The Commission on Higher Education
Teaching Guide for Senior High School
ED
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GENERAL
CHEMISTRY 2
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in collaboration with the Philippine Normal University
SPECIALIZED SUBJECT | ACADEMIC - STEM
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This Teaching Guide was collaboratively developed and reviewed by
educators from public and private schools, colleges, and universities.
We encourage teachers and other education stakeholders to email
their feedback, comments, and recommendations to the Commission
on Higher Education, K to 12 Transition Program Management Unit Senior High School Support Team at k12@ched.gov.ph. We value
your feedback and recommendations.
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Development Team
Team Leader: Myrna S. Rodriguez, Ph.D.,
Writers: Ma. Corazon B. Barrameda, Shirley R.
Jusayan, Ph.D., Veronica C. Sabularse, Ph.D.,
Joseph Carmelo K. San Pascual, Aprhodite M.
Macale
Commission on Higher Education
K to 12 Transition Program Management Unit
Office Address: 4th Floor, Commission on Higher Education,
C.P. Garcia Ave., Diliman, Quezon City
Copy Reader: Kevin Mark R. Gomez
Illustrators: Juan Miguel M. Razon
Cover Artists: Paolo Kurtis N. Tan, Renan U. Ortiz
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Published by the Commission on Higher Education,
2016
Chairperson: Patricia B. Licuanan, Ph.D.
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Technical Editor: Maria Cristina D. Padolina
Senior High School Support Team
CHED K to 12 Transition Program Management Unit
Program Director: Karol Mark R. Yee
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Lead for Senior High School Support:
Gerson M. Abesamis
Consultants
THIS PROJECT WAS DEVELOPED WITH THE PHILIPPINE NORMAL UNIVERSITY.
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University President: Ester B. Ogena, Ph.D.
VP for Academics: Ma. Antoinette C. Montealegre, Ph.D.
VP for University Relations & Advancement: Rosemarievic V. Diaz, Ph.D.
Ma. Cynthia Rose B. Bautista, Ph.D., CHED
Bienvenido F. Nebres, S.J., Ph.D., Ateneo de Manila University
Carmela C. Oracion, Ph.D., Ateneo de Manila University
Minella C. Alarcon, Ph.D., CHED
Gareth Price, Sheffield Hallam University
Stuart Bevins, Ph.D., Sheffield Hallam University
Course Development Officers:
John Carlo P. Fernando, Danie Son D. Gonzalvo,
Stanley Ernest G. Yu
Lead for Policy Advocacy and Communications:
Averill M. Pizarro
Teacher Training Officers:
Ma. Theresa C. Carlos, Mylene E. Dones
Monitoring and Evaluation Officer:
Robert Adrian N. Daulat
Administrative Officers:
Ma. Leana Paula B. Bato, Kevin Ross D. Nera,
Allison A. Danao, Ayhen Loisse B. Dalena
Printed in the Philippines by EC-TEC Commercial, No. 32 St.
Louis Compound 7, Baesa, Quezon City, ectec_com@yahoo.com
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
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Table of Contents
1
DepEd General Chemistry 2 Curriculum Guide . . . . . . . . . . . 5
Chapter 1: Intermolecular Forces and Liquids and Solids
Lesson 14: Colligative Properties of Nonelectrolytes and Electrolyte
Solutions Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
239
Activity Sheet: Acid-Base Titration . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
249
Activity Sheet: Solubility of Salt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
254
Activity Sheet: Determination of Molar Mass by Boiling Point Elevation
260
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lesson 1: Kinetic Molecular Model of Liquids and Solids . . . . . 13
32
Chapter 3: Thermochemistry
65
Lesson 15: Energy Changes in Chemical Reactions . . . . . . . . . . . . . . . .
266
89
Lesson 16: The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . .
286
Lesson 5: Phase Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Lesson 2: Properties of Liquids and Intermolecular Forces. . . .
111
Lesson 17: Thermochemical Equations . . . . . . . . . . . . . . . . . . . . . . . . .
307
Lesson 6: Measuring Viscosity of Liquids . . . . . . . . . . . . . . . . .
128
Lesson 18: Enthalpy and Hess’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . .
323
Lesson 7: Heating and Cooling Curve of a Substance . . . . . . .
140
Lesson 3: Solids and their Properties. . . . . . . . . . . . . . . . . . . . .
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Lesson 4: Phase Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 4: Chemical Kinetics
Lesson 19: Reaction Rates & Collision Theory . . . . . . . . . . . . . . . . . . . .
339
Lesson 8: Types of Solutions and Energy of Solution Formation 159
Lesson 20: Factors that Influence Reaction Rate & Collision Theory . . .
356
Lesson 9: Concentration Units, Mole Fraction, and Molality . . . 180
Lesson 21: Rate of Reaction, Constant, and Concentration of Reactants 372
Lesson 10: Acid-Base Titration & Concentration of Solutions . . 191
Lesson 22: Reaction Rates & the Rate Law . . . . . . . . . . . . . . . . . . . . . . .
387
Lesson 11: Solution Stoichiometry . . . . . . . . . . . . . . . . . . . . . .
201
Activity Sheet: Factors that Affect Reaction Rates . . . . . . . . . . . . . . . . .
405
Lesson 12: Temperature Effect on Solubility . . . . . . . . . . . . . . .
215
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Chapter 2: Physical Properties of Solutions
Lesson 13: Colligative Properties of Nonelectrolytes and
Electrolyte Solutions Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5: Chemical Thermodynamics
227
Lesson 23: Spontaneous Change, Entropy, and Free Energy . . . . . . . .
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
409
Table of Contents
Lesson 24: Enthalpy, Free Energy, and Entropy (Lab) . . . . . . . .
449
Biographical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581
460
Lesson 26: Equilibrium Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . .
476
Lesson 27: Le Chatelier’s Principle . . . . . . . . . . . . . . . . . . . . . .
488
Chapter 7: Acid-Base Equilibria and Salt Solution Equilibria
Lesson 28: Bronsted Acids & Bases and Acid-Base Property of
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Lesson 25: Equilibrium Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . .
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Chapter 6: Chemical Equilibrium
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Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
506
Lesson 30: Strength of Acids and Bases . . . . . . . . . . . . . . . . . .
517
Lesson 31: Buffer Solutions: Henderson Hasselbalch Equation
527
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Lesson 29: Buffer Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lesson 32: Solubility Product Constant . . . . . . . . . . . . . . . . . .
533
Lesson 33: Buffer Solutions and Solutions Equilibria . . . . . . . .
544
Chapter 8: Electrochemistry
Lesson 34: Oxidation-Reduction Reactions . . . . . . . . . . . . . . . . 549
Lesson 35: Redox Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . .
567
Lesson 36: Corrosion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Introduction
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As the Commission supports DepEd’s implementation of Senior High School (SHS), it upholds the vision
and mission of the K to 12 program, stated in Section 2 of Republic Act 10533, or the Enhanced Basic
Education Act of 2013, that “every graduate of basic education be an empowered individual, through a
program rooted on...the competence to engage in work and be productive, the ability to coexist in
fruitful harmony with local and global communities, the capability to engage in creative and critical
thinking, and the capacity and willingness to transform others and oneself.”
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To accomplish this, the Commission partnered with the Philippine Normal University (PNU), the National
Center for Teacher Education, to develop Teaching Guides for Courses of SHS. Together with PNU, this
Teaching Guide was studied and reviewed by education and pedagogy experts, and was enhanced with
appropriate methodologies and strategies.
The SHS for SHS Framework, which stands for “Saysay-Husay-Sarili for Senior High School,” is at the
core of this book. The lessons, which combine high-quality content with flexible elements to
accommodate diversity of teachers and environments, promote these three fundamental concepts:
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SHS for SHS
Framework
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Furthermore, the Commission believes that teachers are the most important partners in attaining this
goal. Incorporated in this Teaching Guide is a framework that will guide them in creating lessons and
assessment tools, support them in facilitating activities and questions, and assist them towards deeper
content areas and competencies. Thus, the introduction of the SHS for SHS Framework.
SAYSAY: MEANING
HUSAY: MASTERY
SARILI: OWNERSHIP
Why is this important?
How will I deeply understand this?
What can I do with this?
Through this Teaching Guide,
teachers will be able to facilitate
an understanding of the value
of the lessons, for each learner
to fully engage in the content
on both the cognitive and
affective levels.
Given that developing mastery
goes beyond memorization,
teachers should also aim for
deep understanding of the
subject matter where they lead
learners to analyze and
synthesize knowledge.
When teachers empower
learners to take ownership of
their learning, they develop
independence and selfdirection, learning about both
the subject matter and
themselves.
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
This Teaching Guide is mapped and aligned to the DepEd SHS Curriculum, designed to be highly
usable for teachers. It contains classroom activities and pedagogical notes, and is integrated with
innovative pedagogies. All of these elements are presented in the following parts:
1. Introduction
• Highlight key concepts and identify the essential questions
• Show the big picture
• Connect and/or review prerequisite knowledge
• Clearly communicate learning competencies and objectives
• Motivate through applications and connections to real-life
2. Motivation
• Give local examples and applications
• Engage in a game or movement activity
• Provide a hands-on/laboratory activity
• Connect to a real-life problem
3. Instruction/Delivery
• Give a demonstration/lecture/simulation/hands-on activity
• Show step-by-step solutions to sample problems
• Give applications of the theory
• Connect to a real-life problem if applicable
4. Practice
• Discuss worked-out examples
• Provide easy-medium-hard questions
• Give time for hands-on unguided classroom work and discovery
• Use formative assessment to give feedback
5. Enrichment
• Provide additional examples and applications
• Introduce extensions or generalisations of concepts
• Engage in reflection questions
• Encourage analysis through higher order thinking prompts
6. Evaluation
• Supply a diverse question bank for written work and exercises
• Provide alternative formats for student work: written homework, journal, portfolio, group/individual
projects, student-directed research project
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Parts of the
Teaching Guide
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
On DepEd Functional Skills and CHED College Readiness Standards
The alignment of both standards, shown below, is also presented in
this Teaching Guide - prepares Senior High School graduates to the
revised college curriculum which will initially be implemented by AY
2018-2019.
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The DepEd articulated a set of 21st century skills that should be
embedded in the SHS curriculum across various subjects and tracks.
These skills are desired outcomes that K to 12 graduates should
possess in order to proceed to either higher education,
employment, entrepreneurship, or middle-level skills development.
On the other hand, the Commission declared the College
Readiness Standards that consist of the combination of knowledge,
skills, and reflective thinking necessary to participate and succeed without remediation - in entry-level undergraduate courses in
college.
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As Higher Education Institutions (HEIs) welcome the graduates of
the Senior High School program, it is of paramount importance to
align Functional Skills set by DepEd with the College Readiness
Standards stated by CHED.
College Readiness Standards Foundational Skills
Produce all forms of texts (written, oral, visual, digital) based on:
Solid grounding on Philippine experience and culture;
An understanding of the self, community, and nation;
Visual and information literacies, media literacy, critical thinking
Application of critical and creative thinking and doing processes;
and problem solving skills, creativity, initiative and self-direction
Competency in formulating ideas/arguments logically, scientifically, and creatively; and
Clear appreciation of one’s responsibility as a citizen of a multicultural Philippines and a
diverse world;
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1.
2.
3.
4.
5.
DepEd Functional Skills
Global awareness, scientific and economic literacy, curiosity,
critical thinking and problem solving skills, risk taking, flexibility
and adaptability, initiative and self-direction
Work comfortably with relevant technologies and develop adaptations and innovations for
significant use in local and global communities
Global awareness, media literacy, technological literacy,
creativity, flexibility and adaptability, productivity and
accountability
Communicate with local and global communities with proficiency, orally, in writing, and
through new technologies of communication
Global awareness, multicultural literacy, collaboration and
interpersonal skills, social and cross-cultural skills, leadership
and responsibility
Interact meaningfully in a social setting and contribute to the fulfilment of individual and
shared goals, respecting the fundamental humanity of all persons and the diversity of
groups and communities
Media literacy, multicultural literacy, global awareness,
collaboration and interpersonal skills, social and cross-cultural
skills, leadership and responsibility, ethical, moral, and spiritual
values
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Systematically apply knowledge, understanding, theory, and skills for the development of
the self, local, and global communities using prior learning, inquiry, and experimentation
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
13.
PY
14.
(LAB) Perform exercises on the
structure of organic compounds using
of models
(LAB) Prepare selected organic
compound and describe their
properties
(LAB) Perform laboratory activities on
enzyme action, protein denaturation,
separation of components in coconut
milk
ED
design a simple
investigation to determine
the effect on boiling point
or freezing point when a
solid is dissolved in water
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Third Quarter – General Chemistry 2
Intermolecular Forces and
1. the properties of
Liquids and Solids
liquids and solids to
1. Kinetic molecular model of
the nature of forces
liquids and solids
between particles
2. Intermolecular Forces
2. phase changes in
3. Dipole-dipole forces
terms of the
4. Ion-dipole forces
accompanying
5. Dispersion forces
changes in energy
6. Hydrogen bonds
and forces between
7. Properties of liquids and
particles
IMF
8. Surface Tension
9. Viscosity
10. Vapour pressure, boiling
point
11. Molar heat of vaporization
12. Structure and Properties of
Water
13. Types and properties of
solids
14. Crystalline and amorphous
solids
15. Types of Crystals – ionic,
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15.
CODE
STEM_GC11OC-IIg-j-96
STEM_GC11OC-IIg-j-97
STEM_GC11OC-IIg-j-98
1.
use the kinetic molecular model to
explain properties of liquids and solids
STEM_GC11IMF-IIIa-c-99
2.
describe and differentiate the types of
intermolecular forces
STEM_GC11IMF-IIIa-c100
3.
predict the intermolecular forces
possible for a molecule
STEM_GC11IMF-IIIa-c101
4.
describe the following properties of
liquids, and explain the effect of
intermolecular forces on these
properties: surface tension, viscosity,
vapor pressure, boiling point, and
molar heat of vaporization
explain the properties of water with its
molecular structure and intermolecular
forces
5.
STEM_GC11IMF-IIIa-c102
STEM_GC11IMF-IIIa-c103
6.
describe the difference in structure of
crystalline and amorphous solids
STEM_GC11IMF-IIIa-c104
7.
describe the different types of crystals
and their properties: ionic, covalent,
molecular, and metallic.
STEM_GC11IMF-IIIa-c105
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 9 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
8.
STEM_GC11IMF-IIIa-c106
9.
interpret the phase diagram of water
and carbon dioxide
STEM_GC11IMF-IIIa-c107
10.
(LAB) Measure and explain the
difference in the viscosity of some
liquids
(LAB) Determine and explain the
heating and cooling curve of a
substance
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properties of solutions,
solubility, and the
stoichiometry of
reactions in solutions
ED
11.
Physical Properties of
Solutions
1. Types of Solutions
2. Energy of solution
formation
3. Concentration Units and
comparison of
concentration units
a. percent by mass, by
volume
b. mole fraction
c. molality
d. molarity
e. percent by volume,
percent by mass, ppm
4. Solution stoichiometry
5. Factors affecting Solubility
6. Colligative Properties of
Nonelectrolyte and
electrolyte solutions
CODE
describe the nature of the following
phase changes in terms of energy
change and the increase or decrease in
molecular order: solid-liquid, liquidvapor, and solid-vapor
PY
covalent, molecular,
metallic
16. Phase Changes
- phase diagrams of water
and carbon dioxide
LEARNING COMPETENCIES
STEM_GC11IMF-IIIa-c108
STEM_GC11IMF-IIIa-c109
1.
describe the different types of
solutions
STEM_GC11PP-IIId-f-110
2.
use different ways of expressing
concentration of solutions: percent by
mass, mole fraction, molarity, molality,
percent by volume, percent by mass,
ppm
STEM_GC11PP-IIId-f-111
3.
perform stoichiometric calculations for
reactions in solution
STEM_GC11PP-IIId-f-112
4.
explain the effect of temperature on
the solubility of a solid and of a gas
STEM_GC11PP-IIId-f-113
5.
explain the effect of pressure on the
solubility of a gas
STEM_GC11PP-IIId-f-114
6.
describe the effect of concentration on
the colligative properties of solutions
STEM_GC11PP-IIId-f-115
7.
differentiate the colligative properties
of nonelectrolyte solutions and of
electrolyte solutions
STEM_GC11PP-IIId-f-116
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 10 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
Calculate boiling point elevation and
freezing point depression from the
concentration of a solute in a solution
STEM_GC11PP-IIId-f-117
9.
calculate molar mass from colligative
property data
STEM_GC11PP-IIId-f-118
10.
(LAB) Perform acid-base titration to
determine concentration of solutions
STEM_GC11PP-IIId-f-119
11.
(LAB) Determine the solubility of a
solid in a given amount of water at
different temperatures
(LAB) Determine the molar mass of a
solid from the change of melting point
or boiling point of a solution
explain the energy changes during
chemical reactions
distinguish between exothermic and
endothermic processes
explain the first law of
thermodynamics
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8.
12.
1.
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energy changes in
chemical reactions
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Thermochemistry
1. Energy Changes in
Chemical Reactions:
exothermic and
endothermic processes
2. First Law of
Thermodynamics
3. Enthalpy of a Chemical
Reaction
- thermochemical equations
4. Calorimetry
5. Standard Enthalpy of
Formation and Reaction
Hess’ Law
Chemical Kinetics
1. The rate of a
1. The Rate of a Reaction
reaction and the
2. Factors that influence
various factors that
reaction rate
influence it
3. The Rate Law and its
2. the collision theory
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
CODE
2.
3.
4.
explain enthalpy of a reaction.
5.
Write the thermochemical equation for
a chemical reaction
Calculate the change in enthalpy of a
given reaction using Hess Law
(LAB) Do exercises on
thermochemical calculations
(LAB)Determine the heat of
neutralization of an acid
describe how various factors influence
the rate of a reaction
write the mathematical relationship
between the rate of a reaction, rate
constant, and concentration of the
6.
7.
8.
1.
2.
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
STEM_GC11PP-IIId-f-120
STEM_GC11PP-IIId-f-121
STEM_GC11TC-IIIg-i-122
STEM_GC11TC-IIIg-i-123
STEM_GC11TC-IIIg-i-124
STEM_GC11TC-IIIg-i-125
STEM_GC11TC-IIIg-i-126
STEM_GC11TC-IIIg-i-127
STEM_GC11TC-IIIg-i-128
STEM_GC11TC-IIIg-i-129
STEM_GC11CK-IIIi-j-130
STEM_GC11CK-IIIi-j-131
Page 11 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
4.
5.
CONTENT STANDARD
PERFORMANCE
STANDARD
components
Collision theory
Catalysis
LEARNING COMPETENCIES
3.
PY
4.
reactants
differentiate zero, first-, and secondorder reactions
write the rate law for first-order
reaction
discuss the effect of reactant
concentration on the half-time of a
first-order reaction
explain the effect of temperature on
the rate of a reaction
explain reactions qualitatively in terms
of molecular collisions
explain activation energy and how a
catalyst affects the reaction rate
cite and differentiate the types of
catalysts
(LAB)Determine the effect of various
factors on the rate of a reaction
5.
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6.
7.
8.
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9.
Chemical Equilibrium
1. The equilibrium condition
prepare a poster on a
specific application of one
of the following:
a. Acid-base
equilibrium
b. Electrochemistry
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Fourth Quarter – General Chemistry 2
Chemical Thermodynamics
spontaneous change,
1. Spontaneous processes
entropy, and free energy
2. Entropy
3. The Second Law of
Thermodynamics
4. Gibbs Free Energy and
Chemical Equilibrium
Chemical equilibrium
and Le Chatelier’s
Include in the poster the
concepts, principles, and
chemical reactions involved,
and diagrams of processes
and other relevant
materials
10.
CODE
STEM_GC11CK-IIIi-j-132
STEM_GC11CK-IIIi-j-133
STEM_GC11CK-IIIi-j-134
STEM_GC11CK-IIIi-j-135
STEM_GC11CK-IIIi-j-136
STEM_GC11CK-IIIi-j-137
STEM_GC11CK-IIIi-j-138
STEM_GC11CK-IIIi-j-139
1. predict the spontaneity of a process based
on entropy
STEM_GC11CT-IVa-b-140
2. determine whether entropy increases or
decreases if the following are changed:
temperature, phase, number of particles
STEM_GC11CT-IVa-b-141
3. explain the second law of
thermodynamics and its significance
STEM_GC11CT-IVa-b-142
4. use Gibbs’ free energy to determine the
direction of a reaction
STEM_GC11CT-IVa-b-143
1. describe reversible reactions
STEM_GC11CE-IVb-e-144
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 12 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
2. Writing the reaction
quotient/equilibrium
constant expression
3. Predicting the direction of a
reaction
4. Significance of the
equilibrium constant
5. Le Chatelier’s Principle
CONTENT STANDARD
PERFORMANCE
STANDARD
Principle
LEARNING COMPETENCIES
CODE
2. explain chemical equilibrium in terms of
the reaction rates of the forward and the
reverse reaction
STEM_GC11CE-IVb-e-145
3. write expressions for the reaction
quotient/equilibrium constants
STEM_GC11CE-IVb-e-146
4. explain the significance of the value of the
equilibrium constant.
STEM_GC11CE-IVb-e-147
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CONTENT
Acid-Base Equilibria and Salt
Equilibria
1. Bronsted acids and bases
2. The acid-base properties of
water
3. pH- a measure of acidity
4. Strength of acids and bases
5. Weak acids/weak bases and
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5. calculate equilibrium constant and the
pressure or concentration of reactants or
products in an equilibrium mixture
6. state the Le Chatelier’s principle and apply
it qualitatively to describe the effect of
changes in pressure, concentration and
temperature on a system at equilibrium
1. acid-base equilibrium
and its applications
to the pH of
solutions and the
use of buffer
solutions
2. solubility equilibrium
and its applications
STEM_GC11CE-IVb-e-148
STEM_GC11CE-IVb-e-149
7. (LAB) Describe the behavior of reversible
reactions
STEM_GC11CE-IVb-e-150
8. (LAB) Describe the behavior of a reaction
mixture when the following takes place:
a. change in concentration of reactants
or products
b. change in temperature
STEM_GC11CE-IVb-e-151
9. (LAB) Perform calculations involving
equilibrium of gaseous reactions
STEM_GC11CE-IVb-e-152
1. define Bronsted acids and bases
STEM_GC11AB-IVf-g-153
2. discuss the acid-base property of water
STEM_GC11AB-IVf-g-154
3. define pH
STEM_GC11AB-IVf-g-155
4. calculate pH from the concentration of
hydrogen ion or hydroxide ions in
aqueous solutions
STEM_GC11AB-IVf-g-156
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 13 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
5. determine the relative strength of an acid
or a base, from the value of the ionization
constant of a weak acid or base
6. determine the pH of a solution of weak
acid or weak base
PY
ionization constants
6. Relationship between the
ionization constants of acids
and their conjugate bases
7. The Common Ion Effect
8. Buffer solutions
9. Solubility equilibria
LEARNING COMPETENCIES
7. explain the Common Ion Effect
Electrochemistry
1. Redox reactions
2. Galvanic cells
3. Standard reduction
potentials
4. Spontaneity of redox
reactions
5. Batteries
6. Corrosion
7. Electrolysis
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EP
ED
C
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8. describe how a buffer solution maintains
its pH
9. calculate the pH of a buffer solution using
the Henderson-Hasselbalch equation
10. explain and apply the solubility product
constant to predict the solubility of salts
11. describe the common ion effect on the
solubility of a precipitate
12. explain the effect of pH on the solubility
of a precipitate
13. (LAB) Determine the pH of solutions of a
weak acid at different concentrations and
in the presence of its salt
14. (LAB)Determine the behavior of the pH
of buffered solutions upon the addition of
a small amount of acid and base
1. define oxidation and reduction reactions
Redox reactions as
applied to galvanic and
electrolytic cells
CODE
STEM_GC11AB-IVf-g-157
STEM_GC11AB-IVf-g-158
STEM_GC11AB-IVf-g-159
STEM_GC11AB-IVf-g-160
STEM_GC11AB-IVf-g-161
STEM_GC11AB-IVf-g-164
STEM_GC11AB-IVf-g-165
STEM_GC11AB-IVf-g-166
STEM_GC11AB-IVf-g-167
STEM_GC11AB-IVf-g-168
STEM_GC11AB-IVf-g-169
2. balance redox reactions using the change
in oxidation number method
STEM_GC11AB-IVf-g-170
3. draw the structure of a galvanic cell and
label the parts
STEM_GC11AB-IVf-g-171
4. identify the reaction occurring in the
different parts of the cell
STEM_GC11AB-IVf-g-172
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 14 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
5. write the half-equations for the reactions
occurring in the electrodes
STEM_GC11AB-IVf-g-173
PY
6. write the balanced overall cell reaction
7. give different examples of galvanic cell
C
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8. define reduction potential, oxidation
potential, and cell potential
9. describe the standard hydrogen electrode
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10. calculate the standard cell potential
11. relate the value of the cell potential to the
feasibility of using the cell to generate an
electric current
12. describe the electrochemistry involved in
some common batteries:
a. leclanche dry cell
b. button batteries
c. fuel cells
d. lead storage battery
STEM_GC11AB-IVf-g-174
STEM_GC11AB-IVf-g-175
STEM_GC11AB-IVf-g-176
STEM_GC11AB-IVf-g-177
STEM_GC11AB-IVf-g-178
STEM_GC11AB-IVf-g-179
STEM_GC11AB-IVf-g-180
13. apply electrochemical principles to explain
corrosion
STEM_GC11AB-IVf-g-181
14. explain the electrode reactions during
electrolysis
STEM_GC11AB-IVf-g-182
15. describe the reactions in some
STEM_GC11AB-IVf-g-183
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 15 of 18
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
CONTENT
CONTENT STANDARD
PERFORMANCE
STANDARD
LEARNING COMPETENCIES
CODE
commercial electrolytic processes
PY
16. (LAB) Determine the potential and
predict the cell reaction of some
assembled electrochemical cells
STEM_GC11AB-IVf-g-185
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C
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17. (LAB) Describe the reactions at the
electrodes during the electrolysis of
water; cite the evidence for your
conclusion
STEM_GC11AB-IVf-g-184
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2 May 2016
This Teaching Guide is a donation by CHED to DepEd. It is for reference purposes only.
Page 16 of 18
Chemistry 2
120 MINS
Kinetic Molecular Model of
Liquids and Solids
LESSON OUTLINE
Communicating learning objectives
15
Motivation
Content Standard
The learners demonstrate an understanding how temperature and pressure affect
Instruction
solubility of solutes (solids and gases) in solvents.
Practice
Performance Standards
Illustration example
20
Discussion and Activity
60
Practice exercises
5
The learners shall design a simple investigation to determine the effect on boiling point
or freezing point when a solid is dissolved in water.
Extension activity
10
Exercises
10
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Introduction
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Enrichment
The learners design a simple investigation to determine the effect of temperature on
solubility of sugar.
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Learning Competencies
Use the kinetic molecular model to explain properties of liquids and solids.
(STEM_GC11IMF-IIIa-c-99)
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Describe and differentiate the types of intermolecular forces. (STEM_GC11IMFIIIa-c-100)
Predict the intermolecular forces that a molecule can possibly form.
(STEM_GC11IMF-IIIa-c-101)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
compare the properties of liquids and solids with those of gases
•
apply the kinetic molecular theory to describe liquids and solids
•
describe the various intermolecular forces and factors that affect their strengths
•
identify the types of intermolecular forces that may operate in a given
Evaluation
Materials
Fo the learner’s activity:
Four different liquid substances: water, ethanol, acetone, butane;
Four Droppers; Eight 1-peso coins; Timer
Resources
(1) Chang, R. (1997). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
(2) Shakhashiri, B. (1989). Chemical demonstrations (3rd ed., pp. 329-332).
Madison, Wis.: Univ. of Wisconsin Pr.
(3) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
Thomson Brooks/Cole.
molecular substance
•
rank substances according to strength of intermolecular forces;
•
illustrate the intermolecular forces between molecules of a compound.
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INTRODUCTION (15 MINS)
1. Communicate to the learners the learning competencies and objectives using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Describe the characteristic properties that differentiate gases, liquids, and solids.
b. Identify the molecular behavior responsible for each property of gases, liquids, and solids.
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c. Describe and differentiate the types of intermolecular forces.
d. Predict the intermolecular forces possibe for a molecule.
e. Rank molecules according to strength of intermolecular forces.
Illustrate the interactions of multiple molecules of a compound.
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f.
2. Present relevant vocabulary that learners should know and will be used in the lesson:
Phase
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A homogeneous part of a system in contact with other parts of the system, but separated from t h e s e
other parts by well-defined boundaries.
Liquids and solids
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Condensed phases
Intramolecular forces and intermolecular forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Teacher Tips:
Some notes on polarity of molecules are
given in Appendix A.
3. Connect the lesson with prerequisite knowledge
A. Recall Kinetic Molecular Theory:
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1. All matter is made of tiny particles.
2. These particles are in constant motion.
3. The speed of particles is proportional to temperature. Increased temperature means greater speed.
PY
4. Solids, liquids, and gases differ in distances between particles, in the freedom of motion of particles,
and in the extent to which the particles interact.
5. For an animation showing the motion of particles in a solid, liquid or gas, the lesson below may be
viewed.
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http://preparatorychemistry.com/KMT_flash.htm
B. Recall Molecular Geometry, Determining Polarity, Bond Dipole, Dipole Moment
Ask the students to
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1. draw the Lewis structures of the following molecules with the correct shape around the central atom;
2. indicate each bond’s polarity by drawing an arrow to represent the bond dipole along each bond;
3. determine the molecule’s polarity and indicate this with an arrow to represent the dipole;
Expected Answers
Cl2
polar or nonpolar
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4. circle their choice in each box to mark the molecule as polar or nonpolar
NH3
CH3Br
CH4
polar or nonpolar
polar or nonpolar
polar or nonpolar
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Roleplay of the three physical states – solid, liquid, gas.
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MOTIVATION (20 MINS)
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The activity involves the students acting as particles (or molecules ) and they will present their behavior and
positions relative to each other in the solid, liquid and gaseous state. Divide the class into three groups. Assign
one group to act out the solid, the second as liquid and the third one as gas. Give the students 3 minutes to
discuss among themselves how to act the assigned state, and 2 minutes to act it. One member of the group will
explain their act.
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Allow the other groups to make comments on the group acts.
INSTRUCTION (60 MINS)
A. Kinetic Molecular Theory of Liquids and Solids
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Small Group Discussion
Group the students with three or four members in each group. One member acts as the recorder and
note-taker.
Teacher Tips:
Questions for Small Group Discussion
•
In liquids, the molecules are so close
together that there is very little empty
space between them. Liquids are much
more difficult to compress and they are
much denser at normal conditions.
•
Molecules in a liquid are held together
by one or more types of attractive forces.
However, the molecules can move past
one another freely. Liquids can flow, can
be poured and assumes the shape of its
container.
•
In a solid, molecules are held tightly in
position with virtually no freedom of
motion. There is even less empty space
in a solid than in a liquid.
•
Solids are almost incompressible and
possess definite shape and volume.
The Condensed State: Liquids and Solids
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Using the roleplays carried out by the class at the motivation part, and the following illustration of solid,
liquid and gas, answer the questions that follow. Prior viewing of the animations suggested in the recall
of requisite knowledge may also be useful in this activity.
Figure 1: Molecular or particle level view of a solid, liquid and a gas
17
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1. Compare the properties of gases, liquids, and solids in terms of distances and arrangement of their
molecules.
a. Compare the distances among molecules in the gas, liquid and solid and rank the phases in increasing
distance between particles.
b. Describe the characteristic movement of the particles of gas, liquid and solid.
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c. How are the molecules of gas, liquid and solid arranged?
d. Arrange the three phases of matter in order of increasing volume of empty space between its molecules.
e. Identify the property of matter that corresponds to the molecular behavior.
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2. Use the table to present the comparison of the properties of gases, liquids and solids.
(Expected answers are given in italics.)
Properties of Matter
Molecular Behavior
liquid
ED
gas
solid
Assumes volume and
shape of container
Fixed volume; assumes
Fixed volume; fixed
shape of occupied part of shape (regardless of size
container.
and shape of container
Density
low
high
Compressibility
Easy to compress
Cannot be appreciably
compressed
Motion of Molecules
Random, fast, cover
large distances
Random, medium speed, Vibration in place
limited distances
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Volume/Shape
high
Cannot be appreciably
compressed
Class Discussion
1. Draw the same table on the board and ask representatives from each group to fill in each box.
2. Use the following illustration to organize the answers of the learners in the preceding questions.
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Teacher Tip:
•
Accept any other appropriate responses. !
•
If there is an internet connection, the
video on factors affecting solubility can
be used for demonstration:!
https://www.youtube.com/watch?
v=OpFW7V_GiUQ!
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Figure 2. Molecular level comparison of gases, liquids and solids. Image obtained from
http://wpscms.pearsoncmg.com/wps/media/objects/3662/3750037/Aus_content_10/Fig10-02.jpg
3. The following table summarizes properties of gases, liquids, and solids and identifies the microscopic
behavior responsible for each property.
4. Put emphasis on the difference in distances of particles in solids and liquids as compared to gases. This is
the reason solids and liquids are called the condensed states. Direct attention to the ability of particles in the
gaseous state to move away from each other. On the other hand, the particles stay close together in the
solid and liquid states.
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Teacher Tip:
For gases:!
An increase in temperature results in
•
increased kinetic energies of gases
dissolved in liquids. This increased
motion enables the dissolved gas to
break intermolecular forces with the
solvent, and escape the solution.!
•
Thus, a warm bottle of carbonated drink/
soft drink does not taste as good as a
cold one, because there is less CO2
dissolved in the warm bottle.!
Ask the following question as jump off point for the lesson on intermolecular forces of attraction.: What
holds the particles in the solid and liquid states?
imagery to your discussion:
http://preparatorychemistry.com/KMT_flash.htm
Hands-on activity
Suggested liquids to use: Water, ethanol, acetone, pentane
Questions to investigate:
How many drops of liquid can a 1-peso coin hold?
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1.
2. How long will it take for one drop of a liquid to evaporate?
Safety Precautions:
•
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The activity should be performed in an airy or well-ventilated room.
Remind the students of the proper handling of the substances they will be using.
•
Prepare the labeled vials with the liquid
before the start of the activity. Four
groups of students can share one set and
take turns in using each liquid. Only the
vials can be shared. Each group should
use their own dropper.
•
Make sure that the students label the
droppers they will be using with the
liquid it is intended for so that
contamination or mixing is avoided.
•
Paper or paper towels are useful in these
types of activities. Instruct the students
to place the coin on top of a sheet of
paper or paper towel.
Avoid contact with the skin and direct inhalation of the vapors of the substances. It is best if the students use
safety gloves, goggles and mask.
1. Tell the students to work in groups of three members. One of the members will act as the recorder of
data.
2. Give each student a data sheet for their results.
3. Check for the availability of the materials for the activity. Each group should have 8 pieces of 1-peso
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If pentane is not readily available, use
butane.
Butane is the fluid used in cigarette
lighters. Acetone and ethanol can
be purchased as aqueous solutions.
Pure or absolute samples are fairly
expensive. The presence of water in
these samples may affect the results,
particularly the evaporation part.
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B. Kinetic Molecular Theory of Liquids and Solids
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5. If internet connection is available, you can use the video found at this site to provide some
coin and 4 droppers.
•
You can try the evaporation part first so
that you can set a maximum time for the
observation of evaporation. It may take
a long time for water to evaporate so
that the students can just use the
maximum time you’ve set. For example,
you set 10 min as the maximum time and
water has not completely evaporated at
this time, the students can record it as
>10 min or more than 10 min.
•
A table should be made on the board
where each group will show their results
for the number of drops and the time of
evaporation.
•
Possible sources of differences in the
results of experiment:
different sides of the coin was used
same member of the group did the
procedures or each member took
turns
surface used is completely flat or not
coin was completely dried or not
before using
old or new coin used
different drop size
technique in dropping: height,
angle, rate, pressure on the dropper
4. Give each group 4 labeled small vials containing each of the liquids.
5. Using the first 4 coins, have the students drop each liquid on a 1-peso coin and count the
number of drops the coin can hold.
PY
6. Then on the next 4 coins, put a drop of the liquid and determine how much time it takes one
drop to evaporate.
Discuss the results of the activity.
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7. Let the students write their results on the board for comparison with the results of the class.
1. Ask the students to share the results of their experiment. Let them compare their results with those of
their classmates for 2 minutes.
2. Ask the students the following questions:
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a. Which molecules can hold more drops on the coin?
b. Which molecules took longer to evaporate?
c. Are the molecules that can hold the lesser number of drops the same as the molecules that took less
time to evaporate?
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d. Based on the formula and geometries of the substances, are the molecules that can hold more drops
on the coin polar or nonpolar? What about those that took longer to evaporate?
3. Discuss the Intermolecular Forces of Attraction between individual particles of a substance in the
condensed states.
Suggested flow of discussion:
a. define intermolecular forces of attraction
Intermolecular forces are attractive forces that act between molecules or particles in the solid or
21
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liquid states. Generally, these attractive forces are much weaker than bonding forces.
b. explain why melting points and boiling points of substances can be used as indicators of strength of
intermolecular forces operating in given solids and liquids
PY
When a solid melts, or a liquid boils, the particles move away from each other. As they do,
intermolecular forces of attraction are broken. The stronger the intermolecular forces to be broken,
the larger the amount of energy needed to break them, hence, the higher the melting point for solid
to liquid transformation, and boiling point for liquid to gas transformation.
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c. describe the different types of intermolecular forces and relate these to the type of molecules that
exhibit them.
The different types of intermolecular forces are the following:
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Dispersion forces – these forces of attraction result from temporary dipole moments induced in
ordinarily nonpolar molecules. These forces are present between all types of molecules due to the
movement of electrons. As electrons move around the nucleus, an uneven distribution causes
momentary charge separations. Slightly positive sides of a molecule are attracted to the slightly negative
sides of the adjacent molecule.
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The extent to which a dipole moment can be induced in a molecule is called its polarisability.
Polarizability of the atom or molecule refers to the ease with which the electron distribution can be
distorted. Generally, the larger the number of electrons and the larger or more diffused the electron
cloud in the atom or molecule, the greater its polarizability. Thus, dispersion forces may be the weakest
of intermolecular forces that can exist between two molecules, but the larger the atoms present, the
stronger the dispersion forces.
For example, F2, the lightest halogen, is a gas, Br2 is a liquid, and the heavier I2 ,is a solid at room
conditions. Further, the more atoms that make up the molecules, the stronger are the dispersion forces.
Methane, CH4, is gaseous, but larger hydrocarbons like butane, C4H10. is liquid, and those with larger
number of carbon atoms, like the waxes, are solids at room temperature.
22
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Teacher Tip
There is a need to guide the students’ group
investigation
An Illustration of London Dispersion Forces using Helium atoms (2 electrons)
Consider atoms of helium. The average distribution of electrons around each nucleus is spherically
symmetrical. The atoms are nonpolar and possess no dipole moment.
•
At a given instant in time, the distribution of electrons around an individual atom, may not be
perfectly symmetrical. Both electrons may be on one side of the nucleus, as shown on the leftmost
atom in the figure below.
•
The atom would have an apparent dipole moment at that instant in time (i.e. a transient dipole).
•
A close neighboring atom, shown on the right, would be influenced by this apparent dipole. The
electrons of the neighboring atom would move away from the negative region of the dipole.Due to
electron repulsion, a temporary dipole on one atom can induce a similar dipole on a
neighboring atom
•
This will cause the neighboring atoms to be attracted to one another. This is called the London
dispersion force (or just dispersion force). It is significant only when the atoms are close together.
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•
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PY
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Figure 3. London dispersion forces between helium atoms. Image obtained from http://
www.mikeblaber.org/oldwine/chm1045/notes/Forces/Intermol/Forces02.htm
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Dipole-dipole forces are attractive forces between polar molecules (molecules that possess dipole
moments). In polar molecules the electrons are unevenly distributed because some elements are more
electronegative than others. The partial negative side of one molecule is attracted to the partial positive
side of another molecule.
This type of force is stronger than the dispersion forces because polar molecules have a permanent
uneven distribution of electrons. The nature of attraction is electrostatic and can be understood in terms
of Coulomb’s law: The larger the dipole moment, the stronger the attraction.
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PY
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Figure 4. Attractive Dipole-Dipole Interactions. Image obtained from http://
www.mikeblaber.org/oldwine/chm1045/notes/Forces/Intermol/Forces02.htm
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The interaction is written as
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Hydrogen bond is a special type of dipole-dipole interaction between the hydrogen atom in a polar
bond, such as N‒H, O‒H, or F‒H, and an electronegative O, N, or F atom. Hydrogen bonds between
water molecules are particularly strong.
A ‒ H ••• B or
A ‒ H ••• A
A and B represent O, N, or F; A ‒ H is one molecule or part of a molecule and A or B is a part of
another molecule; the dotted line represents the hydrogen bond.
Examples of hydrogen bonding in water (H2O), ammonia (NH3) and hydrogen fluoride (HF):
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PY
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Figure 5. Hydrogen bonds in H2O, NH3 and HF. Image obtained from http://
wps.prenhall.com/wps/media/objects/3082/3156196/blb1102.html
d. list the type of intermolecular forces that are expected to operate in the solid or liquid states of simple
molecular substances based on their structures
The following diagram can be used to determine the types of intermolecular forces present in
substances.
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Figure 6. Schematic diagram for determining intermolecular forces in a substance
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Figure 7. Alternative Schematic Diagram for determining Intermolecular Forces in a Substance
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PRACTICE (5 MINS)
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ENRICHMENT (10 MINS)
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Appendix A. Notes on Molecular Polarity
To determine the polarity of a molecule, both the bonds present and the overall shape of the molecule should
be considered. Two or more polar bonds may cancel each other out leading to a nonpolar molecule.
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A molecule will be nonpolar if:
All of the terminal atoms (or groups) are the same
•
All of the terminal atoms (or groups) are symmetrically arranged around the central atom
•
The terminal atoms (or groups) have the same charges
•
Example: CO2
PY
•
A molecule will be polar if:
One or more terminal atoms differ from each other.
•
At least one polar bond is present.
•
The terminal atoms are not symmetrically arranged
•
The molecule has one slightly positive end and one slightly negative end.
•
Example: H2O
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•
ADDITIONAL RESOURCES
(1) Zumdahl, S.S. & S. A. Zumdahl (2012). Chemistry an atoms first approach. United States. Brooks/Cole Cengage Learning Asia Pte.
Ltd. pp. 491 - 495.!
(2) http://www.ck12.org/user:krogers/section/Factors-Affecting-Solubility/ (Retrieved Nov. 5, 2015)
(3) http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_and_Factors_Affecting_Solubility (Retrieved Nov. 5,
2015)
(4) http://chemsense.sri.com/classroom/curriculum/Solubility_Kennedy.pdf (Retrieved Nov. 6, 2015)!
(5) http://www.acs.org/content/dam/acsorg/education/resources/k-8/inquiryinaction/inquiry-in-action.pdf (Retrieved Nov. 2, 2015)
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Chemistry 2
120 MINS
Properties of Liquids and
Intermolecular Forces
LESSON OUTLINE
Motivation
10
Introduction
Content Standard
The learners demonstrate an understanding of the properties of liquids, and the nature of Instruction
forces between particles.
Practice
Communicating learning objectives
5
Lecture/Discussion
85
Practice Exercises
10
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or
freezing when a solid is dissolved in water.
Enrichment
Extension Activity
10
Evaluation
Exercises
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Student Participation/Engagement
Materials
Learning Competencies
Describe the following properties of liquids, and explain the effect of intermolecular
forces on these properties: surface tension, viscosity, vapor pressure, boiling point, and
molar heat of vaporization. (STEM_GC11IMF-IIIa-c-102); and
Resources
Explain the properties of water based on its molecular structure and intermolecular
forces. (STEM_GC11IMF-IIIa-c-103)
(3) McGraw Hill Education,. Retrieved from http://www.mhhe.com/physsci/
chemistry/essentialchemistry/flash/vaporv3.swf
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ED
They also learn how to handle and interpret information to make correct inferences on
how intermolecular forces influence the properties of liquids.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
Illustrations and diagrams, images and objects of liquids
(1) Brown, T. L. et al. (2009). Chemistry: The Central Science (11th ed., pp.
460-511). Pearson Education Inc.
(2) Chang, R. (1997). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
(4) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
Thomson Brooks/Cole.
describe the properties of liquids: surface tension, viscosity, vapor pressure,
boiling point, and molar heat of vaporization;
•
explain the effect of intermolecular forces on these properties; and
•
relate the properties of water to intermolecular forces that operate among its
molecules.
32
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MOTIVATION (10 MINS)
Infer the Topic Protocol
The goal is for learners to infer the topic of the lesson from ideas they obtain from objects or pictures.
Teacher Tips
•
•
e.g. pictures of water boiling, water freezing, glass of water, glass of ice and water, vapor rising from a
liquid, liquids in containers (with labels)
•
PY
1. Prepare objects and pictures that learners will observe. The images and objects to be used should be
related to liquids and their behavior. Some images may need to be labeled, and can range from concrete
to abstract.
e.g. pictures showing properties of liquids like viscous liquids, capillary action, surface tension. in action
C
O
e.g. actual objects that relate to properties of liquids, like thermometer, barometer, sphygmomanometer.
2. Issue at least 6 pictures and/or objects to each group. In one minute or less, learners view an image,
discuss with the group, and record an inference about the upcoming topic of study.
ED
3. After all the images or objects have been seen, each learner in the group share his/her final inference on
the topic of the lesson.
D
EP
4. The group makes a consensus of what they believe is the topic of their lesson, and writes this on a sheet
of Manila paper. The group should be able to make arguments to support their inference based on the
images and objects seen.
5. A representative of each group presents their inferred topic in front of the class for all to see. The teacher
invites a few to explain their inferences about the upcoming topic.
6. After a few have shared, the teacher reveals the topic of study as well as the guide questions and big
ideas. Introduction
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•
•
In order to not to reveal the concepts
yet, the Introduction will be done after
the engaging activity.
Some of the images used in the
motivation may be used in the
discussion and vice versa.
To minimize the movements in a large
class, there may be a need to divide the
interaction into two or three big
groups. This would mean preparing
two or three sets of images and
objects. Each group interacts among
themselves.
Images or pictures may be replaced by
actual objects.
Some application questions may be
assigned as advance reading for the
learners or may be given as further
reading.
7. Debrief: Instruction, Delivery, Practice and Enrichment
Relate the images and objects used by the learners to the larger concept(s).
PY
8. Discuss how learners’ inferences did or didn’t change throughout this protocol. Ask of any lingering
questions about the topic. Evaluation
INTRODUCTION (5 MINS)
C
O
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Describe the properties of liquids: surface tension, viscosity, vapor pressure, boiling point, and molar
heat of vaporization.
ED
b. Explain the effect of intermolecular forces to these properties.
2. Present relevant vocabulary that will be used in the lesson and learners should know.
Fluid
Surface tension
D
EP
A gas or a liquid; a substance that can flow.
The measure of the elastic force in the surface of a liquid. It is the amount of energy required
or increase the surface of a liquid by a unit area.
Capillary action
to stretch
The tendency of a liquid to rise in narrow tubes or to be drawn into small openings.
34
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Teacher Tips
•
•
•
Definitions should remain posted as the
lesson progresses.
More key words may be added to
relevant vocabulary as the need arises.
As the lesson progresses, the list may
include terms like cohesion, adhesion,
evaporation or as the learners figure
them out.
Viscosity
A measure of a fluid’s resistance to flow.
•
Vapor
PY
A gaseous substance that exist naturally as a liquid or solid at normal temperature
Vaporization
The change of phase from liquid to vapor (gaseous phase).
C
O
Vapor pressure of a liquid
The equilibrium pressure of a vapor above its liquid; that is, the pressure exerted by the vapor above
the surface of the liquid in a closed container.
Boiling point
Molar heat of vaporization (ΔHvap)
ED
The temperature at which a liquid boils. The boiling point of a liquid when the external pressure is 1 atm
is called the normal boiling point.
D
EP
The energy (usually in kilojoules) required to vaporize 1 mole of a liquid at a given temperature.
3. Connect the lesson with previous knowledge required.
Recall previous concepts on Kinetic Molecular Theory and intermolecular forces in liquids and solids.
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Some images can be used to recall the
concepts on Kinetic Molecular Theory and
Intermolecular Forces
previously
discussed.
INSTRUCTION (85 MINS)
Teacher Tips
Focus Questions:
•
Discussion of each of the properties
should focus on the behavior of the
liquid and the effect of the
intermolecular forces present among its
molecules.
•
E m p h a s i z e t h e e ff e c t o f c e r t a i n
conditions like temperature and pressure
on these properties.
What are the properties of liquids?
PY
How do intermolecular forces affect the properties of liquids?
A. Class Activity (10 mins)
1. Materials needed:
Liquid samples: water, ethyl alcohol, kerosene
Three burets
Beakers or glass jars with wide mouth
Three pieces blade
C
O
What are the properties of water and how do they relate to its structure and intermolecular forces?
Three droppers
Plastic comb, cloth to rub comb
Glass rod
ED
Small piece of wax paper or plastic sheet
2. Procedure
D
EP
Some newspapers to catch drips and spills
Liquid and charged object
a. Place a liquid sample in each of the buret.
b. Rub the comb on a piece of cloth.
c. Open the buret stopcock and allow a thin steady stream of liquid to flow.
d. Place the rubbed part of the comb close to the stream of liquid. Make sure to rub the comb
with the cloth before using it with the next liquid.
e. Observe and record how the rubbed comb affected the stream of liquid.
36
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Expected observations: The stream of water will bend towards the comb to a large extent; the
stream of ethyl alcohol will bend slightly towards the comb, but the stream of kerosene will be
unaffected by the comb.
Liquid and drops on wax paper
PY
a. Lay a piece of wax paper flat on the surface of the table.
b. Using a dropper, place a drop of a liquid sample on the wax paper. Do the same for the two
other liquids. Make sure a different dropper is used for each of the liquid samples to avoid
contamination.
C
O
c. Observe and record the appearance of the drops of the liquid samples on wax paper.
Liquid and blade on the surface
ED
Expected observations: The drop of water will be of small diameter and almost spherical’ the
drop of ethyl alcohol will have a larger diameter and will spread a little, the drop of kerosene will
be spread out on the wax paper.
a. Place some of the liquid samples in a beaker or a wide mouth glass jar.
b. Carefully place a blade on the surface.
D
EP
c. Did the blade float on the surface of the liquids? Record observations.
Expected observations: The blade will float on water, and may also do so on ethyl alcohol, but
will immediately sink in kerosene.
Discussion on the Properties of Liquid and Intermolecular Forces
The properties of liquids that were observed are consequences of the interactions of particles that
make up the liquid.
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SURFACE TENSION
•
Suggested flow of discussion:
• Define surface tension.
PY
Surface tension is the measure of the elastic force in the surface of a liquid. It is the amount
of energy required to stretch or increase the surface of a liquid by a unit area.
ED
C
O
It is manifested as some sort of skin on the surface of a liquid or in a drop of liquid. Use the
illustrations given below to show manifestations of surface tension.
Figure 1. Examples of how surface tension is manifested.
Illustrate how intermolecular forces can influence the magnitude of surface tension.
D
EP
•
Molecules within a liquid are pulled in all directions by intermolecular forces. Molecules at the
surface are pulled downward and sideways by other molecules, not upward away from the
surface (shown in the diagram below).
These intermolecular forces tend to pull the molecules into the liquid and cause the surface to
tighten like an elastic film or “skin”.
38
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Teacher Tip
•
I f t h e i m a g e s w e re u s e d i n t h e
motivation, you may call on the learners
who held them.
PY
C
O
Figure 2. Intermolecular forces that act on molecules of a liquid. (Image Source:
www.bville.org/.../AP%20Chapter%2011%20Intermolecular%20Forces)
Guide the students to reach the conclusion that liquids that have strong intermolecular forces also
have high surface tension.
•
Use water as an example of a liquid with high surface tension as a result of H-bonds, which are strong
intermolecular forces. Use illustrations, including those used in the activity at the start of this lesson to
show how the high surface tension is manifested in water.
D
EP
ED
•
39
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Figure 3. Water strider walking on the surface of a quiet pond. (Image Source: http://
hyperphysics.phy-astr.gsu.edu/hbase/surten.html)
•
Ask the students why water will form a spherical droplet on a sheet of plastic, but kerosene will spread.
PY
CAPILLARY ACTION
• Define capillary action.
•
C
O
Capillary action is the tendency of a liquid to rise in narrow tubes or be drawn into small openings such
as those between grains of a rock. Capillary action, also known as capillarity, is a result of intermolecular
attraction between the liquid and solid materials.
Describe or show illustrations of examples of capillary action
D
EP
ED
Capillary action is shown by water rising spontaneously in capillary tubes. A thin film of water a d h e r e s
Surface
to the wall of the glass tube as water molecules are attracted to atoms making up the glass (SiO2).
tension causes the film of water to contract and pulls the water up the
tube.
Figure 4. Colored water seen rising up in
glass tubes. (Image Source: http://
water.usgs.gov/edu/capillaryaction.html)
Figure 5. A doctor takes blood sample from
a patient’s finger using a capillary tube.
(Image Source: https://www.colourbox.com/
image/blood-testing-image-6891015)
40
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C
O
PY
•
Explain capillary action in terms of attractive forces formed by molecules of the liquid and those
that make up the tubes.
D
EP
•
ED
Figure 6. Water mixed with food coloring rises up freshly-cut stalks of
celery (Image Source: http://water.usgs.gov/edu/capillaryaction.html)
Two types of forces are involved in capillary action:
Cohesion is the intermolecular attraction between like molecules (the liquid molecules).
Adhesion is an attraction between unlike molecules (such as those in water and in the particles that
make up the glass tube).
These forces also define the shape of the surface of a liquid in a cylindrical container (the meniscus!)
41
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If there is time, this may be done as an
actual demonstration.
When the cohesive forces between the liquid molecules are greater than the adhesive forces between
the liquid and the walls of the container, the surface of the liquid is convex.
Example: mercury in a container
PY
When the cohesive forces between the liquid molecules are lesser than the adhesive forces between the
liquid and the walls of the container, the surface of the liquid is concave.
Example: water in a glass container
D
EP
ED
Example: distilled water in a silver vessel
C
O
When both adhesive and cohesive forces are equal, the surface is horizontal.
Figure 7. Concave and Convex Meniscus. (Image Source: http://
www.diffen.com/difference/Adhesion_vs_Cohesion)
42
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VISCOSITY
• Introduce viscosity with the following activity:
Materials needed:
water, cooking oil, honey, acetone, etc
Test tubes, as many as liquids to test
Droppers
PY
Hard cardboard, tape
Marking pen for labeling
D
EP
ED
C
O
Place two to three drops of each liquid sample in separate and labeled small test tubes. Tape the tubes
upright on a piece of hard cardboard. When the tubes are secure, lay the cardboard and tubes in
horizontal position. Ask the students to observe and time the flow of the contents of the tubes until the
liquid in the tube reaches the edge of the opening of the tube. Start timing by tilting the cardboard
about 4 to 5 inches from the surface to allow the liquid to flow. Which one flowed fastest? Which one
took the longest time? Make the students suggest factors that may influence the ease of flow of the
contents.
Figure 8. Comparison of ease of flow of liquids. (Image Source: http://
www.synlube.com/Viscosity/ColdFlow2.jpg)
Expected answer:
Thick liquid do not flow easily.
43
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•
This activity may be extended to include
various other products familiar to the
learners.
•
(School supplies: paste and glue; Beauty
products: lotion, moisturizers, petroleum
jelly, hair gel, shampoo, conditioner;
Food and cooking products: jams, syrup,
peanut butter, seasoning; Cleaning
materials: liquid detergent, soap, bleach).
Suggested flow of discussion:
• Define viscosity.
Viscosity is a measure of a fluid’s resistance to flow. The greater the viscosity, the slower the
flows.
liquid
PY
Viscosity is expressed in units of centipoise. The table below gives viscosities of liquids of some pure
substances. Water has viscosity of 1 centipoise or 0.001 Pa/s at 20oC.
Liquids
C
O
Substances with lower viscosities include carbon tetrachloride and benzene. Glycerol has a resistance to
flow of more than a thousand times greater than water.
Viscosity (in Ns/m2) at 20oC
3.16 x 10-4
Benzene (C6H6)
6.25 x 10-4
Carbon tetrachloride (CCl4)
9.69 x 10-4
Diethyl ether (C2H5OC2H5)
2.33 x 10-4
Ethanol (C2H5OH)
1.20 x 10-3
Glycerol (C3H8O3)
Mercury (Hg)
1.49
1.55 x 10-3
D
EP
Water (H2O)
ED
Acetone (C3H6O)
1.01 x 10-3
Table 1 . Viscosities of liquids of selected substances
•
Relate viscosity or resistance to flow to strength of intermolecular forces that operate among molecules
of the liquid.
Consider the following examples:
Given molecular structures of water and glycerol, can you tell why glycerol has a higher viscosity than
water?
44
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•
Poise or Stoke is a metric system unit.
The SI unit is Pascal-second (Pa-s) or
Newton-second/meter 2 (N-s/m 2 ). 1
centipoise = 0.001 Pa-s
water
PY
!
!
glycerol
C
O
Expected answer:
The larger number of –OH groups allow glycerol to form more H-bonds with other glycerol molecules,
making its intermolecular forces stronger than those of water, and its resistance to flow greater.
D
EP
ED
Consider the table of viscosities that follow. All the substances in the list are hydrocarbons and nonpolar.
What causes the differences in viscosities of the hydrocarbons in the list?
Table 2. Viscosities of some hydrocarbons. (Image Source: http://
wpscms.pearsoncmg.com/wps/media/objects/3662/3750037/Aus_content_10/Table10-04.jpg)
45
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Expected answer:
The size of the molecules. The larger the molecule, even if it is nonpolar, the stronger the intermolecular
forces and the greater the viscosity compared to nonpolar substances made up of small molecules.
Guide the students to arrive at the following conclusion: Liquids that have strong intermolecular
forces have higher viscosities than those that have weak intermolecular forces.
•
Ask the learners what effect temperature would have on viscosity. Viscosity decreases as
temperature increases: hot molasses flows much faster than cold molasses. The viscosities of some
familiar liquids in the table below were measured at 20 OC, except for lava (ranges between 700 to 1200
O
C.
C
O
PY
•
Ask students to identify familiar liquids with high and low viscosities. Below is a table of
viscosities of some of these.
ED
Viscosity
Liquid
(in centipoise, cps)
Water
1
Blood
3
D
EP
Milk
4 to 10
Castor oil
1000
Latex house paint
1500
Hotcake syrup
5000
Honey
10,000
Hershey’s chocolate syrup
10,000 to 25,000
Ketchup
50,000
Peanut butter
250,000
Lava
≈ 4,300,000
46
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Table 3 . Viscosities of some common liquids
(Image Source: http://
www.wmprocess.com/viscosity-of-commonliquids/ )
VAPOR PRESSURE OF A LIQUID
Introduce this property with the following class activity:
ED
C
O
PY
Ask the students to describe what is happening to the water molecules in the two flasks shown in the
picture.
Figure 9. Evaporation of water in open and in closed containers (Image Source: http://
boomeria.org/physicslectures/heat/equilibrium.jpg)
Expected answer:
D
EP
•
(a) The water molecules in the liquid evaporate and go into the vapor phase. In the open flask, some of
the water molecules in the vapor phase find their way out of the flask are lost to the atmosphere.
(b) When a liquid evaporates to a gas in a closed container, the molecules cannot escape.
47
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•
The learners may be guided to make
comments on the flask on the left first
before comments are given about the
flask on the right.
Gas molecules move in random directions, collide with other gas particles and the walls of the container.
Some will strike the liquid surface and condense back into it. In the closed flask, none of the gas
particles are able to get out of the container, and eventually, the number of molecules that go into the
gaseous state would equal the number of molecules that condense back.
PY
When the rate of condensation of the gas becomes equal to the rate of evaporation of the liquid, the
gas in the container is said to be in equilibrium with the liquid.
liquid ⇋ vapor (gas)
D
EP
ED
C
O
In this condition, the amount of gas and liquid no longer changes.
Figure 10 . Equilibrium between liquid and gas (Image Source: http://
wpscms.pearsoncmg.com/wps/media/objects/3662/3750037/Aus_content_10/Fig10-20.jpg)
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•
Define vapor pressure
Like any gas sample, the molecules in the gaseous state over its liquid create a pressure. The greater the
number of gaseous particles, the greater the pressure exerted by the gas. The pressure exerted by the
gas in equilibrium with a liquid in a closed container at a given temperature is called the equilibrium
vapor pressure or simply vapor pressure of the liquid.
PY
The equilibrium vapor pressure is the maximum vapor pressure of a liquid at a given temperature and
that it is constant at a constant temperature. It increases with temperature.
•
This is one of the relationships between
variables that describe a gas: the
pressure of a gas is directly proportional
to the number of gas particles present.
•
The video clip could be watched if
resources are available.
If not, the
diagram may be printed and posted
instead.
•
C
O
Vapor pressure is independent of the amount of liquid as well as the surface area of the liquid in contact
with the gas.
View animation.
Relate vapor pressure to temperature
D
EP
•
ED
Let the learners view the short clip to get a better image of how equilibrium is achieved at http://
www.mhhe.com/physsci/chemistry/essentialchemistry/flash/vaporv3.swf
Present the following plot of vapor pressure of water as it varies with temperature, and ask the learners
to explain what the plot presents.
49
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ED
C
O
PY
•
Figure 11 . Vapor pressure of water vs. temperature
D
EP
Observation: As the temperature increases, the vapor pressure of water also increases.
When temperature is high, more molecules have enough energy to escape from the liquid. At a l o w e r
temperature, fewer molecules have sufficient energy to escape from the liquid.
Given in the graph below are the vapor pressures for four common liquids: diethyl ether, ethyl a l c o h o l ,
water and ethylene glycol, as a function of temperature. For all four liquids, the vapor pressure
increases as temperature increases.
50
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Emphasize the LIQUID-VAPOR
EQUILIBRIUM in the process.
PY
C
O
ED
D
EP
Figure 12 Vapor pressure of four common liquids, shown as a function of
temperature (Image Source: http://wps.prenhall.com/wps/media/objects/ 3311/3391416/
blb1105.html)
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Relate Vapor Pressure to Strength of Intermolecular Forces
substance
0.71 atm
Acetone
0.28 atm
Ethyl alcohol
0.08 atm
Water
0.03 atm
C
O
Pentane
ED
Observation:
vapor pressure at 25oC
PY
Consider the vapor pressures of the following substances. Relate vapor pressure to strength of
intermolecular forces.
Ethyl alcohol and water have very low vapor pressures. Both liquids have the strong dipole-dipole
interaction called hydrogen bonding. Acetone is polar but does not have H-bonding. Its vapor pressure
is of intermediate value. Pentane is a nonpolar substance, and its vapor pressure is high compared to
those of water and ethyl alcohol.
D
EP
•
When liquids evaporate, the molecules have to have sufficient energy to break the attractive forces that
hold them in the liquid state. The stronger these intermolecular forces are, the greater the amount of
energy needed to break them.
For some substances with weak intermolecular forces, the energy requirement is easy obtained from
collisions with other molecules and absorption of energy from the surroundings. Many molecules can
vaporize, resulting in a high vapor pressure. For molecules with strong intermolecular forces, gathering
enough energy may not be as easy, and register low vapor pressures.
52
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Guide the students to arrive at the following conclusion: The stronger the intermolecular forces of
attraction, the lower the vapor pressure of a liquid.
PY
•
Define molar heat of vaporization.
ED
The molar heat of vaporization (ΔHvap) is the energy required to vaporize 1 mole of a liquid at a
given temperature. H is the symbol for enthalpy, which means heat content at a given standard
condition.
ΔHvap (kJ/
mol)
Boiling Point*
(OC)
6.3
-186
D
EP
•
C
O
MOLAR HEAT OF VAPORIZATION AND BOILING POINT
The relationship between vapor pressure and strength of intermolecular forces is consistent with the
trends in two other properties of liquids, the enthalpy or molar heat of vaporization, and the boiling
point of the liquid.
Substance
Argon (Ar)
Pentane(C5H12)
26.5
36.1
Acetone (CH3COCH3)
30.3
56.5
Ethanol (C2H5OH)
39.3
78.3
Water (H2O)
40.79
100
*Measured at 1 atm
Table 4 . Molar heats of vaporization and boiling points of selected substances
53
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•
Relate molar heat of vaporization to strength of intermolecular forces
The heat of vaporization may be considered a measure of the strength of intermolecular forces in a
liquid. If the intermolecular attraction is strong, it takes a lot of energy to free the molecules from the
liquid phase and the heat of vaporization will be high.
PY
It is easier to vaporize acetone (lower Hvap) than water (higher Hvap) at a given temperature, and more
acetone escapes into the vapor phase at a given temperature. Acetone is a polar substance but has no
H-bonding. It has weaker intermolecular forces than water, and therefore acetone molecules are held
less tightly to one another in the liquid phase.
A practical way to demonstrate differences in the molar heat of vaporization is by rubbing acetone
on your hands. Compare what is felt when water is used. Acetone has a lower ΔHvap than water so that
heat from our hands is enough to increase the kinetic energy of the these molecules and provide
additional heat to vaporize them. As a result of the loss of heat from the skin, our hands feel cool.
•
Define boiling point.
ED
C
O
•
The boiling point of a liquid is the temperature at which the liquid converts into a gas. A more
complete definition includes the vapor pressure, and this is given below.
D
EP
A liquid boils when its vapor pressure equals the pressure acting on the surface of the liquid. The boiling
point is the temperature at which the vapor pressure of a liquid is equal to the external pressure.
The normal boiling point is the temperature at which the liquid converts to a gas when the external
pressure is 1 atm. The normal boiling point of water is 100oC.
The boiling point of a liquid depends on the external pressure. For example, at 1 atm, water boils at
100OC, but if the pressure is reduced to 0.5 atm, water boils at only 82 OC.
54
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•
Ext Pressure: 1 atm
PY
VP of water: 1 atm
C
O
BP of water: 100oC
Ext Pressure: 0.5 atm
VP of water: 0.5 atm
D
EP
ED
BP of water: 82oC
Figure 13 Vapor pressure of four common liquids, shown as a function of temperature
(Image Source: http://wps.prenhall.com/wps/media/objects/3311/3391416/blb1105.html)
•
Relate boiling point to molar heat of vaporization.
The boiling point is related to molar heat of vaporization: the higher ΔHvap , the higher the b o i l i n g
point, as shown in the table.
55
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Emphasize at this point the relationship
among vapor pressure, boiling point and
molar mass of vaporization. Focus on the
energy requirement in the process.
The boiling points of substances often reflect the strength of the intermolecular forces operating among
the molecules. At the BP, enough energy must be supplied to overcome the attractive forces among
molecules before they can enter the vapor phase.
ΔHvap (kJ/mol)
Argon (Ar)
-186
6.3
Benzene (C6H6)
80.1
31.0
Diethyl ether (C2H5OC2H5)
34.6
26.0
Ethanol (C2H5OH)
78.3
Mercury (Hg)
357
Methane (CH4)
-164
Water (H2O)
100
*Measured at 1 atm
C
O
PY
Boiling Point* (OC)
39.3
59.0
9.2
40.79
ED
Substance
Table 4. Boiling points and molar heats of vaporization of selected substances.
It takes more energy
(Hvap) to separate the
molecules of substance A
than molecules of
substance B
D
EP
Molecules of A are held
together by stronger
intermolecular forces than
molecules of B
Boiling point
of A is higher
than that of B
WATER: A VERY UNUSUAL LIQUID
Water is an essential substance to life. It is the most abundant compound on earth, and comprises about
more than 60% of the human body. But it is also one of the most unusual substances on earth.
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Challenge to learners:
•
Can the learners identify substances other than water that can exist naturally in all three states?
Can the learners identify some unusual properties of water?
Introduce the topic of water with the following short activities or demonstrations.
Activity 1
1. Fill a small glass jar all the way to the top with water.
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Let the learners explain their observations using the knowledge they have gained from the earlier part of this
lesson about the structure of water and intermolecular forces that operate between water molecules.
2. What do you think would happen if you were to add twenty-five centavo coins to it?
3. Try adding coins one at a time. What happens to the water in the cup?
Activity 2
ED
4. How many coins can you add without causing the water to overflow?
1. Take some water with a straw and put a few drops on plastic sheet.
(a) What is the shape of the drop?
D
EP
(b) Move a drop around with your straw. Does the drop change?
2. Move one of the drops close to another one with your straw. What happens when two drops meet?
3. Put a small amount of one of the solids (salt, pepper, sugar, talcum powder) on one of the drops. Does
the shape change?
4. Try this again with the other solids.
Activity 3
1. Put some water in your cup.
2. Sprinkle black pepper all over the surface. What does the pepper do? Record your observations.
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Have all the materials ready so that the
activities can be done in a very short
period of time.
3. Add a drop of dish soap to the water. What does the pepper do? Record your observations.
The Unique Properties of Water
•
Water is a good solvent.
Relate this property to the role of water in plant nutrition:
Relate this property to issues of pollution:
ED
Plants are able to absorb nutrient ions dissolved in water.
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A unique property of water is its ability to dissolve a large variety of chemical substances. It
dissolves
salts and other ionic compounds, as well as polar covalent compounds such as a l c o h o l s a n d o r g a n i c
substances that are capable of forming hydrogen bonds with water. Gases like oxygen and carbon dioxide will
dissolve in water meaning that some animals do not need to breathe air in order to respire but they must still be
able to absorb oxygen and excrete carbon dioxide. Water is sometimes called the universal solvent because it
can dissolve so many things.
•
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Issues can be caused however by the ease of which pollutants from farming and industrial plants are
dissolved.
Water has a high specific heat.
Specific heat is the amount of heat or energy needed to raise the temperature of one gram of a
substance by 1oC. The specific heat of water is 1 calorie/g-oC (4.18 J/g-oC), one of the highest for many liquids.
Water can absorb a large amount of heat even if its temperature rises only slightly. To raise the
temperature of water, the intermolecular hydrogen bonds should break. The converse is also true; water can
give off much heat with only a slight decrease in its temperature. This allows large bodies of water to help
moderate the temperature on earth.
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•
Further notes for the teacher: http://
www.britishecologicalsociety.org/wpcontent/uploads/Teachers-notes_Waterchemistry.pdf
Relate this property to changing climate and the capacity of bodies of water to act as
temperature buffer:
Further reference: http://
science.opposingviews.com/large-bodieswater-affect-climate-coastal-areas-22337.html
PY
In summer months this means that water must absorb a great deal of energy in the form of heat from the
sun in order for the temperature to increase. Since most bodies of water are large enough not to be significantly
affected by the heat from the sun, water provides an almost constant temperature for the plants and animals
living there.
•
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It takes about 4.5 times greater amount of energy to heat up water than an equal amount of land.
Hence, large bodies of water heat up and cool down more slowly than adjacent land masses.
The boiling point of water unusually high.
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In the plot on the right, the
broken lines direct one to the
estimated boiling points of HF,
H2O and NH3 if H-bonding was
not present in these three
substances.
ED
Many compounds similar in mass to water have much lower boiling points. The strong
intermolecular forces in water allow it to be a
liquid at a large range of temperatures.
(Image Source: http://
www.reasons.org/Media/
Default/Article/articles/waterdesigned-for-life-part-2-of-7/
part2-2.png)
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Figure 14. Boiling points of Group 14-17
hydrides. The dotted lines direct to the
boiling points of H2O, HF, and NH3 in the
absence of H-bonding.
Relate this property to questions on small water bodies drying up:
Small water bodies like ponds are at risk of drying up in the summer. But since the amount of energy
required to vaporize or evaporate water is so high, this is not expected to happen quickly.
•
Solid water is less dense, and in fact floats on liquid water.
ED
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Unlike all other liquids, the molecules in solid water are actually farther apart than they are in liquid water.
When solid water forms, the hydrogen bonds result in a very open structure with unoccupied spaces, causing
the solid to occupy a larger volume than the liquid. This makes ice less dense than liquid water, causing ice to
float on water.
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Figure 15. The structure of ice. (Image Source: http://media.wiley.com/Lux/
35/168035.image2.jpg)
Relate this property to the survival of aquatic organisms in temperate countries:
Water bodies freeze from the top down. If ice is not able to float, the water bodies would freeze from top
to bottom, and aquatic life will be killed. Because ice floats, aquatic organisms survive under the surface,
which remain liquid. The ice surface also acts as an insulating layer protecting the water beneath from further
freezing, and maintains a temperature adequate for survival. Without this feature, there would be no aquatic life
in temperate and Polar Regions.
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PRACTICE (10 MINS)
Teacher Tips
At 50. ° C the vapor pressure of ethanol is 0.30 atm, acetic acid is 0.08 atm, water is 0.12 atm, and acetone is
0.84 atm.
B. Arrange these substances in order of increasing boiling point temperature.
C. Arrange these substances in order of increasing intermolecular forces.
Expected answer:
•
Why is it necessary for birds to have a
natural coat of wax on their feathers?
•
Why is the application of wax in surfaces
considered as a protection of the
surface?
•
How does the use of soap and
detergents affect the surface tension of
water?
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A. Lowest Rate - acetic acid, water, ethanol, acetone - Highest Rate
PY
A. Arrange these substances in order of increasing rates of evaporation.
Suggested Reading:
B. Lowest Boiling Point - acetone, ethanol, water, acetic acid - Highest Boiling Point
C. Lowest IMF - acetone, ethanol, water, acetic acid – Highest IMF
ENRICHMENT (10 MINS)
ED
Ask the students to share a final inference about the meaning of their images/objects and how they relate to the
larger concept(s).
EVALUATION
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What is the importance/significance/relevance of the object? (industry, school, community, life, health, business,
etc.)
Have the students research areas in the Philippines where faulting and/or folding is present. The students should
submit a short written report identifying the kind of deformation and describing how the deformation has
contributed to the topography of the area.
1. For each of these types of compounds:
a. Predict which will be polar or non-polar
b. List the type or types of intermolecular force(s) present.
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Teacher Tip
Limit the responses here to those images and
objects not yet discussed.
H
C
C
H
O
H
H
H
H
H
H
H
H
Alkane
H
H
C
C
C
H
C
H
H
O
C
C
H
H
H
Ketone
H
H
H
Alcohol
C
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Answers: Alkanes: nonpolar; dispersion
Ketones: polar; dispersion, dipole-dipole
C
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H
Alcohols: polar; dispersion, dipole-dipole, hydrogen bonding
ED
2. Within each group (alkane, ketone, alcohol) in the table given at the end, how does the boiling point
change as the formula mass of that type of compound increases?
Answer: The boiling point increases as the formula mass(FM) of the alkane, ketone, or alcohol
increases.
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3. In terms of intermolecular forces, explain why we see this general trend with formula masses.
Answer: The boiling point increases as the FM increases because the molecule has more protons and
electrons, therefore the dispersion forces increase (induced-dipole – induced dipole interactions).
4. Find an alkane, a ketone, and an alcohol with roughly the almost the same formula mass (within 5g/mol).
Rank these compounds in terms of their relative boiling points.
Answer: Butane, acetone, and 1-propanol all have FM = 60 ±2 g/mole.
The boiling points increase as follows: alkane < ketone < alcohol.
62
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Pentane, 2-butanone, and 1-butanol all have FM = 72±2 g/mole.
The boiling points increase as follows: alkane < ketone < alcohol.
Hexane, 2-pentanone, and 1-pentanol all have FM = 87±2 g/mole.
The boiling points increase as follows: alkane < ketone <alcohol.
PY
5. In terms of intermolecular forces, explain WHY we see this general trend in boiling points, for
compounds of equivalent formula mass. How does the presence of a dipole moment affect the strength
of intermolecular interactions?
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Answer: Since the three compounds have about the same FM, the dispersion forces are about the
same.
ED
Alkanes have only dispersion forces and so fairly weak intermolecular forces, whereas the ketone and
the alcohol have dipole-dipole forces. Only the alcohol has hydrogen bonding — which increases the
intermolecular forces substantially. A dipole moment tends to increase the strength of the intermolecular
forces; alkanes are nonpolar and have the lowest boiling points (for a given FM).
6. Rank these forces in terms of their relative strengths: hydrogen bonding; dipole-dipole; induced dipoleinduced dipole (dispersion), and covalent bonds.
dipole
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Answer: Covalent bonds >>> hydrogen bonding > dipole-dipole > induced dipole-induced
(dispersion)
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ED
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Table 5. Formula Masses and Boiling Points
for Selected Compounds.
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Chemistry 2
60 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
PY
Intermolecular Forces of Liquids
and Solids; Solids and their
Properties
5
Examples and Applications
10
Content Standard
Instruction
The learners demonstrate an understanding of the properties of liquids, and the nature of
Practice
forces between particles.
Lecture/Discussion
25
Test
5
Enrichment
Hands-on Activity
10
Evaluation
Reading Commentary and Observation
5
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Motivation
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or
freezing when a solid is dissolved in water.
Find patterns in the arrangement of particles in a solid and relate them to its properties.
Conduct investigations on the properties of solids using solid samples.
ED
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Describe the different types of crystals and their properties: ionic, covalent, molecular,
and metallic. (STEM_GC11IMF-IIIa-c-105)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
•
Actual sample of solids; magnifying glasses; computer; worksheets
Resources
Learning Competencies
Describe the difference in structure of crystalline and amorphous solids.
(STEM_GC11IMF-IIIa-c-104); and
Materials
(1) Brown, T. L. et al. (2009). Chemistry: The Central Science (11th ed., pp.
460-511). Pearson Education Inc.
(2) Brown, T. L. et al. (2009). Chemistry: The Central Science (11th ed., pp.
460-511). Pearson Education Inc.
(3) Chang, R. (2007). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
(4) Helmenstine, A. (2011). How to Grow Alum Crystals. Retrieved from
https://www.youtube.com/watch?v=sdYS-3J85Pw
(5) McGraw Hill Education,. Retrieved from http://www.mhhe.com/physsci/
chemistry/essentialchemistry/flash/vaporv3.swf
compare the properties of crystalline and amorphous solids;
classify crystals according to the attractive forces between the component atoms, (6) Thechemprofessor’s channel (2009). Lattice Structures Part 1. Retrieved
from https://www.youtube.com/watch?v=Rm-i1c7zr6Q
molecules, or ions (molecular crystals, covalent-network crystals, ionic crystals, and
(7) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
metallic crystals);
Thomson Brooks/Cole.
•
relate the properties of different types of solids to the bonding or interactions among (8) Zumdahl, S., & Zumdahl, S. (2000). Chemistry. Boston: Houghton
Mifflin.
particles in these solids; and
•
predict the strongest force responsible for the formation of a given solid.
65
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Compare the properties of crystalline and amorphous solids;
Teacher Tips
•
•
PY
b. Classify crystals according to the attractive forces among component atoms, molecules, or ions
(molecular crystals, covalent-network crystals, ionic crystals, and metallic crystals);
The different types of crystals: ionic,
covalent, molecular, and metallic are
also referred to as “types of solids.”
We will use the term “crystals” in this
discussion to distinguish from the
general types crystalline and
amorphous solids.
c. Relate the properties of different types of solids to the bonding or interactions among particles in
these solids;
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d. Predict the strongest force responsible for the formation of a solid.
2. Present relevant vocabulary that will be used in the lesson and learners should know.
Fluid
Crystal or crystalline solid
ED
A gas or a liquid; a substance that can flow.
Ion
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A solid material whose components, such as atoms, molecules or ions, are arranged in a highly
ordered microscopic structure.
An atom or group of atoms that has a net positive or negative charge.
Ionic crystal
A solid that consists of positively and negatively charged ions held together by electrostatic
forces.
Electrostatic bonding
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•
More key words will be added to
relevant vocabulary as the need arises.
The attraction between oppositely charged ions in a chemical compound.
Ionic bond
The electrostatic force that holds ions together in an ionic compound.
PY
Network solid or covalent network crystal
A solid that may be a chemical compound (or element) in which atoms are bonded
by covalent bonds in a continuous network extending throughout the material.
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Molecular crystal
A solid composed of molecules held together by van der Waals forces (dispersion force, dipoledipole attraction, hydrogen bonding).
Covalent bond
Dispersion forces
ED
A bond in which one or more pairs of electrons are shared by two atoms.
Dipole-dipole forces
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Interactions that are the result from temporary dipole moments induced in ordinarily nonpolar
molecules.
Attractive forces between polar molecules (molecules that possess permanent dipole
moments).
Hydrogen bond
A special type of dipole-dipole interaction between the hydrogen atom in a polar bond such as N‒H,
O‒H, or F‒H, and any of the electronegative atoms O, N, or F.
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Crystal lattice
The regular repeating structure of a crystalline solid.
Unit cell
PY
The smallest subunit of a crystal lattice that can be repeated over and over to make the entire crystal.
3. Connect the lesson with previous knowledge required. Ask the students the following questions to recall
relevant concepts:
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Recall previous concepts on Kinetic Molecular Theory and intermolecular forces in liquids and solids.
What is a solid?
How is a solid described in terms of the Kinetic Molecular Theory?
b. Distance among particles
c. Arrangement/order of particles
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d. Attractive forces between particles
ED
a. Average kinetic energy
Describe the properties of a solid as a result of the behavior of its particles:
a. Volume/Shape
b. Density
c. Compressibility
d. Motion of molecules
e. Rate of diffusion
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MOTIVATION (10 MINS)
Teacher Tip
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Bring and Show
The teacher brings pictures or actual samples of amorphous and crystalline solids to show to the class. The class
identifies properties or features common to amorphous solids, features common to crystalline solids, and
properties which can distinguish amorphous from crystalline solids. The class can be guided to look at the
rigidity of the solids, behavior on heating, and structural patterns.
INSTRUCTION (25 MINS)
•
Notes
•
A gemstone may be a pure chemical
element (diamond is essentially pure
carbon), a relatively simple chemical
compound (quartz is silicon dioxide,
SiO2), or a more complex mixture of
various compounds and elements (the
garnet family includes a highly variable
mix of iron, magnesium, aluminum, and
calcium silicates). The great majority of
familiar gem materials are oxides or
silicates (e.g., they contain oxygen and
perhaps silicon) and formed as crystals
during the cooling of the earth's crust
over past millennia.
•
Gemstones may be formed in single or
multiple discrete crystals (such as
diamond), in massive collections of
microscopic crystals/ cryptocrystalline
(such
as
chalcedony),or
in amorphous (non-crystalline) masses
(such as opal).
In general, larger crystals were formed in
areas of slow cooling for molten rock,
and smaller crystals in areas of more
rapid cooling. There are several classes
of crystal structure based on symmetry of
the resulting crystals, and there are also
non-crystalline (amorphous) minerals
used as gem materials.
h t t p : / / w w w. t r a d e s h o p . c o m / g e m s /
classify.html
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Focus Questions:
A. What are the two general types of solids? What features can be used to distinguish a
crystalline solid from an amorphous solid?
B. What is the distinguishing feature of crystalline solids? How are the structures of crystals
determined?
ED
C. What are the four types of crystals? What form of unit particles makes up each type of crystal? What
forces bind the unit particles of each type of crystal? What are the properties of each type of crystal?
Lecture Discussion
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A. What are the two general types of solids? What features can be used to distinguish a crystalline
solid from an amorphous solid?
Solids can be categorized into two groups: the crystalline solids and the amorphous solids. The
differences in properties of these two groups of solids arise from the presence or absence of long range order of
arrangements of the particles in the solid.
•
1. Arrangement of particles
The components of a solid can be arranged in two general ways: they can form a regular repeating
three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate
69
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Suggested materials to be brought by
the teacher:
e.g. crystalline solids: salt, sugar,
gemstones, alum (tawas); amorphous
solids: rubber band, plastic,
chocolate bar, glass
•
with no particular long range order,
“shapeless”).
and form an amorphous solid (from the Greek ámorphos, meaning
PY
Crystalline solids are arranged in fixed geometric patterns or lattices. Examples of crystalline solids are
ice and sodium chloride (NaCl), copper sulfate (CuSO4), diamond, graphite, and sugar (C12H22O11). The ordered
arrangement of their units maximizes the space they occupy and are essentially incompressible.
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Amorphous solids have a random orientation of particles. Examples of amorphous solids are glass,
plastic, coal, and rubber. They are considered super-cooled liquids where molecules are arranged in a random
manner similar to the liquid state.
do not have long range order, but may have a limited,
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Amorphous solids (e.g. glass), like liquids,
localized order in their structures.
ED
More than 90% of naturally occurring and artificially prepared solids are crystalline. Minerals, sand, clay,
limestone, metals, alloys, carbon (diamond and graphite), salts (e.g. NaCl and MgSO4), all have crystalline
structures. They have structures formed by repeating three dimensional patterns of atoms, ions, or molecules.
The repetition of structural units of the substance over long atomic distances is referred to as long-range order.
Figure 1. Crystalline and amorphous quartz
70
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Image Sources:
1. http://f.tqn.com/y/chemistry/1/S/a/d/
quartz.jpg
2. https://upload.wikimedia.org/wikipedia/
commons/7/71/SilicaGel.jpg
2. Behavior when heated
•
The presence or absence of long-range order in the structure of solids results in a
difference in the behavior of the solid when heated.
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ED
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The structures of crystalline solids are built from repeating units called crystal lattices.
The surroundings of particles in the structure are uniform, and the attractive forces
experienced by the particles are of similar types and strength. These attractive forces
are broken by the same amount of energy, and thus, crystals become liquids at a
specific temperature (i.e. the melting point). At this temperature, physical properties of
the crystalline solids change sharply.
Figure 2. Examples of crystalline solids (Image Source: http://www.brainfuse.com/
quizUpload/c_83128/crystalline1.GIF)
Amorphous solids soften gradually when they are heated. They tend to melt over a
wide range of temperature. This behavior is a result of the variation in the arrangement
of particles in their structures, causing some parts of the solid to melt ahead of other
parts.
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Figure 3. Examples of noncrystalline solids
Image Sources:
PY
Rubber bands
Glass paper weights
Plastic lunch boxes
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Charcoal
http://1.imimg.com/data/P/9/MY-979264/activated-charcoal_10714160_250x250.jpg
ED
https://upload.wikimedia.org/wikipedia/commons/9/96/Glass_paperweight.jpg
http://healthychild.org/assets/esphoto_plastics_tupperware1-504x334.jpg
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Discussion on the Properties of Liquid and Intermolecular Forces
The properties of liquids that were observed are consequences of the interactions of particles that
make up the liquid.
•
B. What is the distinguishing feature of crystalline solids? How are the structures of crystals
determined?
The Crystal Lattice
•
Crystalline solids are characterized by a regular repeating structure called the crystal lattice.
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There are two main methods for
obtaining salt, evaporation from sea
water and mining salt from the earth.
Most common table salts and salts used
for industrial purposes are obtained
through mining, while specialty
or gourmet salts are still produced via
evaporation of sea water.
http://foodreference.about.com/od/
Ingredients_Basics/a/How-Is-SaltMade.htm
Let the students watch a video clip on growing crystals. This will help visualize how some
crystals are formed. https://www.youtube.com/watch?v=sdYS-3J85Pw .
•
Ask the students if they know how the examples given are formed as you show them, and what
conditions are involved in the formation of these crystals.
Salt crystals
(Image Source: http://
www.gout.com/stages-of-gout)
Amethyst cluster
(Image Source: http://
www.soapgoods.com/images/Brown
(Image Source: https://
en.wikipedia.org/wiki/Amethyst)
ED
(Image Source: http://
www.saltworks.us/salt_info/
si_WhatIsSalt.asp)
Brown sugar crystals
%20Sugar%20-%20Raw%20Demerara
%20Crystals.jpg)
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Salt crystals
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X-Ray Analysis of Solids
Snow fall
(Image Source: https://
n1accord.wordpress.com/
category/natnat/page/2/)
•
•
•
•
•
•
Crystallization in the laboratory
(Image Source: http://
www.reciprocalnet.org/edumodules/
crystallization/)
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•
Cane stalks are shredded and squeezed
to extract its natural juice, which is boiled
until it thickens and molassesrich sugar crystals begin to settle. The
molasses-rich crystals are sent to a
rapidly spinning centrifuge to remove
molasses and leave pure, naturally
white sugar crystals. The sugar crystals
are then dried.
http://www.sugar.org/how-we-get-sugar/
Amethyst is formed in silica-rich liquids
deposited in gas cavities of lava that
occur in crystalline masses. Such cavities
occur in the earth's crust for several
reasons such as gas bubbling in circular
cavities or filling of veins.
academic.emporia.edu/abersusa/go336/
has/
Uric acid is the byproduct of protein
digestion, and among healthy individuals,
it is removed from the blood stream and
excreted by the kidneys. Excess uric acid
is deposited in the joints in crystal form
and creates a painful arthritic condition
known as gout.
http://www.livestrong.com/article/31308uric-acid-formed/
Snow is formed when temperatures are
low and there is moisture - in the form of
tiny ice crystals - in the atmosphere.
When these tiny ice crystals collide they
stick together in clouds to become
snowflakes. If enough ice crystals stick
together, they'll become heavy enough
to fall to the ground.
www.metoffice.gov.uk/learning/snow/
how-is-snow-formed
X-ray Diffraction is a technique used to determine the atomic and molecular structure of a crystal,
wherein atoms cause a beams of incident X-rays to diffract into many specific directions.
ED
C
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PY
A stream of X-rays directed at a crystal diffracts and scatters as it encounters atoms. The scattered rays
interfere with each other and produce a pattern of spots of different intensities that can be recorded on
film, such as that shown in the figure below. X-ray diffraction has provided much of our knowledge about
crystal structure. Below is an image of a diffraction pattern produced by an 8 keV electron beam incident
on a graphite crystal.
Figure 5. An X-ray diffraction pattern of a graphite crystal. (Image Source:
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https://quantumfrontiers.files.wordpress.com/2012/11/graphite2.gif)
C. What are the four types of crystals? What form of unit particles makes up each type of crystal?
What forces bind the unit particles of each type of crystal? What are the properties of each type
of crystal?
The four types of crystals differ in the kind of particles that make up the crystal and the attractive forces
that hold these particles together.
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•
•
Crystallization refers to the formation of
solid crystals from a homogeneous
solution. It is essentially a solid-liquid
separation technique and a very
important one at that.
h t t p : / / w w w. re c i p ro c a l n e t . o rg /
edumodules/crystallization/index.html
1. METALLIC CRYSTALS
Ask the learners to enumerate the properties of metals. Guide the learners on how to make inferences
about the arrangement of atoms in the metallic crystal that are consistent with the properties they listed. The
table below gives a sample of the output of the activity.
C
O
PY
Metallic crystals are made of atoms that readily lose electrons to form positive ions (cations), but no
atoms in the crystal would readily gain electrons. The metal atoms give up their electrons to the whole crystal,
creating a structure made up of an orderly arrangement of cations surrounded by delocalized electrons that
move around the crystal. The crystal is held together by electrostatic interactions between the cations and
delocalized electron. These interactions are called metallic bonds. This model of metallic bonding is called the
“sea of electrons” model.
Inference about the structure
Dense
Atoms are packed close together.
High melting point
Strong attractive forces hold the atoms in the crystal.
Good electrical
conductor
Charged particles move through the crystal.
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ED
Observed property
Good heat conductor
Particles can move through the crystal.
Malleable and ductile
When the crystal is deformed or stress is applied, the attractive
forces are not broken.
Lustrous
Light is easily absorbed and emitted back.
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PY
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Figure 6. Positive ions surrounded by delocalized electrons (Image source:
http://wps.prenhall.com/wps/media/objects/3311/3391416/imag1108/AAAUBAF0.JPG)
Explanation of properties:
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This model is able to explain many physical properties of metals, such as their high melting
points, malleability, ductility, thermal and electrical conductivity, and luster.
High melting point – a large amount of energy is needed to melt the crystal since the forces of
attraction to be broken are numerous and extend throughout the crystal.
•
Dense – atoms are packed closely together. Metals exhibit close-packing structures, a most
economical way by which atoms utilize space.
•
Electrical conductivity – then delocalized electrons move throughout the crystal.
•
Thermal or heat conductor – the delocalized electrons collide with each other as they move
through the crystal, and it is through these collisions that kinetic energy is transferred .
•
Malleability/ductility – when stress is applied to the metal, the metal cations shift in position,
but the mobile electrons simply follow the movement of the cations. The attractive forces
between cations and mobile electrons are not broken.
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•
Luster – the motion and collisions of electrons allow it to gain and lose energy, some of these
in the form of emitted light that is observed as luster.
2. IONIC CRYSTALS
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Ask the learners to enumerate the properties of ionic compounds Guide the learners on how to make
inferences about the arrangement of particles in the ionic crystal that are consistent with the properties they
listed. The table below gives a sample of the output of the activity.
Inference about the structure
Hard
Strong attractive forces hold the crystal together.
High melting point
Strong attractive forces have to be broken to melt the crystal
Poor electrical
conductor in the solid
state
No charged particles move through the crystal
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Observed property
Brittle
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Good electrical
Mobile charged particles are present in the molten state
conductor in the molten
state
Deformation or shift of particles cause attractive forces to be
broken.
Ionic crystals are made of ions (cations and anions). These ions form strong electrostatic interactions
that hold the crystal lattice together. The electrostatic attractions are numerous and extend throughout the
crystal since each ion is surrounded by several ions of opposite charge, making ionic crystals hard and of high
melting points. The figure below shows a model of NaCl crystal, where one Na+ ion is surrounded by six Clions, and a Cl- ion is likewise surrounded by six Na+ ions.
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PY
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Figure 7 . Sodium ion, Na+ and chloride ion, Cl- at lattice points of NaCl crystal (Image
Source: http://chemwiki.ucdavis.edu/Wikitexts/Howard_University/General_Chemistry)
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The energy needed to break the crystal of ionic substances will depend on the magnitude of charges on
the ions (the 2+ and 2- ions attract each other stronger in MgO than 1+ and 1- in NaCl), and the sizes of the ions
(attractions are less between the bigger ions in RbI and as such less heat energy is needed to separate them
than the smaller ions in NaCl).
Ionic substances can conduct electricity in the liquid or molten state or when dissolved in water,
indicating that in these states, charged particles are able to move and carry electricity. However, the solid state
is generally nonconducting since the ions are in fixed positions in the crystal lattice and are unable to move from
one point to another.
Ionic crystals are brittle, and would shatter into small pieces when deformed or when pressure is applied
on the crystal. The shifting of ions cause repulsions between particles of like charges.
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PY
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3. MOLECULAR CRYSTALS
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Figure 8. Shifting of ions cause repulsions in ionic crystal (Image Source: https://
encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcRLoXXkQsqpNSBX3vTHxDaQxfuKUborNlufdFB2Vaoo882KCRe)
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Similar to the starting activity in the discussion of the two earlier types of crystals, ask the students to list
the properties of molecular crystals, and infer from these the arrangements and attractive forces that hold the
particles in the crystal.
Molecular crystals are made of atoms, such as in noble gases, or molecules, such as in sugar, C12H22O11,
iodine, I2, and naphthalene, C10H8. The atoms or molecules are held together by a mix of hydrogen bonding/
dipole-dipole and dispersion forces, and these are the attractive forces that are broken when the crystal melts..
Hence, most molecular crystals have relatively low melting points.
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Inference about the structure
Soft
Weak attractive forces hold the crystal together.
Low melting point
Weak attractive forces are broken when crystals melt
Poor electrical
conductor in the solid
and molten states
No charged particles move through the crystal
Poor heat conductor
No particles can move easily throughout the crystal.
Brittle
Deformation or shift of particles cause attractive forces to be
broken.
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Observed property
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The valence electrons of molecular substances are used in bonding, and cannot move about the crystal
structure. Hence, the crystals are nonconducting. The absence of any mobile particles make molecular crystals
unable to transmit heat fast. The crystals are brittle because the attractive forces that hold the molecules in the
crystal are highly directional and a shift in positions of the molecules would break them.
Figure 9. Arrangement of water molecules in ice crystal (Image Source: http://
chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/)
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4. COVALENT NETWORK CRYSTALS
Using diamond and silicon dioxide as examples, ask the learners to list properties of covalent network
crystals, and infer from these the structure of the crystal.
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Covalent network crystals are made of atoms in which each atom is covalently bonded to its nearest
neighbors. The atoms can be made of one type of atom (e.g. Cdiamond and Cgraphite) or can be made of different
atoms (e.g. SiO2 and BN). In a network solid, there are no individual molecules and the entire crystal may be
considered one very large molecule. Formulas for network solids, like those for ionic compounds, are simple
ratios of the component atoms represented by a formula unit.
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The valence electrons of the atoms in the crystal are all used to form covalent bonds. Because there are
no delocalized electrons, covalent network solids do not conduct electricity. Covalent bonds are the only type
of attractive force between atoms in the network solid. Rearranging or breaking of covalent bonds requires large
amounts of energy; therefore, covalent network solids have high melting points. Covalent bonds are extremely
strong, so covalent network solids are very hard. Generally, these solids are insoluble in water due to the
difficulty of solvating very large molecules. Diamond is the hardest material known, while cubic boron nitride
(BN) is the second-hardest. Silicon carbide (SiC) is very structurally complex and has at least 70 crystalline forms.
Inference about the structure
Hard
Strong attractive forces hold the crystal together.
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Observed property
Very high melting point
Strong attractive forces have to be broken in order to melt
crystals
Poor electrical
conductor in the solid
and molten states
No charged particles move through the crystal
Poor heat conductor
No particles can move easily throughout the crystal.
Brittle
Deformation or shift of particles cause attractive forces to be
broken.
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Diamonds are an example of a covalent network solid in which atoms are covalently bonded with each
other. They tend to be hard and have high melting points.
Silicon dioxide, SiO2 is an example of a covalent network solid in which atoms are covalently bonded to
each other. Notice that each silicon atom is bridged to its neighbors by an oxygen atom.
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Graphite, an allotrope of carbon, differs in properties from other network solids. It is soft and is used as a
solid lubricant. It is also a good conductor of electricity, indicating the presence of charged particles that move
through the crystal.
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Show illustrations of the structure of diamond and graphite. Ask the students to spot any difference in
the bonding behavior of the carbon atoms in the two forms. They should notice that in graphite, each carbon
atom is bonded to only three other carbon atoms, while in diamond, each carbon atom is bonded to four
others. In addition, graphite is made up of layers of rings of carbon atoms. The broken lines connecting the
layers are weak dispersion forces.
Figure 10 . Two allotropes of carbon: graphite and diamond (Image Source: http://
www.designanduniverse.com/articles/images/diamond/atomic_order%20.jpg)
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Each carbon atom has four valence electrons, making it capable of forming four single covalent bonds
with other atoms, like in diamond. In graphite, only three of these four valence electrons are used for bonding,
leaving the fourth electron free. Every carbon atom in graphite has an extra electron that can move about the
layer, allowing graphite to conduct electricity.
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The layers in graphite are held by weak intermolecular forces, and with sufficient pressure, the layers can
slide past one another. When one uses a pencil to write, layers of graphite are transferred to the paper as one
presses the pencil down on the paper.
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Table 1: Comparison of Types of Solids (Image Source: http://wpscms.pearsoncmg.com/
wps/media/objects/3662/3750037/Aus_content_10/Table10-07.jpg)
The previous table presents a summary of component particles, forces involved in the crystal formed, the
general properties of each type of crystal, and some examples.
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PRACTICE (5 MINS)
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Indicate the strongest force holding the crystals together in the following substances by putting a check on the
appropriate box. The first five substances were done for you.
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Expected Answers:
Teacher Tips
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ENRICHMENT (10 MINS)
Observing and Comparing Solids
This is a short activity that can be done in approximately 10 minutes.
Materials needed: Magnifying glasses, sugar, salt, pepper
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Worksheets should be prepared to
include the Practice Test and the
Enrichment Activity.
A sample drill could be done for the first
5 items to remind students of the
different forces and bonds.
Students will observe granules of sugar, salt, and pepper under a magnifying glass, and observe the similarities
and differences between these common materials.
A pinch of each material should be taken from its container and placed on a piece of paper or board paper.
Using different magnifying glasses with different powers, they will examine the grains.
PY
Guide questions for students:
Can you see the difference between the sugar and salt granules? Can you see the different angles?
•
Are there angles in the pepper? Does it look crystalline?
Student Worksheet Questions:
1. Which of the substances appears to be made of crystals?
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2. Defend your answer with evidence from your observation. What did you use to help you decide which
are crystals, and which are not?
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Expected answers:
1. Sugar and salt look like crystals. Pepper doesn’t look like a crystal.
EVALUATION (5 MINS)
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2. The crystals are more regular in shape. Pepper is irregular and all the grains look different.
Higher-Order Thinking Question:
A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its
melting point and found that the material melted over a 12°C range. After it had cooled, she measured the
melting point of the same sample again and found that this time the solid had a sharp melting point at the
temperature that is characteristic of the desired product. Why were the two melting points different? What was
responsible for the change in the melting point?
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Teacher Tips:
•
Different magnifying glasses have
different powers, so have students share.
•
Pepper is ground and not grown from a
liquid.
•
Remind the students that crystals are
very common. They exist in minerals, in
food, in bones and teeth, and elsewhere.
Some of the easiest to see are in sugar
and salt.
•
Guide students as they fill in their
worksheet by asking questions and
engaging in discussion. Point out that
with careful observation and comparison,
they should be able to arrive at some
conclusions.
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Further Evaluation/Enrichment
Teacher Tips:
•
Reading Commentary on Crystal Formation. Read about a crystal of your choice and write a four-six
paragraph essay by answering the following questions. Cite your sources following the APA format.
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1. How is the crystal formed?
2. Is it beneficial or harmful to man or both? Discuss how it is beneficial or harmful to man.
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3. If it is something harmful, what can be done to avoid its formation? If it is something beneficial, how can
its formation be promoted?
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Source: https://teaching.berkeley.edu/sites/teaching.berkeley.edu/files/Rubric%20for
%20Evaluating%20Written%20Assignments%20.pdf
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Suggested crystals for students to work
on:
Oxides of magnesium and calcium
produced from hard water
Calcium oxalate or calcium
phosphate in kidney stones
Uric acid in gout
Barium sulfate in x-ray imaging
Silica in desiccants
Gemstones used for jewelry
Salts in food industry, e.g. table salt,
monosodium glutamate, potassium
nitrate
Metals or alloys used in conductors
-
If time permits, selected number of students
may be asked to share their work to the class.
If not, make a summary of the students’ work
and post for the class to read in their free
time.
The students’ work will be rated whether it
covers the required elements; it presents
information accurately; uses information
creatively; and is evidence- based. A sample
rubric is provided and should be modified
accordingly to suit the teacher’s criteria and
weight for each criterion.
Chemistry 2
90 MINS
Intermolecular Forces of Liquids
and Solids; Phase Changes
LESSON OUTLINE
Communicating learning objectives
5
Engagement and Applications
5
Lecture/Discussion
30
Practice
Problem Exercises
10
Enrichment
Laboratory Experiment
30
Evaluation
Problem Solving
10
Introduction
Phase changes resulting from changes in energy and forces between particles
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Performance Standards
The learners design a simple investigation to determine the effect on boiling point or
freezing when a solid is dissolved in water.
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Content Standards
Motivation
The learners demonstrate an understanding of the properties of liquids, and the nature of
Instruction
forces between particles.
Materials
Learning Competency
Describe the nature of the following phase changes in terms of energy change and the
increase or decrease in molecular order: solid-liquid, liquid vapor, and solid-vapor
Resources
(STEM_GC11IMF-IIIa-c-106)
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Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
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Quantitatively determine energy (heat) change that accompanies phase and temperature
changes.
•
describe the transitions among gas, liquids, and solids in terms of increase or
decrease in molecular order;
•
explain what is happening as a system is heated and relate phase changes to heat
and temperature changes;
•
explain solid-liquid, liquid vapor, and solid-vapor transitions in terms of amount of
energy change; and
•
calculate heat changes in phase and temperature changes.
Illustrations and diagrams; images and objects
(1) Chang, R. (2007). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
(2) Petrucci, R., Harwood, W., & Herring, F. (2007). General Chemistry:
Principles and Modern Applications (9th ed.). Upper Saddle River, NJ:
Pearson!
(3) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
Thomson Brooks/Cole.
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
PY
a. Describe the transitions among gas, liquids, and solids in terms of increase or decrease in molecular
order;
b. Explain what is happening as a system is heated;
c. Relate phase changes to heat and temperature changes;
d. Explain solid-liquid, liquid vapor, and solid-vapor transitions in terms of amount of energy change;
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e. Calculate heat changes in phase and temperature changes.
2. Present relevant vocabulary that will be used in the lesson and learners should know.
Fluid
Phase
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A gas or a liquid; a substance that can flow.
Solid
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A homogeneous part of a system in contact with other parts of the system, but separated by
well-defined boundaries.
A phase of matter with definite shape and volume.
Liquid
A phase of matter with definite volume but no definite shape.
Gas
A phase of matter with no definite shape or volume of its own.
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Teacher Tips
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•
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Most of these terms are recalled from
previous lessons.
Definitions should remain posted as the
lesson progresses.
More key words may be added to
relevant vocabulary as the need arises.
Intermolecular forces
Intermolecular forces are attractive forces between molecules.
Phase changes
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Transformations of matter from one phase to another.
Melting
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A phase change from solid to liquid.
Vaporization
A phase change from liquid to gas.
A phase change from solid to gas.
Condensation
Freezing
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A phase change from gas to liquid.
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Sublimation
A phase change from liquid to solid.
Deposition
A phase change from gas to solid.
Exothermic process
Process that gives off or release heat to the surroundings.
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Endothermic process
Process that absorbs heat from the surroundings.
•
Specific heat of a substance
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The amount of heat needed to raise the temperature of 1 gram of a substance by 1 OC.
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3. Connect the lesson with previous knowledge required.
Recall the three phases of matter and its properties. Use the table comparing the properties of different
phases of matter derived from a previous discussion.
Molecular Behaviour
Volume/Shape
Density
Compressibility
Motion of Molecules
Focus questions:
Liquid
Solid
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Gas
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Properties
What makes a gas different from a liquid or solid?
Why are some substances gases at room temperature, while others are not?
How does the intermolecular force of attraction in a substance relate to its phase?
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It is best if the table filled from the
previous lesson can be used here.
Learner volunteer/s can be called to read
the contents of the table.
MOTIVATION (5 MINS)
Teacher Tip
If available, actual substances showing the
changes should be shown.
Glass of ice water
Solid iodine subliming in a
test tube
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Pan of boiling water
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Focus question:
What phase(s) of matter exist in the following images?
Figure 1: Examples of phase changes
Some answers:
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Ask the students to give other examples of phase changes that they have seen.
cooking oil solidifying during cold mornings, sublimation of dry ice, melting of candle wax.
INSTRUCTION (30 MINS)
Challenge to learners:
What causes the phase change in matter?
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Phase changes are transformations of matter from one physical state to another. They occur when energy
(usually in the form of heat) is added or removed from a substance. They are characterized by changes in
molecular order; molecules in the solid phase have the greatest order, while those in the gas phase have the
greatest randomness or disorder.
What changes in molecular order occur during phase changes?
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The figure below illustrates the difference in molecular order of a substance in the solid, liquid and gaseous
states.
Figure 2. Molecular order in solid, liquid and gas (Image Source: http://
media-2.web.britannica.com/eb-media/65/63065-004-07B69F7B.jpg)
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The next figure shown below summarizes the types of phase changes.
•
•
The reverse change from gas to liquid is condensation, gas to solid is deposition, and liquid
to solid is freezing. These changes give off heat (heat lost) and are exothermic processes.
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The change from solid to liquid is melting, liquid to gas is vaporization, and solid to gas is
sublimation. These changes take place when heat is absorbed (heat gained). They are
endothermic processes.
Figure 3: The different changes in state that matter undergoes (Image Source: http://
www.shmoop.com/matter-properties/test-your-knowledge.html)
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How does a change in energy affect phase changes?
Phase changes occur when heat is added or removed from a substance.
When a substance is heated, the added energy is used by the substance in either of two ways:
a. The added heat increases the kinetic energy of the particles and the particles move faster. The increase
in kinetic energy is accompanied by an increase in temperature.
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b. The added heat is used to break attractive forces between particles. There is no observed increase in
temperature when this happens. Often a change in the physical appearance of the substance is
observed, such as a phase change.
c.
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Conversely, the removal or release of heat results in two ways:
a. A decrease in kinetic energy of the particles. The motion of the particles slow down. A decrease in
temperature is observed.
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b. Forces of attraction are formed, and a phase change may occur. No change in temperature is observed.
Figure 4: Heating curve (Image Source: http://i46.tinypic.com/2rekv40.jpg)
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The change in temperature of a substance as it is being heated can be shown in a graph called the
heating curve, such as the figure shown in the previous page. The heating curve is a plot of
temperature and heat added to the substance. Often, time is used instead of heat added in the
abscissa, because it is assumed that heat is uniformly added per unit time.
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Figure 5. Cooling curve (Image Source: http://www.ausetute.com.au/images/coolcurv.gif)
In both the heating and cooling curves, there are certain portions where the temperature changes as heat is
being added or removed, and portions where the temperature remains constant even if heat is being added or
removed. What is happening at these portions?
1. When heat change is accompanied by a change in temperature, a change in kinetic energies of the
particles in the substance is occurring. The particles are either moving faster or slowing down.
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2. When temperature remains constant during heat change, the particles move at the same speed. The
heat added or removed is involved in breaking or forming attractive forces. A phase change occurs at
this temperature: solid melts or liquid freezes at the melting point,which is also the freezing point; liquid
boils, or gas condenses at the boiling point, which is also the condensation point.
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During phase changes, two physical states of the substance exist at the same time. When addition or removal of
heat is stopped at this temperature, the two physical states will interconvert from one state to the other, and will
be at equilibrium.
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MELTING AND FREEZING: SOLID- LIQUID EQUILIBRIUM
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When a solid is heated, its temperature increases until it reaches its melting point. At this temperature, the
average kinetic energy of the molecules has become sufficiently large to begin overcoming the intermolecular
forces that hold the molecules of a solid state together. The heat absorbed is used to break apart more and
more of the molecules in the solid. The transformation of solid to liquid is called melting, and the reverse
process is called freezing.
During the transition, the average kinetic energy of the molecules does not change, so the temperature
stays constant. The melting point of a solid or the freezing point of a liquid is the temperature at which
solid and liquid phases coexist in equilibrium.
•
Melting points are distinct for each substance. It is dependent on the strength of attractive forces that
hold the particles in the solid. The stronger the attractive forces that hold the particles in the solid, the
higher is the melting point of the substance.
•
The melting (or freezing) point of a substance when the external pressure is 1 atm pressure is called its
normal melting (or freezing) point. For water, this is 0oC.
•
At 0OC and 1 atm, the dynamic equilibrium for water and ice is represented by:
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ice ⇋ water
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A practical illustration of this dynamic equilibrium is provided by a glass of ice water. As the ice cube melts to
form water, some of the water between ice cubes may freeze, thus joining the cubes together.
When heat is added to this system at equilibrium, ice will continue to melt until all have been
transformed to the liquid state. The amount of heat needed to convert the solid to the liquid state at the
melting point is called the heat of fusion of the substance.
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MOLAR HEAT OF FUSION AND MELTING POINT
Heat of fusion is an extensive property. The actual amount of energy involved in the transformation of a
substance from solid to liquid is dependent on the amount of sample used. Thus, this property is often
expressed in terms of molar quantities of sample.
Molar heat of fusion (ΔHfus) is the energy required to melt 1 mole of a solid.
For water, the molar heat of fusion is 6.01 kJ / mol. and its vaporization is 40.7 kJ/mol. If the heat input is
constant, a longer period is needed for one mole of water to evaporate than the time needed for the ice
to melt. An 18-gram sample of ice at 0oC will need 6.01 kJ of energy to be completely transformed into
liquid water, still at 0oC.
•
Like melting points, heats of fusion are influenced by the strength of attractive forces that exist between
particles in the solid. The stronger the attractive forces that hold the particles of the solid together, the
larger is the heat of fusion.
•
The molar heat of fusion is equal to the amount of energy released when one mole or 18 grams of liquid
water at 0oC freezes to ice, still at 0oC.
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Melting Point* (OC)
ΔHfus (kJ/mol)
Argon (Ar)
-190
1.3
Benzene (C6H6)
5.5
10.9
Diethyl ether (C2H5OC2H5)
-116.2
6.9
Ethanol (C2H5OH)
-117.3
-39
Methane (CH4)
-183
Water (H2O)
0
23.4
0.84
6.01
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*Measured at 1 atm
7.61
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Mercury (Hg)
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Substance
Cooling a substance has the opposite effect of heating it, as can be seen from the cooling curve.
If heat is removed from a liquid at a steady rate, its temperature should decrease until the freezing point
is reached. As the solid is being formed, heat is given off by the system, as attractive forces form and
become stronger between particles. Even if heat is being removed, the temperature of the system
remains constant over the freezing period.
•
After all the liquid has frozen, the temperature of the solid drops.
•
The heat change (q) for a given sample during freezing or melting may be calculated using the following
equation: is given by
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q = m ΔHfus
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BOILING AND CONDENSING: LIQUID-VAPOR EQUILIBRIUM
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In the liquid phase, there are still attractions among its particles. The particles are still in contact with each other
but are not locked into fixed positions and are free to move past each other. Although they lack the total
freedom of gaseous molecules, these molecules are in constant motion.
When a liquid is heated, its temperature increases as the kinetic energy of the molecules increases.
When the molecules have sufficient energy to escape from the surface, a phase change occurs.
Evaporation or vaporization is the process in which a liquid is transformed into a gas. The temperature
at which this occurs is the boiling point of the substance. While the liquid vaporizes, the temperature
remains constant.
•
The boiling point is a characteristic of each substance, and is dependent on the strength of attractive
forces that hold the particles or molecules in the liquid state. It is also dependent on the external or
atmospheric pressure. The boiling point of a liquid at 1 atm pressure is called its normal boiling point.
For water, this is at 100oC.
•
The reverse of vaporization or boiling is called condensation, the change from the gas phase to the
liquid phase. Condensation occurs because a molecule strikes the liquid surface and becomes trapped
by intermolecular forces in the liquid. This process occurs at the same temperature when the liquid
vaporizes into the gaseous state. The boiling point can thus be also called condensation point (dew
point), and occur at the same temperature.
•
At the boiling point, both liquid and gaseous states of the substance are present, and the
transformations of liquid to gas and gas to liquid happen at the same time.
•
At 100 OC and 1 atm, the dynamic equilibrium for water and steam is represented by
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•
water ⇋ steam
As heat is absorbed, some water will boil off but the temperature remains at 100 OC (373.15 K) until all
the liquid has vaporized. The amount of heat absorbed by the sample as the liquid transforms into gas is called
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heat of vaporization.
When all of the sample has turned into gas, further heating will cause the temperature of the
increase again.
gas to
PY
MOLAR HEAT OF VAPORIZATION (ΔHvap) AND BOILING POINT
The heat of vaporization is an extensive property and is thus dependent on the amount of sample undergoing
phase change. Hence, published quantities of heats of vaporization specify the amount of substance, and is
often expressed as molar heat of vaporization.
C
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Molar heat of vaporization (ΔHvap) is defined as the energy (usually in kilojoules) required to vaporize 1 mole of
a liquid at a given temperature, usually, at the boiling point. The molar heat of vaporization of water at 100oC is
40.8 kJ/mol.
Boiling Point* (OC)
Argon (Ar)
-186
6.3
80.1
31.0
34.6
26.0
78.3
39.3
357
59.0
-164
9.2
100
40.8
Benzene (C6H6)
Ethanol (C2H5OH)
Mercury (Hg)
Methane (CH4)
Water (H2O)
D
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Diethyl ether (C2H5OC2H5)
ΔHvap (kJ/mol)
ED
Substance
Both the boiling point and molar heat of vaporization of a substance are influenced by the strength of attractive
forces that hold the particles in the liquid state. This can be seen from the table given below.
•
The boiling point is related to molar heat of vaporization: the higher ΔHvap , the higher the boiling point,
as shown in the table:
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PY
SOLID-VAPOR EQUILIBRIUM
In a solid, the particles may be in fixed positions, but they are able to vibrate in place and with increasing
intensity as temperature increases. When particles are able to acquire enough energy to break attractive forces
with adjacent particles, the energetic particles may move into the gaseous state. This phase change is called
sublimation. One of the most familiar examples of sublimation is that of dry ice. The figure below shows iodine
subliming into a purple gas.
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ED
solid ⇋ vapor
C
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Sublimation is the process in which molecules go directly from solid into vapor phase. The reverse process is
called deposition, where molecules make a transition directly from vapor to solid. The process may be
represented by the following equilibrium:
Figure. 6 Sublimation of solid iodine in the bottom of the tube produces a purple gas
that subsequently deposits as solid iodine on the colder part of the tube above. (Image source:
https://courses.candelalearning.com/chemistryformajorsx1xmaster /chapter/phase-transitions-2/
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MOLAR HEAT OF SUBLIMATION
PY
Molar heat of sublimation (ΔHsub) of a substance is the amount of energy that must be added to a mole of
solid at constant pressure to turn it directly into a gas, without passing through the liquid phase. This enthalpy
change associated with sublimation is always greater than that of vaporization even if both sublimation and
evaporation involve changing a substance into its gaseous state because in sublimation, the starting physical
state of the substance is the solid state, which is lower in energy than the liquid state where vaporization starts. ,
D
EP
ED
C
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Sublimation requires that all the forces are broken between the molecules (or other species, such as ions) in the
solid as the solid is converted into a gas.. A comparison of the magnitudes of these thermochemical quantities
can be seen from the heating curve shown below.
Figure 7 . Heating curve showing relative amounts of heats of fusion, vaporization and
sublimation (Image Source: http://chemwiki.ucdavis.edu/@api/deki/files/10006/=IMG.jpg?
revision=1&size=bestfit&width=431&height=273)
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The molar heat of sublimation is generally expressed as Hsub in units of Joules per mole. The sum of the heat
of fusion and the heat of vaporization can give a good estimate of the heat of sublimation of a substance.
HEAT CHANGE WITH CHANGE IN TEMPERATURE
H2O(l) = 4.18 J / g OC
H2O(s) = 2.06 J / g OC
C
O
PY
When a system contains only one phase (solid, liquid, or gas), the temperature will change when it receives
energy during heating or when energy is removed during cooling. The amount of heat received or removed
from the sample to effect a given change in temperature can be calculated using the specific heat of the
substance. This is the amount of heat needed to raise the temperature of 1 gram of a substance by 1OC. It is
The
also equal to the amount of heat lo0st by 1 gram of substance when its temperature drops by 1oC.
specific heat of a substance differs for the solid, liquid, and gaseous statesWater as an example, has the
following specific heat at different phases:
ED
H2O(g) = 2.02 J / g OC
The heat change (q) for this process is given by:
D
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q = m S ΔT
where m = mass of sample in grams
S = specific heat of the sample in the appropriate physical state
T = change in temperature
Sample Problem:
You found a piece of copper metal weighing 3.10 g imbedded in an ice block. How much heat is absorbed by
the piece of metal as it warms in your hand from the temperature of the ice block at 1.5 oC to your body
temperature of 37.0 oC? The specific heat of copper is 0.385 J/g-oC. Assume that the metal is pure copper.
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q = m S ΔT
= (3.10 g)(0.385 J/g-oC)(37.0 oC – 1.5 oC)
= 42.4 J
PY
PROBLEMS INVOLVING CHANGES OF STATE
Use the following examples to show how to deal with problems involving changes of state.
C
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Sample Problem 1: How much energy is required to change 2600 gram of ice at 0˚C into water at the same
temperature?
Solution: Since the problem indicates no change in temperature and involves a solid phase, then the formula
to be used is q = m ΔHfus .
= (2600 g) (6.01 kJ)
= 15,626 kJ
ED
q = m ΔHfus
D
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Sample Problem 2: How much energy is required to change 2600 gram of water at 100˚C into steam at the
same temperature?
Solution: Since the problem indicates no change in temperature and involves a liquid phase, then the formula
to be used is q = m ΔHvap
q = m ΔHvap
= (2600 g) (40.79 kJ)
= 106,054 kJ
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Sample Problem 3: Calculate the amount of energy (in kJ) needed to heat 346 gram of liquid water from 0 OC
to 182 OC. Assume that the specific heat of water is 4.184 J/g OC over the entire liquid range and the specific
heat of steam is 1.99 J/g OC.
PY
Solution: The heat change (q) is calculated for each stage. The calculation is broken down in three steps.
Step 1: Heating of water from 0 OC to 100 OC
q1 = m S Δt
C
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= (346 g) (4.184 J/g OC) (100 OC – 0 OC)
= 1.45 x 105 J
= 145 kJ
q2 = m ΔHvap
= 783 kJ
D
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= (346 g) (40.79 kJ)
ED
Step 2: Evaporating 346 g of water at 100 OC (a phase change)
Step 3: Heating steam from 100 OC to 182 OC.
q3 = m S Δt
= (346 g) (1.99 J/g OC) (182 OC – 100 OC)
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= 5.65 x 104 J
= 56.5 kJ
The overall energy required is given by
PY
qT = q1 + q2 + q3
= 145 kJ + 783 kJ + 56.5 kJ
= 985 kJ
C
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PRACTICE (10 MINS)
Q: Calculate the heat released when 68.0 gram of steam at 124 OC is converted to water at 45 OC.
Expected answer:
ED
Solution: The heat change (q) is calculated for each stage. The calculation is broken down in three steps:
Step 1: Cooling of steam from 124 OC to 100 OC
q1 = m S Δt
D
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= (68.0 g) (2.02 J / g OC) (124 OC – 100 OC)
= 3.30 x 103 J x
1 kJ
1000 J
= 3.30 kJ
Step 2: Condensing 68.0 g of water at 100 OC (a phase change)
q2 = m ΔHvap
= (68.0 g) (40.79 kJ)
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Teacher Tips
Emphasize the step-wise determination of
heat change (q) when both changes of
temperature and phase are observed in the
problem.
= 2770 kJ
Step 3: Cooling liquid water from 100 OC to 45 OC.
q3 = m S Δt
PY
= (68.0 g) (4.184 J/g OC) (100 OC – 45 OC)
= 1.56 x 104 J
The overall energy required is given by
qT = q1 + q2 + q3
= 3.30 kJ + 2770 kJ + 15.6 kJ
ED
= 2790 kJ
C
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= 15.6 kJ
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ENRICHMENT (30 MINS)
Perform Experiment on Heating Curve for Water
EVALUATION (10 MINS)
Solve the following problems systematically and show how you arrived at the final answer.
1. Calculate the amount of heat that must be absorbed by 10.0 gram of ice at 20C to convert it to liquid
water at 60.0C. Given:
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Teacher Tip
Specific heat (ice)
Learners’ answers will be rated according to
correct use and accuracy of data, and
application of concepts.
= 2.1 J/g·C
Specific heat (water) = 4.18 J/g·C
Hfus = 6.0 kJ/mol
PY
2. Calculate the amount of heat needed to melt 2.00 kilogram of iron at its melting point (1,809 K), given
that: Hfus = 13.80 kJ/mol.
H2O(l)
H2O(g)
C
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3. What mass of water would need to evaporate from your skin in order to dissipate 1.7 105 J of heat from
your body? Given:
Hvap = 40.7 kJ/mol
4. How much energy (heat) is required to convert 52.0 gram of ice at 10.0C to steam at
Specific heat (ice):
2.09 J/g·C
Specific heat (water): 4.18 J/g·C
Hfus = 6.02 kJ/mol
Hvap = 40.7 kJ/mol
D
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Specific heat (steam): 1.84 J/g·C
ED
100C? Given:
5. Acetic acid has a heat of fusion of 10.8 kJ/mol and a heat of vaporization of 24.3 kJ/mol. What is the
expected value for the heat of sublimation of acetic acid?
Expected answer:
1. 6,300 J
2. 494 kJ
3. 75.2 g
4. 157.8 kJ
5. 35.1 kJ/mol
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Chemistry 2
60 MINS
Content Standard
The learners demonstrate an understanding of the phase changes in terms of the
accompanying changes in energy and forces between particles.
Learning Competency
Interpret the phase diagram of water and carbon dioxide.
(STEM_GC11IMF-IIIa-c-107)
Communicating learning objectives
5
Motivation
Engagement and Applications
5
Instruction
Lecture/Discussion
20
Practice
Diagram Interpretation
10
Enrichment
Constructing a Diagram
10
Evaluation
Diagram Interpretation
10
Materials
Illustrations of phase diagrams of water and carbon dioxide
Resources
ED
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Introduction
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Performance Standard
The learners design a simple investigation to determine the effect on boiling point or
freezing when a solid is dissolved in water.
LESSON OUTLINE
PY
Intermolecular Forces of Liquids
and Solids; Phase Diagrams
(1) Chang, R. (2007). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
describe the components of a phase diagram;
•
•
use phase diagrams of pure substances to determine its phase at given temperature
and pressure;
(3) Chemwiki.ucdavis.edu,. (2013). Phase Diagrams – Chemwiki. Retrieved
from http://chemwiki.ucdavis.edu/Physical_Chemistry/
interpret the phase diagram of water and carbon dioxide;
•
describe how changes in temperature and pressure can change the state
of matter; and
•
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•
(2) Chemguide.co.uk,. (2016). Phase diagrams of pure substances.
Retrieved from http://www.chemguide.co.uk/physical/phaseeqia/
phasediags.html
Physical_Properties_of_Matter/Phases_of_Matter/Phase_Transitions/
Phase_Diagrams
(4) Petrucci, R., Harwood, W., & Herring, F. (2007). General Chemistry:
Principles and Modern Applications (9th ed.). Upper Saddle River, NJ:
Pearson
construct a phase diagram of a substance from given data.
(5) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
Thomson Brooks/Cole.
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Describe the components of a phase diagram;
PY
b. Use phase diagrams of pure substances to determine its phase at given temperature and pressure;
c. Interpret the phase diagram of water and of carbon dioxide;
d. Describe how changes in temperature and pressure can change the state of matter;
C
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e. Construct the phase diagram of a substance from given data.
2. Present relevant vocabulary that will be used in the lesson and learners should know.
Fluid
Solid
ED
A gas or a liquid; a substance that can flow.
Liquid
D
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A phase of matter with definite shape and volume.
A phase of matter with definite volume but no definite shape.
Gas
A phase of matter with no definite shape or volume of its own.
Vapor
A gaseous substance that exists naturally as a liquid or solid at normal temperature.
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Teacher Tips
•
•
•
Most of these terms are recalled from
previous lessons.
Definitions should remain posted as the
lesson progresses.
More key words may be added to
relevant vocabulary as the need arises.
Melting
A phase change from solid to liquid.
Vaporization
PY
A phase change from liquid to gas.
Sublimation
A phase change from solid to gas.
C
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Condensation
A phase change from gas to liquid.
Freezing
ED
A phase change from liquid to solid.
Deposition
A phase change from gas to solid.
D
EP
Melting (or freezing) curve
The curve on a phase diagram which represents the transition between the liquid and solid
states.
Vaporization (or condensation) curve
The curve on a phase diagram which represents the transition between the gaseous and liquid states.
Sublimation (or deposition) curve
The curve on a phase diagram which represents the transition between the gaseous and solid states.
Triple point
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The point on a phase diagram at which the three states of matter coexist.
Critical point
PY
The point in temperature and pressure on a phase diagram where the liquid and gaseous phases of a
substance merge together into a single phase. The temperature and pressure corresponding to this
are known as the critical temperature and critical pressure.
Normal melting and boiling points
C
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Melting and boiling points when the pressure is 1 atm.
3. Connect the lesson with prerequisite knowledge to recall how heat changes and temperature affect
phase changes.
Focus question:
How can this effect be achieved using CO2 or dry ice?
D
EP
A.
ED
MOTIVATION (5 MINS)
Figure 1: Stage light effects as dry ice sublimes
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Image source: http://www.gettyimages.com/
detail/photo/smoke-beneath-spotlight-highres-stock-photography/200326482-001
Teacher Tip
B.
What does LPG stand for? How can a gas be liquefied? What conditions are needed
to convert a gas into a liquid?
PY
INSTRUCTION (20 MINS)
What is a phase diagram?
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EP
ED
C
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A phase diagram is a graphical representation of the physical states of a substance under different conditions of
temperature and pressure. It gives the possible combinations of pressure and temperature at which certain
physical state or states a substance would be observed. Each substance has its own phase diagram. A typical
phase diagram is shown below.
Because carbon dioxide cannot exist as a
liquid at atmospheric pressure, the dry
ice sublimates and instantly produces a
gas, condensing water vapor, and creating a
thick white fog.
Liquefied petroleum gas or liquid petroleum
gas (LPG or LP gas), also referred to as simply
propane or butane, are flammable mixtures of
hydrocarbon gases used as fuel in heating
appliances, cooking equipment, and vehicles.
https://en.wikipedia.org/wiki/
Liquefied_petroleum_gas
High pressure and low temperature are
needed to liquefy gases.
Teacher Tip
A single phase diagram can be used during
the entire discussion. To allow the addition of
information as the discussion progresses,
place aplastic sheet which could be written
on over the permanent diagram .
The use of colored lines (curves) will be
helpful for the distinction of the processes
involved.
Figure 2: General Phase diagram
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What are the features of a phase diagram?
Teacher tip
Phase diagrams are plots of pressure (usually in atmospheres) versus temperature (usually in degrees Celsius or
Kelvin). The diagram is divided into three areas: solid, liquid and gaseous states. The boundary between the
liquid and gaseous regions stop at point C, the critical temperature for the substance.
•
Emphasize the conversion of units for
temperature and pressure.
•
Units of pressure:
•
•
1 atm = 101325 Pa (pascal)
1 atm = 760 torr (mmHg)
•
Units of temperature: (K = oC + 273)
•
0 oC = 273 K, 100 oC = 373 K
PY
A. The Three Areas
C
O
The three areas are marked solid, liquid, and vapor. Under a set of conditions in the diagram, a substance can
exist in a solid, liquid, or vapor (gas) phase. The labels on the graph represent the stable states of a system in
equilibrium.
D
EP
ED
Suppose a pure substance is found at three different sets of conditions of temperature and pressure
corresponding to A, B, and C as shown in the following diagram:
Figure 3: Phase diagram with three sets of conditions
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Under the set of conditions at A in the diagram, the substance would be a solid as it falls into that
area of the phase diagram. At B, it would be a liquid; and at C, it would be a vapor (gas).
•
B. Three Lines (Curves)
PY
The lines that serve as boundaries between physical states represent the combinations of pressures
and temperatures at which two phases can exist in equilibrium. In other words, these lines define
phase change points.
C
O
1. The green line divides the solid and liquid phases, and represents melting (solid to liquid) and
freezing (liquid to solid) points.
D
EP
ED
Melting (or freezing) curve – the curve on a phase diagram which represents the transition
between liquid and solid states. It shows the effect of pressure on the melting point of the
solid. Anywhere on this line, there is equilibrium between the solid and the liquid.
Figure 4: The freezing
(or melting) curve
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2. The blue line divides the liquid and gas phases, and represents vaporization (liquid to gas) and
condensation (gas to liquid) points.
C
O
PY
Vaporization (or condensation) curve – the curve on a phase diagram which represents the transition
between gaseous and liquid states. It shows the effect of pressure on the boiling point of the liquid. Anywhere
along this line, there will be equilibrium between the liquid and the vapor.
D
EP
ED
Figure 5: The
vaporization or
condensation curve
3. The red line divides the solid and gas phases, and represents sublimation (solid to gas) and deposition
(gas to solid) points.
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C
O
PY
Sublimation (or deposition) curve – the curve on a phase diagram which represents the
transition between gaseous and solid states. It represents the effect of increased temperature
on a solid at a very low constant pressure, lower than the triple point.
D
EP
ED
Figure 6: Sublimation
or deposition curve.
C. Two Important Points
There are two important points on the diagram, the triple point and the critical point.
The triple point
The triple point is the combination of pressure and temperature at which all three phases of matter are at
equilibrium. It is the point on a phase diagram at which the three states of matter coexist. The lines that
represent the conditions of solid-liquid, liquid-vapor, and solid-vapor equilibrium meet at the triple point
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•
It is a unique combination of temperature and pressure where all three phases are in equilibrium
together.
The critical point
PY
The critical point terminates the liquid/gas phase line. It is the set of temperature and pressure on a phase
diagram where the liquid and gaseous phases of a substance merge together into a single phase. Beyond the
temperature of the critical point, the merged single phase is known as a supercritical fluid.
C
O
The temperature and pressure corresponding to this are known as the critical temperature and critical
pressure.
D
EP
ED
If the pressure on a gas (vapor) is increased at a temperature lower than the critical temperature, the liquidvapor equilibrium line will eventually be crossed and the vapor will condense to give a liquid.
Figure 7:
Temperature and
pressure values at the
critical point
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How is the normal melting and boiling points determined in a phase diagram?
PY
The normal melting and boiling points are those when the pressure is 1 atmosphere.
C
O
Figure 8:
D
EP
ED
Locating the normal
melting point and
normal boiling point
These can be found from the phase diagram by drawing a line across pressure at 1 atm.
How does the phase diagram of water look like?
The Phase Diagram for Water
There is only one difference between the phase diagram for water and the other phase diagrams
discussed. The solid-liquid equilibrium line (the melting point curve) slopes backwards rather than forwards.
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For water, the melting point gets lower at higher pressures. This is because solid ice is less dense than
liquid water. This phenomenon is caused by the crystal structure of the solid phase. In the solid forms of water
and some other substances, the molecules crystallize in a lattice with greater average space between molecules,
thus resulting in a solid occupying a larger volume and consequently with a lower density than the liquid. When
it melts, the liquid water formed occupies a smaller volume
PY
ice ⇋ water
ED
C
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An increase in pressure will move the above equilibrium to the side with the smaller volume. Liquid
water is produced. To make the liquid water freeze again at this higher pressure, the temperature should be
reduced. Higher pressures mean lower melting (freezing) points.
D
EP
Figure 9: Phase
diagram for H2O
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Identifying data from the phase diagram of water
PY
Notice that the triple point for water occurs at a very low pressure, 0.006 atm and at 273.2 K temperature. Also
notice that the critical temperature is 647 K (374°C). It would be impossible to convert water from a gas to a
liquid by compressing it above this temperature. The critical pressure is 218 atm.
How does the phase diagram for carbon dioxide look like?
ED
The Phase Diagram for Carbon Dioxide
C
O
The normal melting and boiling points of water are found in exactly the same way as we have already discussed
- by determining where the 1 atm pressure line crosses the solid-liquid, and then the liquid-vapor equilibrium
lines. The normal melting point of water is 273 K (0 oC), and its normal boiling point is 373 K (100 oC).
D
EP
The only thing special about this phase diagram is the position of the triple point, which is well above
atmospheric pressure. It is impossible to get any liquid carbon dioxide at pressures less than 5.2 atmospheres.
At 1 atm pressure, carbon dioxide will sublime at a temperature of 197.5 K (-75.5 °C). This is the reason why
solid carbon dioxide is often known as "dry ice." There is no liquid carbon dioxide under normal conditions only the solid or the vapor.
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PY
PRACTICE (10 MINS)
Interpreting a Phase Diagram
D
EP
ED
C
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Figure 10: Phase
diagrams for CO2
Refer to the following phase diagram of a certain substance to answer the following questions.
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PY
C
O
ED
1. In what phase is the substance at 50 °C and 1 atm pressure?
D
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2. At what pressure and temperature conditions will all three phases of the substance be present?
3. What is the normal melting point of the substance?
4. What phase(s) will exist at 1 atm and 70 °C?
Expected answer:
1. liquid
2. ≈ 0.5 atm and ≈28 °C
3. ≈ 32 °C
4. liquid and vapor (gas)
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ENRICHMENT (10 MINS)
Teacher Tip
Learners may be guided by providing a
scaled graph with pressure and temperature
scales provided.
Constructing a Phase Diagram
PY
Visualize a substance with the following points on the phase diagram: a triple point at 0.05 atm and 150 K; a
normal melting point at 175 K; a normal boiling point at 350 K; and a critical point at 2.0 atm and 450 K. The solid
liquid line is “normal” (meaning positive sloping). For this, complete the following:
C
O
1. Roughly sketch the phase diagram, using units of atmosphere and Kelvin. Label the area 1, 2, and 3, and
points T and C on the diagram.
2. Describe what one would see at pressures and temperatures above 2.0 atm and 450 K.
3. Describe the phase changes from 50 K to 250 K at 1.5 atm.
4. What exists in a system that is at 1 atm and 350 K?
ED
5. What exists in a system that is at 1 atm and 175 K?
Expected answers:
1. 1-Solid, 2-Liquid, 3-Gas, Point T-triple point, Point C-critical point
D
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2. Super-critical fluid
3. Melt at around 180 K and become a liquid at 250 K.
4. Both liquid and vapor exist.
5. Both solid and liquid exist.
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EVALUATION (10 MINS)
Teacher Tip
Learners’ answers will be rated based on
correct use and accuracy of data, and
application of concepts.
Figure 9:
D
EP
ED
C
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PY
Interpreting Phase Diagrams of Water and Carbon Dioxide
Phase diagram of water
Figure 10: Phase diagram of CO2
Based from the phase diagrams of water and carbon dioxide, answer the following questions and justify
your answers:
1. You have ice at 263 K (-10.0 oC) and 1.0 atm. What could you do to make the ice sublime?
2. A sample of dry ice (solid CO2) is cooled to 173 K (-100.0 oC), and is set on a table at room temperature
(298 K; 25 oC). At what temperature is the rate of sublimation and deposition the same (assume that
pressure is held constant at 1 atm)?
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Image sources:
http://chemwiki.ucdavis.edu/@api/deki/
•
f i l e s / 8 7 2 4 / = w a t e r. j p g ?
revision=1&size=bestfit&width=330&hei
ght=320
http://chemwiki.ucdavis.edu/@api/deki/
•
files/8721/=Phase_Diagram_CO2.jpg?
revision=1&size=bestfit&width=322&hei
ght=322
Chemistry 2
90 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Motivation
Engagement and Applications
10
Hands-on Activity
45
Post-lab Discussion
20
Reflection and Application
10
PY
Intermolecular Forces of Liquids
and Solids; Measuring Viscosity
of Liquids
Collect and organize data to determine the viscosity of liquids.
(STEM_GC11IMF-IIIa-c-108)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Poster Presentation
Sphere (marble); liquid to be measured (cooking oil, Karo syrup,
liquid glue, hand sanitizer); balance; graduated cylinder; calculator;
timer or stopwatch; ruler; marker
ED
Learning Competency
Measure and explain the difference in the viscosity of some liquids.
C
O
Instruction
Content Standard
The learners demonstrate an understanding of the properties of liquids, and the nature of Practice
forces between particles.
Enrichment
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or Evaluation
freezing when a solid is dissolved in water.
Materials
measure mass, volume, diameter, length, and time using appropriate instruments;
•
calculate radius, density, and velocity from measured quantities;
•
calculate the viscosity of liquids from data obtained in the experiment; and
•
compare the viscosity of some liquids.
D
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•
Resources
(1) (2016). Retrieved from http://www.udel.edu/pchem/C444/Lectures/
Lecture3.pd
(2) Spacegrant.hawaii.edu,. (2016). Activity: Viscosity. Retrieved from
http://www.spacegrant.hawaii.edu/class_acts/Viscosity.html
(3) Teach like A Champion,. (2012). Take A Stand: Strategy Four. Retrieved
from https://daviskm2psy3010.wordpress.com/about/
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
b. Calculate radius, density, and velocity from measured quantities;
c. Calculate the viscosity of liquids from data obtained in the experiment;
d. Compare the viscosity of some liquids.
PY
a. Measure mass, volume, diameter, length, and time using appropriate instruments;
Viscosity
Measurement of a liquid’s resistance to flow
ED
3. Connect the lesson with previous knowledge/skills required.
C
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2. Present relevant vocabulary that will be used in the lesson and learners should know.
Recall how the quantities (length, mass, volume, time) are measured and check on the skills of learners.
Teacher Tip
•
Relevant vocabulary may still include
the quantities/properties for
measurement (length, mass, volume,
time, velocity, diameter, radius).
•
There may be a need to go over the
skills of using measuring apparatuses
and reporting data in correct significant
figures.
D
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Recall how data are recorded in proper significant figures.
Teacher Tip
MOTIVATION (10 MINS)
•
The list of products/materials provided
may be changed according to the
experiences of learners. Use materials
that are more familiar to the learners.
•
The number of products should be
limited to at least 5-7. The activity can
be done in less than 5 minutes.
Ask learners how they prefer the following products/materials:
(Take a Stand Protocol)
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Product/Materials
THICK
THIN
Milk
Lotion
Toothpaste
PY
Mayonnaise/dressing
Butter/sandwich spread
C
O
1. Post two signs at either end of an imaginary line that goes across the classroom. At one end of the line,
post “THICK.” At the other end, post “THIN.”
2. Tell participants they will use the “Take a Stand Protocol,” wherein they must share and explain their
preferences/choices. After they hear a question, they will move to either side of the imaginary line that
best reflects their choice.
3. Explain the steps of the protocol:
ED
A. The facilitator will ask a question and then participants will move, depending on whether they prefer
the mentioned product/material THICK or THIN, to a side of the imaginary line that goes across the
room. Point out that one side of the room is labeled “THICK,” and the other side labeled “THIN.”
This means that the middle of the line is undecided.
D
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B. After the facilitator asks the question, s/he will pause for the participants to think and will then ask
them to move to the place that best reflects their choice.
C. The facilitator will ask participants to share and justify their preferences, making sure to hear from
people from both sides of the line.
D. If a participant hears an opinion that changes his/her mind, s/he can move quietly to the other part of
the line.
4. Model how the protocol will work. When you ask the question such as, “How do you like your spaghetti
sauce?” show the learners how you would move to reflect your preference. The modelling exercise helps
the participants internalize how to use the invisible line.
5. As you use the protocol, repeat each question twice. Note that you can have participants stand up or sit
down in their places.
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INSTRUCTION (45 MINS)
A. PRE-LAB DISCUSSION:
PY
Make preliminary instructions about the activity.
Questions to investigate:
What is the viscosity of a liquid?
Safety Precautions:
The activity should be performed on a flat surface.
C
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Which of the liquids is most viscous?
ED
Remind the learners of the proper handling of substances they will be using. Avoid contact with the skin and
direct inhalation of the vapors of the substances. It is best if learners use safety gloves, goggles, and masks.
D
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1. Tell the learners to work in subgroups of three members. One of the members will act as the data
recorder. The subgroup should be part of a bigger group of four subgroups, each working with a
different liquid sample.
2. Each subgroup will work with one liquid sample and will ask data from other subgroups to complete the
data required, and be able to compare the viscosities of the four liquids used.
3. Give each learner a data sheet for the results of the experiment.
4. Check the availability of the materials for the activity.
5. Make sure that learners record their data properly and accurately.
6. Allow the learners to show their results on the board for comparison with the results of the class.
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B. LABORATORY PROPER:
Teacher Tip
Learners will perform the activity on Measuring the Viscosity of Some Liquids
Procedure
•
Dropping the sphere into the liquid
numerous times when calculating velocity
ensures an accurate measurement. If time
allows, let the learners do the procedure
at least three to five times to get an
average of the measurements made for
the value of velocity
•
The ball must have a higher density than
the liquid for this process to work.
•
Make sure there is no water or other
liquid in the graduated cylinder when the
activity starts. The presence of another
liquid could throw off measurements.
•
The sphere should be dried before it is
placed in the graduated cylinder. The
liquid should be cleaned or wiped off the
sphere.
•
When filling the graduated cylinder with
liquid, leave sufficient space to prevent it
from overflowing brought by the
displaced liquid.
PY
1. Use a marble for the sphere and one of the liquids for this measurement.
2. Calculate the density of the sphere.
The formula for density is
, so you will need to determine the sphere's mass, ms and volume,
C
O
Vs in turn.
•
Measure mass by placing the sphere on a balance.
•
Determine the volume of the sphere. Volume of a sphere is calculated by using the formula: , where
Vs is the volume and r is the radius of the sphere.
ED
Use two parallel surfaces such as flat boards, to measure the diameter of the sphere. Place the s p h e r e
between two parallel surfaces: if the surfaces are parallel and the sphere is just touching each, the distance
to get the radius, r.
between the surfaces is the diameter, ds of the sphere. Use the formula:
D
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Figure 1:
Determining the
diameter of a
sphere.
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3. Calculate the density of the liquid you are measuring.
•
Measure the mass of the liquid, mL by first weighing the empty graduated cylinder, m1. Pour your
liquid into the graduated cylinder and then weigh it again, m2. Subtract the mass of the empty
cylinder, m1 from that of the cylinder with the liquid, m2 to obtain the mass of the liquid, mL.
PY
mL = m2 - m1
•
To find the volume of the liquid, VL simply determine the amount of liquid you poured into the
graduated cylinder by using the scale on cylinder’s side. Record the volume in cm3 (1 mL = cm3).
•
Use the formula
4. Fill the graduated cylinder with the liquid to be measured.
C
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and your measurements to find the density of the liquid.
ED
When you fill the graduated cylinder with the liquid, make a point not to come too close the top. Leave
sufficient space for the displaced liquid caused by the sphere.
D
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5. Make a mark at a fixed position near the bottom of the graduated cylinder, around 2.5 cm (1 inch) or 5
cm (2 inches) from the bottom. Then make a mark at a fixed position at the top of the cylinder, around
2.5 cm (1 inch) or 5 cm (2 inches) from the top of the liquid.
6. To get distance, dt measure the difference between the top mark and the bottom mark using a meter
stick.
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C
O
PY
•
7. Drop the ball into the liquid.
ED
Figure 2: Distance of fall of the sphere in the cylinder. (Image Source: http://
www.wikihow.com/Measure-Viscosity#/Image:Measure-Viscosity-Step-7.jpg
D
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8. When the bottom of the ball reaches the mark at the top of the cylinder, start the stopwatch.
Then when the ball reaches the mark you made at the bottom of the cylinder, stop the
stopwatch. Determine the time travelled by the sphere, tt.
9. Calculate the velocity, v of the sphere by using your measurements and the formula:
where v is the velocity of the sphere, dt is the distance travelled, and tt is the time it took the
sphere to travel the distance.
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where ∆ρ is the difference between the density of the solid and the liquid
PY
10. Use the given formula for viscosity:
C
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(∆ρ = Ds - Dl ), g is the acceleration of gravity (980 cm/s2), r is the radius of the sphere and v is the
velocity.
ED
PRACTICE (20 MINS)
C. Post-Lab Discussion
D
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1. Have the learners complete the data table for their activity. Calculations should be done and reported in
correct significant digits.
2. Learners share the results of their experiment. In five minutes, let them compare their results with those
of their classmates.
3. Ask the learners the following questions:
a. How does thickness of a material relate to viscosity?
b. From the results of the experiment, rank the liquids in terms of viscosity from the greatest to the
least.
c. What inference can be made from the experiment?
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Teacher Tip
•
To minimize crowding of learners as they
collect data from other subgroups, this
should still be done as a group.
Emphasize the need for each member to
have their own data sheets.
•
Keeping a data table helps keep track of
measurements and be organized.
•
All measurements should be in metric
form.
•
Measured and calculated data should be
reported in correct significant digits.
Quantities
Liquid 1
Liquid 2
Liquid 3
Liquid 4
Cooking oil Karo syrup Liquid glue
Hand
sanitizer
Mass of sphere, ms
PY
Diameter of sphere, ds
Radius of sphere, rs
Volume of sphere, Vs
C
O
Density of sphere, Ds
Mass of cylinder, m1
Mass of cylinder + liquid, m2
Mass of liquid (m2 – m1), mL
ED
Volume of liquid, VL
Density of liquid, DL
Time travelled, tt
Velocity, v
D
EP
Distance travelled, dt
Acceleration of gravity, g
Viscosity of liquid, µ
980 cm/s2
980 cm/s2
980 cm/s2
•
Some units of viscosity:
Poise (symbol: P)
Named after the French physician Jean
Louis Marie Poiseuille (1799–1869), this is
the centimeter-gram-second (cgs) unit of
viscosity, equivalent to dyne-second per
square centimeter (dyne•s/cm2). It is the
viscosity of a fluid in which a tangential
force of 1 dyne per square centimeter
maintains a difference in velocity of 1
centimeter per second between two
parallel planes 1 centimeter apart.
1 dyne = 1 g•cm/s2
Even in relation to high-viscosity fluids,
this unit is most usually encountered as
the centipoise (cP), which is 0.01poise.
Many everyday fluids have viscosities
between 0.5 and 1000 cPs (see table).
For this activity, the unit of viscosity is the
dyne•s/cm2.
Measuring viscosity is an effective way to
determine the state (properties of
matter) or fluidity of a liquid or gas.
980 cm/s2
Expected Result:
Liquid glue, honey, hand sanitizer, glycerin, syrup, cooking oil, water.
*Note: Some variation in the order is predicted depending on the type of each product used. Some
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It plays an important role in the quality
control, various research and
development stages in the laboratory,
processes, health and research
environments as well as a wide range of
industries and applications including
food, chemical, pharmaceutical,
petrochemical, cosmetics, paint, ink,
coatings, oil and automotive.
brands or types may be more viscous than others.
Efficient use of some materials is
dependent on its viscosity.
4. Recall the motivation activity and ask the learners the following questions:
Were your choices the same as your classmates? Why?
Teacher Tip
PY
How do your choices relate to the needs of businesses and industries?
ENRICHMENT (10 MINS)
2. What is the significance of the viscosity of blood thealth?
The enrichment activity will be done as
assigned work since this can be made a
part of the laboratory report as the
“Generalization” portion.
•
Blood viscosity is the thickness and
stickiness of blood. It is a direct measure
of the ability of blood to flow through
the vessels. It is also a key screening test
that measures how much friction the
blood causes against the vessels, how
hard the heart has to work to pump
blood, and how much oxygen is
delivered to organs and tissues.
Importantly, high blood viscosity is easily
modifiable with safe, lifestyle-based
interventions.
•
Blood viscosity is defined as the inherent
resistance of blood to flow. Normal adult
blood viscosity is 40/100, which is read as
“forty over one hundred” and reported
in units of millipoise.
•
Increased blood viscosity is the only
biological parameter that has been
linked with all the other major
cardiovascular risk factors, including high
blood pressure, elevated low density
lipoprotein (LDL) cholesterol, low high
density lipoprotein cholesterol (HDL),
type-II diabetes, metabolic syndrome,
obesity, smoking, age, and gender.
(http://www.bloodflowonline.com/learnabout-blood-viscosity/blood-viscositybasics)
C
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1. Rank the following on the basis of increasing viscosity: ketchup, chocolate syrup, blood, peanut butter,
lava. Use the web to find out.
•
The following table contains some of the common liquids and their viscosity in centipoise (cps).
Viscosity
ED
Liquid
(in centipoise, cps)
1
Milk
3
Blood
4 to 10
Castor oil
D
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Water
1000
Latex house paint
1500
Karo syrup
5000
Honey
Source: http://
www.wmprocess.com/
viscosity-of-common-liquids/
10,000
Hershey’s chocolate syrup
10,000 to 25,000
Ketchup
50,000
Peanut butter
250,000
Lava
≈ 4,300,000
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EVALUATION
Teacher Tip
•
To ensure that all learners will work on
their own project, this can be individual
work or for a small group of students
•
The following elements will be observed
in rating the poster presentation:
(persuasiveness, effective
communication, visual appeal, correct
mechanics).
Poster Presentation: Advertisement of a Product or Sample
Chocolatiers
Carpenters
Cheese makers
Drug manufacturers
Mothers
Car mechanics
Doctors
Chefs
Beautician
C
O
Artist painters
PY
Choose one from the following suggested professions and discuss how the viscosity of an
important product/material is relevant and how it affects their work.
See the next page for the rubric.
ED
The learners can choose other professions/careers not on the list.
D
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(Source: http://rubistar.4teachers.org/index.php?screen=ShowRubric&rubric_id=1357669&)
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3
2
1
Graphics - Relevance
All graphics are related to
the topic and makes it
easier to understand. All
borrowed graphics have a
source citation.
All graphics are related to
the topic and most make it
easier to understand. All
borrowed graphics have a
source citation.
All graphics relate to the
topic. Most borrowed
graphics have a source
citation.
Graphics do not relate to
the topic OR several
borrowed graphics do not
have a source citation.
Graphics - Originality
Several of the graphics
used on the poster reflect
an exceptional degree of
learner creativity in their
creation and/or display.
One or two of the graphics
used on the poster reflect
learner creativity in their
creation and/or display.
The graphics are made by
the learner, but are based
on the designs or ideas of
others.
Required Elements
The poster includes all
required elements as well
as additional information.
All required elements are
included on the poster.
All but one of the required Several required elements
elements is included on
were missing.
the poster.
Attractiveness
The poster is exceptionally The poster is attractive in
attractive in terms of
terms of design, layout
design, layout, and
and neatness.
neatness.
The poster is acceptably
The poster is distractingly
attractive though it may be messy or very poorly
a bit messy.
designed. It is not
attractive.
Grammar
There are no grammatical
mistakes on the poster.
There are 2 grammatical
mistakes on the poster.
ED
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CATEGORY
C
O
5
PY
Sample Rubric - Poster
There is 1 grammatical
mistake on the poster.
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No graphics made by the
learner are included.
There are more than 2
grammatical mistakes on
the poster.
Chemistry 2
90 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Motivation
Engagement activity
5
PY
Intermolecular Forces of Liquids
and Solids; Heating and Cooling
Curve of a Substance
C
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Instruction
Hands-on activity
Content Standards
Reflection and application
The learners demonstrate an understanding of the properties of liquids, and the nature of Enrichment
forces between particles.
Evaluation
Creating a head curve
Phase changes in terms of accompanying changes in energy and forces between
Materials
particles
Beaker (250 mL); crushed ice cubes; thermometer; spatula; timer;
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or
freezing when a solid is dissolved in water.
60
10
10
bunsen burner (or hot plate); iron stand and iron ring (tripod);
thermometer clamp (or cork with boron the middle to fit
thermometer snugly and iron clamp); wire mesh; matches; data
sheets and worksheets; images and diagrams
ED
Collect and organize data needed to construct heating and cooling curves of pure Resources
substances.
(1) (2016). Retrieved from http://www.sausd.us/cms/lib5/CA01000471/
(STEM_GC11IMF-IIIa-c-109)
D
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Learning Competency
(LAB) Determine and explain the heating and cooling curve of a substance.
construct and interpret a heating curve for water;
•
construct heating and cooling curves of a pure substance
(2) Chang, R. (2007). Chemistry (9th ed., pp. 434-485). New York: McGrawHill.
(3) Science.uwaterloo.ca,. (2016). Heating Curve. Retrieved 15 February
2016, from http://www.science.uwaterloo.ca/~cchieh/cact/c123/
heating.html
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
Centricity/Domain/109/HS%20Chemistry%20 Teachers%20Edition
%205.pdf
(4) Whitten, K. (2007). Chemistry (8th ed., pp. 446-499). Belmont, CA:
Thomson Brooks/Cole.
using experimental data; and
•
demonstrate how heat energy can be used to raise the temperature of
a substance and weaken intermolecular forces to cause a phase change.
140
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Construct a heating curve for water;
PY
b. Interpret a heating curve for water;
c. Construct heating and cooling curves of a pure substance using experimental data;
C
O
d. Demonstrate how heat energy can be used to raise the temperature of a substance and weaken
intermolecular forces to cause a phase change.
2. Present relevant vocabulary that will be used in the lesson and learners should know.
Phase
Solid
ED
A homogeneous part of a system in contact with other parts of the system but separated by a
well-defined boundary.
Liquid
D
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A phase of matter with definite shape and volume.
A phase of matter with definite volume but no definite shape.
Gas
A phase of matter with no definite shape or volume of its own.
Phase changes
Transformations from one phase of matter to another.
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Teacher Tip
•
These terms are recalled from previous
lessons on phase changes.
•
Definitions should remain posted as the
lesson progresses.
•
More key words may be added to
relevant vocabulary as the need arises.
Melting
A phase change from solid to liquid.
Vaporization
PY
A phase change from liquid to gas.
Sublimation
A phase change from solid to gas.
C
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Condensation
A phase change from gas to liquid.
Freezing
Deposition
Exothermic process
D
EP
A phase change from gas to solid.
ED
A phase change from liquid to solid.
Process that gives off or releases heat to the surroundings.
Endothermic process
Process that absorbs heat from the surroundings.
Heating curve
A plot of temperature versus time.
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3. Connect the lesson with previous knowledge required.
ED
C
O
PY
Recall how energy is involved in phase changes.
Figure 1: Phase changes that matter undergoes
MOTIVATION (5 MINS)
Focus question:
D
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http://www.shmoop.com/matter-properties/test-your-knowledge.html
Still using the following images from the previous lesson, ask the learners how energy change affects
the phase and temperature of a material?
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Figure 2: Examples of phase changes
A. PRE-LAB DISCUSSION:
Questions to investigate:
D
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Make preliminary instructions about the activity.
Solid iodine subliming
in a test tube
ED
INSTRUCTION (60 MINS)
PY
Glass of ice water
C
O
Pan of boiling water
Teacher Tip
•
This is intended as an exploratory lab
where learners directly see that energy is
absorbed during a phase change, but
does not cause change in temperature.
•
If a thermometer clamp is not available,
one member of the group must always
hold the thermometer when using it. The
thermometer must always stay in the
beaker throughout the lab, so it doesn’t
measure air temperature.
•
Slight changes in temperature during
phase change may be observed but can
be ignored for the purpose of this
experiment. These temperature changes
occur because the thermometer in ice is
also reading the temperature of the air in
between the ice cubes.
What is the viscosity of a liquid?
Which of the liquids is most viscous?
Safety Precautions:
Check if the learners are ready for the experiment. Make sure that learners have read through the purpose,
procedure, and the data table, and that they understand what needs to be recorded during the lab.
Safety precautions:
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Remind the learners of the proper handling of the substances and apparatuses they will be using. It is best if
learners use safety gloves and goggles. Hot pads are also indispensable in this experiment since it involves
heating.
One of the members will act as data
PY
1. Tell the learners to work in groups of three to four members.
recorder.
2. Give each learner a data sheet for the results of the experiment.
3. Check the availability of the materials for the activity.
C
O
4. Make sure that learners record their data properly and accurately.
Allow Each group to present to the class the heating curve that they constructed.
ED
B. LABORATORY PROPER:
Learners will perform the Heating Curve for Water experiment.
Procedure
D
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1. Gather all necessary materials and apparatuses.
2. Set up the beaker on the wire mesh above the Bunsen burner (alcohol lamp) using either the tripod or
iron ring attached to an iron stand. DO NOT light the burner yet.
3. Put about 150 mL of crushed ice cubes into the beaker. Record this temperature at time 0.
Do not let the thermometer rest on the glass.
4. Record the temperature and phases WITHOUT adding heat every minute for five minutes. Use the data
sheet provided for this purpose.
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•
Thermometers used may not be
calibrated accurately, so it may not be
able to read the 0oC temperature of ice.
•
The heating curve will not be perfect,
with more inaccuracies at the beginning
of the curve, as the ice may begin
melting before the learners begin
recording temperature. If possible, keep
ice in a freezer until the start of each
class.
5. Adjust the burner or lamp so medium heat is applied to the beaker with ice. If using a hot
plate, turn the temperature to LOW and stir the ice occasionally with a spatula.
•
6. Record the temperature and phases (solid, liquid, gas) every one minute until the water is
boiling (with lots of bubbles) for 5 minutes.
•
If heat source will provide sufficient
amount of heat, it will take lesser time
until boiling takes place and so will the
observation.
•
Check the graphs made by the learners.
Notice trends in the graph: temperature
increases when only one phase is present
in the beaker; temperature remains
constant when more than one phase is in
the beaker; and constant temperature
represents a phase change.
PY
Remember, there may be more than one phase present. Record all phases present.
Time
Temp oC
Starting Temp
0 oC
Phases present Time
Ice (solid)
1
2
3
Temp
o
C
Phases
present
15
16
ED
0
17
18
19
D
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Data/Observations:
C
O
7. After water has boiled for five minutes, all remaining water can go into the sink. Dry off your
lab table and return all lab materials.
4
20
5
21
6
22
(light the
burner/lamp)
7
23
8
24
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Temp oC
Phases present Time
25
10
26
11
27
12
28
13
29
Phases
present
C
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9
Temp
o
C
PY
Time
14
30
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ED
Graph Making
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•
A system is an imaginary closed container
isolated from its environment. It is
isolated so that we can investigate how
the system changes as it is disturbed
either by transferring mass or energy to
and from it. The existence of the
container is optional in definition, but in
reality a container is used for the isolation.
Use the results to plot your own heating curve for water.
•
•
Phase change between solid and liquid as “A.”
•
Phase change between liquid and gas as “B.”
•
Heating the liquid as “C.”
PY
Label the following points on the graph above:
Processing of Data
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Discuss what are involved in heating and cooling curves.
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When the system is heated, energy is transferred into it. In response to the energy it receives, the system
changes, for example by increasing its temperature. If the temperature of a material is monitored during
heating, it varies with time. A plot of the temperature versus time is called the heating curve. One such
heating curve is shown here.
Figure 3 A heating curve:
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Between A & B, the material is a solid. The heat supplied to the material is used to increase the kinetic
energy of the molecules and the temperature rises.
•
Between B & C, the solid is melting. Heat is still being supplied to the material but the temperature does
not change. Heat energy is not being changed into kinetic energy. Instead, the heat is used to change
the arrangement of the molecules.
•
At point C, all of the material has been changed to liquid.
•
Between C & D, the heat supplied is again used to increase kinetic energy of the molecules and the
temperature of the liquid starts to rise.
•
Between C & D, the liquid is heated until it starts to boil.
•
Between D & E, the liquid is still being heated but the extra heat energy does not change the
temperature (kinetic energy) of the molecules. The heat energy is used to change the arrangement of the
molecules to form a gas.
•
At point E, all of the liquid has been changed into gas.
•
Between E & F, the gas is heated and the heat energy increases the kinetic energy of molecules once
more, so the temperature of the gas increases.
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•
When a system contains only one phase (solid, liquid, or gas), the temperature will increase when it receives
energy. The rate of temperature increase will be dependent on the heat capacity of the phase in the system.
When the heat capacity is large, the temperature increases slowly, because much energy is required to increase
its temperature by one degree. Thus, the slopes of temperature increase for the solid, liquid, and gases are
different.
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•
The heating curve given is sketched
according to the data about water. In a
real experiment, the heat into the system
is hardly transferred at a constant rate
unless the heat source is at a very high
temperature. However, for the sake of
simplicity, assume that heat flows into the
system at a constant rate.
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The figure below shows the heating curve of water. To calculate the total energy change for such a process, all
the steps should be included.
Figure 4. Heating curve for water (Image source: http://wpscms.pearsoncmg.com/wps/media/
objects/3662/3750037/Aus_content_10/Fig10-19.jpg)
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•
Heat capacity of ice (Cice) = 37.6 J/K mol
•
Specific heat of ice (Swater) = 2.03 J/g OC
•
Heat capacity of water (Cwater)= 75.3 J/K mol
•
Specific heat of water (Swater) = 4.184 J/g OC
•
Heat capacity of steam (Csteam)= 35.8 J/K mol (at constant pressure of 1 atm)
•
Specific heat of steam (Ssteam) = 1.99 J/g OC
•
Melting point = 273.15 K (0 OC)
•
Heat of fusion of ice (ΔHfus)= 6.01 kJ/mol
•
Boiling point = 373.15 K.
•
Heat of vaporization (ΔHvap)= 40.79 kJ/mol
•
Heat of sublimation (ΔHsub)= 46.8 kJ/mol
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C. Post-Lab Questions
1. Have the learners complete the data table for their activity.
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Data for water used in the heating curve:
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•
•
Learners may notice other groups
recording different boiling and melting
temperatures, as well as how long it took
for a phase change to occur.
•
Learners should try to account for any
sources of error in their lab and to
suggest modifications on the procedure
to eliminating some of the errors.
•
If time permits, learners may do a re-run
of the experiment, and see how their two
sets of data compare.
2. Have the learners work with their group mates to graph their data.
3. Have the learners compare their graphs with other groups as well.
differences between their graphs.
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4. Ask the learners the following questions:
Have them look closely for
a. What is the chemistry term for a phase change when a solid becomes a liquid?
b. What is the chemistry term for a phase change when a liquid becomes a gas?
c. Describe the phase change that occurs during solidification (freezing).
d. Describe the phase change that occurs during condensation.
e. What happens to the intermolecular forces of attraction inside an ice cube when it melts?
f.
Why did the temperature of the liquid remain unchanged right around 100 oC even though water was
continuously heated?
g. Describe the difference between a phase change and a temperature change.
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D. Extended Understanding of Concept
Construction of the Cooling Curve for Water: A paper exercise
T, oC
Time elapsed,
min
0.0
24.0oC
180.0
210.0
60.0
240.0
90.0
270.0
150.0
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120.0
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30.0
T, oC
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Time elapsed,
min
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1. Using the same groupings in the earlier part of this laboratory session, ask the students to plot the
following data obtained from cooling a sample of pure water in a special cooling equipment from 24oC
to -8oC:
2. Guide the students in explaining the various parts of the cooling curve based on energy changes and
phase changes in the manner that the heating curve was discussed, although the process observed here
is the removal of heat (cooling).
3. Ask the students to draw the missing part of the cooling curve which is the portion that involves cooling
the sample starting from the gaseous to the liquid state.
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•
The cooling curve for water is like a
mirror image of the heating curve,
especially if the same amount of sample
is used to construct the curves.
•
The temperatures at which phase
changes happen are the same in both
curves
ENRICHMENT (10 MINS)
Teacher Tip
•
Graph Analysis
Individual Assessment
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Provide clear and concise explanation for each of the question below:
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Worksheets for both individual and team
assessments should be provided for the
learners.
Teacher Tip
Questions
•
To ensure that all learners will work on
their own project, this can be individual
work or for a small group of students
•
The following elements will be observed
in rating the poster presentation:
(persuasiveness, effective
communication, visual appeal, correct
mechanics).
Answers
1. What phase(s) exist at each of the numbered sections above?
PY
Point 1
Point 2
Point 4
2. At what temperature is this substance condensing?
3. At what temperature is the substance freezing?
4. At which numbered section(s) is/are phase changes occurring?
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Point 3
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5. At which numbered section(s) is/are kinetic energy of the molecules
the greatest?
6. Relate your answer to #5 to the associated intermolecular force of
the molecules.
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7. Evaluate the change in temperature from point A to E, with regard
to heat.
8. Draw the missing section of this heating curve on the graph and
label the phase that best fits. Using the terms temperature and heat,
justify (prove) your chosen phase.
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Teacher Tip
Expected result:
Questions
Answers
1. What phase(s) exist at each of the numbered sections above?
gas
PY
Point 1
liquid/gas
Point 3
liquid
Point 4
solid/liquid
2. At what temperature is this substance condensing?
3. At what temperature is the substance freezing?
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Point 2
60 oC
20 oC
2 and 4
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4. At which numbered section(s) is/are phase changes occurring?
1
6. Relate your answer to #5 to the associated intermolecular force of
the molecules.
In section 1, intermolecular
force is at the lowest, until
they broke apart to allow the
substance to become a gas.
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5. At which numbered section(s) is/are kinetic energy of the molecules
the greatest?
7. Evaluate the change in temperature from point A to E, with regard
to heat.
Heat is lost from A to E as
the substance cools off
(exothermic).
8. Draw the missing section of this heating curve on the graph and
label the phase that best fits. Using the terms temperature and heat,
justify (prove) your chosen phase.
Should extend below E to
represent the solid phase.
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•
To ensure that all learners will work on
their own project, this can be individual
work or for a small group of students
•
The following elements will be observed
in rating the poster presentation:
(persuasiveness, effective
communication, visual appeal, correct
mechanics).
EVALUATION (10 MINS)
Teacher Tip
•
Learners’ answers will be rated for
correct use and accuracy of data, and
application of concepts.
•
A worksheet should be provided for this
assessment.
Creating a Heating Curve: Team Assessment
Boiling point = 60 oC
Melting point = -105 oC
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Background Information on Ethanol:
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Graph the heating curve of ethanol using the information given. Check off each box as you add
additional information to your graph so that none is missed. Each team member must complete his/her own
graph.
Starting temperature = -120 oC
1. After two minutes, frozen cold ethanol starts to melt. It takes two minutes to melt completely.
2. After eight more minutes, it begins to boil. It boils for six minutes.
4. Label “Melting” where this takes place.
5. Label “Vaporization” where this takes place.
ED
3. Heat is added for two more minutes until ethanol reaches 80 oC.
6. Label “Phase Change” where a phase change occurs.
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7. Indicate where ethanol is only a SOLID, only a LIQUID, and only a GAS.
8. Of the three phases, label which phase has: Weakest IMF (intermolecular force), Strongest IMF, and
Medium IMF.
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D
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Sample Heating Curve Template for Ethanol
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Expected result:
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Chemistry 2
60 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
8
Motivation
Demonstration
5
Discussion
35
Group Task/Activity
5
Written Quiz
7
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Physical Properties of Solutions;
Types of Solutions and Energy of
Solution Formation
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Instruction
Content Standard
The learners demonstrate an understanding of the properties of solutions, solubility, and Enrichment
the stoichiometry of reactions in solutions.
Evaluation
Performance Standard
Materials
The learners design a simple investigation to determine the effect on boiling point or For Review:
27 Meta cards (3” x 7”) with the following words written:
freezing when a solid is dissolved in water.
(STEM_GC11PP-IIId-f-110)
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Learning Competencies
Describe the different types of solutions.
Describe the process of solution formation at the atomic/molecular level.
Describe the energy involved during solution formation.
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Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
give examples of the different types of solutions;
•
discuss what happens at the molecular level when a solution forms; and
•
describe energy of solution formation;
Gas Solutions, Liquid Solutions, Solid Solution, Gas in a gas, Air,
Oxygen (gas), Nitrogen (gas), Gas in a liquid, Soda water, Carbon
dioxide (gas), Water (liquid), Liquid in a liquid, Vinegar, Acetic acid
(liquid), Water (liquid), Solid in a liquid, Seawater, Sodium chloride
(solid), Water (liquid), Liquid in a solid, Dental amalgam, Mercury
(liquid), Silver (solid)
Solid in a solid, Steel Carbon (solid), and Iron (solid)
Masking tape ; Manila paper with a blank table
For Demonstration:
Colorless glass; Water; Teaspoon of sugar or sugar cube ; Plastic
stirrer or spoon, teaspoon
For Lecture and Discussion:
Projector; computer; learners’ gadgets like tablets or cell phones (if
available); Manila paper and markers (when projector and laptop are
unavailable) – these will be intended for preparing visual aids in
advance to show the dissolving process and to illustrate how energy
is involved in solution formation
Resources
(1) Brown, T., LeMay, H.E., Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry the CentralScience (11th ed., pp.528-529). Philippines:
Pearson Education South Asia PTE. LTD.
Additional resources at the end of the lesson.
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INTRODUCTION (8 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
a. Discuss what happens to the particles of a solute when a solution forms at the molecular level;
PY
b. Describe energy of solution formation.
Teacher Tip
•
Objectives should be written
prominently on one side of the
chalkboard for learners to refer.
•
The teacher has the option not to
perform the first concept check (#2).
However it is recommended that the
second short activity (#3) be
performed.
•
The meta cards and the Manila paper
must be prepared ahead of the
scheduled lesson.
a. 25 grams of salt dissolved in 95 mL of water;
b. 25 mL of water mixed with 75 mL of isopropyl alcohol;
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2. Check concepts by asking learners to identify the solute and solvent in each of the following solutions and
explain their answers:
c. Tincture of iodine prepared with 0.20 gram of Iodine and 20.0 mL of ethanol.
Expected answers:
a. Salt, the smaller quantity, is the solute; water is the solvent.
ED
b. Isopropyl alcohol, which has the greater volume, is the solvent; water is the solute
c. Iodine, the smaller quantity, is the solute while ethanol is the solvent
3. Short Warm-up Activity
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Learners perform this short activity to recall the different types of solution. Using the meta cards with the
written words, learners fill the blank table by choosing the appropriate answers from among the bunch of
cards.
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PY
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Expected Answer:
(Reference: Timberlake, K.C. (2012). An Introduction to General, Organic and Biological Chemistry, 11th ed.
USA. Prentice Hall. p.247)
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MOTIVATION (5 MINS)
Start the lesson with a short demonstration by preparing a solution by dissolving a teaspoonful of
sugar in a glass half-filled with water. Stir the mixture well. All the time, learners are asked to observe
closely and keenly.
PY
Focus question:
Describe the appearance of the resulting material.
What can be observed? Explain.
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Possible Observations:
After stirring, sugar dissolves completely in water since it could no longer be seen. The mixture
is clear.
Stirring brings undissolved crystals into contact with water, hence sugar dissolved faster in
water.
Possible questions prior the discussion:
ED
From the activity, learners are further led to think beyond what happened during the mixing of sugar
and water.
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Is it possible to picture out or draw what happens when sugar is mixed with water?
What would happen if salt is used instead of sugar?
The questions will be left open for learners to answer. All their responses will be confirmed after they
see the video or visual aids, accompanied by a discussion on the dissolving process.
Mention that water is the solvent for many common solutions, although there are other types as noted
in the warm-up activity.
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Tell that in the succeeding lesson, liquid solutions are the focus of discussion.
Teacher Tip
•
Guide questions can be written in the
chalkboard for reference during the
course of discussion.
•
It is important that the video presentation
be viewed prior to the class discussion.
•
If there is no internet connection, learners’
gadgets can be used, if available.
However, if there is none at all, it is
advisable to prepare visual aids using
Manila paper and markers to show: (1)
dissolving process of sugar and sodium
chloride in water and (2) energy changes
and solution formation.
INSTRUCTION (35 MINS)
http://www.middleschoolchemistry.com/multimedia/chapter5/lesson4
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The Process of Solution Formation
1. Let learners watch the video on how sugar dissolves in water:
a. Describe the sugar before it was mixed with water.
b. What happens to sugar when it is placed in water?
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2. Provide the following guide questions to be answered while watching the video/ showing the visual
aids:
c. Knowing the characteristic (polarity) of water in previous lessons, why does water dissolve sugar?
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d. Describe what happens when water dissolves sugar.
If the video or internet connection is unavailable, the following visual aids with description can be used for
discussion:
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Slide 1 :Sucrose
•
The ball-and-stick, and first space-filling model show that sucrose is a large molecule made up of carbon,
oxygen, and hydrogen.
•
Sucrose has many O–H bonds which are polar.
•
These polar areas are shown with a + near the hydrogen atom, and a − near the oxygen atom.
•
The second space-filling model shows two sucrose molecules held together by their opposite polar
areas.
•
These molecules will separate from each other when sucrose dissolves.
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PY
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Figure 1. Interactions between sugar molecules
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(Image sources: http://www.middleschoolchemistry.com/img/content/
multimedia/chapter_5/lesson_4/sucrose_1.jpg; http://www.middleschoolchemistry.com/img/
content/multimedia/chapter_5/lesson_4/sucrose_2.jpg)
Slide 2: Water Dissolves Sucrose
•
Water molecules arrange themselves around sucrose molecules according to opposite polar areas.
•
The attraction of water molecules and their motion overcome the attraction among sucrose molecules.
•
Sucrose molecules dissolve as they are separated from other molecules and mix into the water.
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C
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•
ED
Figure 2: Separation of sugar molecules during dissolution in water (Image source:
https://dr282zn36sxxg.cloudfront.net/
Expected answers:
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3. After the presentation, have the learners answer the four questions previously given.
a. Sugar particles were crystalline solids before they were added to water.
b. When sugar is placed in water, the sugar is lost, implying it was dissolved.
c. Water dissolves sugar because both are polar molecules. This means that both have areas of
positive and negative charges. The areas of slight positive and negative charges in water are
attracted to the negative and positive areas in sugar molecules. When the attractions between
water and individual sugar molecules overcome the attraction that sugar molecules have for
other sugar molecules, or water with water molecules, then sugar dissolves.
d. The areas of positive and negative charge on a water molecule are attracted to opposite areas
of negative and positive charge in a sugar molecule. As water molecules associate with sugar,
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the attractions between water and individual sugar molecules begin to overcome the attractions that
sugar molecules have for one another. The water pulls away the sugar molecules one by one, eventually
dissolving the sugar.
4. Encourage learners to ask questions during discussion.
What will happen if salt is used instead of sugar?
Expected Answer:
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Just like sugar, salt will also dissolve in water since it is soluble in it.
PY
5. Motivate learners to give a prediction given this question:
Salt will disappear.
Guide Questions
a. Are learners’ predictions correct?
ED
6. Some learners may explain further. To check the predictions, proceed with a similar demonstration, this
time show salt (sodium chloride) being mixed with water.
b. Describe the appearance of salt before and after it was mixed with water.
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c. What happens when salt is added to water? Explain such observation.
d. What kind of compounds are salt and water?
e. Does the kind of solute affect its solubility in water?
7. Accept learners’ appropriate responses after the demonstration.
8. To clarify further how salt dissolves in water, present the next video, or the visual aids with accompanying
discussion:
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Slide 3 :Salt, NaCl Crystals
• The larger green spheres represent the negative chloride ions of the salt, NaCl.
The smaller gray spheres represent the positive sodium ions.
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•
Figure 3: Representation of a NaCl crystal
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Image source: https://online.science.psu.edu/sites/default/files/chem101/
NaClClusterLg.jpg
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Slide 4 :Salt , NaCl, Dissolving in water
• The positive area of water molecules surrounds the negative chloride ions.
•
The negative area of the water molecules surrounds the positive sodium ions.
•
As the attractions from the water molecules and their motion pull the ions apart, the sodium chloride
crystal dissolves.
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Figure 4: NaCl dissolving in water (Image
source:http://butane.chem.uiuc.edu/
pshapley/genchem1/l21/nacl2.gif)
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After the presentation, facilitate the discussion. Encourage learners to justify their speculations based on
their own emerging models of electrostatics that began with elementary science and were again
addressed in Grade 7 Science. Refer to the guide questions given earlier.
•
Have learners look and compare the previous visual aids for sugar with the present for salt.
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Expected answers:
ED
•
a. Our prediction that the salt will dissolve was affirmed.
b. Salt particles were crystalline solids before they were added to water. The crystalline particles dissolved
in water. Nothing is left undissolved.
c. When salt is placed in water, the salt is lost, implying it was dissolved.
Water molecules have areas of slight positive and negative charges (polar), thus they are attracted to ions
which also have positive and negative charges. The areas of slight positive charge in water are attracted to
negatively charged ions, and the areas of slight negative charge in water are attracted to the positively
charged ions.
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•
It is essential to review intermolecular
and intramolecular forces of attraction.
(STEM_GC11IMF-IIIa-c-100 and 102)
Areas of positive and negative charges on a water molecule are attracted to negative and positive ions
that make up salt, NaCl. As water molecules come in contact with the salt crystal, attractions between water
molecules and ions begin to overcome the attractions that salt ions have for one another. Water pulls away the
ions one by one, thereby dissolving the salt.
PY
d. Sugar is a covalent compound and salt is an ionic compound. However, both are soluble in water even if
they are different types of compounds.
C
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e. The kind of solute affects its solubility in water. Water being a polar solvent is capable of dissolving both
ionic and covalent compounds.
Remind learners to recall previous lesson on intermolecular and intramolecular forces of attraction. Salt is
an ionic compound. Point out that ionic bonds between sodium ions and chloride ions are broken due to
the attraction of polar water molecules with oppositely charged ions. Sugar, on the other hand is a
molecular compound. Explain that unlike salt, when sugar dissolves in water, sugar molecules move away
from each other but their individual molecules do not break apart.
•
The different illustrations diagrammatically detail what happens at the molecular level when sugar, a
covalent solid, or salt, an ionic solid dissolves in water.
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ED
•
•
Although the topic on exothermic and
endothermic reactions, as well as enthalpy
of solution (STEM_GC11TC-IIIg-i-1245) is
to be taken the next lesson on
Thermochemistry, it is necessary to
familiarize with these concepts again.
•
Hot packs are used to relax muscles,
lessen aches and cramps. They are
available in sport shops, and drugstores.
Energetics of Solution Formation: Will a solution form?
The formation of solutions of sucrose and water, and of NaCl and water followed very similar processes:
forces of attraction between solvent molecules are broken, forces of attraction between solute particles are
also broken, while forces of attraction are formed between solute and solvent particles are formed. These
processes can happen with any pair of solute and solvent.
•
The major determining factor in solution formation is the relative strengths of intermolecular forces
between and among solute and solvent particles. The extent to which one substance is able to dissolve
in another depends on the relative magnitudes of interactions between: solute-solute, solvent-solvent,
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and solute-solvent in the solution process, and the energy involved in their breaking or formation.
To illustrate this, start with the following demonstration:
PY
Focus questions:
1) What do athletes immediately use if they are injured in their games?
2) Have you tried cold or hot packs?
3) Discuss how cold packs instantly work to treat athletes’ injuries?
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4) How do hot packs work?
Figure 5: Example of an instant hot pack (Image source: http://
www.thermalice.com.au/wp-content/uploads/2012/11/Instant-Hot-Pack-5transparent.jpg
•
Demonstrate how commercial hot and cold packs work. This is done by simply squeezing or kneading
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each pack. The reaction inside both packs shows how energy in the form of heat is transferred or
produced from solution formation.
Allow learners to participate by feeling or touching each pack after each demonstration. Some will be
asked to share their observations.
PY
•
Expected answers:
The hot pack feels hot, while the cold pack feels cold.
C
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I felt there was something inside the hot pack that when squeezed, is mixed and produces heat and caused the
hot sensation in our hands.
Likewise, there was also something inside the cold pack that when squeezed, is also mixed and gave a cold
sensation.
Discuss the different observations.
Instant hot and cold packs are practical aids, which utilize the heat of chemical reactions during solution
formation.
•
Hot packs consist of a pouch of water and a dry chemical, magnesium sulfate (MgSO4) or calcium
chloride. Cold packs on the other hand, are made of water and ammonium nitrate (NH4NO3). When
either pack is squeezed or kneaded, the seal separating the solid compound from the water is broken,
which then mixes with the chemical. Thus, a solution producing instant heat (increase in temperature) or
cold (decrease in temperature) is formed depending on the chemical used as needed (Brown, LeMay,
Murphy & Woodward, 2009, p. 530).
•
To illustrate how energy changes in solution formation, consider how magnesium sulfate (MgSO4) in hot
packs dissolves in water. Use the figure showing the enthalpy of solution below to further facilitate the
discussion. Three steps are carried out in the process:
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•
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PY
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Figure 6: Enthalpy of solution representation. (Image source: http://
img.sparknotes.com/figures/0/07cf18f888c9c21f4b45687743b63ac3/solnform.gif)
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1. Magnesium sulfate (solute – label this A, the darker spheres) breaks apart from ionic bonds that hold the
ions together in the crystal, allowing the ions to separate from each other. The enthalpy (heat absorbed
or absorbed in the reaction) in this process is marked ΔH1. Since this is an endothermic process, energy
is required for this reaction, thus ΔH1 > 0.
2. The second process is very similar to the first step. In the solvent, water (label this B, the lighter shaded
spheres),also needs to overcome the intermolecular forces between molecules and allow them to
separate from each other. The enthalpy of this process is marked ΔH2. This process is also an
endothermic process, where ΔH2 > 0 because energy is required to break the forces between water
molecules.
Let learners visualize what happened so far using the illustration. Solute A has broken the attractive
forces holding it together, and solvent B has broken the intermolecular forces holding it together as well.
Simultaneously happening with these two processes is the third process. Two values, ΔH1 and ΔH2 are
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so far considered. Both values are greater than zero because both processes are endothermic. Energy is
required to separate both solute-solute
particles and solvent-solvent particles.
PY
3. The third process occurs when solute/ magnesium sulfate (A) and solvent/water (B) mix. The solute
molecules and the solvent molecules form attractive forces. The energy involved in formation of
attractive forces between solute and solvent is marked ΔH3. Unlike the earlier processes which involve
breaking of attractive forces and therefore require energy, formation of attractive force is an energyreleasing process or an exothermic process. Thus, ΔH3 < 0.
C
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It is important to note at this point that the energy involved in breaking or forming an attractive force or
a bond depends on the strength of the attractive force. The stronger the attractive force is, the more
energy is needed to break it. The larger also is the amount of energy released during its formation.
The enthalpy of solution can be written as:
ΔHsolution = ΔH1 + ΔH2 + ΔH3
ED
Explain further that the final value for the enthalpy of solution can either be endothermic or exothermic.
Thus, the enthalpy change in forming a solution (ΔHsolution) can either be greater or less than zero,
depending how much energy is required or given off in each step.
D
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If ΔHsolution > 0, the solution formation is endothermic or energy-requiring.
If ΔHsolution < 0, then this solution formation is exothermic or energy releasing.
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PY
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ED
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Figure 7: Energy Diagram for Exothermic Dissolving Process (Image source: http://
chemwiki.ucdavis.edu/@api/deki/files/8782/=Project1AA.png?revision=2
Recall that when the hot pack was touched, it was hot. This illustrates an exothermic process since heat is
given off due to the formation of solution.
Let learners recall their observations when they touched the cold pack and it was cold. The reaction
between water and ammonium nitrate (NH4NO3) is an endothermic process. The energy diagram below
describes the solution process in the cold pack.
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PY
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ED
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Figure 7: Energy Diagram for Endothermic Dissolving Process (Image source:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/
Solutions_and_Mixtures/Solution_Basics/Enthalpy_of_Solution)
In an endothermic solution process, the sum of the strength of the two processes—breaking of forces of
attraction among solute (NH4NO3) particles and among solvent (H2O) particles—is greater than the magnitude
of the force joining the solute and solvent to form the solution. This means that the energy produced from the
formation of attractive forces between solute and solvent is not enough to supply the energy required for
breaking attractive forces. If the amount of energy lacking is small enough and can be absorbed from the
surroundings, the solution forms. However, if the amount of energy lacking is too big, the solution does not
form.
175
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To summarize, introduce the often-used generalization for solubilities, “like dissolves like”.
Polar solutes dissolve in polar solvents. The energy released from the interaction of polar solute and
polar solvent is sufficiently large to provide for the energy required to break dipolar forces in the
solute and dipolar forces in the solvent.
•
Nonpolar solutes dissolve in nonpolar solvents. Nonpolar substances form weak London dispersion
forces (LDF), and release only a small amount of energy, but enough to be used to break weak LDF
between solute particles and between solvent particles.
•
Nonpolar solutes will not dissolve in polar solvents. Nonpolar substances form only weak LDF with
polar substances. Only a small amount of energy is released. This may be sufficient to break LDF
between nonpolar solute particles, but will be too small to break dipolar interactions between polar
solvent particles.
•
Polar solutes will not dissolve in nonpolar solvents for similar reasons as the preceding case.
ED
C
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PY
•
ENRICHMENT (5 MINS)
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Home Experiment: Group learners into five members each to perform the following tasks at home. They will
report the result in the discussion the following day.
1. Each group conducts an investigation to compare how water and isopropyl alcohol (rubbing alcohol –
70% isopropyl alcohol)) dissolve salt.
2. Prepare the setup considering the following variables:
a) Amount of water and isopropyl alcohol
b) Amount of salt added to each liquid
c) Temperature of each liquid
d) Amount of stirring
176
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3. Report the results of the experiment:
a) Comparison of the solubility of salt in water and isopropyl alcohol
b) Discussion of results considering enthalpy of solution
Expected Answers:
PY
There are less undissolved salt (or salt will dissolve totally) in water than the alcohol. More salt is
dissolved in water than in alcohol.
C
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Water dissolves salt better than alcohol. Since water is more polar than alcohol, it attracts positive
sodium and negative chloride ions better than alcohol. This is why water dissolves more salt than alcohol.
Isopropyl alcohol has a polar and a nonpolar part. NaCl will only interact with the polar part of the
molecule and will form less and weaker attractive forces with alcohol.
Essay
D
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EVALUATION (7 MINS)
ED
The magnitude of the energy involved in the two processes: breaking of solute (NaCl) and solvent
(isopropyl alcohol) particles is greater than that produced on joining the solute and solvent to form the solution.
1. Discuss at the molecular level how sugar dissolves in water.
2. Sodium hydroxide (NaOH) reacts rapidly in water and produces heat. Explain the kind of energy
exchange when this base is added to water.
177
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Expected Answers:
1. The discussion presented earlier in the lesson can be used as basis for marking the answers.
2. The process of dissolving sodium hydroxide in water can be broken down into three steps:
a. The solid ionic compound (NaOH) splits into ions. The energy required to do this is equal to the lattice energy.
PY
b. Water particles also separate from each other and also requires energy. The separation creates spaces for ions of NaOH to fit in.
C
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c. The ions dissolve in solution and become hydrated; the energy equal to the hydration energies of the two ions, Na+ and OH - is released to the
surroundings. Since the process is observed to be exothermic, the magnitude of energy in this third process is greater than the sum of the two
processes— separation of the solute (NaOH) and the solvent (H2O). Hence, ΔHsolution < 0..
Rubrics for the Essay
Argument
Meaningfulness
No Answer
1 point
2 points
3 points
4 points
(Not Visible)
(Needs Improvement)
(Meets
Expectations)
(Exceeds
Expectations)
Able to address the
question.
Able to address the
question.
Able to address the
question.
Able to give at least
50% depth of
understanding.
Able to give at least
75% depth of
understanding.
Able to give 100 %
depth of
understanding.
ED
Presentation
0 point
Does not state
relevant arguments.
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Level of
Achievement
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ADDITIONAL RESOURCES
1. Chang, R. (2005). Chemistry (8th ed., p.116). Singapore: McGraw-Hill Companies.
2. Frank, D.V.et al. (2008). California Physical Science, Teacher’s Edition (pp.257-259). Boston, Massachusetts: Pearson Prentice Hall.
3. Middleschoolchemistry.com,. (2016). Multimedia: Why Does Water Dissolve Salt? | Chapter 5, Lesson 3 | Middle School Chemistry. Retrieved 29 July
2015, from http://www.middleschoolchemistry.com/multimedia/chapter5/lesson3
PY
4. Middleschoolchemistry.com,. (2016). Multimedia: Why Does Water Dissolve Sugar? | Chapter 5, Lesson 4 | Middle School Chemistry. Retrieved 29 July
2015, from http://www.middleschoolchemistry.com/multimedia/chapter5/lesson4
5. Nobel.scas.bcit.ca,. (2016). Hot Pack/ Cold Pack. Retrieved 9 August 2015, from http://nobel.scas.bcit.ca/debeck_pt/science/hotColdPack/
pack_p1.htm#Links
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6. Timberlake, K. (2012). An Introduction to General, Organic and Biological Chemistry (11th ed.). USA: Prentice Hall.
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ED
7. UPNISMED Writing Team. (1991). Science and Technology III Textbook for Third Year High School (1st ed., pp. 174-175). Philippines: Book Media
Press.
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Chemistry 2
60 MINS
Physical Properties of Solutions:
Concentration Units, Mole
Fraction, and Molality
LESSON OUTLINE
5
Communicating learning objectives
Motivation
Short Demonstration
5
Content Standards
Instruction
The learners demonstrate an understanding of the properties of solutions, solubility, and
Enrichment
the stoichiometry of reactions in solutions.
The learners demonstrate an understanding of expressing concentration of solutions in Evaluation
Lecture & Seat Work
30
Group Activity
15
Short Quiz
5
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PY
Introduction
mole fraction and molality.
Materials
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or
freezing point when a solid is dissolved in water.
ED
The learners will be able to solve problems on mole fraction and molality of solution
Short Activity: six graham crackers, three marshmallows, two Kisses
chocolate
Lecture/Discussion: projector, computer, scientific calculator, manila paper
& marker (when project & laptop are unavailable
Enrichment Activity: Four Corners Poster (cartolina) to be posted in front
Resources
Learning Competency
(1) Brown, T., LeMay, H.E., Bursten, B., Murphy, C. & Woodward, P. (2009).
Use mole fraction and molality in expressing concentration of solutions (STEM_GC11PPChemistry: the Central Science (11th ed., pp. 149-154). Philippines:
Pearson Education South Asia PTE. LTD.
IIId-f-111)
D
EP
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
express concentration of solutions in mole fraction and molality; and
•
perform calculations for solution concentration given appropriate data.
(2) Masterton, W.L. (2009). Chemistry Principles and Reaction (6th ed., p.
259). Belmont, CA: Brooks/Cole Cengage Learning. p. 259.
(3) Padolina, M.C., Antero, E.S., & Alumaga, M.J.B. (). Conceptual and
Functional Chemistry Modular Approach (pp. 216-219). Manila: Vibal
Publishing House, Inc.
(4) Study.com,. (2016). Molality: Definition & Formula – Video & Lesson
Transcript| Study.com. Retrieved 2 November 2015, from http://
study.com/academy/lesson/molality-definition-formula.html
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INTRODUCTION (5 MINS)
Teacher Tip
PY
In this lesson, students continue to learn about other quantitative ways to express concentration of
solution. Moreover, they continue solve problems with their peer and assess their own progress as they think,
share, and analyze their work for learning to take place.
Common ways of expressing concentrations
have been taken up in Grade 7.!
Communicate learning objectives.
1. Express concentration of solutions in mole fraction and molality;
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2. Perform calculations for solution concentration given molality in terms of mass needed or vice versa.!
Review
MOTIVATION (5 MINS)
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Tang juice as a visual analogy for Molality
ED
Students are asked to recall their previous lesson focusing on the enrichment activity, “Wastewater Four
Corners” and share their experiences when they did the group activity.
1. Show a glass of Tang juice. Ask one student to taste this and let him/her describe the taste after the test.
2. Ask the following questions:
What does Tang juice tastes like if you don’t put much powder?
What does Tang juice tastes like if you put too much powder? Too little?
3. Let students give reasons for their observations. Lead them to discuss the difference in terms of
“concentration:”
They may discuss concentration in terms of molarity. A hint that there are other ways of expressing
concentration of solution can be given.
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Teacher Tip
Students should be informed there are other
ways to present or express concentration of
solution. !
INSTRUCTION (30 MINS)
Begin the short lecture focusing on other ways to express the relative amounts of solute and solvent in a
solution:
Concentration expressions are often based on the number of moles of one or more components of a
solution. The three most commonly used are molarity, which had been discussed earlier, mole fraction,
and molality.
•
Mole Fraction is a way of describing solution composition. It is the ratio of the number of moles of one
component of a mixture to the total number of moles of all components. This is symbolized by the Greek
lowercase letter chi, , with a subscript to indicate the component of interest. It is computed using the
formula:
Mole fraction of component =
Moles of component
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PY
•
Total moles of all components
For example, the mole fraction of NaOH in a sodium hydroxide solution is represented as NaOH.!
ED
!
On the other hand, the mole(s) of a given component can be calculated this way:
Mass of the component in grams
D
EP
Mole of a substance (component) =
Molar Mass of the component in grams/mole
Using board notes, explain how to solve the following problem involving mole fraction:
What is the mole fraction of the solute in a 40% by mass ethanol (C2H6O) solution in water?
The problem asks the mole fraction of the solute (C2H6O), given only the percentage by mass (40%) of the
solute in the solution.
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Teacher tip
•
If computer and LCD are unavailable, it is
important to prepare the problems for
discussion ahead. These could be written
in Manila papers big enough for easy
reading.
Given: 40% by mass ethanol solution
MM ethanol = 46 g/mole
MM water = 18 g/mole
PY
Unknown: mole fraction of solute
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The following steps can be employed to solve the problem:
Step1: In converting concentration units based on the mass or moles of a solute and solvent or
mass percentage, it is useful to assume a certain total mass of solution. Assume there is exactly 100 grams of
solution. Because the solution is 40% ethanol (C2H6O), it contains 40 grams of ethanol and 60 grams of water.
mole ethanol =
40 g
46 g/mol
!
!
!
!
60 g
= 0.87 mol
= 3.33 mol!
D
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mole water =
ED
Step 2: Change the masses of the components ethanol and water to number of moles.
!!!!!!!!18 g/mol
Step 3: Substitute the values obtained in the formula and solve for the mole fraction of
the solute ethanol, and the solvent water.
(x) mole fraction ethanol
=
mole ethanol
mole ethanol +
!
mole water
183
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xethanol =
0.87 mole
= 0.21
The mole fraction of water can be solved using the formula :
xwater
=
mole water
3.33mol water_______ = 0.79
0.87 mol + 3.33 mol
C
O
mole ethanol + mole water
=
PY
0.87 mol + 3.33 mol
Another way to obtain the mole fraction of water is to simply subtract the mole fraction
ED
of ethanol from 1.00 to obtain that of water. This is possible since there are only two
components in the solution.
The mole fractions of all components of a solution (A, B, …..) must add to unity, that is:
D
EP
x A + xB + . . . . . . = 1
Give the second problem to be solved. Instruct students to work alone first, and then share answers with
their seatmate for clarification. After the allotted time, ask a student to solve the problem and explain how he/
she was able to do this.
A 40.0 gram-sample of methanol, CH4O is mixed with 60.0 grams of ethanol, C2H6O. What is
fraction of the methanol?!
the mole
Expected answer:
xmethanol = 0.49
184
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In case the class needs another to solve, give the next problem:
Hydrogen peroxide, H2O2, is used by some water treatment systems to remove the
disagreeable odor of sulfides in drinking water. An aqueous solution of H2O2 prepared in the laboratory was
found to have a concentration of 20.0% by mass. What is the mole fraction of H2O2?
PY
Expected answer:
xH2O2 = 0.117
•
C
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Introduce molality, another important way of expressing concentration of solutions in chemistry. This whole class
discussion uses board notes on molality calculations with examples.
Molality (symbolized by m) is the ratio of the number of moles of solute per kilogram of solvent. It is not
the same as molarity, even if their names are very similar. In molarity, the number of moles of solute is
divided by the volume of the solution, in liters.!
One offshoot of the difference of molality from molarity is that molality does not change with the
solution’s temperature. In molarity, the volume of a solution can change with temperature due to
expansion or contraction, while the mass of solvent in molality does not change with temperature.
•
In equation form:
moles of solute
kilogram of solvent
•
or m = n solute
m = mol/kg
D
EP
m=
ED
•
m solvent!
In problems involving molality, additional formulas are sometimes used to get the final answer. One very
useful formula is that for density:
d=m/v
where d = density,
m = mass
v = volume
185
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•
It is important for students to take the
initiative to “think” or work first alone,
and then pair with their seatmate to
discuss their work.
•
Check how students interpret and
attempt to answer the problem.
Using board notes, explain how to solve the following problem involving molality:
What is the molality of a solution containing 0.75 moles of sodium hydroxide in 500 milliliters of water at 25oC?
The density of water at 25oC is 1.0 gram per milliliter.
PY
Given:
d H2O = 1.0 g/mL
V H2O = 500 mL
n NaOH = 0.75 mol
Unknown:
molality of NaOH solution
C
O
MM H2O = 18 g/mol
ED
The following steps can be followed to solve the problem:
Step1: Determine the mass of water using the density formula. Also convert the mass
d= m/v
D
EP
in grams to kilograms.
mH2O = d H2O x V = 1.0 g/mL x 500 mL = 500 grams x 1 kg/1000 g = 0.50 kg
Step 2: Substitute the given data in the formula to solve for molality.
m=
moles of solute
kilogram of solvent
186
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•
Notice that temperature is indicated in
the problem when density is used. The
density of a substance can change with
temperature.!
m = 0.75 moles NaOH
=
1.5 molal or 1.5 mol/kg
0.50 kg solvent
PY
Think-Pair Share
Give students a short timeframe (1-2 minutes) to think and solve the second problem on their own. After
which, advise them to discuss their work with their seat mate. Ask a student to present and discuss his or her
work to the class.
Expected Answer: !
a. Convert the mass of solute (sucrose) to number of moles
mass C12H22O11
=
75.5 g
= 0.221 mol sucrose!
ED
n=
C
O
What is the molality of a solution containing 75.5 grams sucrose in 400.0 grams water?
molar mass C12H22O11
342.0 g/mol !
D
EP
b. Change the mass of solvent from grams to kilograms.
mass solvent = 400 g x 1 kg/1000 g = 0.400 kg
c. Solve the molality using the formula:
m=
n solute__
m solvent
=
0.221 mol
= 0.553 mol/kg
•
0.400 kg
Give the next problem and ask someone to do the computations on the board and explain how the solution
with the required molal concentration can be prepared.
187
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Check how students are doing their
seatwork.
How many grams of sodium hydroxide (NaOH) are needed to prepare a 0.700 molal solution using 700.0 grams
water?
Expected Answer:
m solution = 0.700 mol/kg
mass H2O = 700.0 g
C
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Unknown: mass of NaOH needed to prepare a 0.700 molal solution
PY
Given:
a. Determine the number of NaOH moles needed to prepare the required concentration.
m=
n solute____ _ thus,
n solute = m x m solvent = (0.700 mol/kg) ( 0.700 kg)
m solvent in kg
ED
n solute = 0.490 mol NaOH!
b. Determine the mass of NaOH from the calculated n of NaOH.
D
EP
Mass NaOH = n x molar mass NaOH!
= (0.490 mol) (40.0 g/mol)
= 19.6 g NaOH!
Hence, to prepare 0.700 m NaOH solution, 19.60 grams of NaOH is needed to be weighed
dissolved in 700.0 grams of water.
and
Ask students if they have questions and problems on the lesson presented.
188
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Let students or volunteers to summarize the gist of the lesson on mole fraction and molality as ways of
expressing concentration of solution.
Teacher tip
The final problem should be given to the
whole class for analysis.
ENRICHMENT (15 MINS)
PY
Assign students to do the following problem set which will be submitted the following day.
Problem Set:
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1. Calculate the mole fraction of the solute in the following solutions:
a. 100.0 grams C2H6O in 100.0 grams H2O
b. 30% HCl solution by mass
Expected answers:
D
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1. a. mole fraction of C2H6O = 0.2811
ED
2. A solution is prepared by mixing 1.00 gram of ethanol (C2H6O) with 100.0 gram water to give a final
volume of 101 mL. Calculate the mole fraction for the solute and solvent, and the molality of ethanol in
the solution.
b. mole fraction of HCl = 0.22
2. a. mole fraction
nC2H6O = 0.00389
nH2O = 1 - 0.00389 = 0.9961
189
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b. molality of C2H6O = 0.2217 m
EVALUATION (5 MINS)
PY
Written Test( Quiz)
A. Describe mole fraction and molality as methods of expressing concentration of solution.
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B. Solve the given problem:
A solution of phosphoric acid was made by dissolving 10.0 g H3PO4 in 100.0 mL of water. The resulting
volume was 104 mL. Calculate the mole fraction (solute and solvent) and molality of
the solution. Assume
3
water has a density of 1.00 g/cm .
ED
Expected Answers:
B.
D
EP
A. Mole fraction is the ratio of the number of moles of one component in a solution to the total number
of moles of all the components in a solution. Molality on the other hand, is defined as the number of
moles of solute per kilogram of solvent.
mole fraction of H3PO4 = 0.0180
mole fraction of H2O = 0.9820
molality of solution = 1.02 mol/kg
190
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Chemistry 2
120 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives and
Review
20
PY
Physical Properties of Solutions:
Acid-Base Titration &
Concentration of Solutions
C
O
Motivation
Short Demonstration
10
Content Standards
Group Activity and Class Discussion
70
The learners demonstrate an understanding of the properties of solutions, solubility, and Instruction
the stoichiometry of reactions in solutions.
Enrichment
Group Activity
10
The learners demonstrate an understanding of the quantitative properties of solution
Evaluation
Quiz and Discussion of Lab Results
10
through acid-base titration to determine concentration of solutions.
Materials
Performance Standards
Review: masking tape; Manila paper with written review problems
The learners design a simple investigation to determine the effect on boiling point or
Demonstration: colored powdered juice; three transparent drinking
glasses half-filled with water: first glass with 1 tbsp. of powdered juice,
freezing point when a solid is dissolved in water.
ED
The learners design a simple procedure to determine the percentage of acetic acid,
HC2H3O2 in a sample of vinegar.
(STEM_GC11PP-IIId-f-119)
D
EP
Learning Competency
Perform acid-base titration to determine concentration of solution.
Use different ways of expressing concentrations of solutions: molarity.
(STEM_GC11PP-IIId-f-111)
second glass with 2 tbsps. of powdered juice powder, and third glass with
3 tbsps. of powdered juice
Discussion: projector; computer; calculator for students (if available);
Manila paper and markers (when projector and laptop are unavailable)
Group Activity: iron stand; 10 mL pipette; two beakers (50 mL); burette
clamp; base burette; aspirator; one graduated cylinder (25 mL or 50 mL);
three Erlenmeyer flasks or beakers (250 mL); medicine dropper for
indicator; phenolphthalein indicator; 250 mL distilled water; 50 mL of 0.1M
or mol/L sodium hydroxide in a clean, dry labeled glass container (titrant);
50 mL hydrochloric acid (analyte)
Resources
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
use experimental results to determine the concentration (molarity) of an acid or base
by titration;
•
solve problems involving molar concentration of solution; and
•
carry out a titration procedure to determine the percentage of acetic acid (HC2H3O2)
in sample vinegar.
(1) Brown, T., LeMay, H.E., Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry the Central Science (11th ed., pp. 542-526). Philippines:
Pearson Education South Asia PTE. LTD.
(2) Committee on General Chemistry (1986). Laboratory Manual and
Workbook for Chemistry (Revised Ed. Part 1., pp. 101-103) Quezon
City: Ken Incorporated.
Additional resources at the end of the lesson
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INTRODUCTION (20 MINS)
Teacher Tip
•
If there is a need to review students on
the following ways of expressing
concentration: percent by mass,percent
by volume, parts per million, and mole
fraction, then they should be given
problems to solve before proceeding to
the present lesson.
•
The objectives should be clearly written
on one side of the chalkboard for
students to see.
•
The following formulas can be used for
the review questions:
PY
There are several quantitative ways to express the relative amounts of solute and solvent in a solution.
One is by determining the number of moles of solute per liter of solution or its molarity. The concept is
introduced using a neutralization reaction data obtained through acid-base titration. Titration is a technique
which makes use of a solution of known concentration to determine the concentration of a substance of interest
present in an unknown solution.
Other ways of expressing concentration such as percent (%) by mass and volume, and parts per million
(ppm) were already taken in Grade 7 (Quarter I), while mole fraction was taken in Grade 9 (Quarter II).
Communicate the learning objectives:#
2. Solve problems involving molar concentration of solution;#
C
O
1. Use experimental results to determine the concentration (molarity) of an acid or base by titration;#
ED
3. Design a simple investigation to determine the percentage of acetic acid (HC2H3O2) in a sample
vinegar.
Review (10 mins)
Ask the learners to answer the following questions to check their prior knowledge on the concepts presented in
each of the query. Let a learner explain how he/she was able to attain the correct response.
D
EP
1. Calculate the percentage concentration of a solution that contains 20 grams of sodium hydroxide,
NaOH,. in a 500 mL solution.
Teacher Tip
The following formulas can be used for the
review questions:
2. Calculate the percentage concentration of ethyl alcohol in a solution of 10 mL ethyl alcohol in 40 mL
water.#
% (m/m) =
3. In the United States, drinking water cannot contain more than 5 x 10-8 gram of arsenic per gram of
water, according to law. Express this concentration of arsenic in ppm.
mass solute____
total mass of solution
% (v/v) =
volume solute
x 100
total volume of solution
4. Differentiate an acid from a base.
ppm =
192
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mg solute
liter of solution
x 100
Expected answers:
mole fraction solute =
mole solute
mole solute + mole solvent#
1. 4%
2. 20%
mole fraction solvent =
mole solvent
mole solute + mole solvent#
3. 0.05 ppm#
PY
4. An acid is a substance that is capable of donating a proton to another substance, while a base is a
substance that accepts protons (Bronsted-Lowry).
Acid
-
Analyte
-
Base
-
Equivalence point
-
Indicator
-
Molarity
-
Neutralize
-
Titrant
-
Titration
D
EP
ED
-
C
O
List the following science terms that the learners will encounter in the course of the lesson:
MOTIVATION (10 MINS)
Teacher’s Demonstration
#
Show three glasses with varying amount of powdered orange juice. Ask students to smell, observe, and
compare the different solutions. Ask which one contains the most powdered juice? Ask the learners how they
arrived at their answer.#
193
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•
These terms could be given a day before
the actual lesson. This will enable
learners to study in advance what they
can expect in the new lesson. The terms
acid and base are still included since
students will work with these substances.#
Expected Answer: !
!
The darkest colored juice is probably the sweetest among the three glasses. Perhaps it contained the
most amount of powdered juice.
Teacher tip
INSTRUCTION (70 MINS)
PRE-LABORATORY DISCUSSION (20 minutes)
Divide the class into 6 groups. The group’s composition will depend on class size and availability of
chemicals and equipment.#
2.
Distribute Student Activity Sheet (SAS) 1: Acid-Base Titrations. Give five to ten minutes for reading
and discuss some important ideas before the activity.
ED
1.
Focus Questions:
D
EP
A.
C
O
PY
From the learner’s observation, discuss that the differences in properties of the samples may be
explained by knowing the amount of orange powder present in each of the prepared solutions. There are
several ways of expressing the concentration of a solution, and one of these, molarity, will be the focus of the
day’s lesson.
a.
Describe a standard solution. How can we prepare standard solution?
b.
What is titration?
c.
Differentiate titrant from analyte.
d.
In titration, when is an acid-base reaction completed? What material can be used to help detect
the end point of titration?
e.
What is molarity of a solution?
Expected Answers: #
194
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Reduce the amount of chemicals when
necessary. Micro scale experiment can also be
applied.
A standard solution is one whose concentration has been accurately determined. It is prepared by
accurately weighing a pure solute, called a primary standard, and dissolving it to a specific volume.
b.
Titration is the process of determining the volume of one solution required to react quantitatively
with a given volume of another in which one solution is added to the other, a small amount at a
time until just sufficient has been added to complete the reaction. #
c.
In titration, the solution with known concentration is the titrant (delivered from a burette), while the
analyte is the solution of unknown concentration.
d.
The completion of an acid- base reaction is indicated by a distinct change in color of an indicator.
This is the equivalence point (end point) or the completion of an acid - base reaction, when all
the acid has been neutralized by the base.
C
O
PY
a.
At the equivalence point, the stoichiometric amounts of acid and base should have been mixed to
achieve neutralization, based on the balanced equation.#
#
e.
ED
An indicator is a substance that changes color when a reaction is completed.In the present activity,
phenolphthalein is utilized as an indicator, a substance that is pink in basic solutions, and clear in acidic
ones.
Molarity is one way to express the concentration of a solution in moles of solute present in one (1)
liter, L, of solution. It can be used to convert between moles of solutes and volumes of their
solutions.
Safety precautions :!
D
EP
#
- Never let students stir solutions with their fingers. They should use a glass rod.#
-
If either acid or base solution gets into a student’s eyes, flush them with water until help arrives. If
either solution gets on their skin, this can be washed off with soap and plenty of water after
examining they have no serious burns.#
-
Acids and bases should be stored in separate container.
-
Remind students to wash their hands after working and ensure they don’t rub any acid or base into
195
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Teacher tip
It is important that the base (NaOH) be
prepared and standardized before the actual
laboratory activity.
their eyes.
-
•
It is important for students to take the
initiative to “think” or work first alone,
and then pair with their seatmate to
discuss their work.
•
Guide questions can be written in the
chalkboard for reference during the
discussion,
•
The investigation of the different groups
should be done under the guidance of
the teacher.#
PY
LABORATORY ACTIVITY PROPER (30 minutes)
B.
Distribute the materials to the different groups. #
2.
Check other required materials to be brought by each group.
3.
Let the different groups perform Activity 01 (SAS 01) #
4.
Learners are expected to fill the table (Data for Acid-Base Titration) and answer the different questions
in the activity sheet.
C
O
1.
POST-ACTIVITY (20 minutes)
ED
Ask among the groups to show and discuss their filled table. Encourage them to show their
computations and explain how they obtained the concentration, molarity, of HCl.
Let the groups compare and discuss their obtained results.
D
EP
C.
All waste materials should be poured into a large beaker together and treated before being thrown
into waste.
Discuss other questions in SAS 01. Emphasize the following questions :
a.
What are the reactants in this reaction?#
b.
What are the products of this reaction?#
c.
Describe what would have happened in one of the titrations if phenolphthalein was not added to
the sample flask.
Expected answers:
a.
Hydrochloric acid (HCl) and sodium hydroxide (NaOH) are the reactants.
b.
The products of the neutralization reaction are sodium chloride and water:#
196
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HCl
—>
NaCl
+ H2O
The equivalence point cannot be determined if the indicator is not added to the HCl solution.
Provide another sample problem on acid-base reaction to solve molarity.
Example:
PY
c.
+ NaOH
C
O
If 20.0 mL of HCl solution of unknown concentration requires 35.0 mL of NaOH solution with 0.25 mol/L
concentration to reach the end point in a titration, what is the HCl solution’s concentration?
Expected Answer: #
M HCl = 0.44 mol/L
D
EP
ENRICHMENT (10 MINS)
ED
Let students share some science ideas they have learned from the lesson.
Using the same groupings, assign students to perform an investigation determining the percentage of acetic
acid (HC2H3O2) in vinegar samples, to be performed during their vacant time in school.
Each group will use a different local brand and compare which among the samples they will recommend
customers to use. They will show and discuss their results the following day.
197
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EVALUATION (10 MINS)
Titration involves the gradual addition of a solution (___________________) of known concentration
to a known volume of a second solution (______________________) until the
__________________________ is attained.
2.
The end point is indicated by a color change of the _____________________.
3.
H2O +
Based on the balanced equation for the neutralization, HC2H3O2 + NaOH —>
NaC2H3O2, when the number of moles of acetic acid = number of moles of sodium hydroxide, the
________________________________ is attained.
Solve the problems and show your computations.
1.
What is the molarity of a sodium hydroxide solution if 9.0 mL of the solution is titrated to the end
point with 10.0 mL of standard 0.20M hydrochloric acid?
2.
What is the molarity of a sodium hydroxide solution if 4.50 mL of the solution is titrated to the end
point, with 5.00 mL of acetic acid with 0.0830M?
Expected Answers:
A.
C
O
PY
1.
ED
B.
Fill in the missing words:
D
EP
A.
1. Titrant; analyte; equivalence point
2. Indicator
3. Equivalence point
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B.
1)
Balanced equation:
NaOH
mol HCl used:
+ HCl
—>
H2O + NaCl
0.010 L x 0.20 mole HCl
=
0.0020 moles HCl
=
0.0020 moles NaOH
mol NaOH reacted:
0.0020 mol HCl x
1 mol NaOH
PY
L
Molarity of the NaOH solution:
0.0020 mol NaOH
0.0090 L
NaOH
+
HC2H3O2
mol HC2H3O2 : 0.0830 mol HC2H3O2
L
mol NaOH: 0.000415 mol HC2H3O2
—>
0.22 M NaOH
H2O +
ED
Balanced equation:
=
x 0.00500 L
x
NaC2H3O2
= 0.000415 mol HC2H3O2
1 mol NaOH
D
EP
2)
C
O
1 mol HCl
= 0.000415 moles NaOH
1 mol HC2H3O2
M NaOH solution:
0.000415 moles NaOH =
0.0922 M NaOH
0.00450 L
199
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ADDITIONAL RESOURCES:
Deft Studios,. (2016). Lab Activity 20: ACID_BASE TITRATION LAB. Retrieved 27 September 2015, from http://
www.deftstudios.com/webchem/pdf/lab20sm.pdf#
Masterton, W.L. (2009). Chemistry Principles and Reaction (6th ed., pp. 259-262). Belmont, CA:
PY
Brooks/Cole Cengage Learning.
Mendoza, E. E . & Religioso, T. F. (1997). You and the Natural World Science and Technology Series,
C
O
Chemistry (pp.231-239). Quezon City: Phoenix Publishing House, Inc. #
Sharp, D.W.A. (1987). Dictionary of Chemistry (p.400). Great Britain: Richard Clay Ltd.Bungay.
Stoker, H.S. (2008). Exploring General, Organic, and Biological Chemistry (pp. 268-276). Philippines:
ED
Cengage Learning Asia Pte. Ltd. pp. 268 – 276.#
D
EP
https://s3.amazonaws.com/el-gizmos/materials/TitrationSE.pdf (Retrieved September 25, 2015)#
200
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Chemistry 2
60 MINS
Physical Properties of Solutions:
Solution Stoichiometry
LESSON OUTLINE
Introduction
15
Short Activity
5
PY
Motivation
Content Standards
The learners demonstrate an understanding of the properties of solutions, solubility, and Instruction
the stoichiometry of reactions in solutions.
Enrichment
The learners demonstrate an understanding of stoichiometry of reactions in solutions.
Evaluation
Performance Standards
Materials
The learners design a simple investigation to determine the effect on boiling point or
Communicating learning objectives
The learners use stoichiometry to solve problems, including those involving real life
situations.
understand solution stoichiometry and its usefulness in relation to real world
application; and
D
EP
•
ED
Learning Competency
Perform stoichiometric calculations for reactions in solution. (STEM_GC11PP-IIId-f-112)
•
30
Group Activity
5
Homework: Problem-Solving
5
C
O
freezing point when a solid is dissolved in water.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Pair Activity and Discussion
perform stoichiometric calculations for reactions in solution using model on how to
set up the problem and the steps to solve it.
Short Activity: six graham crackers, three marshmallows, two Kisses
chocolate
Lecture/Discussion: projector, computer, scientific calculator, manila paper
& marker (when project & laptop are unavailable
Enrichment Activity: Four Corners Poster (cartolina) to be posted in front
Resources
(1) Brady, J.E. & Holum J.R. (1988). Fundamentals of Chemistry (3rd ed.,
pp. 130-133). United States of America: John Wiley & Sons.
(2) Brown, T., LeMay, H.E., Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry: the Central Science (11th ed., pp. 149-154). Philippines:
Pearson Education South Asia PTE. LTD. !
(3) Chemed.chem.purdue.edu,. (2016). Acid-Base Titration 1. Retrieved 5
September 2015, from http://chemed.chem.purdue.edu/genchem/
probsolv/stoichiometry/acid-base1
(4) Masterton, W.L. (2009). Chemistry Principles and Reaction (6th ed., p.
259). Belmont, CA: Brooks/Cole Cengage Learning. p. 259.
(5) Mendoza, E. E . & Religioso, T. F. (1997). You and the Natural World
Science and Technology Series, Chemistry (pp. 237-238). Quezon City:
Phoenix Publishing House, Inc.
(6) https://oise-is-chemistry-2011-2012.wikispaces.com/file/view/
Turner_Ross_Chemistry_LP.pdf
(7) Turner_Ross_Chemistry_LP.pdf (Retrieved September 15, 2015)
201
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INTRODUCTION (15 MINS)
Teacher Tip
PY
As students build their knowledge on concentration of solution, they will be lead to the concept of
stoichiometry as it will be related to the chemistry of solution. As a topic, stoichiometry is often a source of
weariness among learners in as much as mathematical skills is necessary in solving stoichiometric problems.
Thus to make this subject more understandable a model will be used and examples are to be performed at the
board as a class. Students will also have an opportunity to do problems with a peer and assess their own
progress as they think and share their learning as a pair. (3 minutes)
Communicate learning objectives.
C
O
1. Understand solution stoichiometry and its usefulness in relation to real world application;!
2. Perform stoichiometric calculations for reactions in solution using model on how to set up the
problem and the steps to solve it.
ED
Review (12 minutes)
Give the review question on how to compute for the molarity of solution in an acid-base titration. Ask
someone to do the solution in the board:!
D
EP
Calculate the molarity of an acetic acid solution if 34.57 mL of this solution is needed to neutralize 25.19
mL of 0.1025 M sodium hydroxide.!
CH3COOH (aq) + NaOH (aq)!!
Expected Answer:
—>
NaCH3COOH (aq) + H2O (l)
Convert the volume of both acid and base from mL to L.!
202
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Objectives should be clearly written on one
side of the chalkboard for reference when
presenting the learning objectives.
25.19 mL
X
1L
=
0.02519 L
=
0.03457 L
1000 mL
34.57 mL
X
1L
Calculate the number of moles of sodium hydroxide used up in the reaction.
Teacher tip
•
This portion can be given after the topic
on acid-base titration depending on the
students’ assimilation of the lesson.
However, if there is a need to take this in
the present lesson, it is strategic to give
this before solution stoichiometry.!
•
Some students need to be encouraged
to show their computations. Give them
enough time to calculate first then ask for
volunteers.!
C
O
moles NaOH = 0.1025 mol x 0.02519 L = 0.002582 mol NaOH
PY
1000 mL
L!
In the balanced chemical equation, note that 1 mole of acetic acid needs 1 mole of sodium hydroxide for
complete neutralization.
ED
Hence, to calculate the number of moles of CH3COOH present in the solution, !
Present the idea that a solution can also be prepared to a specified molarity by:!
Weighing the calculated mass of solute in the laboratory and dissolving it in enough solvent to form the
desired volume of solution:!
molarity (M) =
D
EP
•
moles solute
=
volume of solution (L)
•
mass solute/molar mass solute!
volume of solution (L)
Starting with a more concentrated solution, dilute with water to give a solution of the desired molarity:!
203
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n solute (concentrated solution) = n solute (dilute solution)
Mc Vc
=
Md Vd!
PY
Where subscripts c and d stand for concentrated and dilute solutions respectively.
Use the following sample problems for board work and discussion. Ask some volunteers:
C
O
1. If 149.1 g of KCl is dissolved in water to make 500 mL solution, what is the molarity of the solution? !
Expected answers:
ED
2. Copper sulfate is widely used as a dietary supplement for animal feed. A laboratory technician prepares
a 1.0 M “stock” solution of CuSO4 . An experiment requires 1.5 L of a 0.10 M solution of CuSO4.
Describe how the solution is to be prepared.!
Question number 2 requires finding out the volume of the stock 1.0 M CuSO4 that should be
to give 1.5 L of 0.10 M CuSO4 solution. The dilution equation can be used here. !
=
Md Vd!
D
EP
Mc Vc
Vc = Md X Vd
Mc
= 0.10 M X 1.5 L
diluted
= 0.150 L = 150 mL!
1.0 M
Thus, to prepare the solution, measure 150 mL of the 1.0 M CuSO4 stock solution and dilute
enough water to form 1.5 L of 0.10 M CuSO4 solution.!
with
204
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MOTIVATION (5 MINS)
Teacher tip
“What is the composition of SM Choco?” !
•
Materials for the motivation activity
“What is the composition of SM!Choco?”
should be prepared in front of the
students.!
•
It is important to move around to check
how students interpret and attempt to
answer the questions.!
•
Stoichiometry is a scary topic, according
to some students. It is therefore
important to introduce the module to
serve as guide in solving stoichiometric
problems involving concentration of
solutions. !
•
The topic on stoichiometry has been
covered in CHEM 1 under LCs #38-41. !
1. Show the ingredients of SM Choco:
!
Three Marshmallow (M)!
!
Two Kisses chocolate (KC)
PY
Six Graham Crackers (GC); !
!
Two Graham Crackers (GC)!
!
One Marshmallow (M)!
!
One Kisses chocolate (KC)
C
O
2. Inform students that one SM Choco is composed of:!
Focus Questions!
!
ED
3. Instruct students to make more SM Choco using the ingredients presented. Let them write their own
chemical equation.
How might “we write that in a chemical formula format?”
a. What would the reactants be?!
D
EP
b. What is the product? How many SM Choco can be prepared?!
c. What is the ratio of the ingredients to form SM Choco? !
Expected Answers:!
2 Graham Crackers (GC) + 1 Marshmallow (M) + 1 Kisses chocolate (KC)
Reactants
——>
1 SM Choco
Product!
205
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The ingredients can only form 2 SM Choco. The Kisses chocolate lacks to come up with
another SM Choco. The ratio of the ingredients:
PY
2 GC : 1 M : 1KC = 1 SM Choco
INSTRUCTION (30 MINS)
C
O
Present solution “stoichiometry” as how it is simulated in the motivating activity. This is illustrated in the way the
number of ingredients is combined to produce an SM Choco. !
!
Stoichiometry!
!
Mole!
!
Mole to mole ratio!
!
Molar mass!
ED
Recall the following terms that were met in previous lessons and would be used in the course of discussion.
Stoichiometry !
D
EP
Expected Answers:
is the relationship between the relative quantities of substances taking part in a reaction or
formation of a compound. !
!
!
Mole !
is the SI unit for amount of substance.!
!
!
!
Mole to mole ratio!
206
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!
is the quantitative relationship between the amounts of reactants and/or products in a given chemical
equation.!
Molar mass is the mass of one mole of substance, usually measured in g/mol.!
PY
This is clearly simulated in the activity, such that you can say:
The reactants: 2 moles of GC need 1 mole of M and 1 mole of KC to produce 1 mole of SM Choco!
C
O
The present lesson focuses on solution chemistry such that problems presented will incorporate concentration. It
is important that correct conversions are observed, thus labeling units used is a must.!
D
EP
ED
Present board notes, examples, and use a model like the one shown below to set a given problem and the steps
to solve it. !
Present example problems for board work:!
1. Nitric acid reacts with sodium hydroxide in solution to give sodium nitrate and water.!
HNO3(aq)
1 mole
+
NaOH(aq)
1 mole
—>
NaNO3(aq)
1 mole
+
H2O(l)
1 mole
207
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Let students check if the reaction is balanced. If yes, proceed to the next step.!
Ask: “How many moles of water are formed when 25.0 mL of 0.100 M HNO3 completely reacts with NaOH?”!
Relate the situation to the model earlier presented: !
—>
mol A
—>
mol B
—>
QB
PY
QA
In the given chemical equation, there is HNO3. How many moles of this material are reacting to produce water?!
!
!
!
V HNO3 = 25.0 mL X 1L/1000 mL = 0.025 L.
Thus:!
!
moles HNO3 = M HNO3 X V HNO3 = 0.100 mol/L
!
C
O
!
X
0.025 L
= 0.0025 mol
Computing for the moles of water formed:
ED
According to the balanced chemical reaction, for every mole of HNO3 reacted, one mole of H2O is formed. The
mole: mole ratio is 1:1.!
D
EP
moles H2O = mol HNO3 X mole ratio of H2O to HNO3
2. What volume of a 0.470M HCl reacted with enough Zn metal to produce 1.50g ZnCl2 based on the following
reaction?
Zn(s)
+
2 HCl(aq)
—>
Molar mass of ZnCl2 = 136.29 g/mol
ZnCl2(aq)
+
H2(g)
!
!
208
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Teacher tip
The teacher should roam around to check and
assess the students’ work. This will also clarify
some misconceptions.!
!
$
1 mol
& = 0.0110 mol ZnCl
2
#136.29g ZnCl &
"
2%
V
=
HCl
0.0220 mol HCl
= 0.0468 L = 46.8 mL
0.470 mol
L
C
O
!
2 mol HCl $&
#
mol HCl = 0.0110 mol ZnCl2
= 0,0220 mol HCl
#1 mol ZnCl &
"
%
2
PY
mol ZnCl2 =1.50g ZnCl2 #
3. Let students perform a team work - Think and Solve-Pair Share to answer the problem. (5 minutes) !
Ca(OH)2(s)
+ 2 HCl(aq)
—>
ED
Consider the following equation:!
CaCl2(aq)
+ 2 H2O(l)!
D
EP
a) How many liters of 0.100 M HCl is required to completely react with 5.00 grams of calcium hydroxide?
b) If 15.0 grams of calcium hydroxide is combined with 75.0 mL of 0.500 M HCl, how many grams of
calcium chloride would be formed?
Give students a short and specific timeframe (1-2 minutes) to think and solve freely based on their
understanding of the given problem.!
Students in each pair then share their computations with their classmates for another short and specific
timeframe (e.g. 1 minute each).!
209
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Finally, the teacher leads the whole class to solve the problem.!
Expected answers:!
b) Mol Ca(OH)2 in 15.0 g
!
Mol HCl used
:
0.135 mol HCl needed to neutralize 5.00g Ca(OH)2
:
0.202 mol Ca(OH)2 !
:
0.0375 mol HCl
PY
a) Mol HCl needed
C
O
Mol Ca(OH)2 that will react with 0.0375 mol HCl:
ED
! 1mol Ca(OH)2 $
mol Ca(OH)2 = 0.0375 mol HCl #
& = 0.01875 mol Ca(OH)2
" 2mol HCl %
This amount of Ca(OH)2 for reaction is 0.202 mol. This means Ca(OH)2 is in excess and
not all will react. HCl is limiting, and all HCl available will be used up in the reaction.
present.
D
EP
The amount of CaCl2 produced in the reactionwill be calculated from the amount of HCl
! 1mol CaCl2 $
mol CaCl2 = 0.0375 mol HCl #
& = 0.01875 mol CaCl2
" 2mol HCl %
210
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ENRICHMENT (30 MINS)
Teacher tip
•
The Four Corner Poster should be
presented in front of the class to facilitate
the group task.!
•
If possible, prepare an activity sheet
depicting the four corner poster. !
•
It is important to go around the different
groups to check their outputs. If there is
a group that needs further help, a group
discussion might be helpful.
Retrieved from Turner_Ross_Chemistry_LP
(retrieved September 21, 2015)
PY
1. Introduce the Enrichment Activity, the “Wastewater Four Corners.”
2. Group students into four subgroups.
3. Give the following introduction and instruction for the activity:
ED
C
O
One of the main reasons for treating wastewater is to remove the organic carbon and o t h e r
nutrients from the water. If these nutrient sources are not removed, when the water
reaches lakes or other
bodies of water, it will affect that body of water’s ecosystem. For
example, if the waste material is
excess nitrogen, this is digested by algae that causes
excessive algae growth. The algae uses up
oxygen from the water to grow and to digest the nutrients found in the wastewater. The depletion of oxygen
in the water kills other life such as fish and other marine organisms that is dependent on it. Such will result to
an alarming state
that will eventually destroy marine life.
Instruction:
D
EP
In each corner of the room (referring to the Four Corner Poster), there are equations
involving
the breakdown of wastewater material. In every corner, a group will be assigned to
determine the amount
of carbon dioxide created by the reaction specific to their assigned
place. The four groups will converge
after four minutes and compare their results with the
other groups. As a class, you will determine which
reaction produces the most carbon dioxide.
Corner1: Estrogen is found in many pharmaceuticals and in biologically enhanced foods.
Estrogen: C18H24O2 + 23 O2
—>
18CO2 + 12H2O
Start with 20g Estrogen.
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Corner 2: Casein is a protein found in milk products and digested.
C8H12N203
+ 3 O2
—>
C5H7O2N
+
NH3
+
3CO2
Start with 30g Casein.
Corner 3: Urea is a product found in urine.
H2O
+
7H
—>
3NH4
+
CO2
ED
+
C
O
C8H12N203 is Casein
NH2COH2N
H2O
PY
C5H7O2N is Bacterial Cell
+
Turning urea to ammonium, start with 60g urea.
6NO-3 +
D
EP
Corner 4:
5CH3OH
—>
3N2
+
5CO2
+
7H2O
+
6OH-
Turning nitrate into nitrogen gas.
Start with 20 g nitrate.
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Expected Answers:
Corner 2) 21.49 g CO2
PY
Corner 1) 58.15 g CO2 - Estrogen produces the most amount of CO2
Corner 3) 43.96 g CO2 - The reaction of urea produces the second largest amount of CO2
C
O
Corner 4) 11.83 g CO2 - Nitrate reaction produced the least amount of CO2
Scoring:
ED
Each group will be graded according to the quality of their concerted effort. A correct answer with a very
thorough explanation will be given 10 points; correct answer with explanation will have 7 points; only a correct
answer y will have 5 points; and 2 points for the group who tried to answer though incorrect.
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EVALUATION (3 MINS)
The assignment serves as the evaluation of the lesson taken. Each student will submit their answers the
following day for marking.
Solve the following stoichiometry problems:
1. Using the following equation:
2 NaOH
+
H2SO4
—>
2 H2O
+
Na2SO4
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How many grams of sodium sulfate will be formed, if you start with 1.25 L of a 4.0 M solution of sodium
hydroxide?
Ca(OH)2 (s)
—>
Expected Answers:
+ Ca(NO3)2 (aq)
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ED
1. 355.3 grams of Na2SO4
2. 0.0926 grams of Ca(OH)2
H2O(l)
C
O
2HNO3 (aq) +
PY
2. How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of 0.100M HNO3? The reaction proceeds
as follows:
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Chemistry 2
60 MINS
Physical Properties of Solutions:
Temperature Effect on Solubility
LESSON OUTLINE
Introduction
Communicating learning objectives
40
5
10
C
O
PY
Instruction
Discussion and Demonstration
Content Standard
The learners demonstrate an understanding how temperature and pressure affect
Enrichment
Group task: Investigation Design
solubility of solutes (solids and gases) in solvents.
Evaluation
Quiz
Performance Standards
Materials
The learners shall design a simple investigation to determine the effect on boiling point
Fo the Assignment (Answers to the problem set on mole fraction
or freezing point when a solid is dissolved in water.
5
The learners design a simple investigation to determine the effect of temperature on
solubility of sugar.
ED
Learning Competencies
Explain the effect of temperature on the solubility of a solid and a gas. (STEM_GC11PPIIId-f-113)
Explain the effect of pressure on the solubility of a gas. (STEM_GC11PP-IIId-f-114 )
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
explain the effect of temperature on the solubility of a solid and a gas; and
•
explain the effect of pressure on the solubility of a gas.!
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•
and molality): manila paper with problem set; masking tape
For Demonstration:
a. Temperature affects the solubility of a solid
20 mL hot water; 20 mL cold water; Two (10) grams white sugar preweighed using weighing scale in two separate plastic cups and
properly labeled; Two plastic spoons; Two beakers or plastic bottles;
Graduated cylinder
b. Temperature affects the solubility of gases
Carbonated drink (preferably colored);"Hot water;!Cold water;!Four
clear plastic cups
c. Pressure affects solubility of a gas dissolved in a liquid
Canned soft or cola drink
Resources
(1) Brown, T.,Le May, H.E.,Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry the central science, 11th ed. Philippines. Pearson Education
South Asia PTE. LTD. pp. 539 - 541.!
(2) Masterton, W.L. (2009). Chemisty principles and reaction, 6th ed.
Belmont, CA 94002.3098 USA. Brooks/Cole Cengage Learning. pp. 264
– 267.!
(3) Padolina, M.C, E.S. Antero, & M.J.B. Alumaga.
Conceptual and
functional chemistry modula approach. Manila. Vibal Publishing House,
Inc. pp. 208 – 210.
Additional resources at the end of the lesson
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INTRODUCTION (5 MINS)
Teacher Tip
Give the learning objectives:!
C
O
PY
Overview !
If a teaspoon of sand is added to water, it will not dissolve since by nature, it is insoluble in water. If the same
amount of sugar or salt is added to water, the result will be different since both sugar and salt are soluble in
water. But there is a limit on how much of sugar or salt can dissolve in a given amount of water. This is the
concept of solubility, defined as the maximum amount of a substance that dissolves in a given volume of solvent
at a given temperature. It could be deduced that temperature directly affects solubility. Pressure too, is another
factor to consider. Temperature affects the solubility of both solids and gases. However, pressure only affects
gases solubility. In this lesson, how these two factors affect the solubility of gases and solids will be explored.
a. Explain the effect of temperature on the solubility of a solid and a gas;
b. Explain the effect of pressure on the solubility of a gas.!
ED
Review (5 minutes)!
Ask learners to answer the following questions to check their prior knowledge on solubility and factors affecting
solubility:!
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Suppose you have two glasses half-filled with water and 1 teaspoon each of sand and salt.
What will happen if you add the sand and salt separately in each of the glass with water? Explain your
predictions.
!
Possible Answers:!
a. The sand will not dissolve in water even if you stir the mixture.
b. The salt will dissolve in water even if the mixture is not stirred.
c. Sand will not dissolve because it is insoluble in water. Salt, on the other hand will dissolve since it is
soluble in water. The nature of solute and solvent is a factor that affects solubility of solute (sand and salt)
in water.
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•
•
This topic on factors affecting solubility
was first taken in Grade 7 (Quarter I).
Learners need to be reviewed on
solubility from their G 7 Science, the
following concepts can be presented
prior to this lesson’s discussion or during
review of the previous lesson:!
•
•
Solubility is the maximum amount of
solute that can be dissolved in a
given quantity of solvent at a
particular temperature.
The rate of dissolution is affected by
several factors: particle size, stirring,
and application of heat.
INSTRUCTION (40MINS)
1. Ask students to define solubility again, assuming they have learned this term in Grade 7."
PY
2. Discuss that there are two other important factors that affect the solubility of solute in a given solvent, aside
from the nature of solute and solvent. How these factors directly affect solubility of solids and liquids will be
explored and further discussed.
3. Pose these questions:
a. What other factors affect the solubility of solids and gases?
C
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b. Can temperature affect solubility? How about pressure?
c. Can you cite instances or situations you have experienced that would support these claims?
Possible answers:
a. Temperature and pressure are factors that affect solubility of solutes in solvent.
b. Yes, both temperature and pressure can affect solubility.
ED
c. For example, it is easier to dissolve powdered chocolate in hot water than in iced water, thus
temperature affects the solubility of a chocolate drink.
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Solubility is affected by temperature and pressure. How would this be possible? The proceeding demonstration
activities “Temperature affects the solubility of a solid (sugar) and of a gas” will show proof to the matter. The
result in both demonstrations will serve as a springboard for later discussion.
Question to investigate (for the first demonstration):
Where does sugar dissolve better, in hot or cold water?
4. Demonstrate the activity with the help of a student:!
a. Pour 10 mL of hot water into one beaker or clean plastic bottle, while one student pours 10 mL of cold
water into another.
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b. With the help of another student, pour 10 grams of sugar into each plastic bottle at the same time.
c. Again with the help of a student, stir the contents of each plastic bottle with a plastic spoon. The teacher
and the student should try to stir in the same way, force, and length of time.
d. Show that the contents will be allowed to settle and let students observe what happens.
PY
e. Instruct students to record the time for the sugar to dissolve in water in both gerber bottles.
5. Ask learners to take note of all their observations and possible explanations. They will still be observing the
effect of temperature on solubility of a gas.
C
O
6. Proceed with the next demonstration on “Temperature Affects the Solubility of a Gas”
Question to be investigated:
Does temperature have an effect on how quickly dissolved gas escapes from a soda?!
ED
7. Demonstrate the activity with the help of a student:"
a. In front of the learners, open a bottle of club soda or soft drink.
b. With the help of a student, fill two clear cups or transparent plastic bottles in half with the soda.
c. Fill one empty cup with about 1/3 of ice cold water, and another empty cup about 1/3 of tap water.
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d. Again with the assistance of a student, place one of the soda glass in the cold water and the other into
the hot water.
e. Ask the learners to watch and observe the surface of the soda in both cups.
8. Instruct students to also take note of all their observations and possible explanations.
9. Lead the discussion on the effect of temperature on solubility of solid and gas in a liquid. Initiate the
discussion by making students answer the following questions.
Guide Questions:"
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Teacher Tip:
Students should be given precautionary
measures especially when handling hot water.!
a. Where does sugar dissolve better, in hot or cold water?!
b. Does temperature affect the solubility of sugar? How do you know?
•
In the second demonstration, two cups of
carbonated drink is needed. One cup
should be heated and the other one
cooled. However, one simple way to heat
and cool the cups is to use a hot and
cold water bath.
•
Guide questions can be written in the
chalkboard for reference during the
course of discussion.!
c. Does temperature have an effect on how quickly dissolved gas escapes a soda/cola drink?
d. Can you tell if there is a difference in the amount of gas escaping from each sample of soda? What did
you observe that made you think so?
PY
e. Based on your observation in the second experiment, why do you think people store soda or soft drinks
in refrigerators?
Expected Answers:!
C
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a. There is a small amount of sugar left at the bottom of the glass with the hot water. There is significantly
more sugar in the bottom of the glass with cold water.
b. Temperature affects solubility of sugar. The higher the temperature, the more sugar is dissolved.
c. The way gas escapes in the form of bubbles from soda is affected by temperature.
d. More bubbles form and rise to the surface in the soda placed in hot water than in cold water. Dissolved
gas comes out of solution faster in warm water than cold water.
ED
e. People store soft drinks in refrigerators to prevent gas (CO2), which gives the drink its biting taste, from
easily escaping the solution.
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10. Provide additional explanation on the effect of temperature on solubility of solids and gases."
"
A. Effect of Temperature on Solubility !
"
"
Temperature changes have a direct effect on solubility of solids and gases.
Solids!
!
For most solids like sugar and salt, an increase in temperature means an increase in solubility. Everyday
experiences like the one observed in the first demonstration activity may lead one to think that solubility always
increases with temperature. This is not the case, however. The dissolving of a solid occurs more quickly at higher
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ED
C
O
PY
temperature, but the amount of solid that can be dissolved may increase or decrease with increased
temperature. The effect of temperature on the solubility of several solids in water (aqueous solution) is shown in
the figure below.
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Figure 1: Solubility of Several Inorganic and Organic Solids in Water as a Function of
Temperature"(http://chemwiki.ucdavis.edu/Wikitexts/University_of_California_Davis)!
!
It can be noted that compounds such as glucose and sodium acetate (CH3CO2Na) exhibit an increase in
solubility with increasing temperature. Sodium chloride (NaCl) and potassium sulphate (K2SO4) on the other
hand, exhibit a little variation and still others like lithium sulphate (Li2SO4) become less soluble with increasing
temperature.
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Teacher Tip:
•
Accept any other appropriate responses. !
•
If there is an internet connection, the
video on factors affecting solubility can
be used for demonstration:!
https://www.youtube.com/watch?
v=OpFW7V_GiUQ!
PY
Notice that there are only very few solids that decrease in solubility with increase in temperature. While
the endothermicity or exothermicity of dissolution may have some effect on the solubility as temperature
changes, the main effect of added heat is an increase in kinetic energies of particles, allowing solute particles to
break free from each other, and disperse into the solvent. Even CaCl2, whose dissolution is highly exothermic,
shows an increase in solubility with increasing temperature, as seen in the Figure. In general, solids will increase
in solubility as temperature increases.!
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Gases
Gases behave quite differently from solids. It is typical for gases dissolved in aqueous solutions at
ordinary pressures to exhibit decreasing solubility with increasing temperature.
For gases, an increase in temperature results in increased kinetic energies of gas particles dissolved in
liquids. This increased motion enables the dissolved gas to break intermolecular forces with the solvent, and
escape the solution.
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!!!!!B.!Effect!of!Pressure!on!Solubility!
ED
Thus, a warm bottle of carbonated drink/soft drink does not taste as good as a cold one, because there
is less CO2 dissolved in the warm bottle. Even newly-boiled water tastes flat because there is less oxygen gas
dissolved in it.!
Discuss that pressure has a significant effect on solubility only for gases in a liquid system, but negligible
effects on the solubility of solids in liquids. Such observation is exemplified in the next short demonstration.
Show the effect of pressure on solubility with the help of a student. That is when pressure is applied to a
gas above the solvent like water, the gas will move into the solvent and occupy some of the spaces between the
particles of the solvent. "
1. Show an unopened canned soft drink. Explain that in the preparation of the soft drink, pressure is
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Teacher Tip:
For gases:!
An increase in temperature results in
•
increased kinetic energies of gases
dissolved in liquids. This increased
motion enables the dissolved gas to
break intermolecular forces with the
solvent, and escape the solution.!
•
Thus, a warm bottle of carbonated drink/
soft drink does not taste as good as a
cold one, because there is less CO2
dissolved in the warm bottle.!
applied to force the carbon dioxide (CO2) molecules into the soda. This gas gives the biting taste in
the soft drink. But when the gas pressure is decreased, the solubility of CO2 is also decreased.
PY
Direct the students’ attention again to the can of soft drink. Ask them to give a prediction and an
explanation of what will occur once the can is opened.!
2. Try to open the can this while students are observing and noting some possible changes.
C
O
Possible answers:
Bubbles will come out from the opened can. The bubbles carry with them the gases stored in the liquid
soft drink solution.!
ED
When a can of carbonated soft drink is opened, the pressure in the soft drink is lowered, hence t h e g a s
starts to leave the solution immediately. This is manifested by the release of stored CO2 through fizzing, which
could be seen on the surface of the liquid. Once this gas is released, the beverage becomes flat due to the loss
of CO2.!
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Explain further that at low to moderate pressure, gas solubility is directly proportional to pressure.
If there is a lot of gas on the space above the solvent (the more gas, the higher the pressure of the gas),
then there would be more gas particles colliding with the surface of the liquid. The more collisions with the
solvent, the greater the possibility of dissolving. Hence, solubility of gas increases with increased pressure over
the liquid. The mathematical relationship that describes this is Henry’s Law, that states that aat a given
temperature, the solubility of a gas in a liquid is proportional to the partial pressure of the gas above the
liquid.
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Ask students to answer the questions to apply concepts learned.
•
a. Consider the following system that is in equilibrium:
CO2 (g)
+
H2O(l)
H2CO3 (aq)
ΔHsoln < 0"
I.
PY
What will happen to the solubility of carbon dioxide if:
Temperature is increased?
II. Pressure is increased?
C
O
b. Danielle has always wanted to start her own carbonated company. Just recently she opened her
bottling company to produce her drinks. She wants her product to “out-fizz” all other competitors.
She wants to maximize the solubility of the gas in her drink. What conditions would best allow her to
achieve her goal to put her company on top?
Possible answers:
ED
c. CJ is trying to increase the solubility of salt in some water. He begins to anxiously stir the mixture.
Should he continue stirring? Why?
!
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a. i. An increase in temperature means a decrease in solubility of the gas.!
!!!!!!ii. An increase in pressure results in more gas particles dissolving or entering the liquid to decrease
the partial pressure. Consequently, the solubility of the gas would increase.
b. Danielle would be able to maximize the solubility of CO2 in her drink, out-fizzing the others, if she
increases the pressure and lower the temperature during soft drink production.!
c. CJ should stop stirring since this will only affect how fast the salt will dissolve, but not the solubility of
salt in water. He might as well heat the mixture since increasing the solution’s temperature will also
increase the solubility of the salt.
223
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Let students summarize or give a synthesis of their learnings on the effect of temperature and pressure on
solubility of solids and gases.
Two factors that affect the degree of solubility of a solute in a solvent
As pressure increases,
solubility increases.
As temperature
increases, solubility
increases
(endothermic
reaction) and
solubility decreases
(exothermic
reaction)
As temperature
increases, solubility
decreases
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ENRICHMENT (5 MINS)
No effect
PY
Temperature
Gas in a Liquid
C
O
Pressure
Solid in a Liquid
ED
Factor Affecting Solubility
Inquiry in Action
1. Form groups of five students each. "
2. Each group will design an experiment to compare how well sugar and salt dissolve in hot and cold water.
They will report the results of their investigation.
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Teacher Tip
There is a need to guide the students’ group
investigation
EVALUATION (10 MINS)
Problems on conceptual understanding :
PY
a. One manufacturer’s instructions for setting up an aquarium specify that if boiled water is used, the water
must be cooled to room temperature and allowed to stand overnight before fish are added. Why is it
necessary for the water to stand for such period of time?
ED
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b. Using Figure 2 below, compare the solubilities of potassium nitrate and cesium sulfate.
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1.
Figure 2: Solubilities of several ionic solids as a function of temperature.
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Possible answers:
PY
c. When water is boiled, all of the dissolved oxygen (DO) and nitrogen are removed. Oxygen in particular is
needed by fish to live. When the water is cooled to room temperature, it initially contains very little DO.
But allowing the water to stand overnight, when temperature is low, allows oxygen gas in the air to
dissolve faster, preventing the fish to suffocate and thus stay alive in the water.
ADDITIONAL RESOURCES
ED
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d. As temperature increases, solubility of potassium nitrate also increases. Conversely, cesium sulfate’s
solubility decreases with increasing temperature.
(1) Zumdahl, S.S. & S. A. Zumdahl (2012). Chemistry an atoms first approach. United States. Brooks/Cole
Cengage Learning Asia Pte. Ltd. pp. 491 - 495.
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(2) http://www.ck12.org/user:krogers/section/Factors-Affecting-Solubility/ (Retrieved Nov. 5, 2015)
(3) h t t p : / / c h e m w i k i . u c d a v i s . e d u / P h y s i c a l _ C h e m i s t r y / E q u i l i b r i a / S o l u b i l t y /
Solubility_and_Factors_Affecting_Solubility (Retrieved Nov. 5, 2015)
(4) http://chemsense.sri.com/classroom/curriculum/Solubility_Kennedy.pdf (Retrieved Nov. 6, 2015)
(5) http://www.acs.org/content/dam/acsorg/education/resources/k-8/inquiryinaction/inquiry-in-action.pdf
(Retrieved Nov. 2, 2015)
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Chemistry 2
60 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Instruction
Discussion
40
Group task: laboratory investigation
5
Quiz
10
PY
Colligative Properties of
Nonelectrolytes and Electrolyte
Solutions
Content Standards
Enrichment
The learners demonstrate an understanding of how concentration of solute affects the
Evaluation
colligative properties of nonelectrolyte and electrolyte solutions.
C
O
The learners will be able to calculate boiling point elevation and freezing point Materials
Laptop; LCD projector; manila paper; masking tape; graphs to show
depression given appropriate concentration for various solutions.
Performance Standard
The learners design a simple investigation to determine the effect of an electrolyte and
nonelectrolyte on the boiling point of water.
ED
The learners perform calculations on problems involving Boiling Point elevation and
Freezing Point Depression of Solution
the effects of adding solute on the vapor pressure, boiling point, and
freezing point of solutions
Resources
(1) Boiling Point Elevation. Retrieved from http://www.chem.purdue.edu/
gchelp/solutions/eboil.html!
(2) Brown, T., Le May, H.E.,Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry the central science, 11th ed. Philippines. Pearson Education
South Asia PTE. LTD. pp. 546 – 556).
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Learning Competencies
Describe the effect of concentration on the colligative properties of solutions. (3) Colligative Properties of Solutions. Retrieved from http://
www.chem1.com/acad/webtext/solut/solut-3html!
(STEM_GC11PP-IIId-f-115), Differentiate the colligative properties of nonelectrolyte
(4) Masterton, W.L. & Hurley, C. N. (2009). Chemisty principles and
solutions and of electrolyte solutions. (STEM_GC11PP-IIId-f-116 ), Calculate boiling
reaction, 6th ed. Belmont, CA 94002.3098 USA. Brooks/Cole Cengage
point elevation and freezing point depression from the concentration of a solute in a
Learning. pp. 267 – 271.
solution. (STEM_GC11PP-IIId-f-117)
(5) Padulina, M.C, E.S. Antero, & M.J.B. Alumaga. (2010) Conceptual and
functional chemistry modular approach. Manila: Vibal Publishing House,
Inc. pp. 221 – 228.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
describe the effect of solute concentration on the colligative properties of solutions;
(6) Vapor Pressure Lowering. Retrieved from http://www.chem.purdue.edu/
gchelp/solutions/colligv.html
•
differentiate the colligative properties of nonelectrolyte solutions and of electrolyte
solutions; and
•
calculate or solve problems involving boiling point elevation and freezing point
depression from the concentration of a solute in a solution.
(7) Zumdahl, S.S. & S. A. Zumdahl (2012). Chemistry an atoms first
approach. United States. Brooks/Cole Cengage Learning Asia Pte. Ltd.
pp. 502 - 504.
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INTRODUCTION (5 MINS)
PY
Overview
Often times a solution is described in terms of concentration of one or more solutes present in it. However,
there are some important physical properties of solution which are more directly dependent on the
concentration of solute particles. Such properties are called colligative (Latin, coligare – which means “tied
together”) properties which means, they depend on the collective effect of the concentration of solute particles
present in the solution. These properties include: (1) vapor pressure lowering, (2) boiling point elevation, (3)
freezing point depression, and (4) osmotic pressure.
C
O
1. Communicate learning objectives.
Describe the effect of solute concentration on the colligative properties of solutions;
•
Differentiate the colligative properties of nonelectrolyte solutions and of electrolyte solutions; and !
•
Design a simple investigation to determine the effect of an electrolyte and nonelectrolyte on the
boiling point of water.!
ED
•
2. Review: Ask learners to say something regarding the following science ideas (as jumping board for the
new lesson)
Solution, Solute, Solvent, and Concentration
•
Ionic and Covalent Compounds
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•
INSTRUCTION (40 MINS)
Motivation
• Ask learners to picture this out:” It is a hot summer day and you have a picnic at the park or beach front
with your classmates, friends or relatives with watermelon and “dirty ice cream”. Mmmmmm….. tastes
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Teacher Tip
Let learners recall the ideas learned earlier
from solution.
good… refreshing…. The ice cream is an old-fashioned homemade kind ice cream. The kind of where
the maker has a tub full of mix of ingredients immersed in a bigger tub filled with ice and salt. But wait a
minute, why salt? Why the ice cream vendor does add salt to the ice?
The answer will be withheld until the end of the class. The next lesson, Colligative properties of solution
will be able to explain the answer.
PY
•
•
Present the following guide queries as bases for the lecture.
Questions:
1) What is colligative property?
C
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The salt is added because it lowers the freezing temperature of the ice and makes the ice cream colder
faster. The salt added to the ice or water slush is a solute that has affected the property of the solution it was
added to. This property is a colligative property of solution.
ED
2) Identify the different colligative properties of solutions.
3) Describe the effect of solute concentration on the colligative properties of solutions.
•
Colligative Properties
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4) Differentiate between the effects that an electrolyte and the of nonelectrolyte solutions and of
electrolyte colligative properties solutions. !
Colligative properties are properties of a solution that depend only on the number and not on the
identity of the solute particles. Thus, these depend on the collective effect of the concentration of solute
particles present in an ideal solution. Because of their direct relationship to the number of solute particles, the
colligative properties are very useful for characterizing the nature of a solute after it is dissolved in a solvent and
for determining the molar masses of substances. The latter will be discussed in the next lesson.
•
The Different Colligative properties of Solution
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Teacher Tip
Learners may give other observations
regarding homemade ice cream. It is
important to lead them through questions
that will lead them to the present topic.
Colligative properties include the following: (1) vapor pressure lowering; (2) boiling point elevation; and
(3) freezing point depression.
•
Effect of solute concentration on the colligative properties of solutions!
Effects of electrolyte and nonelectrolyte on colligative properties solutions.
C
O
•
PY
!
The concentration or amount of nonvolatile solute (i.e., a solute that does not have a vapor pressure of
its own) in the solution has an effect on the colligative properties of solutions. The effect would depend on the
ratio of the number of particles of solute and solvent in the solution and not on the identity of the solute.
However, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte.
1) Vapor Pressure Lowering
ED
Vapor pressure is a direct measure of escaping tendency of molecules. A pure liquid (solvent) in a closed
container will establish equilibrium with its vapor. And when that equilibrium is reached, the pressure exerted by
the vapor is called the vapor pressure. A substance that has no measurable vapor pressure is nonvolatile, while
one that exhibits a vapor pressure is volatile.
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When a liquid evaporates easily, it will have a large number of its molecules in the gas phase resulting to
a high vapor pressure. The picture (Fig. 1) in the left shows a surface entirely occupied by liquid molecules,
some of which evaporated and form a vapor. On the right, a nonvolatile solute like salt or sugar has been
dissolved into the solvent, having the effect of diluting the water. The addition of a nonvolatile solute resulted to
a lowering of the vapor pressure of the solvent. The lowering of the vapor pressure depends on the number of
solute particles that have been dissolved. The chemical nature of the solute is not considered because vapor
pressure is merely a physical property of the solvent and does not undergo a chemical reaction with the solvent
and does not itself escape into the gas phase.
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Teacher tip
•
If internet connection is available learners
will be allowed to view the video on
Boiling Point Elevation. retrieved from
http://www.chem.purdue.edu/gchelp/
solutions/eboil.html!
PY
C
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Figure 1: Vapor Pressure of Liquids
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ED
It is important to note that the reduction in the vapor pressure of a solution of this example is
directly proportional to the fraction of the volatile molecules in the liquid, which is the mole fraction of the
solvent. This reduced vapor pressure can be determined using Raoult’s Law (1886).
Fig. 2: Graph showing the relationship between vapor pressure and mole
fraction of water (Image source: http://chem.libretexts.org/@api/deki/files/64408/
=H2O_Raoult.png?revision=1)
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Where:
C
O
PY
•
Recall from the definition of mole fraction that in a two component solution (a solvent and a
single solute), X solvent = 1 – X solute. !
NaCl(s)
D
EP
ED
While the chemical nature of the solute is not a factor to consider, it is important to take into
consideration whether the solute is an electrolyte or nonelectrolyte. Ionic compounds like sodium chloride,
NaCl, are strong electrolytes that dissociate into ions when they dissolve in solution results in a larger number
of dissolved particles. Consider two different solutions of equal concentration: one is made from ionic
compound NaCl, while the other is made from the molecular compound glucose (C6H12O6). The equations
below show what happens when these solutions dissolve :
C6H12O6 (s)
———>
Na+ (aq)
+
————->
C6H12O6 (aq)
Cl" (aq)
2 dissolved particles
1 dissolved particle
The sodium chloride, NaCl dissociates into 2 ions, while glucose does not dissociate. Thus, equal
concentrations of each solution will result in twice as many dissolved particles as in the case of NaCl. The vapor
pressure of the solvent in NaCl solution (electrolyte) will be lowered twice as much as that of the solvent in the
glucose (nonelectrolyte) solution. Since the surface now of salt solution is covered by more solute particles,
232
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At this point mathematical calculations
involving, vapor pressure, boiling point
elevation and freezing point depression
may not be included. Since this will be
taken up in the next lesson. However, the
formulas can be presented for each of
the concepts presented.
there is less room for solvent molecules to evaporate lowering the vapor pressure of the solvent, water.
2) Boiling Point Elevation
PY
The addition of a nonvolatile solute lowers the vapor pressure of the solution; consequently the
temperature must be raised to restore the vapor pressure of the solution to the value conforming to the pure
solvent. Specifically, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling
point by an amount known as the boiling point elevation.
D
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ED
C
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Figure 3 below shows the phase diagram of a solution and the effect that the lowered vapor
pressure has on the boiling point of the solution compared to the solvent. In this case the sucrose solution has
a higher boiling point than the pure solvent. Since the vapor of the solution is lower, more heat must be
supplied to the solution to bring its vapor pressure up to the pressure of the external atmosphere. The boiling
point elevation is the difference in temperature between the boiling point of the pure solvent and that of the
solution.
Figure 3: The lowering of the vapor pressure
in a solution causes the boiling point of the
solution to be higher than pure solvent
•
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Discussion could be enriched by referring
to the www.ck12.org (pdf).!
ED
C
O
PY
Figure 4: Normal boiling point for water
(solvent) as a function of molality in several
solutions containing sucrose (a non-volatile
solute).
D
EP
For dilute solution the elevation of the boiling point is directly proportional to the molal
concentration of the solute:
the solvent.
The molal boiling point elevation constant, Kb, has a specific value depending on the identity of
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•
For more discussion refer to http://
www.chem.purdue.edu/gchelp/solutions/
colligv.html
•
3) Freezing Point Elevation
PY
The freezing point of a substance is the temperature at which the solid and liquid forms
can coexist indefinitely, at equilibrium. Under these conditions molecules pass between the 2
phases at equal rates because their escaping tendencies from the two phases are identical.
D
EP
ED
C
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Figure 5 below shows the phase diagram for a pure solvent and how it changes when a solute is
added to it. The solute lowers the vapor pressure of the solvent resulting in a lower freezing point for the
solution compared to the pure solvent. The freezing point depression is the difference in temperature between
the freezing point of a pure solvent and that of a solution. On the graph, Tf represents the freezing point
depression.
Figure 3: The lowering
of the vapor pressure in
a solution causes the
boiling point of the
solution to be higher
than pure solvent
(purple). As a result, the
freezing point of a
solvent decreases when
any solute is dissolved
into it.
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At a given temperature, if a substance is added to a solvent like water, the solute-solvent
interactions prevent the solvent from going into the solid phase, requiring the t e m p e r a t u re t o d e c re a s e
further before the solution will solidify. Meaning, more energy must be removed from the solution in order to
freeze it and the freezing point of the solution is
power than that of the pure solvent.
PY
solution. Thus:
The magnitude of the freezing point depression is directly proportional to the molality of the
Tf = Kf m
Where:
C
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Kf – is the molal freezing - point depression constant., a constant that is equal to the
change in the freezing-point for a 1 molal solution of a nonvolatile molecular
solute
Tf – freezing point depression
Inquiry in Action:
•
D
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ENRICHMENT (5 MINS)
ED
M – molality of solute
Ask learners to give some practical applications of the lesson.
Some possible questions:
1. People who live in colder climates have seen the trucks put or sprinkle salt on the roads when
snow or ice is forecast. Why do they do that?
2. When planes fly in cold weather, the planes need to be de-iced before liftoff. Why is that done?
•
For an investigation group learners into 5 groups to design a simple investigation to determine the
effect of an electrolyte and nonelectrolyte on the boiling point of water.
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•
SUMMARY
Ask learners what have they learned from the lesson such as:
•
colligative properties of solution and their examples;
•
the effect of solute concentration on the colligative properties of solutions; and
•
the effect of electrolyte and nonelectrolyte on colligative properties of solutions
•
PY
•
EVALUATION (10 MINS)
C
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Formative Assessment
Problems on Conceptual Understanding :
1. Which would increase more the boiling point of water: salt or sugar? Why?
2. Identify which of the following statements are true. When a solute is added to a solvent forming a
solution: (Explain your answer)
(II) the boiling point decreases
(III) the freezing point increases
A. i and ii are true
B. i and iv are true
C. ii and iv are true
D. ii and iii are true
Expected Answers:
D
EP
(IV) the freezing point decreases
ED
(I) the boiling point increases,
1. Salt will increase more the boiling point of water. The increase in boiling point depends on the number
of molecules added to water. Salt is a very small molecule. In addition it splits into two particles when in
water, the Na+ ion and the Cl- ion. In numbers: if suppose you add 6g of salt into water you add about
4,400,000,000,000,000,000,000 (4.4 * 1022) particles to the water. On the other hand, sugar has a
molecular weight that is 3 times larger than that of salt. It does not split up in different particles when in
water. So adding 6 g sugar into water you add around 700,000,000,000,000,000,000 particles (7.3 * 1021)
237
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Learners may ask for help in their
investigation, thus this could be done
during their vacant period as a group
performance output.!
Answers will vary among groups. Allow
the different groups to compare results.
to the water. Still a huge number, but considerably less than with salt. To get the same effect with sugar
that you get with salt, you will have to use about 6 times as much sugar as salt.
The same is true in principle with lowering the freezing point of liquids. That is the reason why we use
salt in winter on our streets and not sugar - as we would need 6 times as much for the same effect. But it
would work with sugar too, if we use just enough of it.
PY
2. Both may occur if a nonvolatile solute is added to the solvent. When solute is added to a solvent like
water, its boiling point increases. However, the effect is opposite as regards freezing point.
C
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APPENDIX A
D
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ED
Concept Map on Colligative Properties of solution. This could be given at the end of the lesson as a wrapping
up activity
(Image source: http://www.chem1.com/acad/webtext/solut/solut-3.html#SEC2)
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Chemistry 2
120 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
7
Instruction
Group Activity and Discussion
35
PY
Colligative Properties of
Nonelectrolytes and Electrolyte
Solutions
Motivation
Content Standards
Enrichment
The learners demonstrate an understanding of how temperature and pressure affect
Evaluation
solubility of solutes (solids and gases) in solvents.
Demonstration
Take Home Calculation Activity
Laboratory Activity
C
O
The learners demonstrate an understanding on how the molar mass of a nonvolatile solid Materials
Projector; computer; calculator; periodic table of elements
is determined from the change of boiling point of a resultant solution.
Performance Standard
The learners design a simple investigation to determine the effect of an electrolyte and
nonelectrolyte on the boiling point of water.
ED
The learners shall be able to perform an investigation to determine the molar mass of a
nonvolatile solute from the change of boiling point of a resultant solution.
Learning Competencies
Calculate molar mass from colligative property data. (STEM_GC11PP-IIId-f-118)
For Group Activity 3: “Determination of Molar Mass by Boiling Point
Elevation of Urea Solution”
Triple beam balance; SAS 03 Activity sheet; 250 mL distilled water;
ethanol; urea; sugar (unknown solute)
Resources
(1) Brown, T., Le May, H.E.,Bursten, B., Murphy, C. & Woodward, P. (2009).
Chemistry the central science, 11th ed. Philippines. Pearson Education
South Asia PTE. LTD. pp. 546 – 556).
(2) Elvins, C, et al. (1991). Heineman Chemistry in Context: Chemistry one.
Australia: Dah Hua Printing Press. p. 193 – 195. (retrieved October 30,
2015)
D
EP
Determine the molar mass of a solid from the change of melting point or boiling point of
a solution. (STEM_GC11PP-IIId-f-121)
(3) Lab Report on Molecular Mass Determination by Boiling Point Elevation
Method. Retrieved (08/122/2015) http://www.art-xy.com/2010/11/labreport-on-molecular-mass.html
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
(4) Masterton, W.L. (2009). Chemisty principles and reaction (6th ed).
Belmont, CA 94002.3098 USA. Brooks/Cole Cengage Learning. p. 265.
•
calculate molar mass from a colligative property data; and
•
determine the molar mass of a solid from the change of boiling point of a solution.
(5) Madu, C.E. & Attili, B. (2012). Determination of molar mass by boiling
point elevation of urea solution. Retrieved (10/12/2015) from http://
www.collin.edu/chemistry/Handouts/1412/Determination%20of
%20Molar%20Mass%20By%20Boiling%20Point%20Elevetion%20BA
%20Jan_1.pdf
Additional resources at end of the lesson
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INTRODUCTION (3 MINS)
Learning objectives:
At the end of the lesson, the learners should be able to:
1. Calculate the molar mass from a colligative property data; and
C
O
PY
Overview (2 minutes)
When a non-volatile solute is added to a pure solvent, the resultant solution would have a higher boiling point
than the pure solvent. The boiling point of a solution is a colligative property (Atkins, 2010: 170 – 171) and is
dependent on the concentration of the solute in the solution, but not on the kind of solute and solvent. The
boiling point can be measured by an ebullioscope, an instrument for observing the boiling point of liquids,
especially for determining the alcoholic strength of a mixture by the temperature at which it boils
(www.thefreedictionary.com).
Review (5 minutes)
Ask learners to answer the following questions:!
ED
2. Determine the molar mass of a solid from the change of boiling pint of a solution.
D
EP
a. Calculate the molality of a solution that contains 1.875 grams of potassium chloride (KCl) (MM= 74.55 g/
mol) in 175 grams water.!
b. Determine the mass of NaCl (MM = 58.44 g/mol) needed to prepare a 1.0 molal solution using 25 grams
of water as the solvent:!
Expected Answers:
a. 0.143 mol/kg KCl solution!
b. 1.5 g NaCl!
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Teacher Tip
Let learners recall the ideas learned earlier
from solution.
INSTRUCTION (70 MINS)
Teacher Tip
PY
Pre-Laboratory Discussion (15 minutes)
1. Start the discussion with the following introductory question regarding their take home activity: What is
the effect of salt and sugar separately on the boiling point of water? This leads them to what they could
expect from the present lesson. Divide the class into groups.
2. Distribute SAS 03: Determination of Molar Mass by Boiling Point of Urea Solution.
a. Present the objective of the laboratory activity. !
b. Review safety precautions:!
I.
For working around open flames, tie back long hair;
II. Wearing of laboratory gown;
C
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3. Give five minutes for reading, and discuss some important details before the activity:
III. Using of test tube holder when removing tubes from the hot water bath;
ED
IV. Not placing chemicals directly on a balance pan;
V. Ethanol is flammable; hence it should be kept away from open flames;
VI. Dispose excess material in the appropriate waste container. Do not throw any solid material in
the trash.
D
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c. Questions to be answered in the course of the activity!
4. Distribute materials to the different groups.
5. Check other required materials assigned like box of matches, rags, and distilled water.
6. Remind learners to answer the questions and fill the table as required in the laboratory activity.
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Learners may give other observations
regarding homemade ice cream. It is
important to lead them through questions
that will lead them to the present topic.
Laboratory Activity Proper (40- 50 minutes)!
PY
1. Let the groups perform Activity 3 (Refer to SAS 3). Learners record their observations and discuss results
among themselves. They will be graded for their laboratory performance.
2. Let them discuss the detailed calculations from their data.
C
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Post-Activity (25 minutes)!
1. Let different groups present data table that includes: molal concentration, boiling point and Kb
(ebullioscopic constant) of ethanol, and the change in boiling point of ethanol.
3. Integrate a short lecture while discussing the answers to questions in SAS 3 (Appendix A).
4. Discuss possible sources of errors.
5. Learners should be able to realize that they can also determine the mass of an unknown substance from
a colligative property of solution.
Additional Questions for SAS 03:
1. What is the unknown solute?!
ED
6. Ask learners to share their experiences.
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2. How did the solute elevate the boiling point of ethanol?
3. What are the possible sources of errors and explain how these affected the result.
Possible answers:
1. Sugar, sucrose, C12H22O11 (or the substitute provided)!
2. In any solution, like the ethanol solution, the mole of a component decreases with the addition of
another component. Thus, when the unknown solute was added into the ethanol solvent, the number of
components (solvent and solute) in the mixture increased, hence the mole fraction of ethanol decreased.
As a result, the vapor pressure of the pure ethanol solvent decreased because the less volatile solute
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Teacher tip
•
If internet connection is available learners
will be allowed to view the video on
Boiling Point Elevation. retrieved from
http://www.chem.purdue.edu/gchelp/
solutions/eboil.html!
molecule present at the solution surface layer partially blocked the evaporation of the solvent molecules.!
PY
Moreover, when a solute dissolves in a solvent, the initial solvent-solute interaction is broken or c u t a n d
replaced by a stronger solvent-solute interaction. Hence, more energy is required to overcome the
stronger forces of attraction between the solvent and solute, accounting for the elevation or rise in the
boiling point of ethanol when the unknown solute was added.
3. Possible sources of errors:!
C
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a. The ethanol solution may not be pure as the presence of impurities can also increase the boiling
point of ethanol;!
ED
b. The mass of solvent used is very crucial as it is involved in the calculation of the molecular mass of
the unknown sample. Any spills or vaporization resulting to the loss of solvent will directly cause a
deviation on the empirical molecular mass from its literature value. Hence, any steps involving the
transfer of ethanol has to be handled with great care, as ethanol is a volatile liquid.!
D
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c. Temperature of the water bath must be controlled and not be too high. If the temperature of the
water bath is much higher than the boiling point of ethanol, the ethanol sample will vaporized
excessively, leading to a loss of the results’ accuracy.!
Use the following information during the discussion!
Data and Results:!
Unknown:
______________________________________________
Volume of ethanol used:
Density of ethanol:
0. 785 g/mL
Molar Mass of urea:
60.06 g/mol
__________________________________ mL
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Ebullioscopic (Kb) constant of ethanol:
______________________0C/m
Weight of watch glass with unknown (W1):
______________________ g
Weight of empty watch glass (W2):
____________________________ g
Weight of Unknown (W1 - W2):
____________________________ g
Temperature of boiling ethanol (Tb (solvent)):
______________________0C
Molar Mass of unknown solute:
____________ 0C
PY
Temperature of boiling unknown in ethanol (Tb (solution)):
____________________________ g/mol
Percent (%) Error [(TV-EV/TV) x100]
______________________ %
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Table 1: Data in determining the Ebullioscopic (Kb) constant of ethanol
Grams of Urea, (NH2)2 CO
Trial
To make 50 mL of 1.0 molal ethanol
Boiling Temp. of Ethanol solution (0C)
3
Average
Generalization (8 minutes)
D
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2
ED
1
For concept attainment, let learners provide ideas learned from the activity.!
Possible answer:!
The molecular mass of an unknown sample is determined to be _______(will depend on the result of the
activity) by the boiling point elevation method. The rise in the boiling point of the solution is a common
phenomenon observed when a solute is dissolved in a pure solvent.
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•
At this point mathematical calculations
involving, vapor pressure, boiling point
elevation and freezing point depression
may not be included. Since this will be
taken up in the next lesson. However, the
formulas can be presented for each of
the concepts presented.
ENRICHMENT (7 MINS)
The following take home problems are enrichment activities to be performed individually.!
PY
Questions:
C
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1. Why does the concept of colligative properties apply only to dilute solutions?!
2. Eugenol is the active ingredient in the oil of cloves used to relieve toothache. Calculate the boiling
point of a solution in which 0.17 grams of eugenol, (C10H12, O2) is dissolved in 10.0 grams of
benzene. (Kb benzene= 2.53 0C/m; Tb(solvent) = 80.100C)
D
EP
Expected Answers:
ED
3. A solution was prepared by dissolving 18.00 grams glucose in 150.0 grams water. The resulting
solution was found to have a boiling point of 100.34 0C. Calculate the molar mass of glucose. (Kb
(water) = 0.510C kg/mol; Tb(water) = 80. 00C)
1. A dilute solution is one that has a small quantity of solute dissolved in a relatively large amount of
solvent. The solute particles are quite far apart from each other in the solution, creating an ideal
environment wherein solute particles experience little intermolecular interaction. This assumption
thus works well for a lot of dilute solutions, particularly solution of molecular compounds.!
Figure 3: The lowering of the vapor pressure
in a solution causes the boiling point of the
solution to be higher than pure solvent
•
On the other hand, as the concentrations increase, there will be stronger solute-solute interactions
and the nature of the solute particles become important. Now there are more solute particles, and
these particles are closer together as such, intermolecular interactions are no longer negligible. The
properties of solution will no longer be colligative since it is affected by the identity of the solute too.
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Discussion could be enriched by referring
to the www.ck12.org (pdf).!
Therefore, colligative properties do not apply to concentrated (more of solute than solvent) solutions.
Figure 4: Normal boiling point for water
(solvent) as a function of molality in several
solutions containing sucrose (a non-volatile
solute).
2. Tb (solution) = 80.37 0C
PY
3. MM glucose = 180 g/mol
EVALUATION
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Appendix A (Attachment)
SAS 03: “Determination of Molar Mass by Boiling Point Elevation of Urea Solution”
Appendix B
ED
Rubric for Laboratory Performance
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This rubric generally describes the performance level of learners and is meant to give generalized,
meaningful feedback. The score is not based on getting a number of points, but instead correlates a level of
performance to a corresponding point grade. A 5-point rubric will be used.
5 – Exceeds expectations
4 – Meets expectations
3 – Partial accomplishment/Adequate
2 – Minimal accomplishment
1 – Below expectations
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•
For more discussion refer to http://
www.chem.purdue.edu/gchelp/solutions/
colligv.html
Score
3
2
1
•
Multiple data points were collected and some thought was given regarding the number of points needed to
increase the validity of results.
•
The analysis of the data included a tabular representation.
•
The tabular analysis was directly used in the conclusion to respond to the objective and explain how these were
met.
•
The conclusion included a discussion of other factors that may have influenced the data. Each of these factors was
fully explained and its possible effect evaluated. The learner’s explanation indicated a superior understanding of
the concepts and/or a deeper thought process than was required in the completion of the activity.
•
The learner was a full and active participant in the laboratory activity.
•
Multiple data points were collected.
•
The analysis of the data included a tabular representation.
•
The tabular analysis was directly used in the conclusion to respond to the objective and explain how these were
met.
•
The conclusion included a discussion of other factors that may have influenced the data. Each of these factors is
fully explained and its possible effect evaluated.
•
The learner participated fully in the laboratory activity.
•
A sufficient amount of data was collected to perform an analysis.
•
The analysis was completed and included a tabular representation of the data.
•
The data was used to form a conclusion that responds to the objective.
•
The learner showed a general understanding of the concepts involved and is able to answer any analysis questions.
•
The student participated fully in the laboratory activity.
•
Data is recorded on the data table and some attempt is made at analyzing the data was made.
•
The lab report included conclusion statements.
•
The student was present for the lab, participated, and recorded data.
C
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The learner assumed a leadership role in the laboratory activity and/or was instrumental in the group effort toward
a successful completion.
ED
4
•
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5
Description
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Appendix C
Experimental Techniques for SAS 03
C
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PY
In the setup of the experiment (Figure 1), the glass tubing serves as a means for escape for the gas
molecules present in the boiling tube. This ensures that a closed rigid environment is not established during the
experiment, since boiling does not occur in a closed rigid environment. In a closed environment, while the
vapour pressure of the liquid ethanol rises, the extra vapor that evaporates increases the pressure upon the
liquid. The vapor pressure therefore, never reaches the same value as the environmental pressure, because
both values increase as a result of heating. This problem does not occur in an open environment (that is with the
presence of the glass tubing), because the vapor that evaporates is free to disperse thus, is able to establish
pressure equality between the vapor and the surroundings, when boiling can happen.
ED
Besides providing an open environment where boiling can occur, the glass tubing also allows g a s e o u s
molecules to escape into the environment. This prevents a dangerous build-up of pressure within the boiling
test tube, which may cause the glassware to crack or shatter. Also, the glass tubing acts as a condense with its
cool internal surface, it will cause gaseous ethanol molecules, which have a low boiling point at around 780C –
to condense and prevent them from escaping into the environment. This is important as the number of ethanol
molecules directly affect the mass of the solution, and consequently the molality of solution.
Teacher Tip
Let learners recall the ideas learned earlier
from solution.
D
EP
In addition, the experiment was carried out in a hot water bath. This allowed the temperature of the
solution to increase gradually, which in turn allows the boiling points of ethanol with and without the unknown
sample to be more accurately determined. A glass chip or capillary tube is placed also inside the test tube to
avoid abrupt change in temperature and breakage.
Reference: Lab Report on Molecular Mass Determination by Boiling Point Elevation Method. Retrieved
(08/122/2015) http://www.art-xy.com/2010/11/lab-report-on-molecular-mass.html
ADDITIONAL RESOURCES
Padulina, M.C, E.S. Antero, & M.J.B. Alumaga. (2010) Conceptual and functional chemistry modular
approach. Manila: Vibal Publishing House, Inc. pp. 222 – 223.
Zumdahl, S.S. & S. A. Zumdahl (2012). Chemistry an atoms first approach. United States. Brooks/Cole Cengage Learning Asia Pte. Ltd. pp. 502 - 503.
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GENERAL CHEMISTRY 2
STUDENT ACTIVITY SHEET
SAS 01
Learning Competency
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(LAB) Perform acid-base titration to determine concentration of solutions
(STEM_GC11PP-IIId-f-119)
PY
Acid-Base Titration
Standard Solution
Molarity
ED
A solution whose concentration or strength has been correctly established is a standard solution. This can be prepared by the accurate
weighing of a pure solute and dissolving to the correct final volume of the solution. However, this is seldom done except in such cases where the
desired chemicals can be obtained in very pure form. In most cases, the concentration of the solution is determined by the standardization process.
This is done by measuring the volume of the solution that will react with either a known weight of a pure substance or a primary standard, or a
known volume of a standard solution. In this activity, the standardized sodium hydroxide (NaOH) solution has been prepared to be used in the
analysis.
Acid-Base Titration
D
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The concentration of a solution can be measured by molarity (M). Molarity or molar concentration is defined as the number of moles of the
solute dissolved in 1 liter of solution.
Titration is the process of determining the volume of a standard solution that will react completely with a given weight or volume of a
sample. Acid-base titration can be used to determine the concentration of an acid or base by measuring the amount of a solution with a known
concentration (titrant) that reacts completely with a solution of unknown concentration (analyte). The point at which this occurs is called the
equivalence point or the end point of titration, and this is determined by the use of an indicator which changes color at the desired point. There are
many kinds of indicators, and in the present activity, phenolphthalein will be used.
Objectives:
1. Determine the molar concentration of hydrochloric acid (HCl) using a standard sodium hydroxide solution (NaOH).
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2. Use the titration equation (MA)(VA) = (MB)(VB) ,where MA is the molar concentration of the acid, VA is the volume of the acid sample you use in
the activity, MB is the base molarity, and VB is the base volume.
Materials:
Iron stand
10 mL pipette
Burette clamp
Base burette
Aspirator
One (25 mL or 50 mL) graduated cylinder
One (50 mL) beaker
Three (250 mL) Erlenmeyer Flasks or beakers
Medicine dropper for the indicator
•
•
•
Phenolphthalein indicator
250 mL distilled water
50 mL hydrochloric acid (to be obtained from the teacher)
•
50 mL of 0.1 M or mol/L sodium hydroxide in a clean, dry labeled glass container (to be prepared and standardized by the teacher)
ED
C
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•
•
•
•
•
•
•
•
•
Procedure:
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SAFETY: laboratory gown/aprons and gloves
1. Wash the glasswares (Erlenmeyer flasks, beakers, pipette) with dishwashing liquid. Rinse these thoroughly with tap water, and finally with
distilled water.
2. Invert the glasswares to drain the distilled water. Leave these inverted until use.
Preparing the Burette for Titration
a. Wash the burette with dishwashing liquid using a long-handled brush. Make sure to clean the tip of the burette. Empty the burette into
the sink.
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b. Using a beaker, place some distilled water into the base burette and rinse it thoroughly. Drain any liquid inside the burette by inverting it
on the burette clamp, while the pinchcock is released.
PY
c. Using the small beaker, rinse the burette with about 5 mL freshly prepared NaOH standard solution. Make sure to drain the base through
the tip of the burette. Do this twice and discard the washing in the sink.
d. Secure the pinchcock of the burette, making sure that no liquid flows through the tip. Fill this with NaOH standard solution until the 50
mL-mark, though it does not have to be exact.
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e. Let the solution flow through the tip of the burette, such that no air space remains in it, including the part after the pinchcock. Collect the
solution using a clean beaker. This solution can still be used for titration.
Hint: Close and open the pinchcock alternately to remove air spaces. A continuously smooth flow of the liquid out of the burette
means there is no more air space.
If there is more air space, fill the burette with NaOH standard solution up to the 50 mL-mark. Make sure that the lower meniscus is at the
50 mL mark of the burette.
ED
f.
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g. The titrant is ready for use in the analysis.!
Titration Setup
(Image source: https://
water.me.vccs.edu/courses/
ENV211/lesson12_print.htm)
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PY
Titrating the Hydrochloric Acid
1. Using the 10 mL pipette, add 10 mL of HCl to a clean Erlenmeyer flask.
2. Using the graduated cylinder, add 25 mL of distilled water to the acid.
3. Add two drops of phenolphthalein indicator and swirl the solution. Place the flask under the burette, and titrate with the standard solution
while shaking the flask, until the appearance of a very pale pink color that persists for about 20 seconds. Record the final volume of the base
to 2 decimal places.
4. Do two to three trials of the titration, Fill the table below with your results:
Table1: Data for acid-base titration.
Final Burette
Reading (mL)
Volume of Base
(mL)
(Final – Initial)
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Initial Burette
Reading (mL)
Molarity of Base
(mole/L)
ED
Trial
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Average
Questions:
1. What is a standard solution?
_____________________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________________
2. When does the completion of the acid-base reaction occur? What is this condition?
_____________________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________________
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3. What are the reactants in this reaction?
_____________________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________________
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4. What are the products of this reaction?
_____________________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________________
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5. Describe what would have happened in one of your titrations if you had forgotten to add phenolphthalein to the sample flask.
_____________________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________________!
ED
Calculate the concentration of the base (sodium hydroxide) in mol/L (M).
Conclusion:
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What is the average concentration (M) of the hydrochloric acid solution?
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
_
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GENERAL CHEMISTRY 2
STUDENT ACTIVITY SHEET
SAS 02
PY
SOLUBILITY OF SALT
Learning Competency
(LAB) Determine the solubility of a solid in a given amount of water at different temperatures (STEM_GC11PP-IIId-f-120)
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The solubility of a pure substance in a particular solvent is the quantity of that substance that will dissolve in a given amount of the solvent. Solubility
varies with the temperature of the solvent. Thus, solubility must be expressed as a quantity of solute per quantity of solvent at a specific
temperature. For most ionic solids in water, especially salts, solubility varies directly with temperature.
ED
In this laboratory activity, the solubility of potassium nitrate (KNO3) in water will be studied. Several quantities of this salt will be dissolved in a given
amount of water at a temperature close to the water’s boiling point. Each solution will be observed as it cools, and the temperature at which
crystallization of the salt occurs will be noted and recorded. The start of crystallization indicates that the solution contains the maximum quantity of
solute that can be dissolved in that amount of solvent.
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After collecting solubility data for several different quantities of solute, plot these on a graph. Construct a solubility curve for KNO3 by connecting
the plotted points.
Objective
To collect the experimental data necessary to construct a solubility curve for potassium nitrate (KNO3) in water.
Materials
A.
•
•
•
•
Equipment
Triple beam balance
Alcohol lamp or burner
Spatula
Iron stand
•
•
•
•
One graduated cylinder (10 mL)
Stirring rod
Iron ring
Wire gauze
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Test tube holder
Test tube rack
Thermometer
One 50 mL beaker
Four test tubes (18 x 150 mm)
One beaker (400 mL)
•
•
•
•
•
•
Utility clamp or iron clamp
Clamp holder
Pencil
Graphing paper
Ruler
Laboratory gown
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•
•
•
•
•
•
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B.
Reagent
•
Distilled water
•
Potassium nitrate (KNO3)
Safety
•
Use laboratory gown/aprons and gloves when working.
•
Tie back long hair, and secure loose clothing when working with an open flame.
•
Be sure to use a test tube holder when removing test tubes from the hot water bath.
ED
Procedure
While one laboratory partner carries out steps 1 through 4, the other group member should go on to step 5.
1. Using a marking pencil or marker, number four test tubes 1 through 4. Place these in a test tube rack.
2. On the balance, measure exactly 2.0 grams of potassium nitrate (KNO3). Pour the salt into test tube 1.
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3. Repeat step 2 for the following masses of KNO3. Add each quantity to the corresponding test tube each with 5.0 mL of distilled water:
Test tube #
Mass (grams) of KNO3
Volume (ml) of distilled H2O
1
2.0
5
2
4.0
5
3
6.0
5
4
8.0
5
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ED
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4. Fill a 400-mL beaker about three –fourths full of tap water. This will be used as a water bath. Using the water bath and test tube #1, prepare the
setup as shown below (Figure 1). Heat the water to 900C and adjust the flame to maintain the water at about this temperature.
Figure 1: Setup to Determine the Solubility of Salt
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5. Stir the KNO3 – water mixture with a glass stirring rod until the KNO3 is completely dissolved. Remove the stirrer and rinse it off. Loosen the
clamp and use a test tube holder to remove the tube.
6. While the first laboratory partner repeats step 5 for test tube #2, the other partner should place a warm thermometer (dipped into the hot water
bath) into the solution in test tube #1. Hold the test tube up to the light and watch for the first sign of crystallization in the solution. At the instant
crystallization starts, observe and record the temperature. Should crystallization start too quickly because of a cold temperature, redissolve the
solid in the hot-water bath and repeat the step.
7. Steps 5 and 6 should be followed for all four test tubes. One laboratory partner should stir the KNO3 until it dissolves, and the other partner
should record the temperatures of crystallization. Record all temperatures in Table 1: Observations and Data.
8. If any doubtful results are obtained, repeat the procedure by redissolving the salt in the hot-water bath and allowing it to recrystallize.
Table1: Observation and Data
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Grams of KNO3/5.0 mL H2O
1
2.0
2
4.0
3
6.0
4
8.0
Crystallization temperature (0C)
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Test tube #
Calculations:
1. Using proportions, convert the experimental mass/volume ratios to equivalent mass/100-mL ratios.
ED
2.0 g/5.0 mL = _______________________ g/100 mL
4.0 g/5.0 mL = _______________________ g/100 mL
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6.0 g/5.0 mL = _______________________ g/100 mL
8.0g/5.0 mL = _______________________ g/100 mL
2. Plot your experimental data on the grid provided. Plot mass of solute per 100 mL of water on the y-axis and temperature on the x-axis (refer
Figure 2 below).
3. Construct a solubility curve by connecting the plotted points on your graph.
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ED
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Figure 2: Solubility Curve of potassium nitrate (KNO3)
Questions
1. How many grams of KNO3 can be dissolved in 100 mL of H2O at the following temperatures?
a. 300C _________________________________
b. 600C __________________________________
c. 700C __________________________________
2. Define the terms saturated, unsaturated, and supersaturated.
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____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
PY
3. Classify the following KNO3 solutions as saturated, unsaturated, or supersaturated. Explain your answers.
a. 75 gKNO3/100 mL H2O at 400C
b. 60 gKNO3/100 mL H2O at 400C
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____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
4. Do the solubilities of all ionic solids increase as temperature increases? Explain your answer.
Conclusion
ED
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
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____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
Reference
http://www.cpet.ufl.edu/wp-content/uploads/2013/03/%E2%80%98Solubility-of-a-Salt%E2%80%99- (Constructing-a-Solubility-Curve-for-PotassiumNitrate-in-Water.pdf)
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GENERAL CHEMISTRY 2
SAS 03
PY
STUDENT ACTIVITY SHEET
DETERMINATION OF MOLAR MASS BY BOILING POINT ELEVATION OF UREA SOLUTION
Learning Competency
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(LAB) Determine the molar mass of a solid from the change of melting point or boiling point of a solution (STEM_GC11PP-IIId-f-121)
A solution describes a system in which one or more substances, the solutes, are homogeneously dissolved in another substance, the solvent. The
proportion of the solute and the solvent in the solution varies with the solvent usually in greater quantity.
ED
Physical properties are divided into two categories namely (1) extensive properties, such as mass and volumes which depend on the size of the
sample, and (2) intensive properties such as density and concentration, which are characteristic properties of the substance that do not depend on
the size of the sample. There is a third category of property, a subset of a system’s intensive properties and is called colligative properties applicable
only to solutions.
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When a nonvolatile solute is dissolved in a solvent, the freezing point of the resulting solution becomes lower while the boiling point of the solution
becomes higher than those of the pure solvent. The degree to which the freezing point is lowered, or the boiling point elevated depends on the
concentration of the solute particle. Properties that depend on the concentration, but not the identity of the solute in solution are called colligative
properties.
Objective
In this experiment, the following will be determined:
1. The boiling point of pure ethanol and its Kb (boiling point elevation constant) using urea solution; and
2. The molar mass of an unknown by measuring the boiling point elevation of the unknown/ethanol.
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*Note
From the difference in the boiling point of pure ethanol and the boiling point of the urea solution, the boiling point elevation constant of ethanol
can be calculated. The magnitude of boiling point elevation (ΔTb) is related to the molality of the solution (m) as shown in the equation below:
ΔTb =T(solution) − T(solvent) = Kbm
m =
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Where:
Kb = boiling point elevation constant, which is the function of the solvent and not the solute
m = molality of the solution calculated as follows:
moles of solute , (mol)
__________________________________
mass of solvent, (kg)
[mass of solute, (g)] [1 mol of solute/molar mass of solute, (g)]
_______________________________________________________
mass of solvent, (kg)
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m (mol/Kg) =
ED
Expressing the number of mole in terms of its mass and molar mass, the above equation can be rewritten in the following manner:
Then use the boiling point elevation measurement to determine the molar mass of an unknown solute.
Materials
A.
•
•
•
•
Equipment
Triple beam balance
Alcohol lamp or burner
Laboratory Thermometer (110 – 1200C)
Iron stand
•
•
•
•
Spatula
Wire gauze
Three (8 x 1—in) test tubes
One test tube rack
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One (400 mL) beaker
One (600 mL) beaker
One (100 mL) graduated cylinder
Two watch glasses
Two holed rubber stoppers
Iron ring
Clamp holder
•
•
•
•
•
•
•
One thermometer (0C - 1100 scale)
Stirring rod
Test tube holder
8-inch long glass tubing
Iron clamp
Laboratory gown
Small capillary tube (3-4” long) or
broken glass chip
B.
•
•
Reagent
250 mL distilled water
Urea
•
•
Ethanol
Unknown solute (to be obtained from
the teacher)
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•
•
•
•
•
•
•
Procedure
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ED
Safety
•
Use laboratory gown/aprons and gloves when working.
•
Tie back long hair, and secure loose clothing when working with an open flame.
•
Be sure to use a test tube holder when removing test tubes from the hot water bath.
•
Avoid placing chemicals directly on a balance pan.
•
ETHANOL IS FLAMMABLE, KEEP AWAY FROM OPEN FLAMES.
1. First prepare a 900C water bath by heating 400 mL of water in a 600 mL beaker.
2. Determine the mass of urea (60.06 g/mol) necessary to make 50 mL of 1.0 molal solution in ethanol (d = 0.785 g/mL). Measure three urea
samples of the calculated mass. Learners show their calculations to the teacher before continuing.
3. Measure 50.00 mL of ethanol and place it into a clean, dry (8 x 1 in) test tube. Place a small capillary tube (about 3-4 inches long) open end
down, or broken glass chip in the ethanol. The test tube is then fitted with a two holed rubber stopper with an 8-inch long glass tubing, and
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thermometer inserted so that its tip is about an inch below the surface of ethanol. The test tube is then clamped and immersed in the hot water
bath, and heated gently.
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ED
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4. Use the setup in Figure 1 to determine the boiling point of pure ethanol.
Figure 1: Experimental setup for the determination of the molecular mass by Ebullioscopic Method
5. Use the same ethanol and determine the elevated boiling point when 1.0, 2.0, and 3.0 molals of urea is added (measured in step 2).
6. Record observations in Table 1.
7. Get an unknown sample sample from the teacher. To a new batch of 50.0 mL ethanol, add a mass of the unknown equal to the mass of one
molal of urea as calculated in step 2. Use the procedures above (step 1 to 4) to measure the boiling point elevation of the unknown/ethanol
solution. From the data, calculate the molar mass of the unknown.
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8. Given the density of ethanol (d = 0.785 g/mL), the mass of ethanol and that of the unknown in 1 kg of solvent may be found. From these data
and molality of the solution, the molar mass of the unknown sample may be determined.
Trial
PY
Table1: Data in Determining the Ebullioscopic (Kb) constant of Ethanol
Grams of Urea, (NH2)2 CO
to make 50 mL of 1.0 molal ethanol
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1
2
3
!
Volume of ethanol used:
!
!
!
_____________________________________
__________________________________ mL#
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Unknown:! !
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Average
Fill the needed data:
Boiling Temperature of Ethanol
solution (00C)
Density of ethanol :!0. 785 g/mL
Molar mass of urea:!60.06 g/mol
Boiling point elevation constant, Kb, of ethanol!!_________________________________0C/m#
Weight of watch glass with unknown (W1):!!
____________________________________g#
Weight of empty watch glass (W2):!!
____________________________________g#
!
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Weight of Unknown (W1 - W2):!! !
!
Boiling point of ethanol (Tb (solvent)):
____________________________________g#
____________________________________0C#
Boiling point of solution of Unknown in
#
____________________________________0C#
PY
Ethanol (Tb (solution)):## #
Change in Boiling Point [ΔTb =T(solution) -T(solvent)] ____________________________________0C#
_________________________________g/mol#
Percent (%) Error [(TV-EV/TV) x100]
___________________________________ %
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Molar Mass of Unknown Solute:
Questions
1. What is the unknown solute?
ED
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
2. How did the solute elevate the boiling point of ethanol?
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____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
3. What are the possible sources of errors and explain how these affected the result.
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________
Reference
Lab Report on Molecular Mass Determination by Boiling Point Elevation Method. Retrieved
(08/122/2015) http://www.art-xy.com/2010/11/lab-report-on-molecular-mass.html
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Chemistry 2
220 MINS
Content Standard
The learners demonstrate an understanding of energy changes in chemical reactions.
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Motivation
Real-life application
10
PY
Thermochemistry: Energy
Changes in Chemical Reactions
Learning Competencies
Explain the energy changes during chemical reactions.
lecture materials to be presented; materials for demonstration (see
Appendix B)
Resources
ED
(STEM-GC11TC-IIIg-i-122)
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Instruction
85
Discussion & Demonstration
Performance Standards
The learners design a simple investigation to determine the effect on boiling point or Enrichment
100
Laboratory exercise
freezing when a solid is dissolved in water.
Evaluation
20
Laboratory exercise
The learners demonstrate an understanding of exothermic and endothermic processes,
the first law of thermodynamics and the use of calorimetry in the determination of heats Materials
Computer or overhead projector; projector screen; transparencies of
of reaction
Distinguish between exothermic and endothermic processes.
(STEM-GC11TC-IIIg-i-123)
(STEM-GC11TC-IIIg-i-124)
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Explain the First Law of Thermodynamics.
Specific Learning Outcome
At the end of the lesson, the learners will be able to:
•
describe and explain how heats of reactions are determined by calorimetry.
(1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
New York: John Wiley & Sons.
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
(3) Padolina, MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
Inc.
(4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
Co., Inc.
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INTRODUCTION (5 MINS)
Communicating Learning Objectives
1. Communicate the learning competencies and objectives to the learners using any of the suggested
protocols. (verbatim, own words, read-aloud)
PY
2. Present a list of terms that learners may have encountered in previous lessons (in lower grade levels), and
those that will be introduced and defined as they come up in this lesson
a. Energy
b. Work
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c. Kinetic energy
d. Potential energy
e. Heat energy
f.
Exothermic change
i.
Specific heat
j.
Calorimetry
k. Heats of reaction
l.
Enthalpy
m. Thermodynamics
n. System
o. Surroundings
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h. Heat capacity
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g. Endothermic change
p. First Law of Thermodynamics – Law of Conservation of Energy
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Teacher Tip
Project/display the objectives all at once or
show each one at a time.
MOTIVATION (10 MINS)
Physical and chemical changes are accompanied by energy change.
PY
Give some examples of familiar physical and chemical changes that are obviously accompanied by energy
change. Ask the learners to name one or two more. Burning of fuels releases energy. Boiling of water or cooking
of food require energy. Photosynthesis need energy from the sun in order to happen.
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Inform the learners that the lesson is about thermochemistry, the study of heat and energy changes that
accompany physical and chemical processes.
INSTRUCTION (85 MINS)
1. Review
Ask learners to define energy.
Ask them how they recognize objects with energy, perhaps by giving specific examples.
2. Lecture Proper
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Energy and the first law of thermodynamics
ED
Ask them if energy can take different forms and give examples of these.
Ask the learners to give examples of instances or activities that require or produce energy that are
familiar to them
Ask them where their energy to exercise, dance, or walk to school come from.!
Ask them what happens to energy used in these activities? Is energy used up? Does it disappear?
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Teacher Tip
•
Energy is the capacity to do work.
•
Matter has energy that can make
things happen. Objects can affect other
objects when they have energy. For
example, a car running at a certain
speed can do work when it collides
with a parked car by moving the latter
a certain distance.
From learners’ responses, state that:
•
Energy is not seen, but its effects are obvious.
•
Matter has energy.!
•
Energy can take different forms.
PY
Proceed to discuss the interconversion of energy from one form to another.
Give examples of energy conversions (chemical to electrical energy in batteries; water in a dam
being converted to electrical energy – hydroelectric power).
•
Emphasize to the learners that in this interconversions, no energy is lost, as such, energy is
conserved.
•
State the First Law of Thermodynamics, the Law of Conservation of Energy.
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•
ED
The Law of Conservation of Energy states that energy can neither be created nor destroyed. It can
only be transformed from one kind to another. The energy of the universe is constant.
Exothermic and Endothermic Reactions
• Classify the changes, processes and reactions named in the earlier part of the lesson into two
groups: those that release energy, and those that require energy.
Start the discussion by stating that in every chemical reaction, energy is either absorbed or
released.
•
Define exothermic and endothermic change.
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•
If energy is released to the surroundings when a change occurs, such change is described as an
exothermic change. The heat released by an exothermic reaction often results to an increase in the
temperature of the reaction mixture and the reaction vessel, and possibly the air surrounding the vessel. !
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•
The interconversion of energy from one
form to another has been taken up in
earlier grades.
• The Law of Conservation of Energy
controls energy changes that occur when
chemical or physical changes take place.
On the other hand, a change that involves absorption of energy from the surroundings is said to
be an endothermic change. When an endothermic change occurs, the temperature of the reaction
mixture decreases since part of the kinetic energies of particles in the surroundings are absorbed for use
in the reaction.!
PY
Give a demonstration of exothermic and endothermic reactions (Appendix A).
Demo 1
1. Tell learners to write down their observations. Ask:
b. Why did the lid or container’s cover pop?
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a. Is the reaction exothermic or endothermic? Why?
c. Why did the container pop as soon as the lighted matchstick was introduced into the
container?
Sample responses:
ED
a. The reaction is exothermic because heat was produced. The container felt hot/warm when
touched.
b. The reaction that occurred produced a gas, which expands because of the heat produced
by the reaction.
c. The reaction was very rapid because by shaking the container after putting a few drops
vaporizes the alcohol and hence would react very quickly with oxygen in the air.
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•
2. Lead the discussion towards what happens when a chemical change occurs, wherein bonds
break and form (Energy is required to break bonds and released when bonds are formed).
Also, allow learners to recall the Law of Conservation of Energy and show how this relates to
their observations of the demo.
a. Where did the heat produced come from?
b. What does this tell you about the potential energy of the products compared to that of
reactants?
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Teacher Tip
•
Guide questions can be written in the
chalkboard for reference during the
course of discussion.
•
It is important that the video presentation
be viewed prior to the class discussion.
•
If there is no internet connection, learners’
gadgets can be used, if available.
However, if there is none at all, it is
advisable to prepare visual aids using
Manila paper and markers to show: (1)
dissolving process of sugar and sodium
chloride in water and (2) energy changes
and solution formation.
Sample response:
a. When a chemical reaction occurs, bonds are broken and formed. The amount of heat
released when chemical bonds of the products are formed is greater than the energy
required to break the chemical bonds of the reactants.
b. The potential energy of the products is less than that of the reactants.
PY
Demo 2
1. Tell students to write down their observations. Ask:
b. Why is the container cold when touched?
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a. Is the reaction exothermic or endothermic? Why?!
c. Where does the energy absorbed from the surroundings go?
d. How would you compare the potential energy of the products with that of the reactants?
Sample response:
ED
a. The reaction is endothermic. The container felt cold when touched.
b. Part of the energy needed for the reaction to happen was obtained from the immediate
surroundings of the reactants, including the plastic cup container. Because heat from the
plastic cup was absorbed by the reaction, the container felt cold.
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c. It goes into the formation of chemical bonds. A smaller amount of energy is released when
chemical bonds are formed than required to break chemical bonds of reactants.
d. The potential energy of the products is greater than the potential energy of the reactants.
Again consider the Law of Conservation of Energy.
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Calorimetry: measuring energy changes in chemical reactions
•
Burning gasoline is a highly exothermic reaction. The total amount of heat obtained,
which is actually the energy change when the reaction occurs , is called the heat of reaction. !
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A concept related to heat of reaction is enthalpy change, H, which simply stated,
refers to the heat transferred by a process that occurs at constant pressure. This property is
important since many reactions that are often studied are constant pressure processes,
including reactions in test tubes and beakers, and reactions in biological systems. Thus, for
most reactions, heats of reactions are good approximations of the enthalpy changes and are
often referred to interchangeably.!
Units of expressing heat
ED
The measurement of the amount of heat evolved or absorbed when a process or
chemical reaction takes place is called calorimetry.
Pose the question: How is energy expressed in measurements?
•
Remind the learners that energy is measured as heat released or absorbed
•
Discuss the units used to express heat – calorie, kilocalorie, joule, kilojoule, and the
conversion of one unit to another:
D
EP
•
1 cal = 4.184 J
1 kcal = 4.184 kJ
A device called a calorimeter may be used to measure the heat released from a chemical
reaction. Basically, what it does is measure the change in temperature of the system when a
reaction takes place.
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ED
C
O
PY
Describe a bomb calorimeter, show the learners an illustration, and explain how the heat of a reaction is
measured or determined. !
D
EP
•
Figure 2: Diagram of a bomb calorimeter (Image source: https://www.learner.org/
courses/chemistry/images/lrg_img/BombCalorimeter.jpg)
A bomb calorimeter is commonly used to determine the heat of combustion of a compound, that is,
the heat released when a particular quantity of a compound is burned in oxygen. A more detailed
description is given in Appendix B.
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•
When gasoline burns in a car engine, the
energy derived from the reaction can do
many things: make the car run, cause the
horn to sound, operate the radio and
other devices. Part of these is lost as
heat.
•
It is possible to burn gasoline without
deriving any work from it. We can make
all the energy derived from it appear as
heat, and we can measure this amount of
heat.
In a general chemistry laboratory, a simpler calorimeter, called a coffee cup calorimeter, is
often used in determination of heats of reactions. Tell the learners that they will do an
experiment in the laboratory using a calorimeter. Show the learners an actual or illustration of a
coffee cup calorimeter (Appendix C).
Discuss heat capacity and specific heat or specific heat capacity. Tell learners to recall intensive
and extensive properties, and let them distinguish between heat capacity and specific heat
according to these properties.
PY
•
C
O
Substances vary in their ability to absorb heat. Some substances require a large amount of heat to raise i
ts temperature by one degree Celsius; others may require only a small amount. The general term for
this property is heat capacity, which is the amount of heat required to raise the temperature of a given
amount of substance one degree Celsius. It has the unit J/oC. This is an extensive property (since its
magnitude is dependent on the amount or size of the substance).
q
where: q = amount of heat absorbed
-mΔT
m = mass of substance
D
EP
c=
ED
An intensive property related to heat capacity is specific heat or specific heat capacity. Specific heat,
c, is defined as the amount of heat required to raise the temperature of one gram of a substance one
degree Celsius. It is expressed as:
ΔT = change in temperature
The specific heat of water is 4.18 J/g-oC (4.18 J g-1 oC-1) while for copper it is 0.382 J/g- oC, for iron,
0.446 J/g- oC and for aluminum 0.900 J/g- oC. Note that the specific heat of water is much larger than
that of the metals.
•
Show the relevant equation for the calculation of the heat involved when the temperature of the coffee
cup calorimeter changes due to heat released by a reaction. Note that the Law of Conservation of
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Energy should apply.
The equation shows that the total amount of heat evolved is equal to the total amount of heat
absorbed by calorimeter contents.
ΔH + qmix + qcal = 0
Where:
ΔH - heat of reaction
C
O
Or
PY
ΔH = qreaction = - (qmix + qcal)
qmix - heat absorbed by reaction mixture
qcal – heat absorbed by calorimeter
ED
qmix may be written as:
qmix = mmix cmix ΔT
D
EP
where mmix is the mass of reaction mixture, cmix the specific heat
of the reaction mixture and ΔT the change in temperature.
The heat absorbed qcal, by the calorimeter, may be shown as
qcal = mcal ccal ΔT
Usually mcal and ccal are taken together as a product, instead of measuring these values
separately. This product is called the calorimeter constant Ccal.
Ccal = mcal ccal
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Thus,
qcal = Ccal ΔT
PY
The overall equation may then be written as:
ΔH + mmix cmix ΔT + Ccal ΔT = 0
or
ED
qwarm water = -(qcold water + qcal)
C
O
The calorimeter must first be calibrated before it is used to determine the heat of reactions. A s p e c i f i c
quantity of cold water is placed in the calorimeter, after which a measured amount of warm water is
added to the cold water. The cold water gains the heat lost by the warm water because energy is
conserved.
qwarm water + qcold water + qcal = 0
D
EP
Heat released by the warm water may be represented as:
qwarm water = mwarm water cwarm water ΔT
Heat absorbed by the cold water may be written as:
Qcold water = mcold water ccold water ΔT
Heat absorbed by the calorimeter may be shown as:
qcal = Ccal ΔT
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Therefore, the overall equation is:
PY
mwarm water cwarm water ΔT + mcold water ccold water ΔT + Ccal ΔT = 0
Give sample problems that involve the calculation of the calorimeter constant and determine the heat of
reaction. Sample problems are given in Appendix D.
•
At the end of the lesson, refer back to the list of words that were introduced and ask learners to define
them. They could write this on paper or give their definitions verbally. If stated verbally, any wrong
answers or misconceptions may be clarified with other learners participating in the discussion.
C
O
•
ED
ENRICHMENT (100 MINS)
Exercise 1: Coffee-cup Calorimetry- Determining heat of solution and heat of neutralization
D
EP
1. One week before this laboratory session, task the learners to read and study the exercise. Hand out
copies of the experiment if leaners do not have a laboratory manual. Instruct them to write the (1)
objectives of the experiment and (2) the necessary data tables in their laboratory notebooks.
2. Have a pre-laboratory discussion of calorimetry. Review what was discussed in the lecture about
calorimetry.
3. Ask learners to state the objectives of the exercise.
4. Describe what they are to do when they perform the exercise. Show them the materials that will be used
in the exercise. Demonstrate how they will set the calorimeter.
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5. Check their data tables and make corrections when needed.
PY
The exercise involves determining the calorimeter constant; the heat of solution of a salt and the heat of
neutralization when given amounts and specific concentrations of NaOH and HCl are mixed.
EVALUATION (20 MINS)
2
(Not Visible)
(Needs Improvement)
ED
1
3
4
(Meets Expectations)
(Exceeds Expectations)
D
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Appendix A
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For the laboratory experiment, instruct the learners to prepare their data tables prior to coming to class. Check these tables before
they do the experiment. Check the data they collected and let them answer the questions in the exercise. They could do this as a
group where they discuss their answers among themselves. The learners’ answers to the questions will indicate their understanding
of the concept of calorimetry.
Demonstrations on Exothermic and Endothermic reactions:
Demo 1: Pringle Pop
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Materials:
•
Potato chips container complete with plastic cover – the type lined with aluminum foil (Example: Pringles
potato chips container)
•
Dropper with pointed tip
•
Matches
•
Ethanol
Teacher Tip
PY
•
Procedure:
The reaction that occurs is the
combustion of ethanol:
CH3CH2OH(g) + 3O2(g)
—> H2O +
CO2 + heat
C
O
1. Poke a hole about 1½ inches from the bottom of the potato chips container using a nail. The hole should
be big enough to fit a matchstick.
2. Squirt a few drops of alcohol into the potato chips container with its cover through the hole.
3. Shake the container vigorously and immediately introduce a lighted match into the hole.
Demo 2: Vinegar and Baking Soda
Materials:
ED
4. Let the learners touch the container.
One clear plastic cup
•
Measuring spoon (one tablespoon, one teaspoon)
•
Thermometer
•
Vinegar
•
Baking soda
Procedure:
D
EP
•
1. Measure about four tablespoons of vinegar into the plastic cup.
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•
The reaction of baking soda and acetic
acid:
CH3COOH + NaHCO3 —>
NaCH3COO + H2O + CO2
2. Tell learners to record the temperature of the vinegar in the plastic cup.
3. Add two teaspoons of baking soda and stir.
4. Tell learners to record the temperature of the mixture.
5. Let them touch the cup.
PY
Source: http://highschoolenergy.acs.org/content/hsef/en/how-can-energy-change/exothermicendothermic-chemical-change.html
D
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ED
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Appendix B
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Diagram of a bomb calorimeter
PY
A bomb calorimeter is usually used to measure heats of reactions of exothermic reactions. It is made of a strong
steel container called the bomb, where the reactants are placed. The bomb is placed into an insulated water
bath, which has a stirrer and a thermometer. The initial temperature of the water is recorded, and igniting the
reactants in the bomb with a small heater starts the reaction. The heat released by the reaction is absorbed by
the water, which causes the water’s temperature to rise. By measuring the change in temperature and knowing
the heat capacity of the calorimeter, we can calculate the energy released when the reaction takes place.
C
O
The insulation of the calorimeter prevents the flow of heat to and from the surroundings. Exothermic changes
make the insulated system warmer, while endothermic changes make it cooler.
D
EP
ED
Appendix C
A basic Coffee Cup Calorimeter
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Appendix D
1. The following illustrates the calculation of the calorimeter constant.
PY
Problem:
Sample Answer:
C
O
Fifty (50) milliliters of water at 60 oC is added to 75 mL of water at 30 oC in a coffee cup calorimeter. The
resulting mixture was observed to have a temperature of 40 oC. What is the calorimeter constant?
Assume that heat cannot flow in or out of the calorimeter. Thus we write:
ED
qwarm water + qcold water + qcal = 0
Consider that the density of water is 1 g/mL and its specific heat is 4.18 J/oC-g. Since the mass has to be in
grams, do the following calculations:
D
EP
50 mL(1.0 g/mL) = 50 g water at 60 oC
75 mL (1.0 g/mL) = 75 g water at 30 oC
Thus, the equation is:
50 g (4.18 J/oC-g) (40 oC – 60 oC) + 75 g (4.18 J/oC-g) (40 oC -30 oC)+ Ccal (40 oC -30 oC) = 0
-4180 + 3135 + Ccal(10) = 0
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Ccal = (4180 - 3135) / 10
Ccal = 104.5 J/oC
PY
The calorimeter constant is 104.5 J/oC.!
2. The following problem illustrates the determination of the heat of reaction of chemical reactions.
C
O
A student transferred 50.0 mL 1.00 M HCl into a coffee-cup calorimeter, which had a temperature of 25.5 oC.
He then added 50.0 mL 1.00 M NaOH, which also had a temperature of 25.5 oC, and stirred the mixture
quickly. The resulting solution was found to have a temperature of 32.5 oC. The calorimeter constant for the
coffee-cup calorimeter used was 15.0 J/ oC. Calculate the heat of reaction.
ED
Sample Answer:
Consider that the density of water is 1 g/mL and its specific heat is 4.18 J/oC-g. Assume that the density of
the solutions used is close to that of water to have the same density and specific heat.
D
EP
Use of the following equation to calculate for heat of reaction:
ΔH + mmix cmix ΔT + Ccal ΔT = 0
Since the amount of each reagent is given in units of volume, change this to grams. Therefore,
(50.0 mL + 50.0 mL)(1 g/mL) = 100.0 g reaction mixture
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Substitute:
ΔH + 100.0 g (4.18 J/ oC-g)(32.5 oC - 25.5 oC ) + 15.0 J/ oC (32.5 oC - 25.5 oC ) = 0
PY
ΔH + 2884.2 J + 103.5J = 0
ΔH = - 2987.7 J or 29.88 kJ
50.0 mL x 1.00 mol/1000 mL = 0.0500 mol HCl
ED
The same quantity of NaOH was used, 0.0500 mol.
C
O
Thus, the heat released when the given amounts of HCl and NaOH react is 29.88 kJ since ΔH or change in
enthalpy is expressed in kJ/mol. The calculated value is for the reaction of 50.0 mL HCl and 50.0 mL NaOH
expressed in kJ/mol. The number of mole of HCl used in the reaction must be obtained.
Calculate ΔH in J/mol by dividing the value obtained above by the number of moles of HCl used.
= - 59.8 kJ/mol
3. Another sample problem
D
EP
ΔH = - 29.88 kJ/0.0500 mol
Five (5.00) grams of an unknown compound was dissolved in 75.0 mL of water with an initial temperature of
22.5 oC contained in a coffee-cup calorimeter. Upon the compound’s dissolution, the final temperature of
the resulting solution was 33.2 oC. The calorimeter used had a calorimeter constant of 200 J/ oC. What is
the heat of solution of the compound? Assuming the compound is sodium hydroxide (NaOH), what is the
heat of solution in kJ per mol?
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Answer:
Assume the density and specific heat of the resulting solution is close to that of pure water; density is 1.0 g/
mL and specific heat is 4.18 J/ oC.
PY
mwater = 75.0 mL (1.0 g/mL)
= 75.0 g
C
O
ΔHsoln + mmix cmix ΔT + Ccal ΔT = 0
ΔHsoln + 75.0 g (4.18 J/ oC) (33.2 oC – 22.5 oC) + 200 J/ oC (33.2 oC – 22.5 oC) = 0
ΔHsoln = - 5629 J or 5.629 kJ
ED
ΔHsoln + 3488.6 + 2140 = 0
Assuming the compound is NaOH (Molar mass = 40.0 g/mol),
D
EP
mol compound = 5.0 g/40.0 g/mol
= 0.125 mol
Therefore in kJ per mol:
ΔHsoln = -5.629 kJ/ 0.125 mol
= 45.0 kJ/mol
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Chemistry 2
210 MINS
Content Standard
The learners demonstrate an understanding of energy changes in chemical reactions.
Learning Competencies
Calculate boiling point elevation and freezing point depression from the concentration of
a solute in a solution. (STEM_GC11PP-IIId-f-117)
Explain the energy changes during chemical reactions
ED
(STEM-GC11TC-IIIg-i-122)
Distinguish between exothermic and endothermic processes
(STEM-GC11TC-IIIg-i-123)
Introduction
Communicating learning objectives
5
Motivation
Short class activity
5
Instruction
Discussion and Demonstration
80
Enrichment
Laboratory Exercise
120
C
O
Performance Standard
The learners shall demonstrate an understanding of exothermic and endothermic
processes, the first law of thermodynamics and the use of calorimetry in the
determination of heats of reaction
LESSON OUTLINE
PY
Thermochemistry: the First Law
of Thermodynamics
Explain the first law of thermodynamics (STEM-GC11TC-IIIg-i-124)
D
EP
Describe and explain how heats of reactions are determined by calorimetry
Determine the heat of neutralization of an acid (STEM_GC11TC-IIIg-i-129)
Evaluation
Laboratory Exercise
Materials
Computer or overhead projector; projection screen; transparencies
of lecture materials to be presented; materials for demonstration
(See Appendix B).
Resources
(1) Brady EB. 1990. General Chemistry – Principles and Structure. New
York: John Wiley & Sons. 852 p.
(2) Masterson WL, Hurley CN, Neth EJ. 2012. Chemistry: Principles and
Reactions. California, USA: Brooks/Cole, Cengage Learning. 774 p.
(3) Padolina MCD, Sabularse VC, Marquez LA. 1995. Chemistry for the
21st Century. Makati, Philippines: Diwa Scholastic Press, Inc. 340 p.
(4) Silberberg MS. 2007. Chemistry – The Molecular Nature of Matter and
Change. International Edition. New York: McGraw-Hill Co., Inc. 1088 p.!
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
explain the first law of thermodynamics;
•
describe and explain how heat of reactions are determined by calorimetry; and
•
determine the heat of neutralisation of an acid.
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Teacher Tip
INTRODUCTION (5 MINS)
Project/display the objectives all at once or
show each one at a time.
PY
1. Introduce the learning objectives using any of the suggested protocols (verbatim, own words, readaloud)
C
O
2. Introduce terms that learners may have encountered in previous lessons (in lower grade levels) and those
that will be introduced in this lesson.
Energy
-
Work
-
Kinetic energy
-
Potential energy
-
Heat energy
-
Exothermic change
-
Endothermic change
-
Heat capacity
-
Specific heat
-
Calorimetry
-
Heats of reaction
-
Enthalpy
-
Thermodynamics
-
System
-
Surroundings
-
First law of thermodynamics – Law of conservation of energy
D
EP
ED
-
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Teacher Tip
Learners may give other observations
regarding homemade ice cream. It is
important to lead them through questions
that will lead them to the present topic.
MOTIVATION (5 MINS)
Ask students - why are they able to do physical exercise.
- why are they able to get up in the morning and come to school.
C
O
Students respond – they are able to all these activities because they have energy.
INSTRUCTION (80 MINS)
Review
Ask students how they would define energy.
ED
1.
PY
Tell students to stand up and do physical exercise.
Ask students how they recognize objects with energy. Perhaps they could give specific
examples.
2.
Lecture Proper
D
EP
Ask students to differentiate between potential and kinetic energy.
Energy and the first law of thermodynamics
Ask students where the energy they have to exercise, dance or walk to come to school come from.
Ask students to give examples familiar to them that require or produce energy.
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From students responses state that
Energy is not seen but its effects are obvious.
•
Matter has energy
•
Energy is produced from chemical reactions
•
Atoms and molecules have energy; they possess kinetic and potential energy
PY
•
Proceed to discuss kinetic and potential energy of atoms and molecules
Atoms or molecules of a solid, liquid or gas possess kinetic energy because they are in constant motion;
they collide with one another and bounce off from each other.
•
Atoms and molecules possess potential energy because of attractions and repulsion of these particles.
The nuclei and electrons in atoms are electrically charged and therefore would attract and repel each
other. Potential energy of chemical substances is sometimes referred to as chemical energy.
C
O
•
Teacher tip
•
ED
Consider the interconversion of kinetic energy to potential energy.
D
EP
Show a diagram where a wheelbarrow filled with rocks is being pushed up a hill – kinetic energy is changed to
potential energy as wheelbarrow is pushed up a hill; and potential energy is changed to kinetic energy when the
wheelbarrow rolls down the hill. A similar diagram is shown in Appendix A.
Ask students where the energy required to push the wheelbarrow or to peddle the bicycle up the hill comes
from. Likewise, ask them where the energy they expended to do exercise or dance comes from.
Possible response: The energy comes from the conversion of some of the chemical (potential) energy in the
food (like rice) to kinetic energy.
Emphasize to the students that in this interconversion of energy no energy is lost, that is, energy is conserved.
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•
Energy comes from the complex
chemical reactions involved in the
metabolism of nutrient molecules in the
body. Various chemical reactions in an
individual’s body produce energy that
allows one to do work, that is exercise or
walk to school.
The law of conservation of energy states
that energy can be neither created nor
destroyed. It can only be transformed
from one kind to another. The energy of
the universe is constant. The law of
conservation of energy controls energy
changes that occur when chemical or
physical changes take place.!
State the first law of thermodynamics, the law of conservation of energy.
PY
Tell students to give other examples of energy conversions (chemical to electrical energy in batteries; water in a
dam being converted to electrical energy – hydroelectric power).
Exothermic and Endothermic Reactions
•
Energy is required to break chemical bonds
•
Energy is released when bonds are formed
C
O
Start the discussion by stating that in every chemical reaction, there is either absorption or release of energy.
State that when a chemical reaction occurs:
ED
Consider the changes in energy when reactions occur in an insulated system and an open system. Define
exothermic and endothermic change.
Demo 1
D
EP
Give a demonstration of exothermic and endothermic reactions (Appendix B).
Tell students to write down their observations.
Ask the students
•
Is the reaction exothermic or endothermic? Why?
Answer: The reaction is exothermic because heat was produced. The container felt hot/warm
to the touch.
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•
Why did the lid or cover of the container pop?
Answer: The reaction that occurred produced a gas, which expands because of the heat
produced by the reaction.
Why did the container pop as soon as the lighted matchstick was introduced into the container?
PY
•
Answer: The reaction was very rapid because by shaking the container after putting a few drops
vaporizes the alcohol and hence would react very quickly with oxygen in the air.
•
C
O
Lead the discussion towards what happens when a chemical change occurs, that is, bond breaking and
bond formation. (Energy is required to break bonds and energy is released when bonds are formed). Also, allow
students to recall the law of conservation of energy and show how this relates to the observations of the demo.
Where did the heat produced come from?
What does this tell you about the potential energy of the products compared to that of the reactants?
D
EP
•
ED
Answer: When a chemical reaction occurs bonds are broken and formed. The amount of heat released
when chemical bonds of the products are formed is greater than the energy required to break the chemical
bonds of the reactants.
Answer: The potential energy of the products is less than that of the reactants.
Demo 2
Tell students to write down their observations.
Ask the students
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•
Is the reaction exothermic or endothermic? Why?
•
Why is the container cool to the touch?
•
Where does the energy absorbed from the surroundings go?
C
O
Answer: Energy (heat) was absorbed by the system from the surroundings.
PY
Answer: The reaction is endothermic. The container felt cool to the touch.
Answer: It goes into the making of the chemical bonds. A smaller amount of energy is released when
chemical bonds are formed than the energy required to breaking chemical bonds of the reactants.
How would you compare the potential energy of the products with that of the reactants?
ED
•
Answer: The potential energy of the products is greater than the potential energy of the reactants.
Units of expressing heat
D
EP
Again consider the Law of conservation of energy.
Pose the question: How do chemists measure energy?
Then proceed to state that chemists measure energy as heat.
Discuss the units used to express heat – calorie, kilocalorie, joule and kilojoule, and the conversion of one unit to
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another.
1 cal = 4.184 J
Calorimetry: how do chemists measure energy changes in chemical reactions?
Introduce the terms thermochemistry and heat of reaction. Define these terms.
PY
1 kcal = 4.184 kJ
C
O
Discuss calorimetry. Describe a bomb calorimeter and show the students an illustration of a bomb calorimeter
and explain how the heat of a reaction is measured or determined. (Appendix C).
ED
Discuss heat capacity and specific heat or specific heat capacity. Tell students to recall extensive and extensive
properties and let them distinguish between heat capacity and specific heat according to these properties.
D
EP
Define specific heat and write the equation that expresses it (Appendix D). Give the specific heat of water and
that of other materials like certain specific metals (Appendix D). Ask students to explain the differences in
values. Ask students of what importance is the large specific heat of water in their daily lives. This will allow
you to ascertain their understanding of specific heat.
Consider the coffee cup calorimeter. Show the students an actual or an illustration of coffee cup calorimeter
(Appendix E) and describe it. Tell the students that they will be doing an experiment in the laboratory using this
calorimeter.
Show the relevant equation for the coffee cup calorimeter that applies the law of conservation of energy.
The equation shows that the total amount of heat evolved is equal to the total amount of heat absorbed by
calorimeter contents.
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•
Or
ΔH + qmix + qcal = 0
Where
ΔH - heat of reaction
PY
ΔH = qreaction = - (qmix + qcal)
qmix - heat absorbed by reaction mixture
qcal – heat absorbed by calorimeter
C
O
qmix may be written as:
qmix = mmix cmix ΔT
ED
where mmix is the mass of reaction mixture, cmix the specific heat of the reaction mixture and ΔT the change
in temperature.
The heat absorbed by the calorimeter, qcal may be shown as
D
EP
qcal = mcal ccal ΔT
Usually, mcal and ccal are taken together as a product instead of measuring these values separately. This product
is called the calorimeter constant, Ccal.
Ccal = mcal ccal
Thus,
qcal = Ccal ΔT
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The overall equation may then be written as:
•
ΔH + mmix cmix ΔT + Ccal ΔT = 0
PY
The calorimeter must be first calibrated before it is used for determining heats of reactions. A specific quantity
of cold water is placed in the calorimeter after which a measured amount of warm water is added to the cold
water. The cold water gains the heat lost by the warm water because energy is conserved.
or
qwarm water + qcold water + qcal = 0
The heat released by the warm water may be represented as
ED
qwarm water = mwarm water cwarm water ΔT
C
O
qwarm water = -(qcold water + qcal)
The heat absorbed by the cold water may be written as
D
EP
Qcold water = mcold water ccold water ΔT
The heat absorbed by the calorimeter may be shown as
qcal = Ccal ΔT
Therefore the overall equation is
mwarm water cwarm water ΔT + mcold water ccold water ΔT + Ccal ΔT = 0
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Give sample problems that illustrate calculating for the calorimeter constant and determining the heat of
reaction. Sample problems are given in Appendix F.
PY
At the end of the lesson list words that were introduced in the lesson and ask students to define them. They
could write this down on paper or give their definitions orally. If stated orally, any wrong answers or
misconceptions may be cleared up with other students participating in the discussion.
ENRICHMENT (120 MINS)
C
O
Exercise 1: Coffee-cup Calorimetry: Determining heat of solution and heat of neutralization
One week before this laboratory session assign students to read and study this exercise. Hand out copies of the
experiment if the students do not have a laboratory manual. Assign them to write in their laboratory notebooks
the (1) objectives of the experiment and (2) the necessary data tables.
Ask students to state the objectives of the exercise.
ED
Have a pre-laboratory discussion of calorimetry. Review what was discussed in the lecture about calorimetry.
D
EP
Describe to the students what they will be doing when they perform the exercise. Show the students the
materials that will be used in the exercise. Demonstrate how they will set up the calorimeter.
Check their data tables and make corrections when needed.
The exercise will involve determining the calorimeter constant; the heat of solution of a salt and the heat of
neutralization when given amounts and specific concentrations of NaOH and HCl are mixed.
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EVALUATION
2 (NEEDS
IMPROVEMENT)
3 (MEETS
EXPECTATIONS)
4 (EXCEEDS
EXPECTATIONS)
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1 (NOT VISIBLE)
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For the laboratory portion, instruct the students to prepare their data tables prior to coming to class. Check these tables before
they do the experiment. Check the data they had collected. Let them answer the questions in the exercise. They could do this
as a group where they discuss their answers among themselves. The students’ answers to the questions will indicate their
understanding of the concept of calorimetry.
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ED
APPENDIX A
Figure A. Interconversion of kinetic
energy and potential energy. (http://
www.chem1.com/acad/webtext/
energetics/CE-1.html#SEC1)
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Teacher tip
•
CH3CH2OH(g) + 3O2(g)
H2O + CO2 + heat
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APPENDIX B
Demonstrations+on+Exothermic+and+Endothermic+reactions:+
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Demo 1
Pringle Pop
Materials:
(Example: Pringles potato chips container)
Dropper with a pointed tip
Procedure:
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Matches
ED
Potato chips container complete with plastic cover – the type lined with aluminum foil
Ethanol
The reaction that occurs is the
combustion of ethanol:
•
Poke a hole about 1½ inches from the bottom of the potato chips container using a nail. The hole
should be big enough to fit a matchstick.
•
Squirt a few drops of alcohol into the potato chips container with its cover through the hole.
•
Shake the container vigorously and immediate introduce a lighted match into the hole.
•
Let the students touch the container.
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—>
Teacher tip
•
Demo 2
CH3COOH + NaHCO3
—>
NaCH3COO + H2O + CO2
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Vinegar and Baking Soda
Materials:
1 clear plastic cup
C
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Measuring spoon (1 Tablespoon, 1 teaspoon)
Thermometer
Vinegar
Procedure:
ED
Baking soda
Measure about 4 tablespoons of vinegar into the plastic cup.
•
Tell students to record the temperature of the vinegar in the plastic cup
•
Add 2 teaspoons of baking soda and stir.
•
Tell students to record the temperature of the mixture.
•
Let students touch the cup.
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•
Source: !
The reaction of baking soda and baking
soda:
http://highschoolenergy.acs.org/content/hsef/en/how-can-energy-change/exothermic-endothermicchemical-change.html
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PY
C
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APPENDIX C
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ED
Appendix Figure C.
Diagram of a bomb
calorimeter.
Calorimetry is the measurement of the amount of heat evolved or absorbed when a process or chemical
reaction takes place. A device called a calorimeter is used to measure the heat released from a chemical
reaction. Basically, what it does is measure the change in temperature of the system when a reaction takes
place. One such apparatus is called a bomb calorimeter. This type of calorimeter is usually used to measure
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PY
heats of reactions of exothermic reactions. It is made up of a strong steel container called the bomb, where the
reactants are placed. The bomb is placed into an insulated water bath, which has a stirrer and a thermometer.
The initial temperature of the water is recorded and igniting the reactants in the bomb with a small heater starts
the reaction. The heat released by the reaction is absorbed by the water causing the temperature of the water
to rise. By measuring the change in temperature and knowing the heat capacity of the calorimeter, we can
calculate the energy released when the reaction takes place.
The insulation of the calorimeter does not allow heat to flow to and from the surroundings. Exothermic
changes make the insulated system warmer while endothermic changes make it cooler.
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A bomb calorimeter is commonly used to determine the heat of combustion of a compound, that is, the
heat released when a particular quantity of a compound is burned in oxygen.
c=
q
-mΔT
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ED
APPENDIX D
Specific heat or specific heat capacity is an intensive property that can be used to identify or determine the
purity of a substance. Specific heat, c, is defined as the amount of heat required to raise the temperature of
one gram of a substance one degree Celsius. It is expressed as:
where:
q = amount of heat absorbed
m = mass of substance
ΔT = change in temperature
When the temperature of a system of given mass, m and specific heat capacity is raised from T1 to T2, the
amount of heat, q, is given by the equation:
q = mc(T2 – T1) = mc ΔT
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water is 4.18 J/g-oC (4.18 J g-1
it is 0.382 J/g- oC, for iron,
aluminum 0.900 J/g- oC. Note
of water is much larger than
PY
The specific heat of
o -1
C ) while for copper
0.446 J/g- oC and for
that the specific heat
that of the metals.
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ED
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APPENDIX E
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Appendix Figure E. A simple coffee cup calorimeter.
1)
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APPENDIX F
The following illustrates the calculation of the calorimeter constant.
C
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Problem:
Fifty (50) milliliters of water at 60 oC is added to 75 mL of water at 30 oC in a coffee cup calorimeter. The
resulting mixture was observed to have a temperature of 40 oC. What is the calorimeter constant?
ED
Answer:
We assume that heat cannot flow in or out of the calorimeter. Thus we write
D
EP
qwarm water + qcold water + qcal = 0
We consider that the density of water is 1 g/mL and its specific heat is 4.18 J/oC-g. Since we have to have mass
in grams we do the following calculations:
50 mL(1.0 g/mL) = 50 g water at 60 oC
75 mL (1.0 g/mL) = 75 g water at 30 oC
Thus, the equation is
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50 g (4.18 J/oC-g) (40 oC – 60 oC) + 75 g (4.18 J/oC-g) (40 oC -30 oC)+ Ccal (40 oC -30 oC) = 0
-4180 + 3135 + Ccal(10) = 0
10
Ccal = 104.5 J/oC
2)
C
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The calorimeter constant is 104.5 J/oC.
PY
Ccal = 4180 - 3135
The following problem illustrates the determination of the heat of reaction of chemical reactions.
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Answer:
ED
A student transferred 50.0 mL 1.00 M HCl into a coffee-cup calorimeter, which had a temperature of 25.5 oC.
He then added to this 50.0 mL 1.00 M NaOH which also had a temperature of 25.5 oC and stirred the mixture
quickly. The resulting solution was found to have a temperature of 32.5 oC. The calorimeter constant for the
coffee-cup calorimeter used was 15.0 J/ oC. Calculate the heat of reaction.
We consider that the density of water is 1 g/mL and its specific heat is 4.18 J/oC-g and assume that the density
of the solutions used is close to that of water to have the same density and specific heat.
We make use of the following equation to calculate for heat of reaction.
ΔH + mmix cmix ΔT + Ccal ΔT = 0
Since the amount of each reagent is given in units of volume we have to change this to grams. Therefore,
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(50.0 mL + 50.0 mL)(1 g/mL) = 100.0 g reaction mixture
Substituting
ΔH + 100.0 g (4.18 J/ oC-g)(32.5 oC - 25.5 oC ) + 15.0 J/ oC (32.5 oC - 25.5 oC ) = 0
PY
ΔH + 2884.2 J + 103.5J = 0
ΔH = - 2987.7 J or 29.88 kJ
The same quantity of NaOH was used, 0.0500 mol.
ED
50.0 mL x 1.00 mol/1000 mL = 0.0500 mol HCl
C
O
Thus, the heat released when the given amounts of HCl and NaOH react is 29.88 kJ. Since ΔH or change in
enthalpy is expressed in kJ/mol. The value we calculated is for the reaction of 50.0 mL of HCl and 50.0 mL of
NaOH we much express this this in kJ/mol. We must obtain the number of mole of HCl used in the reaction.
D
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We calculate ΔH in J/mol by dividing the value we obtained above by the number of moles of HCl used.
ΔH = - 29.88 kJ/0.0500 mol
= - 59.8 kJ/mol
3)
Another sample problem
Five (5.00) grams of an unknown compound was dissolved in 75.0 mL of water with an initial temperature of 22.5
o
C contained in a coffee-cup calorimeter. Upon dissolution of the compound the final temperature of the
resulting solution was 33.2 oC. The calorimeter used had a calorimeter constant of 200 J/ oC. What is the heat
of solution of the compound? Assuming the compound is sodium hydroxide, NaOH, what is the heat of
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solution in kJ per mol?
Answer:
PY
W assume the density and specific heat of the resulting solution is close to that of pure water; density is 1.0 g/
mL and specific heat is 4.18 J/ oC.
mwater = 75.0 mL (1.0 g/mL)
= 75.0 g
C
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ΔHsoln + mmix cmix ΔT + Ccal ΔT = 0
ΔHsoln + 75.0 g (4.18 J/ oC) (33.2 oC – 22.5 oC) + 200 J/ oC (33.2 oC – 22.5 oC) = 0
ΔHsoln = - 5629 J or 5.629 kJ
ED
ΔHsoln + 3488.6 + 2140 = 0
D
EP
Assuming that the compound is NaOH (Molar mass = 40.0 g/mo)
mol compound = 5.0 g/40.0 g/mol
= 0.125 mol
Therefore in kJ per mol:
ΔHsoln = -5.629 kJ/ 0.125 mol
= 45.0 kJ/mol
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Chemistry 2
150 MINS
Thermochemistry:
Thermochemical Equations
LESSON OUTLINE
Communicating learning objectives
5
Instruction
Discussion & Demonstration
70
PY
Content Standard
The learners demonstrate an understanding of energy changes in chemical reactions.
Introduction
(STEM-GC11TC-IIIg-i-126)
Calculate the change in enthalpy of a given reaction using Hess Law.
(STEM-GC11TC-IIIg-i-127)
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Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
10
(1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
New York: John Wiley & Sons.
ED
Write the thermochemical equation for a chemical reaction.
65
C
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Enrichment
Laboratory exercise
Performance Standard
The learners design a simple investigation to determine the effect on boiling point or Evaluation
Quiz
freezing when a solid is dissolved in water.
Materials
Computer or overhead projector; projector screen; transparencies of
Learning Competencies
lecture materials to be presented.
Explain enthalpy of a reaction.
Resources
(STEM-GC11TC-IIIg-i-125)
•
define enthalpy of reaction;
•
write thermochemical equations;
•
interpret thermochemical equations;
•
state Hess’ Law in own words; and
•
determine enthalpy of reactions from known thermochemical data using Hess’ Law.
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
(3) Padolina, MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
Inc.
(4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
Co., Inc.
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INTRODUCTION (5 MINS)
Teacher Tip
1. Introduce the learning objectives using any of the suggested protocols (Verbatim, Own words, Readaloud).
Enthalpy
•
Heat content
•
Thermochemical equations
•
Hess Law
•
Heat of reaction
•
Standard heat of formation
C
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•
PY
2. Post a list of terms that learners may have encountered in previous lessons (in lower grade levels and),
and those that will be introduced in this lesson.
INSTRUCTION (70 MINS)
ED
A. Recall of concepts learned from previous lesson
First Law of Thermodynamics
•
Exothermic and endothermic reactions
•
Heat of reaction
•
Calorimetry Thermochemistry and Enthalpy
D
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•
B. A closer look at Enthalpy Change,
1. Define enthalpy and heat content.
2. Write the equation ΔH = Hproducts - Hreactants. Explain what the equation means.
3. Ask learners to deduce the sign of ΔH for an exothermic and endothermic reaction at constant
pressure.
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Project/display the objectives all at once or
show each one at a time.
An exothermic reaction at constant pressure has a negative change in enthalpy – ΔH, while an
endothermic reaction at constant pressure has a positive change in enthalpy + ΔH.
PY
For an exothermic reaction, the heat content of reactants is greater than the heat content of the
products since heat is released. The value of ΔH < 0.
ΔH = Hproducts - Hreactants
ΔH < 0
ED
Hproducts
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Hreactants
For an endothermic reaction, the heat content of the products is greater than the heat content of the
reactants because heat is absorbed. The value of ΔH > 0.
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Hproducts
ΔH = Hproducts - Hreactants
ΔH < 0
Hreactants
4. State that enthalpy is a state function. Define state function.
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PY
The change in enthalpy ΔH is a state function, which refers to a quantity whose value depends on the
current state of the system and not on what has previously occurred. For example, the temperature
of a sample of water is 25 oC. This temperature does not depend on its previous temperature. Its
current value is 25 oC. Thus, ΔH being a state function depends on the state of each reactant and
product. To explain further, the value of ΔH for a reaction depends only on the conditions defining
the state of the reactants and products and not on the path it takes from reactants to products.
Thermochemical Equations
C
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1. State that the heat released or absorbed when a reaction takes place is an important and integral
part of the reaction, and could be indicated in the chemical equation. An equation which shows the
heat involved is called a thermochemical equation.
C6H12O6 (s) + 6 O2 (g)
—>
ED
2. Show the learners how to write thermochemical equations. An example is the thermochemical
equation for the combustion of glucose written as:
6 CO2 (g) + 6 H2O (l)
ΔH = -2805 kJ/mole
3. Ask them: a. What does the negative sign of ΔH indicate?
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b. If the reaction is exothermic what happens to heat when the reaction takes
place?
Sample response:
a. The negative sign of ΔH indicates that the reaction is exothermic.
b. Heat is released when the reaction occurs.
State that since the reaction is exothermic, heat could be written as a product, as shown below
C6H12O6 (s) + 6 O2 (g)
—>
6 CO2 (g) + 6 H2O (l) + 2805 kJ
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4. Give sample problem and ask the learners to write the corresponding thermochemical equation.
Sample exercise:
When one mole of nitrogen gas, N2, reacts with one mole of oxygen gas, O2, two moles of nitric
oxide, NO, are formed. In the process, 180.5 kJ of heat are required.
PY
5. Discuss the properties of heat of reaction ΔH, and give examples to illustrate these properties
The heat of reaction ΔH has the following properties:
C
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a. ΔH is expressed in units of kJ/mole. The value of ΔH depends on the amount of material. It is an
extensive property. Thus, if 2 moles of glucose are burned, then 5610 kJ of heat will be
produced.
ED
b. The value of ΔH for a given reaction depends on the physical state of each component. Thus, the
state of every reactant and product participating in the reaction must be indicated.
Therefore,
D
EP
c. Considering the first law of conservation of energy, the amount of heat released when 1 mole of
glucose is burned in oxygen producing carbon dioxide and water, is the same amount of heat
required for 6 moles of carbon dioxide, and 6 moles of water to react to form 1 mole of glucose.
Note that the value of ΔH is the same for the reverse reaction but has an opposite sign.
6 CO2 (g) + 6 H2O (l)
Or
—>
C6H12O6 (s) + 6 O2 (g)
6 CO2 (g) + 6 H2O (l) + 2805 kJ
—>
ΔH = 2805 kJ/mole
C6H12O6 (s) + 6 O2 (g)
In the second equation above, the amount of heat is written as one of the reactants.
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The thermochemical equation for the combustion of benzene may be written as:
C6H6 (l) + 7 ½ O2 (g)
—>
3 H2O (l ) + 6 CO2 (g)
ΔH = - 3269 kJ/mole
—>
3 H2O (l ) + 6 CO2 (g) + 3269 kJ
C6H6 (l) + 7 ½ O2 (g)
The reaction is exothermic since heat is produced by the reaction.
PY
Or
NH3 (g)
—>
3/2 H2 + 1/2 N2 (g)
ΔH = 29.61 kJ
Or
—>
3/2 H2 + 1/2 N2 (g)
ED
NH3 (g) + 29.61 kJ
C
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The thermochemical equation for the decomposition of ammonia, NH3 may be written as:
D
EP
The decomposition of ammonia is an endothermic reaction, as such requires the supply o f h e a t
to decompose ammonia. Note that 29.61 kJ of heat are required to decompose 1 mole of
ammonia. The reaction is endothermic. The heat content of the products is greater than the heat
content of the reactants.
The heat content of chemical substances depends on temperature and pressure. By convention,
ΔH values are generally reported at 25 oC (298 K) and standard atmosphere pressure (1 atm.)
Hess’s Law
1. Reiterate that enthalpy is a state function, that the magnitude of ΔH does not depend on the path
reactants take to form products. This means that chemical reactions can be carried out in one or
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several steps.
example.
In both cases, the net change is the same.
Illustrate this by giving an
•
—>
Or we can do it in two steps:
carbon dioxide.
CO2 (g)
ΔH = - 393.509 kJ
carbon-to-carbon monoxide, then carbon monoxide to
C
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C (s) + O2(g)
PY
We can burn carbon directly to carbon dioxide.
—>
CO (g)
ΔH = - 110.524 kJ
CO (g) + ½ O2 (g)
—>
CO2(g)
ΔH = - 282.985 kJ
C (s) + O2 (g)
—>
CO2 (g)
ΔH = - 393.509 kJ/mol
ED
C (s) + ½ O2 (g)
Cancel CO because it appears on both sides of the equation.
D
EP
From this example, the overall change is the net result of a series of steps, and the net
value of ΔH for the overall reaction is just the sum of all the enthalpy changes of the
different steps.
Note that in the above reaction the heats of reaction of the individual steps involved are
added algebraically to obtain the overall heat of reaction.
2. Write or show the following statement of Hess’ Law on the board:
Hess’s Law states that the change in enthalpy for any chemical reaction is
constant, whether the reaction occurs in one or several steps.
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Thermochemical calculations
1. Discuss the properties of thermochemical equations. Give examples that illustrate calculating ΔH
based on these properties .
Thermochemical equations possess two properties:
PY
a. They may be reversed.
Consider these properties of thermochemical equations.
A. Reversing thermochemical equations
C
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b. They may be treated as algebraic expressions, and therefore, maybe added, subtracted,
multiplied, or divided by a factor.
H2 (g) + ½ O2 (g)
H2O (l)
—>
—>
ED
When an equation is reversed, the sign of ΔH is also reversed.
H2O (l)
H2 (g) + ½ O2 (g)
ΔH = -286 kJ
ΔH = +286 kJ
D
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This means that the heat involved in the formation of 1 mole of H2O (l), is equal to the amount of
heat required to decompose 1 mole of liquid water.
B. Thermochemical equations may be added or subtracted as though these are algebraic equations.
Consider this problem:
Calculate the heat of formation of methane (CH4).
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The equation involved is C (s) + 2 H2 (g)
—>
CH4 (g)
ΔH = ?
The thermochemical equations for the combustion of these species are;
a. CH4 (g) + 2 O2 (g)
—>
CO2 (g) + 2H2O (l)
b. 2 H2 (g) + O2 (g)
—>
2H2O (l)
c. C (s) + O2 (g)
ΔH = - 890.4 kJ
ΔH = - 571.5 kJ
CO2 (g)
ΔH = - 393.7 kJ
C
O
—>
PY
It is impossible to measure this change directly. However, the heats of combustion of CH4
(g), C (s), and H2 (g) can be measured directly.
Solution:
ED
To solve the problem, combine equations (a), (b), and (c) so that when added, everything cancels out
except the formulas in the desired equation, the formation of methane.
Note that in the desired chemical equation, CH4 is on the product side, thus equation (a) needs to be
reversed. When done, the sign of ΔH must also be reversed.
CH4 (g) + 2 O2 (g)
ΔH = 890.4 kJ
D
EP
CO2 (g) + 2H2O (l)
Nothing needs to be done with equations (b) and (c) since C(s) and H2 (g) are on the reactant side. In
the desired equation, thexd are also on the reactants side.
Add the three equations, cancelling terms that appear on both sides:
CO2 (g) + 2H2O (l)
—>
CH4 (g) + 2 O2 (g)
ΔH = 890.4 kJ
2 H2 (g) + O2 (g)
—>
2H2O (l)
ΔH = - 571.5 kJ
C (s) + O2 (g)
—>
CO2 (g)
ΔH = - 393.7 kJ
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C (s) + 2 H2 (g)
—>
CH4 (g)
ΔH = - 74.8 mol/kJ
PY
Note that the net equation obtained is the desired equation above for which heat of reaction is
calculated (the heat of formation of CH4 (g)). This example illustrates Hess’s Law, in which the heat of
reaction ΔH depends only on the final products and initial reactants, and is independent of the
reaction (how the chemical change was carried out, in one or several steps).
C
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2. Summarize your discussion after giving examples on how to calculate ΔH using thermochemical
equations.
To summarize how to calculate for the ΔH of a specific reaction:
1. Look for the formulas that appear only once among the equations, and place it in the right place
just as in the desired equation.
ED
2. Note the number of moles of each reactant and products in the desired equation.
3. Manipulate the equations with known ΔH values so that the number of moles of reactants and
products are on the correct sides as in the desired equation. Do not forget to:
a. Change the sign of ΔH when the equation is reversed.
D
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b. Multiply/divide the number of moles and ΔH by the same factor.
4. Add the manipulated equations, cancelling terms that are common to both sides of the equation
to obtain the desired equation. Algebraically add the ΔH values to get the final ΔH, or change in
enthalpy of the desired equation.
Heats of Formation
1. Define heat of formation/standard heat of formation ΔHfo.
2. Discuss standard state and give examples (Padolina et al., 1995).
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3. State that values of standard states of formation are important because these are conveniently used
to calculate changes in enthalpy, ΔH of many reactions.
4. Give examples that illustrate the use of standard heats of formation in calculating changes in
enthalpy or ΔH of reactions.
PY
Calculating changes in enthalpy or ΔH of reactions.
1. Calculate the heat of hydrogenation of ethane (C2H4) given the following thermochemical equations:
—>
C2H6 (g)
ΔHfo = - 84.5 kJ/mol
b. 2 C (graphite) + 2 H2 (g)
—>
C2H4 (g)
ΔHfo = 52.3 kJ/mol
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a. 2 C(graphite) + 3 H2 (g)
Answer:
The desired equation is:
—>
C2H6 (g)
ΔH = ?
ED
C2H4 (g) + H2 (g)
Applying Hess’s Law,
C2H4 (g)
Add equation (a):
D
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Reverse equation (b) and change the sign of ΔH:
—>
2 C(graphite) + 3 H2 (g)
Overall equation:
C2H4 (g) + H2 (g)
2 C (graphite) + 2 H2 (g)
—>
C2H6 (g)
—>
C2H6 (g)
ΔH = - 52.3 kJ/mol
ΔH = - 84.5 kJ/mol
ΔH = - 136.8 kJ/mol
Examining this problem closely, the heat of reaction can be calculated by subtracting the sum of the
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heats of formation of reactants from the sum of the heats of reaction of products.
ΔHrxn = Σ ΔHf products - Σ ΔHf reactants
Therefore, the equation:
PY
Using ΔHfo to calculate the heat of hydrogenation of ethane:
ΔHrxn = ΔHfo C2H6 (g) – [ΔHfo C2H4 (g) + ΔHfo H2 (g)]
(- 84.5 kJ/mol) - (52.3 kJ/mol + 0)
C
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ΔHrxn =
= - 136.8 kJ/mol
ED
Note that compounds are the only one assigned heats of formation values while elements are not
(the ΔHf of elements is zero).
2. Calculate the ΔH for the reaction: CS2 (l) + 2 O2 (g)
ΔHfo CO2 (g)
= - 393.5 kJ/mol;
ΔHfo SO2
= -296.8 kJ/mol;
ΔHfo CS2 (l)
Answer:
CO2 (g) + 2 SO2 (g)
D
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Given:
—>
=
87.9kJ/mol
Use the equation: ΔHrxn = Σ ΔHf products - Σ ΔHf reactants
Write the equation for the problem:
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ΔHrxn = [ΔHfo CO2 (g) + 2ΔHfo SO2 (g)] - [ΔHfo CS2 (l) + 2 ΔHfo O2 (g)]
Substitute:
ΔHrxn = [-393.5kJ/mol + 2(-296.8 J/mol)] - [87.9 J/mol+ 2(0 J/mol)]
C
O
= - 1075.0 kJ
ENRICHMENT (65 MINS)
Laboratory exercise
ED
•
PY
The ΔHfo of SO2 (g) and the ΔHfo of O2 (g) are multiplied by 2, because there are two moles each of
these species in the desired equation.
Exercise 2.
D
EP
Prepare an exercise on solving problems involving thermochemical equations (See attached Laboratory
Exercise 2 for General Chemistry 2 Thermochemistry).
Learners do their calculations during the laboratory period. They may do this as a group or in pairs. Ask
them to show their calculations on the board.
•
Other notes for enrichment
Energy changes in chemical reactions can also take place in forms other than heat, such as light,
electricity, mechanical energy, etc. Some common examples are the burning of petrol or diesel in motor
engines of cars, trucks, tractors or buses produces mechanical energy, which is used in running these
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vehicles. The reaction involved in this example is
C
O
PY
Chemical reactions taking place in batteries produce electrical energy to run transistors, radios, torches
and watches, etc. For example, in a Daniel cell, the chemical reaction between zinc metal and copper
sulphate solution is accompanied by electrical energy.
Photosynthesis
D
EP
ED
In this process, chlorophyll in green plants converts carbon dioxide and water into glucose and oxygen.
Energy is provided by the energy of sunlight,
Visit also http://chemistry.tutorvista.com/physical-chemistry/endothermic-reaction.html
SEE ALSO ANIMATION:
http://www.tutorvista.com/chemistry/animations/endothermic-and-exothermic-reactionsanimation
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EVALUATION (10 MINS)
Quiz
Multiple choice. Write the letter corresponding to the best answer.
PY
1. A reaction is allowed to take place in an insulated container containing 100 mL of water. If the
reaction is exothermic, what happens to the temperature of water?
a. The temperature of the water goes down.
b. The temperature of the water goes up.
C
O
c. The temperature of the water does not change.
2. The thermochemical equation showing the formation of ammonia (NH3) from its elements is:
—>
2 NH3 (g)
This equation shows that 92 kJ of heat is:
ΔH = -92 kJ
ED
N2 (g) + 3 H2 (g)
a. Lost to the surroundings when one mole of hydrogen is used up in the reaction.
b. Absorbed from the surroundings when one mole of nitrogen reacts.
c. Absorbed from the surroundings when one mole of ammonia is formed.
D
EP
d. Lost to the surroundings when 2 moles of ammonia is formed.
3. Given the hypothetical thermochemical equation:
A + B
—>
C + D
ΔH = - 430 kJ
Which among the following statements is correct about this reaction?
a. The reaction is endothermic.
b. The equation may be written as A + B +
430 kJ
C + D
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c. The heat content of C and D is greater than the heat content of A and B.
d. The heat content of A and B is greater than the heat content of C and D.
C2H2 (g) + 3/2 CO2 (g)
—>
PY
4. An oxyacetylene torch is a tool that mixes and burns oxygen and acetylene to produce an extremely
hot flame. This tool is used to cut steel or weld iron and other metals. The temperature of the film
can reach 3480 oC. The burning of acetylene is given by the thermochemical equation:
CO2 (g) + H2O (l)
ΔH = - 1301.1 kJ
2CO2 (g) + 2H2O (l)
—>
C
O
For the reaction,
2 C2H2 (g) + 3 CO2 (g)
what is the ΔH for the reaction?
a. ΔH =
1301.1 kJ
b. ΔH = - 1301.1 kJ
2602.2 kJ
ED
c. ΔH =
Evaluation
D
EP
d. ΔH = - 2602.2 kJ
Learners’ understanding of the concepts may be assessed from their answers to the exercise in the laboratory.
A quiz may also be given during the lecture session to assess their understanding of the concepts (See suggested quiz above.)
1
2
3
4
(Not Visible)
(Needs Improvement)
(Meets Expectations)
(Exceeds Expectations)
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Chemistry 2
150 MINS
Content Standard
The learners demonstrate an understanding of energy changes in chemical reactions.
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Instruction
Discussion and Demonstration
85
Laboratory Exercise
60
PY
Thermochemistry: Enthalpy and
Hess’s Law
Enrichment
Performance Standard
The learners shall demonstrate an understanding of energy changes in chemical reactions Evaluation
Materials
C
O
Learning Competencies
Explain enthalpy of a reaction (STEM-GC11TC-IIIg-i-125)
Write the thermochemical equation for a chemical reaction (STEM-GC11TC-IIIg-i-126)
Computer or overhead projector; projection screen; transparencies
of lecture materials to be presented; materials for demonstration
(See Appendix B).
Calculate the change in enthalpy of a given reaction using Hess Law (STEM-GC11TC-IIIgResources
i-127)
(1) Brady EB. 1990. General Chemistry – Principles and Structure. New
York: John Wiley & Sons. 852 p.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
ED
Do exercises on thermochemical calculations (STEM_GC11TC-IIIg-i-128)
explain the enthalpy of a reaction;
•
write the thermochemical equation of a given reaction;
•
determine the change in enthalpy of a given reaction using Hess’ Law; and
•
perform calculations involving thermochemical equations
D
EP
•
(2) Masterson WL, Hurley CN, Neth EJ. 2012. Chemistry: Principles and
Reactions. California, USA: Brooks/Cole, Cengage Learning. 774 p.
(3) Padolina MCD, Sabularse VC, Marquez LA. 1995. Chemistry for the
21st Century. Makati, Philippines: Diwa Scholastic Press, Inc. 340 p.
(4) Silberberg MS. 2007. Chemistry – The Molecular Nature of Matter and
Change. International Edition. New York: McGraw-Hill Co., Inc. 1088 p.
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INTRODUCTION (5 MINS)
1. Introduce the learning objectives using any of the suggested protocols (verbatim, own words, readaloud)
Enthalpy
-
Heat content
-
Thermochemical equations
-
Hess Law
-
Heat of reaction
-
Standard heat of formation
D
EP
INSTRUCTION (80 MINS)
ED
-
C
O
PY
2. Introduce terms that learners may have encountered in previous lessons (in lower grade levels) and those
that will be introduced in this lesson.
Thermochemistry and Enthalpy
Define thermochemistry.
Define enthalpy and heat content.
Write the equation ΔH = Hproducts - Hreactants. Explain what the equation means.
Ask students to deduce the sign of ΔH for an exothermic and endothermic reaction at constant pressure.
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Teacher Tip
Project/display the objectives all at once or
show each one at a time.
To further discuss the sign of ΔH for an exothermic and endothermic reaction you may use the diagrams given
in Appendix A.
State that enthalpy is a state function. Define state function (Appendix A).
PY
Thermochemical Equations
Show the students how to write thermochemical equations.
C
O
State that the heat released or absorbed when a reaction takes place is an important and integral of the reaction
and that this could be indicated in the chemical equation. The equation showing the heat involved is called a
thermochemical equation.
An example is the thermochemical equation for the combustion of glucose which could be written as:
6 CO2 (g) + 6 H2O (l)
ΔH = -2805 kJ/mole
C6H12O6 (s) + 6 O2 (g)
6 CO2 (g) + 6 H2O (l) + 2805 kJ
ED
Or
C6H12O6 (s) + 6 O2 (g)
D
EP
Ask students: What does the negative sign of ΔH indicate?
Answer: The negative sign of ΔH indicates that the reaction is exothermic.
Ask students: If the reaction is exothermic what happens to heat when the reaction takes place?
Answer: Heat is released when the reaction occurs.
State that since the reaction is exothermic then heat could be written as a product just like in the second
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equation above.
Ask students: Since the reaction is exothermic what can you say about the heat content of the reactants
compared to the products?
PY
Answer: The heat content o the products is less than the heat content of the reactants.
Give an example of a thermochemical equation for an endothermic reaction. Have the students interpret it.
•
C
O
Discuss the properties of ΔH, heat of reaction and give examples illustrating these properties (Appendix B).
Hess’ Law
D
EP
Write on the board or show the statement:
ED
Repeat the statement that enthalpy is a state function; that is, the magnitude of ΔH does not depend
upon the path the reactants take to form the products. What this means is that chemical reactions can be carried
out in one step or in several steps. In both cases, the net change is the same. Illustrate this by giving an example
(Appendix C).
Hess’s Law states that the change in enthalpy for any chemical reaction is constant, whether the
reaction occurs in one step or in several steps.
Thermochemical calculations
Discuss the properties of thermochemical equations (Appendix D). Give examples that illustrate calculating for
ΔH based on these properties (Appendix D).
After giving examples on how to calculate ΔH using thermochemical equations you may summarize your
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discussion (Appendix D).
Heats of Formation
Define heat of formation/standard heat of formation, ΔHfo.
Discuss standard state and give examples (Padolina et al., 1995).
PY
State that values of standard states of formation are important because they are convenient to use in calculating
changes in enthalpy, ΔH of many reactions.
C
O
Give examples that illustrate the use standard heats of formation in calculating changes in enthalpy or ΔH of
reactions (Appendix E).
ENRICHMENT (120 MINS)
Exercise 2.
ED
Laboratory exercise
Prepare an exercise on solving problems involving thermochemical equations (See attached Laboratory Exercise
2 for General Chemistry 2 Thermochemistry).
EVALUATION
D
EP
Students do their calculations during the laboratory period. They may do this as a group or in pairs. Call upon
students to show their calculations on the board.
Quiz
Multiple choice. Write the letter corresponding to the best answer.
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1. A reaction is allowed to take place in an insulated container containing 100 mL of water. If the reaction is
exothermic, what happens to the temperature of water?
a. The temperature of the water goes down.
b. The temperature of the water goes up.
PY
c. The temperature of the water does not change.
2. The thermochemical equation showing the formation of ammonia, NH3 from its elements is:
—>
2 NH3 (g)
ΔH = -92 kJ
C
O
N2 (g) + 3 H2 (g)
This equation shows that 92 kJ of heat is __.
a. lost to the surroundings when one mole of hydrogen is used up in the reaction.
b. Absorbed from the surroundings when one mole of nitrogen reacts.
c. Absorbed from the surroundings when one mole of ammonia is formed.
ED
d. Lost to the surroundings when 2 moles of ammonia is formed.
3. Given the hypothetical thermochemical equation:
—>
C + D
ΔH = - 430 kJ
D
EP
A + B
Which among the following statements is correct about this reaction?
a. The reaction is endothermic.
b. The equation may be written as A + B +
430 kJ
—>
C + D
c. The heat content of C and D is greater than the heat content of A and B.
d. The heat content of A and B is greater than the heat content of C and D.
4. An oxyacetylene torch is a tool that mixes and burns oxygen and acetylene to produce an extremely hot
328
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flame. This tool is used to cut steel or weld iron and other metals. The temperature of the film can reach
3480 oC. The burning of acetylene is given by the thermochemical equation:
C2H2 (g) + 3/2 CO2 (g)
—>
CO2 (g) + H2O (l)
ΔH = - 1301.1 kJ
2CO2 (g) + 2H2O (l)
—>
2 C2H2 (g) + 3 CO2 (g)
C
O
what is the ΔH for the reaction?
a. ΔH =
PY
For the reaction:
1301.1 kJ!
b. ΔH = - 1301.1 kJ!
ED
Student understand of the concepts may be assessed from their answers to the exercise in the laboratory.
A quiz may also be given during the lecture session to assess their understanding of the concepts. (See suggested quiz above)
c. ΔH =
2602.2 kJ!
d. ΔH = - 2602.2 kJ!
2 (NEEDS
IMPROVEMENT)
D
EP
1 (NOT VISIBLE)
3 (MEETS
EXPECTATIONS)
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4 (EXCEEDS
EXPECTATIONS)
APPENDIX A
An exothermic reaction at constant pressure has a negative change in enthalpy, - ΔH and an endothermic
reaction at constant pressure has a positive change in enthalpy, + ΔH.
!
C
O
Hreactants
PY
For an exothermic reaction, the heat content of the reactants is greater than the heat content of the products
heat is released. The value of ΔH < 0.
ΔH = Hproducts - Hreactants
ΔH < 0
Hproducts
ED
For an endothermic reaction, the heat content of the products is greater than the heat content of the reactants
heat is absorbed. The value of ΔH > 0.
Hproducts
D
EP
!
ΔH = Hproducts - Hreactants
ΔH > 0
Hreactants
The change in enthalpy, ΔH is a state function. What does this mean? A state function is a quantity whose
value depends on the current state of the system and not on what has previously occurred in the system. For
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PY
example, the temperature of a sample of water is 25 oC. This temperature does not depend on its previous
temperature. Its current value is 25 oC. Thus, ΔH being a state function depends on the state of each reactant
and product. To explain further, the value of ΔH for a reaction depends only on the conditions defining the
state of the reactants and products and not on the path it takes from reactants to products.
APPENDIX B
C
O
The heat of reaction, ΔH has the following properties:
1. ΔH is expressed in units of kJ/mole. The value of ΔH is dependent on the amount of material. It is an
extensive property. Thus, if 2 moles of glucose are burned then 5610 kJ of heat will be produced.
2. The value of ΔH for a given reaction depends on the physical state of each component. Thus, the state of
every reactant and product participating in the reaction must be indicated.
6 CO2 (g) + 6 H2O (l)
Or,
D
EP
Therefore
ED
3. Considering the first law of conservation of energy, the amount of heat released when one mole of glucose
is burned in oxygen producing carbon dioxide and water is the same amount of heat required for 6 moles of
carbon dioxide and 6 moles of water to react to form one mole of glucose. Note that the value of ΔH is the
same for the reverse reaction but is opposite in sign.
—>
6 CO2 (g) + 6 H2O (l) + 2805 kJ
C6H12O6 (s) + 6 O2 (g)
—>
ΔH = 2805 kJ/mole
C6H12O6 (s) + 6 O2 (g)
In the second equation above, the amount of heat is written as one of the reactants.
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•
•
The thermochemical equation for the combustion of benzene may be written as
C6H6 (l) + 7 ½ O2 (g)
—>
3 H2O (l ) + 6 CO2 (g)
ΔH = - 3269 kJ/mole
C6H6 (l) + 7 ½ O2 (g)
—>
3 H2O (l ) + 6 CO2 (g) + 3269 kJ
C
O
The reaction is exothermic, heat is produced by the reaction.
PY
Or,
The thermochemical equation for the decomposition of ammonia, NH3 may be written as
Or,
NH3 (g) + 29.61 kJ
3/2 H2 + 1/2 N2 (g)
ΔH = 29.61 kJ
ED
—>
3/2 H2 + 1/2 N2 (g)
D
EP
NH3 (g)
The decomposition of ammonia is an endothermic reaction, that is, heat must be supplied for ammonia to
decompose. Note that 29.61 kJ of heat are required to decompose one mole of ammonia. The reaction is
endothermic. The heat content of the products if greater than the heat content of the reactants.
The heat content of chemical substances depends upon temperature and pressure. By convention, ΔH values
are generally reported at 25 oC (298 K) and standard atmosphere pressure (1 atm.)
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APPENDIX C
We can burn carbon directly to carbon dioxide.
ΔH = - 393.509 kJ
PY
!!!!!!!! C (s) + O2(g) !!!!!!!—>!!!!!!!!!!!!CO2 (g)
Or, we can do it in two steps: carbon to carbon monoxide then carbon monoxide to carbon dioxide.
CO (g) + ½ O2 (g)
—>
—>
CO (g)
ΔH = - 110.524 kJ
CO2(g)
ΔH = - 282.985 kJ
C
O
!!!!!!!!! C (s) + ½ O2 (g)
______________________________________________________________________________
C (s) + O2 (g)
—>
CO2 (g)
ΔH = - 393.509 kJ/mol
ED
We cancel CO because it appears on both sides of the arrows.
From this example we see that the overall change is the net result of a series of steps, and the net value of ΔH
for the overall reaction is the just the sum of all the enthalpy changes of the different steps.
APPENDIX D
D
EP
Note that in the above reaction the heats of reaction of the individual steps involved are added algebraically to
obtain the overall heat of reaction.
Thermochemical equations possess two properties:
1. They may be reversed.
2. They may be treated as algebraic expressions. Therefore, they may be added, subtracted, multiplied by a
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factor or divided by a factor.
Let us consider these properties of thermochemical equations.
When we reverse an equation the sign of ΔH is also reversed.
H2 (g) + ½ O2 (g)
—>
H2O (l)
ΔH = -286 kJ
H2 (g) + ½ O2 (g)
ΔH = +286 kJ
C
O
H2O (l)
—>
PY
1. Reversing thermochemical equations
This means that the heat involved in the formation of one mole of H2O (l) is equal to the amount of heat
required to decompose one mole of liquid water.
ED
2. Thermochemical equations may be added or subtracted as though they are algebraic equations.
Consider this problem:
D
EP
Calculate the heat of formation of methane, CH4.
The equation involved is C (s) + 2 H2 (g)
—>
CH4 (g)
ΔH = ?
It is impossible for us to measure this change directly. However, the heats of combustion of CH4 (g), C (s)
and H2 (g) can be measured directly.
The thermochemical equations for the combustion of these species are;
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(a) CH4 (g) + 2 O2 (g)
—>
CO2 (g) + 2H2O (l)
ΔH = - 890.4 kJ
(b) 2 H2 (g) + O2 (g)
—>
2H2O (l)
ΔH = - 571.5 kJ
(c) C (s) + O2 (g)
—>
CO2 (g)
ΔH = - 393.7 kJ
PY
Solution:
To solve the problem we have to combine equations (a), (b) and (c) so that when they are added
everything cancels out except the formulas in the desired equation, that is the formation of methane.
CO2 (g) + 2H2O (l)
—>
CH4 (g) + 2 O2 (g)
C
O
Note that in the desired chemical equation CH4 is on the product side, thus we need to reverse eq. (a).
When we do so we also have to reverse the sign of ΔH.
ΔH = 890.4 kJ
ED
We do not need to do anything with equations (b) and (c) since C(s) and H2 (g) are on the reactant side.
In the desired equation, they are also on the reactants side.
CO2 (g) + 2H2O (l)
2 H2 (g) + O2 (g)
C (s) + O2 (g)
C (s) + 2 H2 (g)
D
EP
So we add the three equations, cancelling terms that appear on both sides:
—>
—>
—>
—>
CH4 (g) + 2 O2 (g)
2H2O (l)
CO2 (g)
CH4 (g)
ΔH = 890.4 kJ
ΔH = - 571.5 kJ
ΔH = - 393.7 kJ
ΔH = - 74.8 mol/kJ
Note that the net equation we obtain is the desired equation above for which we calculated its heat of reaction
(the heat of formation of CH4 (g). This example illustrates Hess’s Law, from which we see that the heat of
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reaction, ΔH depends only on the final products and initial reactants, and is independent of the of the reaction
(how the chemical change was carried out, in one step or in several steps).
To summarize how to calculate for the ΔH of a specific reaction:
2. Note the number of moles of each reactant and products in the desired equation.
PY
1. Look for the formulas that appear only once among the equations and place it in the right place just as that
in the desired equation.
• Change the sign of ΔH when the equation is reversed.
C
O
3. Manipulate the equations with known ΔH values so that the number of moles of reactants and products are
on the correct sides as in the desired equation. Do not forget to
• Multiply/divide the number of moles and ΔH by the same factor.
APPENDIX E
1.
D
EP
Calculating changes in enthalpy or ΔH of reactions.
ED
4. Add the manipulated equations, cancelling terms that are common to both sides of the equation to obtain
the desired equation. Algebraically add the ΔH values to get the final ΔH or change in enthalpy of the
desired equation.
Calculate the heat of hydrogenation of ethane, C2H4 given the following thermochemical equations:
(1) 2 C(graphite) + 3 H2 (g)
—>
C2H6 (g)
ΔHfo = - 84.5 kJ/mol
(2) 2 C (graphite) + 2 H2 (g)
—>
C2H4 (g)
ΔHfo = 52.3 kJ/mol
Answer:
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The desired equation is:
C2H4 (g) + H2 (g)
—>
C2H6 (g)
ΔH = ?
PY
We apply Hess’s Law.
Reverse equation 2 and change the sign of ΔH:
—>
2 C (graphite) + 2 H2 (g)
Add equation 1: 2 C(graphite) + 3 H2 (g)
Overall equation:
C2H4 (g) + H2 (g)
—>
ΔH = - 52.3 kJ/mol
C2H6 (g)
C2H6 (g)
ΔH = - 84.5 kJ/mol
C
O
C2H4 (g)
ΔH = - 136.8 kJ/mol
We, therefore have the equation:
ED
If we examine this problem closely, we will see that the heat of reaction can be calculated by subtracting the
sum of the heats of formation of reactants from the sum of the heats of reaction of products.
ΔHrxn = Σ ΔHf products - Σ ΔHf reactants
D
EP
So using ΔHfo to calculate for the heat of hydrogenation of ethane:
ΔHrxn = ΔHfo C2H6 (g) – [ΔHfo C2H4 (g) + ΔHfo H2 (g)]
ΔHrxn =
(- 84.5 kJ/mol) - (52.3 kJ/mol + 0)
= - 136.8 kJ/mol
Note that compounds are the only one assigned heats of formation values while elements are not (the ΔHf of
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elements is zero).
2. Calculate the ΔH for the reaction: CS2 (l) + 2 O2 (g)
CO2 (g) + 2 SO2 (g)
Answer:
C
O
We make use of the equation: ΔHrxn = Σ ΔHf products - Σ ΔHf reactants
PY
Given: ΔHfo CO2 (g) = - 393.5 kJ/mol; ΔHfo SO2 = -296.8 kJ/mol; ΔHfo CS2 (l) = 87.9kJ/mol
We write the equation for the problem:
ΔHrxn = [ΔHfo CO2 (g) + 2ΔHfo SO2 (g)] - [ΔHfo CS2 (l) + 2 ΔHfo O2 (g)]
D
EP
Substituting:
ED
The ΔHfo of SO2 (g) and the of ΔHfo O2 (g) are multiplied by 2 because in the desired equation there are two
mols each of these species.
ΔHrxn = [-393.5kJ/mol + 2(-296.8 J/mol)] - [87.9 J/mol+ 2(0 J/mol)]
= - 1075.0 kJ
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Chemistry 2
210 MINS
Chemical Kinetics: Reaction
Rates & Collision Theory
Introduction
1. the rate of a reaction and the various factors that influence it; and
2. the Collision Theory.
Identifying reactions
5
Instruction
Discussion
82
Enrichment
Laboratory exercise
100
Laboratory exercise question & answer
20
C
O
D
EP
ED
Learning Competencies
Describe how various factors influence the rate of a reaction. (STEM_GC11CK-IIIi-j-130),
Explain the effect of temperature on the rate of a reaction,(STEM_GC11CK-IIII-J-135),
Explain reactions qualitatively in terms of molecular collisions, (STEM-GC11CK-IIIi-j136),
Explain activation energy and how a catalyst affects the reaction rate (STEM-GC11CK-IIIij137), Cite and differentiate the types of catalysts, (STEM-GC11CK-IIIi-j138), (LAB)
Determine the effect of various factors on the rate of reaction. (STEM-GC11CK-IIIi-j139)
3
Motivation
Performance Standard
Evaluation
The learners shall be able to explain the effects of various factors on the rates of chemical
Materials
reactions and describe the collision theory.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Communicating learning objectives
PY
Content Standards
The learners demonstrate an understanding of
LESSON OUTLINE
Computer or overhead projector; projector screen; transparencies of
lecture materials to be presented; materials for demonstration
(Appendix A)
Resources
(1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
New York: John Wiley & Sons.
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
(3) Padolina, MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
Inc.
•
define rate of reaction;
•
identify factors that affect the rate of a reaction;
•
describe how each factor can affect the rate of a reaction;
•
state the premises of the Collision Theory;
•
explain the effect of each factor on the rate of a reaction using the Collision Theory;
•
define activation energy; and
•
explain the effect of a catalyst on the rate of a reaction by providing a different reaction path.
(4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
Co., Inc.
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INTRODUCTION (3 MINS)
Teacher Tip
•
1. Communicate learning objectives.
•
•
Chemical kinetics
•
Rate of reactions
•
Catalyst
•
Activation energy
PY
2. Introduce terms that the learners will encounter.
C
O
MOTIVATION (5 MINS)
Ask the learners to name some chemical process happening around them that they think happens too
fast and they want to be slowed down. Alternatively, ask for processes that are too slow that they want to
happen faster. Some possible answers are aging, ripening of fruits, spoilage of food, etc.
ED
Pose the question on possible ways of changing the rate at which some reactions occur. One example is
storing food at low temperatures (inside a refrigerator!) to slow down spoilage.
D
EP
INSTRUCTION (82 MINS)
A. Recall of concepts learned from previous lesson
•
First Law of Thermodynamics
•
Exothermic and endothermic reactions
•
Heat of reaction
•
Calorimetry Thermochemistry and Enthalpy
1. Tell learners to recall certain reactions previously discussed which occur at different rates.
340
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Project/display the objectives all at once
or show each one at a time.
Write terms on the board or show them
one at a time as you go along.
Ask them to recall the previous lesson on heat changes (enthalpy changes). State that neither the heat of
reaction nor the balanced equation can give any indication of how fast a reaction will take place..
PY
Define “rate of reaction” as a measure of how fast a reaction takes place. The rate of a reaction is often
expressed as a change in amount or concentration of a substance (reactant or product) per unit time. Chemical
kinetics is the study of rates of reactions and factors that affect them.
C
O
2. Discuss the importance of knowing rates of reactions (Why learners should be interested in studying
rates of reactions).
3. Factors that affect rates of reactions
ED
Many chemical industries make use of chemical reactions whose rates should be fast enough to be
economically viable but slow enough to allow some control. Action of drugs or medicines is an important
consideration in medicine.
D
EP
Ask learners the following questions from which the teacher could deduce the depth of their
understanding of factors:
a. How long does it take an iron nail exposed to the rain to rust?
b. Compare the rusting of iron to how fast milk curdles when an acid like vinegar or calamansi juice is
added to it.
c. Which has a more rapid reaction, the burning of liquid gasoline in air, or gasoline in a car engine that
is first vaporized, then mixed with air?
d. Do you think you could light a log with a single matchstick?
e. How about twigs or smaller pieces of wood?
341
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Sample response:
a. It takes a long time for an iron nail to rust.
b. Milk curdles as soon as acid is added to it. Curdling of milk is occurs faster than the rusting of an iron
nail.
c. Vaporized gasoline in car engines readily ignites and burns very rapidly compared to liquid gasoline.
PY
d. No.
e. Yes
f.
C
O
(questions, continued)
Baking powder or sodium bicarbonate (NaHCO3), is used in baking to make cakes rise because the
carbon dioxide gas produced when baking powder reacts with water in the cake batter. Compare
the volume of a cake prepared with the right amount of baking powder, with that of just about half
the amount needed.
g. Why do we keep food in the refrigerator?
Sample response:
The cake with the right amount of baking powder is expected to have a larger volume because more
carbon dioxide gas will be produced.
D
EP
f.
ED
h. How do particles move at high temperatures compared at low temperatures?
g. To keep it from spoiling, which can happen faster at room temperature.
h. Particles move faster at higher temperatures than at lower temperatures.
Perform a demonstration that shows the difference in the rate of reaction of an uncatalyzed and
catalyzed reaction (See Appendix A for this). Let learners describe what they see.
342
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Sample response: Bubbling and steam is produced.
Which reagent caused the reaction to proceed?
PY
Sample response: Baker’s yeast or potassium iodide—depending on which demonstration the t e a c h e r
will use (Learners identify the reagent which acts as a catalyst).
C
O
While doing the demonstration, allow the learners to touch the reaction vessel, ensuring that they do
not come in contact with the reacting materials. Let them recall endothermic and exothermic reactions, and later
identify whether the reaction is endothermic or exothermic.
From learners’ answers, the teacher could have a discussion with them on the factors that affect rates of
reactions.
Nature of reactants – state that substances vary in their chemical reactivity. Chemical reactivity is a
major factor that determines the rate of a reaction. Give examples of substances of varying chemical
reactivity and reactions where these are involved.
ED
•
Ability of reactants to meet - consider surface area of contact, classification of reactions as
homogeneous and heterogeneous reactions
•
Concentration of reactants - This could be likened to a hallway with just a few learners to a crowded
hallway where they are likely to bump into each other. Give an application of this i.e., prescription of
medicines by doctors.
•
Temperature of reaction system
•
Presence of a catalyst
4. Collision theory
D
EP
•
Teacher Tip
•
This portion may be discussed after the
laboratory experiment has been carried
out. The Collision Theory is a way of
explaining the effect of the factors
investigated in the experiment.
•
Other activities relating to factors that
affect reaction rates:
http://www.teachsecondary.com/mathsand-science/view/lesson-plan-ks4science-rates-of-reaction-in-chemistry
Ask learners their ideas on how they would explain the effects of the different factors on reaction rates at
the molecular level. From their answers, proceed to a discussion of the Collision Theory and emphasize that for
343
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•
Lesson plan: KS4 science – rates of
reaction in chemistry
a reaction to take place, reactant particles must have activation energy or the minimum amount of energy
required and the right orientation for effective collision to bring about a chemical change.
Show illustrations showing (Appendix B):
b. Slow moving particles approach each other then eventually fly apart.
PY
a. Two reactant particles approach each other then fly apart because they are not correctly oriented for
a reaction to occur.
c. Fast moving, energetic molecules correctly oriented forming new substances.
C
O
Ask learners to give their explanation on the basis of Collision Theory how reaction rates are affected by
various factors.
In your discussion of how reaction rates are affected by various factors consider the following:
Temperature – show a plot of the fraction of molecules having a kinetic energy at two different
temperatures (Appendix E). Ask learners to interpret the graph. The graph would show that at
higher temperatures, the total fraction of molecules having the required energy is greater than that at
lower temperatures.
A greater number of reactant molecules are more energetic at higher
temperatures than at lower temperatures, thereby making collisions more effective for products to
form.
•
Concentration – Present to learners an illustration showing a few reactant molecules and another with
more reactant molecules. The more number of reactant molecules, the greater probability for
effective collisions to form products. Compare these to scenarios where there are just two or three
learners in a hallway, and a crowded one. Learners are more likely to bump into each other in the
crowded hallway than with just a few ones.
D
EP
ED
•
5. Potential energy diagrams
Start your discussion of potential energy diagrams with the question:
344
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•
Possible illustrations for factors that
affect reaction rates:
http://www.freezeray.com/flashFiles/
RatesOfReactionSurfaceArea.swf
http://www.freezeray.com/flashFiles/
RatesOfReactionConc.swf
http://www.freezeray.com/flashFiles/
RatesOfReactionTemp.swf
What happens after reactant molecules collide?
•
Sample response:
PY
If a reaction occurs, during the collision, the particles that separate are different from those that collide.
When the particles collide, the molecules slow down. Thus, the total kinetic energy (K.E.) they possess
decreases. Because energy cannot disappear, this means that the total potential energy (P.E.) of the particles
must increase.
D
EP
ED
C
O
Further state that the relationship between activation energy and total potential energy of reactants and
products may be expressed graphically through a potential energy diagram.
Figure 1. A potential energy diagram for a reaction. (Image source: http://
schools.birdville.k12.tx.us/cms/lib2/tx01000797/centricity/domain/912/chemlessons/Lessons/
Energy/image001.jpg)
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Point out that the activation energy appears as a “hill,” which is referred as the potential energy barrier
between reactants and products. Reactant particles must go over this hill, the minimum energy required for a
reaction to occur, to form the products.
C
O
PY
Show a figure (Appendix C) where the teacher likens activation energy to the energy needed by the
person to push the ball over the hill. If the person does not push the ball hard enough, it rolls back down just
as reactant particles that do not have the minimum amount of energy required. If the ball is given enough
energy, it will go over the hill as reactant particles with the required amount of energy would go over the
potential energy barrier to form the products.
D
EP
ED
Activation energy is the minimum energy required for the electron clouds and nuclei of reactant particles
to overcome the repulsions and form bonds.
Figure 2: Energy diagram for a reaction showing H and Activation Energy (Image
source: www.gcsescience.com)
Describe/define the activated complex or the transition state.
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•
•
Show learners which portion of the potential energy diagram corresponds to the heat of
reaction, ΔHrxn.
PY
How does a catalyst increase the rate of a reaction?
Tell them to recall the demonstration that showed the increase in the rate of a reaction with the
addition of a catalyst to the reaction mixture.
C
O
Ask learners of their idea about how a catalyst increases the rate of a reaction.
Show a potential energy diagram for a catalyzed and uncatalyzed reaction (Appendix D). Ask learners to
interpret the diagram.
ED
Point out that the catalyst provides a path for the reaction with a lower activation energy, wherein the
catalyst participates in the reaction by changing its mechanism.
D
EP
Discuss the different types of catalysts (homogeneous and heterogeneous catalysts). Give examples. Ask
learners to recall enzymes and what type of catalyst is an enzyme.
ENRICHMENT (100 MINS)
Exercise 1: !
Factors affecting rates of reactions
1. One week before this laboratory session, task learners to read and study this exercise. Hand out
copies of the experiment if they do not have a laboratory manual. Instruct them to write the (1)
objectives of the experiment and (2) the necessary data tables in their laboratory notebooks.
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2. Have a pre-laboratory discussion reviewing the factors that affect reaction rates discussed in the
lecture.
•
3. Ask learners to state the objectives of the exercise.
4. Describe what they are to do when they perform the exercise. Show them the materials that will be
used in the exercise. Demonstrate how they will mix the reactant solutions and set the water bath.
PY
5. Check their data tables and make corrections when needed.
C
O
The exercise involves determining the effect of reactant concentration, temperature, and the presence of
a catalyst on the rate of reactions.
EVALUATION (20 MINS)
ED
Evaluation
Check the data learners collected when performing the laboratory experiment. Let them answer the questions in the exercise.
They could do this as a group where they discuss their answers among themselves. They should write down their answers in their
notebooks. The learners’ answers to the questions will reflect their understanding of how certain factors affect the rate of a
chemical reaction.
2
3
4
(Needs Improvement)
(Meets Expectations)
(Exceeds Expectations)
D
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1
(Not Visible)
APPENDIX A
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Demonstration procedure
Teacher Tip
The teacher may use any of the following depending on available materials.
Materials:
PY
Demo 1*
One empty 500 mL soda plastic bottle or a 500 mL plastic water bottle
•
½ cup 3% or 6% hydrogen peroxide (the former is available in drug stores, the latter may be available in
beauty supply stores)
•
Dish washing liquid solution or any soap solution
•
Baker’s active yeast (available in supermarkets or bakery supply stores)
•
Food coloring (optional)
ED
Procedure:
C
O
•
1. Dissolve one teaspoon or one packet of active yeast in a small amount of warm water. Keep still for
about 5 minutes.!
D
EP
2. Dilute a small amount of dishwashing liquid in about ¼ cup of water, or dissolve soap in water.
3. Place about ¼ cup of the dishwashing liquid solution or soap solution into the plastic bottle. Two to
three drops of food color can be added and mixed.
4. Add ½ cup of hydrogen peroxide to the soap solution.
5. Ask learners their observation.
6. Add the yeast to the mixture in the bottle.
7. Ask the learners to describe what they see and explain the phenomenon. Ask them to identify which
reagent caused the reaction to proceed.
8. Let them determine if the reaction is exothermic or endothermic.
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•
The teacher can give a background of
the reaction that proceeds.
•
The chemical reaction involved produces
a foamy soap, which froths out of a
bottle or a graduated cylinder. The
reaction involved is the catalytic
decomposition of hydrogen peroxide,
H2O2. The reaction involved is:
2H2O2 (l) —> 2H2O + O2 (g) + heat
•
In Demo 1, the catalyst used is Baker’s
Yeast.
•
Tell the learners that Baker’s yeast
(Saccharomyces cerevisiae) has many
uses. For example, it is used in the
preparation of bread like pan de sal, as
well as in fermentation processes – in
wine making and brewing (beer making).
•
Yeast contains the enzyme catalase. This
enzyme brings about the decomposition
of hydrogen peroxide to water and
oxygen. The oxygen gas that produced
in the reaction causes soap bubbles to
form, and learners observe the foam
coming out of the bottle. The same
reaction occurs in our body. Small
amounts of hydrogen peroxide are
formed in our bodies by metabolic
reactions. Hydrogen peroxide is harmful
to the body if it accumulates in body
cells. However, its build up is prevented
by the presence of the enzyme catalase
produced in these very cells by
decomposing it to water and oxygen.
•
*Source: http://www.coolscience.org/CoolScience/KidScientists/h2o2.htm
Step 1:
H2O2 (aq) + I- (aq) —> H2O (l) + OI- (aq)
Demo 2*
•
•
10 mL graduated cylinder
•
Goggles
•
Rubber gloves
•
Plastic tray or basin
•
30% hydrogen peroxide
•
Dishwashing liquid or soap that readily forms bubbles
•
2 M potassium iodide solution
•
Food coloring (optional)
C
O
500 mL graduated cylinder or 1.5 liter soda bottle
ED
•
PY
Materials:
D
EP
Procedure:
1. Place the graduated cylinder or soda bottle on a plastic tray or basin.
2. Measure 20 mL 30% hydrogen peroxide and place this in the graduated cylinder or bottle.
3. Add 5 mL of dishwashing detergent or soap solution.
4. Ask learners to describe what they see.
5. Add 10 mL of potassium iodide solution.
6. Ask students to describe what they see and explain the phenomenon. Ask them to identify which
reagent caused the reaction to proceed.
7. Let them determine if the reaction is exothermic or endothermic.
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Step 2:
H2O2 (aq) + OI- (aq) —> H2O (l) +
O2(g) + I-(aq)
•
Note that when common terms on both
sides of the arrow cancel, the overall
reaction si obtained. The iodide ion does
not appear in the overall reaction
2H2O2 (l) —> 2H2O + O2 (g) + heat
•
•
Again, the oxygen gas form causes soap
bubbles to form.
•
Safety precautions: 1. Handle hydrogen peroxide with care. It is severely corrosive to the skin, eyes,
and the respiratory tract. In case of contact, flush with water for 15 minutes.
If eyes are affected, get medical attention. !
2. Steam and oxygen gas form very quickly. Stand back from the reacting
vessel. Do not stand over the reaction.
PY
3. Use gloves and goggles during the demonstration
*Source: http://cldfacility.rutgers.edu/content/catalytic-decomposition-hydrogen-peroxide-
APPENDIX B
Collision
After collision
D
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ED
Before collision
C
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potassium-iodide
Figure B1. Two reactant particles approach each other then fly apart because they are not correctly
oriented for a reaction to occur.
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•
Collision
After collision
C
O
PY
Before collision
Figure B2. Slow moving particles approach each other then eventually fly apart.
Collision
After Collision
D
EP
ED
Before collision
Appendix Figure B3. Fast moving, energetic molecules correctly oriented form new substances.
APPENDIX C
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PY
C
O
ED
D
EP
A catalyst provides a path for the reaction with a lower activation energy.
Source: Bloomfield MM. 1980.
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D
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ED
C
O
PY
APPENDIX D.
Comparison of reaction rate between a catalyzed and uncatalyzed reaction.
Source: http://www.chem.uiuc.edu/rogers/text13/tx133/tx133p2.GIF
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D
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ED
C
O
PY
APPENDIX E.
When a catalyst is present, more molecules possess the minimum amount of energy needed
effective collisions.!
!
!
!
Source: Brady, 1990.
for
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Chemistry 2
210 MINS
Content Standards
The learners demonstrate an understanding of:
Introduction
Communicating learning objectives
5
Instruction
Discussion and Activity
20
Motivation
Demonstration
55
Enrichment
Laboratory Exercise
120
Evaluation
Processing of Laboratory Exercise
10
C
O
1. the rate of a reaction and the various factors that influence it; and
LESSON OUTLINE
PY
Chemical Kinetics: Factors that
Influence Reaction Rate and
Collision Theory
2. the Collision Theory.
Materials
Performance Standard
Computer or overhead projector; projection screen; transparencies
The learners shall be able to explain the effects of various factors on the rates of chemical for lecture materials to be presented; materials for demonstration
(See Appendix A)
reactions and describe the collision theory.
D
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ED
Learning Competencies
Describe how various factors influence the rate of a reaction (STEM_GC11CK-IIIi-j-130).
Explain the effect of temperature on the rate of a reaction (STEM_GC11CK-IIII-J-135).
Explain reactions qualitatively in terms of molecular collisions (STEM-GC11CK-IIIi-j136).
Explain activation energy and how a catalyst affects the reaction rate (STEM-GC11CK-IIIij137). Cite and differentiate the types of catalysts (STEM-GC11CK-IIIi-j138). (LAB)
Determine the effect of various factors on the rate of reaction. (STEM-GC11CK-IIIi-j139)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
describe the factors that influence the rate of reaction;
•
explain reactions qualitatively using molecular collisions;
•
explain activation energy and how a catalyst affects reaction rates; and
•
cite and differentiate the types of catalysts.
Resources
(1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
New York: John Wiley & Sons.
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
(3) Padolina, MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
Inc.
(4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
Co., Inc.
356
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INTRODUCTION (5 MINS)
1. Communicate learning objectives
Teacher Tip
2. Present relevant vocabulary that learners will encounter:
•
Phase
•
Rate of Reaction
PY
Catalyst
Activation Energy
C
O
INSTRUCTION (20 MINS)
1. Tell learners to recall certain reactions previously discussed which occur at different rates.
ED
Ask them to recall the previous lesson on heat changes (enthalpy changes). State that heat changes do not tell
how fast a reaction takes place nor does a balanced chemical equation.
2. Lecture Proper
D
EP
Discuss the importance of knowing rates of reactions (Why learners should be interested in studying rates of
reactions).
Factors that affect rates of reactions
Ask learners the following questions from which the teacher could deduce the depth of their understanding of
factors:
a. How long does it take an iron nail exposed to the rain to rust?
b. Compare the rusting of iron to how fast milk curdles when an acid like vinegar or calamansi juice is
added to it.
c. Which has a more rapid reaction, the burning of liquid gasoline in air, or gasoline in a car engine that is
357
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Project/display the objectives all at once
or show each one at a time.
Write terms on the board or show them
one at a time as you go along.
first vaporized, then mixed with air?
d. Do you think you could light a log with a single matchstick?
e. How about twigs or smaller pieces of wood?
Sample response:
PY
a. It takes a long time for an iron nail to rust.
b. Milk curdles as soon as acid to it. Curdling of milk is faster than the rusting of an iron nail.
c. Vaporized gasoline in car engines readily ignites and burns very rapidly compared to liquid gasoline.
d. No.
C
O
e. Yes
(questions, continued)
Baking powder or sodium bicarbonate (NaHCO3), is used in baking to make cakes rise because the
carbon dioxide gas produced when baking powder reacts with water in the cake batter. Compare the
volume of a cake prepared with the right amount of baking powder, with that of just about half the
amount needed.!
g. Why do we keep food in the refrigerator?
ED
f.
Sample response:
f.
D
EP
h. How do particles move at high temperatures compared at low temperatures?
The cake with the right amount of baking powder is expected to have a larger volume because more
carbon dioxide gas will be produced.
g. To keep it from spoiling.
h. Particles move faster at higher temperatures than at lower temperatures.
358
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MOTIVATION (55 MINS)
1. Perform a demonstration that shows the difference in the rate of reaction of an uncatalyzed and catalyzed
reaction (See Appendix A for this). Let learners describe what they see.
PY
Sample response: Bubbling and steam is produced.
Which reagent caused the reaction to proceed?
C
O
Sample response: Baker’s yeast or potassium iodide—depending on which demonstration the teacher will
use (Learners identify the reagent which acts as a catalyst).
2. While doing the demonstration, allow the learners to touch the reaction vessel, ensuring that they do not
come in contact with the reacting materials. Let them recall endothermic and exothermic reactions, and later
identify whether the reaction is endothermic or exothermic.
ED
From learners’ answers, the teacher could have a discussion with them on the factors that affect rates of
reactions.
Nature of reactants – state that substances vary in their chemical reactivity. Chemical reactivity is a major
factor that determines the rate of a reaction. Give examples of substances of varying chemical reactivity
and reactions where these are involved.
•
Ability of reactants to meet - consider surface area of contact, classification of reactions as homogeneous
and heterogeneous reactions
•
Concentration of reactants - This could be likened to a hallway with just a few learners to a crowded
hallway where they are likely to bump into each other. Give an application of this i.e., prescription of
medicines by doctors.
•
Temperature of reaction system
•
Presence of a catalyst
D
EP
•
Allow the learners to give other examples of chemical reactions they meet in every day life that would illustrate
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the effects of these various factors.
Other activities relating to factors that affect reaction rates:
PY
http://www.teachsecondary.com/maths-and-science/view/lesson-plan-ks4-science-rates-of-reaction-inchemistry
Lesson plan: KS4 science – rates of reaction in chemistry
C
O
Possible illustrations for factors that affect reaction rates:
http://www.freezeray.com/flashFiles/RatesOfReactionSurfaceArea.swf
http://www.freezeray.com/flashFiles/RatesOfReactionConc.swf
http://www.freezeray.com/flashFiles/RatesOfReactionTemp.swf
ED
Collision theory
D
EP
1. Ask learners their ideas on how they would explain the effects of the different factors on reaction rates at the
molecular level. From their answers, proceed to a discussion of the Collision Theory and emphasize that for
a reaction to take place, reactant particles must have activation energy or the minimum amount of energy
required and the right orientation for effective collision to bring about a chemical change.
2. Show illustrations showing (Appendix B):
a. Two reactant particles approach each other then fly apart because they are not correctly oriented for a
reaction to occur.
b. Slow moving particles approach each other then eventually fly apart.
c. Fast moving, energetic molecules correctly oriented forming new substances.
360
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3. Ask learners to give their explanation on the basis of Collision Theory how reaction rates are affected by
various factors.
In your discussion of how reaction rates are affected by various factors consider the following:
Temperature – show a plot of the fraction of molecules having a kinetic energy at two different
temperatures. Ask learners to interpret the graph. The graph would show that at higher temperatures,
the total fraction of molecules having the required energy is greater than that at lower temperatures. A
greater number of reactant molecules are more energetic at higher temperatures than at lower
temperatures, thereby making collisions more effective for products to form.
•
Concentration – Present to learners an illustration showing a few reactant molecules and another with
more reactant molecules. The more number of reactant molecules, the greater probability for effective
collisions to form products. Compare these to scenarios where there are just two or three learners in a
hallway, and a crowded one. Learners are more likely to bump into each other in the crowded hallway
than with just a few ones.
ED
C
O
PY
•
Potential energy diagrams
D
EP
1. Start your discussion of potential energy diagrams with the question:
What happens after reactant molecules collide?
Sample response:
If a reaction occurs, during the collision, the particles that separate are different from those that collide. When
the particles collide, the molecules slow down. Thus, the total kinetic energy (K.E.) they possess decreases.
Because energy cannot disappear, this means that the total potential energy (P.E.) of the particles must increase.
2. Further state that the relationship between activation energy and total potential energy of reactants and
361
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products may be expressed graphically through a potential energy diagram.
3. Show the learners potential energy diagrams of an exothermic and endothermic reaction.
differentiate between the two plots. Make them interpret the diagrams.
Ask them to
PY
4. Point out that the activation energy appears as a “hill,” which is referred as the potential energy barrier
between reactants and products. Reactant particles must go over this hill, the minimum energy required for
a reaction to occur, to form the products.
C
O
5. Show a figure (Appendix C) where the teacher likens activation energy to the energy needed by the person
to push the ball over the hill. If the person does not push the ball hard enough, it rolls back down just as
reactant particles that do not have the minimum amount of energy required. If the ball is given enough
energy, it will go over the hill as reactant particles with the required amount of energy would go over the
potential energy barrier to form the products.
D
EP
6. Describe/define the activated complex.
ED
Activation energy is the minimum energy required for the electron clouds and nuclei of reactant particles to
overcome the repulsions and form bonds.
7. Show learners which portion of the potential energy diagram corresponds to the heat of reaction, ΔHrxn.
How does a catalyst increase the rate of a reaction?
8. Tell them to recall the demonstration that showed the increase in the rate of a reaction with the addition of a
catalyst to the reaction mixture.
9. Ask learners of their idea about how a catalyst increases the rate of a reaction.
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10. Show a potential energy diagram for a catalyzed and uncatalyzed reaction (Appendix D). Ask learners to
interpret the diagram.
PY
11. Point out that the catalyst provides a path for the reaction with a lower activation energy, wherein the
catalyst participates in the reaction by changing its mechanism.
12. Show a plot of number of molecules vs. kinetic energy (Appendix E). Call on the learners to explain this plot.
What does it show?
C
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13. Discuss the different types of catalysts (homogeneous and heterogeneous catalysts). Give examples. Ask
learners to recall enzymes and what type of catalyst is an enzyme.
Exercise 1!
Factors affecting rates of reactions
ED
ENRICHMENT (120 MINS)
D
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1. One week before this laboratory session, task learners to read and study this exercise. Hand out copies of
the experiment if they do not have a laboratory manual. Instruct them to write the (1) objectives of the
experiment and (2) the necessary data tables in their laboratory notebooks.
2. Have a pre-laboratory discussion reviewing the factors that affect reaction rates discussed in the lecture.
3. Ask learners to state the objectives of the exercise.
4. Describe what they are to do when they perform the exercise. Show them the materials that will be used in
363
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the exercise. Demonstrate how they will mix the reactant solutions and set the water bath.
•
5. Check their data tables and make corrections when needed.
PY
The exercise involves determining the effect of reactant concentration, temperature, and the presence
of a catalyst on the rate of reactions.
C
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EVALUATION (10 MINS)
Check the data learners collected when performing the laboratory experiment. Let them answer the questions in the exercise.
They could do this as a group where they discuss their answers among themselves.
2 (NEEDS
IMPROVEMENT)
3 (MEETS
EXPECTATIONS)
D
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1 (NOT VISIBLE)
ED
They should write down their answers in their notebooks. The learners’ answers to the questions will reflect their understanding
of how certain factors affect the rate of a chemical reaction.
364
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4 (EXCEEDS
EXPECTATIONS)
Appendix A
Teacher Tip
•
The teacher can give a background of
the reaction that proceeds.
•
The chemical reaction involved produces
a foamy soap, which froths out of a
bottle or a graduated cylinder. The
reaction involved is the catalytic
decomposition of hydrogen peroxide,
H2O2. The reaction involved is:
Demonstration procedure
PY
The teacher may use any of the following depending on available materials.
Demo 1*
C
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Materials:
One empty 500 mL soda plastic bottle or a 500 mL plastic water bottle
•
½ cup 3% or 6% hydrogen peroxide (the former is available in drug stores, the latter may be available in
beauty supply stores)
•
Dish washing liquid solution or any soap solution
•
Baker’s active yeast (available in supermarkets or bakery supply stores)
•
Food coloring (optional)
ED
•
D
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Procedure:
1. Dissolve one teaspoon or one packet of active yeast in a small amount of warm water. Keep still for about 5
minutes.!
2. Dilute a small amount of dishwashing liquid in about ¼ cup of water, or dissolve soap in water.
3. Place about ¼ cup of the dishwashing liquid solution or soap solution into the plastic bottle. Two to three
drops of food color can be added and mixed.
4. Add ½ cup of hydrogen peroxide to the soap solution.
5. Ask learners their observation.
6. Add the yeast to the mixture in the bottle.
7. Ask the learners to describe what they see and explain the phenomenon. Ask them to identify which
365
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2H2O2 (l)
—>
2H2O + O2 (g) + heat
In Demo 1, the catalyst used is Baker’s Yeast.
•
Tell the learners that Baker’s yeast
(Saccharomyces cerevisiae) has many
uses. For example, it is used in the
preparation of bread like pan de sal, as
well as in fermentation processes – in
wine making and brewing (beer making).
•
Yeast contains the enzyme catalase. This
enzyme brings about the decomposition
of hydrogen peroxide to water and
oxygen. The oxygen gas that is formed
causes soap bubbles to form, and
learners observe the foam coming out of
the bottle. The same reaction occurs in
our body. Small amounts of hydrogen
peroxide are formed in our bodies by
metabolic reactions. Hydrogen peroxide
is harmful to the body if it accumulates in
body cells. However, its build up is
prevented by the presence of the
enzyme catalase produced in these very
cells by decomposing it to water and
oxygen.
reagent caused the reaction to proceed.
•
8. Let them determine if the reaction is exothermic or endothermic.
In Demo 2, potassium iodide acts as the
catalyst. In the presence of potassium
iodide, the decomposition of hydrogen
peroxide takes place in two steps:
*Source: http://www.coolscience.org/CoolScience/KidScientists/h2o2.htm
H2O2 (aq) + I- (aq) —>
PY
Demo 2*
500 mL graduated cylinder or 1.5 liter soda bottle
•
10 mL graduated cylinder
•
Goggles
•
Rubber gloves
•
Plastic tray or basin
•
30% hydrogen peroxide
•
Dishwashing liquid or soap that readily forms bubbles
•
2 M potassium iodide solution
•
Food coloring (optional)
D
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ED
•
C
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Materials:
Procedure:
Step 1:
1. Place the graduated cylinder or soda bottle on a plastic tray or basin.
2. Measure 20 mL 30% hydrogen peroxide and place this in the graduated cylinder or bottle.
3. Add 5 mL of dishwashing detergent or soap solution.
4. Ask learners to describe what they see.
5. Add 10 mL of potassium iodide solution.
6. Ask students to describe what they see and explain the phenomenon. Ask them to identify which reagent
366
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H2O (l) + OI- (aq)
Step 2:
H2O2 (aq) + OI- (aq) —>
+ I-(aq)
•
Note that when common terms on both
sides of the arrow cancel, the overall
reaction si obtained. The iodide ion does
not appear in the overall reaction
2H2O2 (l)
•
H2O (l) + O2 (g)
—>
2H2O + O2 (g) + heat
Again, the oxygen gas form causes soap
bubbles to form.
caused the reaction to proceed.
•
7. Let them determine if the reaction is exothermic or endothermic.
Safety precaution:
PY
1. Handle hydrogen peroxide with care. It is severely corrosive to the skin, eyes, and the respiratory tract. In
case of contact, flush with water for 15 minutes. If eyes are affected, get medical attention. !
Steam and oxygen gas form very quickly. Stand back from the reacting vessel. Do not stand over the
reaction.
3.
Use gloves and goggles during the demonstration
C
O
2.
*Source: http://cldfacility.rutgers.edu/content/catalytic-decomposition-hydrogen-peroxide- potassium-iodide
D
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ED
APPENDIX B
Figure B1. Two reactant particles approach each other then fly apart because they are not correctly
oriented for a reaction to occur.
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C
O
PY
•
D
EP
ED
Figure B2. Slow moving particles approach each other then eventually fly apart.
Figure B3. Fast moving, energetic molecules correctly oriented form new substances.
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APPENDIX C
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ED
C
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PY
•
A ball given enough energy will go over the hill, as reactant particles with the required amount of
energy would go over the potential energy barrier to form products.
Source: Bloomfield MM. 1980.
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D
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ED
C
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PY
APPENDIX D
A catalyst provides a path for the reaction with a lower activation energy.
Source: Bloomfield MM. 1980.
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D
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ED
C
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PY
APPENDIX E
When a catalyst is present, more molecules possess the minimum amount of energy needed for effective collisions.
Source: Brady, 1990.
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Chemistry 2
150 MINS
LESSON OUTLINE
Introduction
Communicating learning objectives
5
Instruction
Group Activity and Discussion
15
PY
Chemical Kinetics: Rate of
Reaction, Rate Constant, and
Concentration of Reactants
C
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Motivation
Demonstration and Discussion
Content Standards
The learners demonstrate an understanding of he rate of a reaction and the various Enrichment
Laboratory Exercise
factors that influence it; and the Collision Theory.
Evaluation
Seat Work
Performance Standard
Materials
Computer or overhead projector; projection screen; transparencies
Learning Competencies
Write the mathematical relationship between the rate of a reaction, rate constant, and of lecture materials to be presented
concentration of the reactants (STEM_GC11CK-IIIi-j-131)
Resources
70
30
30
ED
State the order of the reaction with respect to each reactant and the overall order of the (1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
New York: John Wiley & Sons.
reaction. (STEM_GC11CK-IIIi-j-132)
Write the rate law for first-order reaction. (STEM_GC11CK-IIIi-j-133)
Discuss the effect of reactant concentration on the half-life of a first-order reaction rate.
(STEM_GC11CK-IIIi-j-134)
D
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(LAB) Write mathematical expression of rate of reaction and solve problems involving
half-life.
(Lab) Perform an exercise illustrating half-life.
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
(3) Padolina, MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
Inc.
(4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
Co., Inc.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
calculate molar mass from a colligative property data; and
•
determine the molar mass of a solid from the change of boiling point of a solution.
372
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INTRODUCTION (5 MINS)
Teacher Tip
•
1. Communicate learning objectives.
•
-
Chemical kinetics
-
Rate of reaction
-
Reaction mechanism
-
Rate law
PY
2. Introduce terms that the learners will encounter:
C
O
INSTRUCTION (15 MINS)
1. Let learners recall the factors that affect rates of reactions. Briefly review these factors.
Expressing rates of reactions
ED
2. Lecture proper:
D
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The rate of a chemical reaction tells how fast a given amount of a reactant or product changes with time.
It can be expressed either as the disappearance of a reactant or the appearance of the product.
How are rates of reactions measured?
Ask learners give their ideas of how one could measure how fast a reaction takes place.
Choosing one of the reactants or products and measuring its change in concentration with time
determine the rate of a reaction. The change in concentration may be monitored by change in pH, color,
pressure, or electrical conductivity.
373
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Project/display the objectives all at once
or show each one at a time.!
Write terms on the board or show them
one at a time as you go along.
MOTIVATION (70 MINS)
1. Demonstrate measuring rates of reaction using a dye that decolorizes with the addition bleach (See
Appendix A). !
PY
2. Tell learners to write down their observations and explain these (Difference in the length of time the dyes
changed color).
How is the rate of a reaction defined?
C
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The rate of a chemical reaction is defined as the number of moles of reactant consumed per unit time, or
the decrease in concentration of reactant with time.
Also, it could be defined as the number of moles of product formed per unit time, or the increase in
concentration of product with time.
ED
3. Relate rate of reaction to the speed of a car. The speed of a car gives its rate of travel:
Rate of travel = change in position = kilometers
So,
hour
D
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Time
Rate of reaction = change in concentration
Time
= moles/liter
= mol/L
second
sec
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4. Give a sample reaction and discuss how the rate of the reaction may be measured and expressed. State that
the rate of a reaction may be determined by measuring the change in concentration of reactants or products
with time; reactant concentrations decrease and product concentrations increase as the reaction proceeds.
Writing expressions of rate of reaction
C
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1. Show the learners how to write the expression for the reaction’s rate of reaction.
PY
5. Show a graph that illustrates the change in reactant and product concentrations with time (Appendix B).
For the general reaction:
A + B —> C + D
Rate = - D[A]
=
Δt
=
- D[C]
=
Δt
-D[D]
Δt
D
EP
Δt
- D[B]
ED
The rate of the reaction may be expressed in several ways:
The symbol Δ denotes change; t is time and [ ] indicates molar concentration. Since rate is a positive
quantity, the concentrations of the products C and D increase with time while that of the products decrease. A
negative sign is placed before the change in concentration of reactants, indicating that the concentration of the
reactant decreases with time. A negative sign is always used whenever reactants express the rate of a reaction.
Putting a negative sign will result in rate with a positive sign.
2. Give an example of a reaction and write the expression for the rate of the reaction. The following are
examples that may be used:
1. H2 + I2
—>
2HI
375
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Teacher tip
During the laboratory period the students
could do an exercise on writing expressions
of rate of reactions.!
2. N2(g)
+ 3H2 (g) —> 2NH3(g)
3. Write a general reaction with species having different coefficients, and write its rate of reaction expression.
Rate = - D[A]
aD
=
- D[B]
=
- D[C]
bDt
=
PY
aA + bB —> cC + dD
-D[D]
cDt
dDt
C
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4. Give other examples of reaction and ask learners to write the expression for the rate of the reaction. This
could be board work or seat work.
ED
5. Learners may be given an exercise on writing expressions for rates of reaction (Exercise 1 Part A for this unit)
during the laboratory period.
Mathematical relationship between the rate of a reaction, rate constant, and concentration of
reactants
Rate law
D
EP
Make learners recall that the concentration of reactants influences the rate of chemical reactions.
1. Discuss the rate law by considering a general equation and mathematically show the direct relationship
between rate of reaction and reactant concentrations. Consider the following equations:
aA + bB —> products
Rate α [A]a [B]b
376
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By changing the proportionality sign to an equality sign a term, k is introduced. This is the rate constant.
Rate = k[A]a [B]b
PY
2. Define the rate law and state the significance of the value of the rate constant. From the exponents of the
concentrations in the rate equation, define the order of the reaction with respect to each reactant as well as
the overall order of the reaction.
C
O
3. Give an example of a reaction. Ask learners to go to the board and write the rate expression of the reaction.
Tell them to state the order of the reaction with respect to each reactant, and also the overall order of the
reaction. Give other reactions.
Relation of concentration and time
—> products
D
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aA
ED
1. State that an important relationship in the study of kinetics is the dependence of reactant concentration with
time. For this discussion, consider reactions involving a single reactant:
2. 2. Proceed to state that for a first order reaction, rate depends on the concentration of the single reactant
where the exponent is one. Write the expression of the rate law for this reaction:
Rate = k[A]
3. 3. State that the rate equation, by using calculus, was transformed to integrated rate law, which expressed
the relationship of concentration with time. Write the equation in terms of logarithm to the base 10:
377
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log [A]o =
kt .
[A]t
2.303
PY
Half-life and radioactivity
1. Prior to this lesson, give task learners to look for news reports regarding radioactive materials or nuclear and
radiation incidents, or perhaps the role of radioisotopes in every day life.
C
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The teacher may also assign learners to bring pictures that show the use of radioactivity (radioisotopes)
in various fields like medicine, agriculture and food, archeology, etc. Let them show these pictures to the class
and allow them to discuss it.
ED
2. Incorporate whatever the learners report in the discussion on radioactive materials. This should then lead to a
discussion on unstable isotopes and their half-lives. At this point, show how the half-life of a radioisotope is
calculated and how much remains of a radioactive substance after a given period of time.
3. Tell the learners to recall their lesson on atomic structure, subatomic particles, and isotopes.
D
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4. Give a background on radioactivity. Describe how Henri Becquerel discovered radioactivity in 1896. Define
radioactivity. Briefly discuss why certain atoms are unstable (unstable isotopes or radioactive isotopes).
5. Illustrate what is meant by half-life and let learners define the term.
6. Show them a figure similar to Appendix Figure _ to illustrate the meaning of half-life.
7. Ask them if there will come a time when a radioisotope will have zero atoms. Let them explain their answer.
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8. Give examples of radioisotopes of varying half-lives. Indicate that some radioisotopes have very long halflives, while others have short. Show the learners a table of radioisotopes pointing out those with very long halflives (thousands of years), and those with very short (seconds).
PY
9. From the definition of half-life, show them the necessary equations to arrive to the equation for calculating
half-life (Appendix D). Indicate this is related to integrated rate law for a first order reaction.
10. Illustrate how to calculate half-life using sample problems (Appendix E).
C
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11. In the laboratory, learners work on a problem set on calculations involving half-life (Exercise 2 Part B). Also,
they can perform an exercise that illustrates half-life (Exercise 3).
First Laboratory Class for this unit:
Laboratory Exercises
ED
ENRICHMENT (30 MINS)
D
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Exercise 2: Prepare a set of questions on writing rate of reaction expressions, and problem sets involving
calculations of half-life.
Learners will write rates of reactions expressions, and solve problems involving half-life. They then show
their solutions on the board and discuss their answers.
Exercise 3: Exercise that illustrates half-life. !
!
!!!!!!See http://www.nuclearconnect.org/in-the-classroom/for-teachers/half-life-of-paper-mmspennies-or-puzzle-pieces or
http://www.google.com.ph/url?url=http://portal.utpa.edu/portal/page/portal/
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1AEBCCCA4D323F62E054000E7F4F739C&rct=j&q=&esrc=s&sa=U&ved=
0ahUKEwjc84aUpbLJAhXJOJQKHWa7Ad8QFgg1MAY&sig
PY
Guide learners as they perform the exercise.
EVALUATION (30 MINS)
C
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Seat Work - this may be conducted during the laboratory period (Exercise 2). Learners are rated based on their
solutions to these problems.
Give them chemical reactions, and instruct them to write the reaction rate expression and order of reaction with
respect to each reactant. Give the overall reaction order.
ED
Learners work on calculating problems relating to half-life.
D
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Laboratory on Half-life (Exercise 3). Learners’ understanding of half-life may be assessed from their answers to
questions in the exercise.
1 (NOT VISIBLE)
2 (NEEDS
IMPROVEMENT)
3 (MEETS
EXPECTATIONS)
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4 (EXCEEDS
EXPECTATIONS)
Appendix A
Demonstration Procedure
Demo 1** (This demo may take about 5 minutes)
PY
Materials:
Household bleach (sodium hypochlorite)
•
Food dyes (yellow, red, blue or green)
•
Large glass containers
•
250- mL beakers or equivalent glass containers (peanut butter or jam jars)
•
Droppers
C
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•
ED
Procedure:
1. Prepare dye solutions (this could be done before the class period; the teacher may choose to use only two dyes. Suggestion: use the yellow
dye and any other color).
D
EP
2. Place about 250 mL of water in separate containers. Add two drops of food dyes.!
3. Measure 100 mL of the resulting solutions and place in 250-mL beakers.!
4. Add 1-2 droppers full of household bleach into one of the dye solutions and start timing till the dye changes color. !
5. Repeat this with other dye solutions.
6. Tell learners to write down their observations.!
381
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7. Ask them to explain their observations (Difference in the length of time the dyes changed color).
**Source: https://www.teachchemistry.org/content/dam/AACT/middle-school/reactions/reaction-rate/
secure/Demo_SimpleKinetics.pdf
—>
2HI
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H2 + I2
PY
Appendix B
For the reaction:
D
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ED
the concentrations of the reactants H2 and I2 decrease with time; the concentration of the product HI
increases. The concentration of reactants and products could be plotted with time to obtain a graph similar to
that shown below.
Change in concentration of I2 and HI with time
—>
2HI
as the reaction. H2 + I2
proceeds.
382
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The t1/2 of nitrogen-13 = 10 minutes.
C
O
PY
Appendix C
ED
After one half-life, one-half of the sample decays to a new element (carbon-13) and one-half remains.
After two half-lives, one-fourth of the original sample remains. After three half-lives, one-eight of the original
sample remains.
D
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Consider the isotope of nitrogen, nitrogen-13, which is used in positron emission tomography. The halflife of nitrogen-13 is 10 minutes. If you have a 10-gram sample of nitrogen-13, after 10 minutes or one half-life,
5 grams of the sample would have decayed and 5 grams of nitrogen-13 remains. After another 10 minutes or 2
half-lives, half of the 5 grams of nitrogen-13 or 2.5 grams would have decayed and 2.5 grams remains.
383
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Appendix D
What does half-life mean?
PY
Half-life is the length of time required for the concentration of a reactant to decrease by half its original
or initial concentration. At this time, t = t1/2:
For a first order reaction, substitute in equation 3:
[A]o =
kt1/2
½ [A]o
2.303
Solving for t1/2,
log 2 =
kt1/2
D
EP
2.303
ED
log
C
O
[A]t = ½ [A]o
t1/2 = 2.303 log 2
k
t1/2 = 0.693
Equation 4
k
This equation tells that for a first-order reaction, t1/2 is independent of the initial concentration. It
depends on the value of k, which is constant throughout the reaction.
384
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Appendix E
Consider problems illustrating how to calculate the half-life of radioisotopes, and how to calculate the amount of
a radioisotope that remains after a given period of time.
PY
Example 1. Iodine-131 is commonly used to treat throat cancer. The rate constant for decay of iodine-131 is
0.0864/day. What is the half-life of iodine-131?
To calculate half-life, use the expression:
t1/2 = 0.693
k
0.693
ED
t1/2 =
C
O
Solution:
0.0864/day
D
EP
t1/2 = 8.021 days
The half-life of iodine-131 is 8.021 days
Example 2. The half-life of sodium-24, a radioisotope used as a tracer for the cardiovascular system, is 14.951
hours. How many grams of sodium-24 will remain after 24.0 hours in a sample initially containing 100 mg?
Solution:
First, calculate k. Use the expression:
t1/2 = 0.693
k
385
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k =
0.693
.
14.951 hr
PY
k = 0.0464/hr
log [A]o =
kt .
[A]t
2.303
C
O
Then use the equation:
Initial amount = 100 mg.
Amount after 24 hrs = x
ED
In the equation, quantities are in molar concentrations. However, number of moles is
proportional to mass, thus:
log 100 = (0.0464/hr) (24 hrs)
2.303
D
EP
x
log 100 – log x = 0.483
log x = log 100 – 0.483
x = 32.9 mg
The amount of sodium-24 remaining after 24 hours is 32.9 mg.
386
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Chemistry 2
150 MINS
Chemical Kinetics: Reaction
Rates & the Rate Law
Introduction
Communicating learning objectives
3
Motivation
Discussion
82
Instruction
Demonstration
5
Enrichment
Laboratory exercise
100
Evaluation
Laboratory exercise question & answer
20
PY
Content Standards
The learners demonstrate an understanding of
LESSON OUTLINE
1. the rate of a reaction and the various factors that influence it; and
2. the Collision Theory.
C
O
The learners demonstrate an understanding of the mathematical relationships between
reaction rate, rate constants, and concentration of reactants.
Performance Standard
Materials
Computer or overhead projector; projector screen; transparencies of
lecture materials to be presented;
ED
Learning Competencies
Write the mathematical relationship between the rate of a reaction, rate constant, and Resources
(1) Brady, EB. (1990). General Chemistry – Principles and Structure (p. 852).
concentration of the reactants. (STEM_GC11CK-IIIi-j-131)
New York: John Wiley & Sons.
Give the order of the reaction with respect to each reactant and the overall order of the
(2) Masterson, WL., Hurley, CN., & Neth, EJ. (2012). Chemistry: Principles
reaction. (STEM_GC11CK-IIIi-j-132)
and Reactions (p. 744).
California, USA: Brooks/Cole, Cengage
Learning.
Write the rate law for first-order reaction. (STEM_GC11CK-IIIi-j-133)
D
EP
MCD., Sabularse, VC., & Marquez LA. (1995). Chemistry for
Discuss the effect of reactant concentration on the half-life of a first-order reaction rate. (3) Padolina,
the 21st Century (p. 340). Makati, Philippines: Diwa Scholastic Press,
(STEM_GC11CK-IIIi-j-134)
Inc.
(LAB) Write mathematical expression of rate of reaction and solve problems involving (4) Silberberg, MS. (2007). Chemistry – The Molecular Nature of Matter
and Change (International Edition, p. 1088). New York: McGraw-Hill
half-life.
Co., Inc.
(LAB) Perform an exercise illustrating half-life.
Specific Learning Outcomes
387
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INTRODUCTION (3 MINS)
Teacher Tip
•
1. Communicate learning objectives.
•
Chemical kinetics
•
Rate of reactions
•
Reaction mechanism
•
Rate law
C
O
•
PY
2. Introduce terms that the learners will encounter.
Project/display the objectives all at once
or show each one at a time.
Write terms on the board or show them
one at a time as you go along.
INSTRUCTION (5 MINS)
1. Let learners recall the factors that affect rates of reactions. Briefly review these factors.
Expressing rates of reactions
ED
2. Lecture proper:
D
EP
The rate of a chemical reaction tells how fast a given amount of a reactant or product changes with time.
It can be expressed either as the disappearance of a reactant or the appearance of the product.
How are rates of reactions measured?
Ask learners give their ideas of how one could measure how fast a reaction takes place.
Choosing one of the reactants or products and measuring its change in concentration with time
determine the rate of a reaction. The change in concentration may be monitored by change in pH, color,
pressure, or electrical conductivity.
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Teacher Tip
•
Emphasize that rate is always a positive
value.
MOTIVATION (82 MINS)
1. Demonstrate measuring rates of reaction using a dye that decolorizes with the addition bleach (See
Appendix A). !
PY
2. Tell learners to write down their observations and explain these (Difference in the length of time the dyes
changed color).
How is the rate of a reaction defined?
C
O
The rate of a chemical reaction is defined as the number of moles of reactant consumed per unit time, or
the decrease in concentration of reactant with time.
ED
It can also be defined as the number of moles of product formed per unit time, or the increase in
concentration of product with time.
3. Relate rate of reaction to the speed of a car. The speed of a car gives its rate of travel:
Rate of travel = change in position = kilometers
So,
hour
D
EP
Time
Rate of reaction = change in concentration
Time
= moles/liter =
mol/L
second
sec
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4. Give a sample reaction and discuss how the rate of the reaction may be measured and expressed. The
rate of a reaction may be determined by measuring the change in concentration of reactants or products
with time; reactant concentrations decrease and product concentrations increase as the reaction
proceeds.
PY
5. Show a graph that illustrates the change in reactant and product concentrations with time (Appendix B).
C
O
Writing expressions of rate of reaction
1. Show the learners how to write the expression for the reaction’s rate of reaction.
A + B
—>
ED
For the general reaction:
C + D
Rate =
D
EP
The reaction may be expressed in several ways:
Teacher tip
•
The symbol Δ denotes change; t is time and [ ] indicates molar concentration. Since reaction rate is always
expressed as a positive quantity, a negative sign is placed before the change in concentration of reactants,
indicating that the concentration of the reactant decreases with time. Putting a negative sign will result in rate
with a positive value.
2. Give an example of a reaction and write the expression for the rate of the reaction. The following are
examples that may be used:
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During the laboratory period the
students could do an exercise on writing
expressions of rate of reactions.
1. H2 + I2
2. N2(g)
—>
2HI
+ 3H2 (g) —> 2NH3(g)
PY
In working with each of the above examples, where the reactants and products of the reactions have
different coefficients in the balanced equation, the rate of reaction that should be obtained is the same
regardless of how the rate was expressed – whether in terms of reactants and products.
C
O
3. Ask the learners to write a general expression for rate of reaction for reactions whose balanced equations
have different coefficients,
4. Give other examples of reactions and ask learners to write the expression for the rate of the reaction.
This could be board work or seat work.
ED
5. Learners may be given an exercise on writing expressions for rates of reaction (Exercise 1 Part A for this
unit) during the laboratory period.
D
EP
Mathematical relationship between the rate of a reaction, rate constant, and concentration of reactants !
Make learners recall that the concentration of reactants influences the rate of chemical reactions. The
effect of concentration of reactants on the rate of reaction can be seen quantitatively using the rate law for the
reaction.
Rate law
1. The rate law for a reaction is an expression that gives the mathematical relationship of the rate of a
reaction and the concentration of reactants.
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Considering a general equation
aA + bB
—>
products
The rate law for the reaction is
PY
Rate = k[A]m [B]n
C
O
k is the rate constant, and is dependent on temperature. The exponents m and n indicate the order of
the reaction with respect to the corresponding reactant and are experimentally determined.
For example, the rate of the gas-phase decomposition of dinitrogen pentoxide
2N2O5 → 4NO2 + O2
has been found to be directly proportional to the concentration of N2O5:
ED
rate = k [N2O5]
D
EP
This equation is called the rate equation or the rate law for the reaction. Based on this e q u a t i o n , t h e
reaction is first order in N2O5. If the concentration of N2O5 is doubled, the rate of this reaction will be twice as
fast. The reaction will slow down by half if the starting concentration of N2O5 is reduced to half.
The expression for the rate law generally bears no necessary relation to the stoichiometric coefficients in
the balanced equation for the reaction, and must be determined experimentally.
Example: Rate data were obtained for following reaction:
A + 2B —> C + 2D
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•
The rate law
rate = k [N2O5]
is actually
rate = k [N2O5]1
indicating first order.
Teacher tip
Initial A
(mol/L)
Initial B
(mol/L)
Rate of Formation
of C (M min-1)
1
0.10
0.10
3.0 x 10-4
2
0.20
0.10
1.2 x 10-3
3
0.10
0.30
3.0 x 10-4
4
0.20
0.40
For rate laws where the exponents to
which the concentrations of reactants are
raised are equal to the coefficients of the
reactants in the balanced equation, this is
purely coincidental. The values of the
exponents or the order of the reaction
are determined experimentally.
•
What about a third order reaction with
respect to a certain reactant? If
?
C
O
What is the rate law for this reaction?
•
PY
Experiment No.
Answer: To get the rate law, the order of the reaction with respect to each reactant has to be
determined from the experimental data . To do this:
ED
a. Compare experiments 1 and 3. The concentration of Reactant A is kept constant, while [B] is tripled
in Expt. 3. However, the observed rates of reaction are the same in both experiments. Conclusion:
Changing the concentration of B does not affect the rate of reaction. B is not in the rate law
expression.
D
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b. Compare experiments 1 and 2. When the concentration of A is doubled, the rate increases by four
times. Conclusion: The reaction is second order with respect to A.
The rate law for the reaction is
rate = k[A]3
rate = k[A]2 [B]0 = k[A]2
then if [A] is doubled, the reaction will be
8x faster, or the rate will increase by 8.
c. Can the learners predict the rate in experiment 4?
rate = k (2[A])3 = k (8)[A]3
2. Discuss the significance of the value of the rate constant. From the exponents of the concentrations in
the rate equation, define the order of the reaction with respect to each reactant as well as the overall
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order of the reaction.
•
PY
3. Give an example of a reaction. Ask learners to go to the board and write the rate expression of the
reaction. Tell them to state the order of the reaction with respect to each reactant, and also the overall
order of the reaction. Give other reactions.
C
O
Relation of concentration and time
The rate law gives the rate of the reaction when the concentrations of reactants are known, such as at the
start of the reaction. However, as the reaction proceeds, the reactants decrease in concentrations as they are
used up in the reaction.
1. State that an important relationship in the study of kinetics is the dependence of reactant
concentration with time. For this discussion, consider reactions involving a single reactant:
—>
products
ED
aA
2. Assume that this is a first order reaction and therefore the rate depends on the concentration of the
single reactant where the exponent is one. Write the expression of the rate law for this reaction:
D
EP
rate = k [A]
3. State that the rate equation, by using calculus, can be transformed to the integrated rate law, which
expresses the relationship of concentration with time. Write the equation in terms of logarithm to the
base 10:
log [A]o =
kt
[A]t
2.303
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4. An important event in a reaction is the time when half of the original amount of reactant
has been used up in the reaction. This event is called the half-life of the reaction, t½. At t½ ,
or
[A]o = 2[A]t
The integrated rate law for a first order reaction indicating the half-life is
[A]o
=
½[A]o
log 2
kt½
2.303
=
C
O
log
PY
[A]t = ½[A]o
kt½
2.303
ED
A step by step transformation of this equation is given in Appendix D.
Application of rate laws and integrated rate equation
D
EP
Half-life and radioactivity
1. Prior to this lesson, give task learners to look for news reports regarding radioactive materials or
nuclear and radiation incidents, or perhaps the role of radioisotopes in every day life.
The teacher may also assign learners to bring pictures that show the use of radioactivity
(radioisotopes) in various fields like medicine, agriculture and food, archeology, etc. Let them show
these pictures to the class and allow them to discuss it.
2. Incorporate whatever the learners report in the discussion on radioactive materials. This should then
lead to a discussion on unstable isotopes and their half-lives. At this point, show how the half-life of
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•
a radioisotope is calculated and how much remains of a radioactive substance after a given period of
time.
3. Tell the learners to recall their lesson on atomic structure, subatomic particles, and isotopes.
C
O
5. Illustrate what is meant by half-life and let learners define the term.
PY
4. Give a background on radioactivity. Describe how Henri Becquerel discovered radioactivity in 1896.
Define radioactivity. Briefly discuss why certain atoms are unstable (unstable isotopes or radioactive
isotopes).
6. Show them a figure similar to Appendix Figure _ to illustrate the meaning of half-life.
ED
7. Ask them if there will come a time when a radioisotope will have zero atoms. Let them explain their
answer.
D
EP
8. Give examples of radioisotopes of varying half-lives. Indicate that some radioisotopes have very long
half-lives, while others have short. Show the learners a table of radioisotopes pointing out those with
very long half-lives (thousands of years), and those with very short (seconds).
9. From the definition of half-life, show them the necessary equations to arrive to the equation for
calculating half-life (Appendix D). Indicate this is related to integrated rate law for a first order
reaction.
10. Illustrate how to calculate half-life using sample problems (Appendix E).
11. In the laboratory, learners work on a problem set on calculations involving half-life (Exercise 2 Part B).
Also, they can perform an exercise that illustrates half-life (Exercise 3).
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•
ENRICHMENT (100 MINS)
!
!
!
First Laboratory Class for this unit:
PY
Laboratory Exercises
Exercise 2: Prepare a set of questions on writing rate of reaction expressions, and problem sets involving
calculations of half-life.
C
O
Learners will write rates of reactions expressions, and solve problems involving half-life. They then show
their solutions on the board and discuss their answers.
Exercise 3: Exercise that illustrates half-life. !
!
See the following links:
ED
1. http://www.nuclearconnect.org/in-the-classroom/for-teachers/half-life-of-paper-mmspennies-or-puzzle-pieces
2. http://www.google.com.ph/url?url=http://portal.utpa.edu/portal/page/portal/
1AEBCCCA4D323F62E054000E7F4F739C&rct=j&q=&esrc=s&sa=
D
EP
U&ved=0ahUKEwjc84aUpbLJAhXJOJQKHWa7Ad8QFgg1MAY&sig
Guide learners as they perform the exercise.
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EVALUATION (20 MINS)
•
Evaluation
PY
Seat Work - this may be conducted during the laboratory period (Exercise 2). Learners are rated based on their solutions to these
problems.
Give them chemical reactions, and instruct them to write the reaction rate expression and order of reaction with respect to each
reactant. Give the overall reaction order.
Learners work on calculating problems relating to half-life.
2
(Not Visible)
(Needs Improvement)
3
4
(Meets Expectations)
(Exceeds Expectations)
ED
1
D
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APPENDIX A
Demonstration procedure
C
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Laboratory on Half-life (Exercise 3). Learners’ understanding of half-life may be assessed from their answers to questions in the
exercise.
Demo 1** (This demo may take about 5 minutes)
Materials:
•
Household bleach (sodium hypochlorite)
•
Food dyes (yellow, red, blue or green)
•
Large glass containers
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•
250- mL beakers or equivalent glass containers (peanut butter or jam jars)
•
Droppers
•
Procedure:
PY
1. Prepare dye solutions (this could be done before the class period; the teacher may choose to use only two
dyes. Suggestion: use the yellow dye and any other color).
C
O
2. Place about 250 mL of water in separate containers. Add two drops of food dyes.!
3. Measure 100 mL of the resulting solutions and place in 250-mL beakers.!
5. 5. Repeat this with other dye solutions.
D
EP
6. Tell learners to write down their observations.!
ED
4. Add 1-2 droppers full of household bleach into one of the dye solutions and start timing till the dye changes
color. !
7. Ask them to explain their observations (Difference in the length of time the dyes changed color).
**Source: https://www.teachchemistry.org/content/dam/AACT/middle-school/reactions/
reaction-rate/secure/Demo_SimpleKinetics.pdf
APPENDIX B
For the reaction:
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H2 + I2
—>
2HI
•
D
EP
ED
C
O
PY
The concentrations of the reactants H2 and I2 decrease with time; the concentration of the product HI increases.
The concentration of reactants and products could be plotted with time to obtain a graph similar to that shown
below.
Change in concentration of I2 and HI with time as the reaction.
H 2 + I2
—>
2HI proceeds.
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•
C
O
PY
APPENDIX C
The t1/2 of nitrogen-13 = 10 minutes.
ED
After one half-life, one-half of the sample decays to a new element (carbon-13) and one-half remains.
After two half-lives, one-fourth of the original sample remains. After three half-lives, one-eight of the original
sample remains.
APPENDIX D.
D
EP
Consider the isotope of nitrogen, nitrogen-13, which is used in positron emission tomography. The halflife of nitrogen-13 is 10 minutes. If you have a 10-gram sample of nitrogen-13, after 10 minutes or one half-life,
5 grams of the sample would have decayed and 5 grams of nitrogen-13 remains. After another 10 minutes or 2
half-lives, half of the 5 grams of nitrogen-13 or 2.5 grams would have decayed and 2.5 grams remains.
What does half-life mean?
Half-life is the length of time required for the concentration of a reactant to decrease by half its original
or initial concentration. At this time, t = t1/2:
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•
[A]t = ½ [A]o
For a first order reaction, substitute in equation 3:
[A]o =
kt1/2
½ [A]o
PY
log
2.303
log 2 =
C
O
Solving for t1/2,
kt1/2
2.303
k
t1/2 = 0.693
Equation 4
D
EP
k
ED
t1/2 = 2.303 log 2
This equation tells that for a first-order reaction, t1/2 is independent of the initial concentration. It
depends on the value of k, which is constant throughout the reaction.
APPENDIX E
Consider problems illustrating how to calculate the half-life of radioisotopes, and how to calculate the
amount of a radioisotope that remains after a given period of time.
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•
Example 1. Iodine-131 is commonly used to treat throat cancer. The rate constant for decay of
iodine-131 is 0.0864/day. What is the half-life of iodine-131?
PY
Solution:
To calculate half-life, use the expression:
k
t1/2 =
0.693
0.0864/day
C
O
t1/2 = 0.693
ED
t1/2 = 8.021 days
The half-life of iodine-131 is 8.021 days
Solution:
D
EP
Example 2. The half-life of sodium-24, a radioisotope used as a tracer for the cardiovascular
system, is
14.951 hours. How many grams of sodium-24 will remain after 24.0 hours in a sample initially containing 100
mg?
First, calculate k. Use the expression:
t1/2 = 0.693
k
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•
k =
0.693
14.951 hr
PY
k = 0.0464/hr
Then use the equation:
log [A]o =
2.303
C
O
[A]t
kt .
Initial amount = 100 mg.
Amount after 24 hrs = x
ED
In the equation, quantities are in molar concentrations. However, number of moles is
proportional to mass, thus:
log 100 = (0.0464/hr) (24 hrs)
2.303
D
EP
x
log 100 – log x = 0.483
log x = log 100 – 0.483
x = 32.9 mg
The amount of sodium-24 remaining after 24 hours is 32.9 mg.
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GENERAL CHEMISTRY 2
LABORATORY EXERCISE 1
FACTORS THAT AFFECT REACTION RATES*
PY
Learning Competency
(LAB) Determine the effect of various factors on the rate of reaction (STEM_GC11CK-IIIi-j-139)
ED
C
O
Chemical reactions take place at different rates. Some take a very long time while others occur instantaneously. Sodium metal reacts instantaneously
with oxygen in the air, while iron takes a long time to rust when exposed to the atmosphere. Combustion reactions like the burning of liquid
petroleum gas (LPG) in air or oxygen are very fast reactions. Explosions are very fast chemical reactions. Dynamite, when ignited, produces powerful
explosions in a fraction of a second. Interest in how fast reactions take place are important because these impact on economic, health, and safety
issues. To the community in general, fine powders like flour dust or the presence of methane gas could be detrimental because of the possibility of
huge explosions occurring. Hence, it is important to know and understand the various factors that affect or alter the rates of chemical reactions.
Studies of chemical kinetics have recognized that reactant concentration, temperature, and the presence of a catalyst influence the rate of reaction.
Use the “clock” reaction in the study of the effect of concentration and temperature on the rate of a reaction. The following is a simplified sequence
of reactions proposed:
Iodate ions (IO3-) react with bisulfite ions (HSO3-) to form iodide ions (I-)
IO3- + HSO3-
2.
—>
I- + 3SO4- + 3H3O+
Iodide ions then react with iodate ions to form molecular iodine (I2).
I- + IO3-
3.
D
EP
1.
+
H3O+
—>
3I2 + + 9H2O
Molecular iodine reacts with starch to produce the dark blue starch-iodine complex.
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Study the effect of a catalyst on a chemical reaction using the reaction between magnesium and water with phenolphthalein as indicator. The
reaction involved is:
Mg + 2H2O
—>
Mg2+ + 2OH- + H2
PY
Objective
C
O
At the end of the exercise, learners will be able to state the effect of concentration, temperature, and the presence of a catalyst on the rate of
reactions.
Materials
M KIO3 (Solution A)
0.014 M HSO3- (Solution B)
Phenolphthalein
Magnesium ribbon
NaCl
Starch
Ice
B. Apparatus
•
•
•
•
•
•
•
D
EP
•
•
•
•
•
•
•
ED
A. Reagents
10-mL volumetric flask
Ten test tubes
Test tube rack
250-mL beaker
10-mL graduated cylinder
Burner or hot plate
Thermometer
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Procedure
PY
This lab exercise has three main parts. For each part, learners will determine the effect of concentration, temperature, and the presence of a catalyst
on the rate of a chemical reaction.
A. Effect of concentration
C
O
1. From a 0.01 M KIO3 stock solution, prepare in test tubes 5-mL KIO3 solutions with the following concentrations: 0.01 M, 0.008 M, 0.006 M,
0.004 M, 0.002 M. Refer to these as Solution A.
2. In another set of five test tubes, place 5 mL of 0.014 M HSO3- and mark as Solution B.
3. Pour each of the prepared Solution A, one concentration at a time, into Solution B, then transfer the mixture back and forth from one test
tube to the other to ensure uniform mixing.
4. Note the time elapsed from the instant the two solutions come in contact with each other until the first sign of a blue color is detected.
5. Record this time in the table prepared prior to coming to class.
ED
a. Calculate the concentration of iodate ions in moles per liter in each of the reaction mixtures. Learners will have to consider the final
volume of the reaction mixture after adding Solution B. Write the values obtained in the table prepared.
b. Plot a graph of reaction time versus concentration of iodate. What effect does concentration have on the rate of the reaction?
D
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c. Why does increasing the concentration of a reactant increase the rate of reaction?
B. Effect of temperature
1. Prepare five test tubes each containing 10 mL of solution A (0.01 M KIO3), and five test tubes each containing 10 mL of Solution B (0.014 M
HSO3-).
2. Place a pair of test tubes of solution A and solution B in a water bath to bring it to the required temperature.
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3. Mix the two solutions as described above and record the time elapsed from the instant the two solutions came in contact until a blue color is
detected.
4. Record data in the table prepared prior to coming to class.
PY
5. Repeat the experiment at five different temperatures within the range of 10 – 65 oC.
a. Plot a graph of temperature against time. Write a generalization on the effect of varying temperatures of the reaction mixture on the rate
of a reaction.
C
O
b. Why does increasing the temperature of the reactants increase the rate of a reaction?
C. Effect of a catalyst
1. Place three test tubes on a test tube rack. Add 5.00 mL distilled water into each test tube and add the following:
ED
Test tube 1: 1 drop phenolphthalein
Test tube 2: 1 drop phenolphthalein and 1 cm Mg ribbon
Test tube 3: 1 drop phenolphthalein, 1 cm Mg ribbon, and 0.05 gram NaCl
a.
b.
c.
d.
D
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2. Observe what happens for a period of five minutes.
Which test tube exhibited the fastest reaction? How did you arrive at this conclusion?
Which substance served as a catalyst in the experiment?
How does a catalyst affect the rate of a chemical reaction?
How does a catalyst work?
*Adapted (with modifications) from General Chemistry and Chemical Division, Institute of Chemistry, UP Los Baños (2006). Laboratory Manual for
Chemistry 16 General Chemistry I.
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Chemistry 2
330 MINS
Spontaneous Change, Entropy,
and Free Energy
LESSON OUTLINE
Introduction
Communicating learning objectives
15
Demonstration Activity
10
Lecture
215
Practice
Sample question
20
Enrichment
Experiment
60
Learning Competencies
Predict the spontaneity of a process based on entropy. (STEM_GC11CT-IVa-b-140)
Evaluation
Written homework
10
Use Gibbs’ free energy to determine the direction of a reaction.
Transparent cups; Coffee granules; Hot and cold water; Diagrams,
Worksheets
C
O
Performance Standards
The learners shall find patterns and make predictions on heat and energy involved in
chemical reactions
ED
Determine whether entropy increases or decreases if the following are changed:
temperature, phase, number of particles. (STEM_GC11CT-IVa-b-141)
Explain the second law of thermodynamics and its significance
(STEM_GC11CT-IVa-b-142)
Materials
Resources
(1) Chang, R. (2007) Chemistry, 9th Ed. McGraw-Hill, Inc., USA.
(2) Whitten, K.W., et al (2007) Chemistry, 8th Ed. Thomson-Brooks/Cole,
USA.
(3) Zumdahl, S.S. and Zumdahl, S.A. (2000) Chemistry, 5 ed., Houghton
Mifflin.
D
EP
Use Gibbs’ free energy to determine the direction of a reaction
(STEM_GC11CT-IVa-b-143)
PY
Content Standard
Motivation
The learners demonstrate an understanding of spontaneous change, entropy, and free
Instruction
energy.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
identify spontaneous processes;
•
predict the spontaneity of a process based on entropy and Gibbs free energy;
•
state the effect of change in temperature, phase, number of particles and size of
molecules to entropy;
•
use Gibbs free energy to determine the direction of a reaction.
•
determine entropy change and Gibbs free energy change using thermodynamic
data.
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INTRODUCTION (15 MINS)
1. Communicate learning competencies and objectives to the learners using any of the suggested protocols
(Verbatim, Own words, Read-aloud).
a. I can identify spontaneous processes.
PY
b. I can predict the spontaneity of a process based on entropy and Gibbs free energy.
Teacher tip
•
Presenting the relevant vocabulary and
concepts may already serve to connect
the current lesson with prerequisite
knowledge. It is best if it is done
chronologically with the intention of
allowing students’ to ask questions for
clarification.
•
Use diagrams to facilitate the
introduction part.
c. I can state the effect of change in temperature, phase, number of particles and size of molecules to
entropy.
d. I can use Gibbs free energy to determine the direction of a reaction.
C
O
e. I can determine entropy change and Gibbs free energy change using thermodynamic data.
2. Present relevant vocabulary the students should know that will be used in the lesson.
Exothermic process
Process that gives off heat to the surroundings
ED
Endothermic process
Process that absorbs heat from the surroundings
D
EP
Thermodynamics
A scientific discipline that deals with the interconversion of heat and other forms of energy
Enthalpy, H
A thermodynamic quantity used to describe heat changes taking place at constant pressure
Enthalpy of reaction, ΔHrxn
The difference between the enthalpies of the products and the enthalpies of reactants
Spontaneous process
A physical or chemical change that occurs by itself. A process that takes place without energy from an external
source
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Image source: http://
chemed.chem.purdue.edu/genchem/
topicreview/bp/ch5/work.html
Entropy, S
A thermodynamic quantity that is a measure of how spread out or dispersed the energy of a system is among
the different possible ways that system can contain energy
PY
Absolute Entropy
The absolute value of entropy of a substance
Standard entropy, S°
The absolute entropy of a substance at 1 atm and 25 °C
C
O
Gibbs Free Energy, G
Energy available to do useful work. Used to express the spontaneity of a reaction more directly
ED
Standard Free Energy, ΔGrxn°
The free energy change for reaction when it occurs under standard state conditions, when reactants in their
standard states are converted into products in their standard states
Equilibrium constant, K
A number equal to the ratio of the equilibrium concentrations of the products to the equilibrium concentrations
of reactants, each raised to the power of their stoichiometric coefficients
D
EP
State functions
Properties that can be expressed as (final – initial) states
3. Connect the lesson with prerequisite knowledge
Recall from previous lessons.
a. What is thermodynamics?
b. How do endothermic processes differ from exothermic processes?
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Image source: http://
employees.csbsju.edu/hjakubowski/
classes/ch331/lipidstruct/systemsurr.gif
c. State the First Law of Thermodynamics. How does it relate to the Law of Conservation of Energy?
The First Law of Thermodynamics:
“Energy of the universe is constant.”
PY
The Law of Conservation of Energy :
“Energy can be converted from one form to another but cannot be created or destroyed.”
Teacher tip
•
C
O
This may be a very general law but it is a very important one. It helps us understand the type of change
that can occur in our universe.
d. What are the parts of the universe of interest? How do they relate to each other?
•
the system (part we are investigating)
•
the surroundings (everything else)
ED
For thermodynamic studies we need to divide the universe into two parts:
D
EP
e. How is system changes treated in the study of thermodynamics?
System Change is based on going from initial state to final state. It is and ALWAYS written
* What is important in this process is to get
the correct + or – sign.
•
It is important in thermodynamics, as
elsewhere in chemistry that the right
sign, value, and units are used.
•
Some of these properties were discussed
in previous lessons, others will be
discussed in this lesson.
•
Important State Functions are: ΔT, ΔH,
ΔE, ΔS, and ΔG
•
Important Path Functions are: q and w
Systemfinal ‒ Systeminitial and using the symbol Δ for change.
Change (Δ) = (final – initial)
Change in temperature is written:
ΔT = Tf – Ti
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For example ‒ if you are on a ladder at
height 5ft and then you climb up to 12 ft
the change in height is
Δh = hf - hi !
Δh = 12ft – 5ft !
Δh = +7ft
!
Example: Hot coffee cools from 55oC to 28oC so
ΔT = Tf – Ti
= 28 oC – 55oC
= –27oC
PY
Change in internal energy (E) is written:
ΔE= Ef - Ei
What are state functions? What are path functions?
C
O
f.
Properties that can be expressed as (final – initial) and we write with Δ such as ΔE = Ef – Ei are called
state functions.
MOTIVATION (10 MINS)
ED
Properties that you cannot calculate by just knowing final and initial states but must know how process
occurred are called path functions.
D
EP
Demonstration Activity. Using two cups of water, one hot and one cold, show how coffee granules will behave
when added to the water.
•
Ask the students what will happen if granules of coffee are added to the water in a cup.
•
Let them make their observations and compare what happens in the two cups.
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Teacher Tip:
•
C
O
PY
•
D
EP
INSTRUCTION (215 MINS)
ED
Diffusion of coffee granules in hot and cold water. (Image obtained from http://
i.ytimg.com/vi/zg9bUoMcrzI/maxresdefault.jpg)
Lesson Proper. Lecture Discussion on Spontaneous Process, Entropy, the Second Law of Thermodynamics and
Gibbs Free Energy and Equilibrium
Focus Question
Will a reaction occur? Is the reaction spontaneous?
The Three Laws of Thermodynamics
State the three laws of thermodynamics.
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•
•
Emphasize the difference in temperature
of the water in the cup. A third cup with
warm water may be added to the
demonstration.
Transparent glass is best for the
demonstration so that the students will
be able to see the mixing of the coffee
with water.
Coffee is used for better visibility of
mixing solid and liquid. The granular
ones will be better. Five to ten granules
are enough. Add the coffee granules to
the water at the same time.
Don’t stir.
1st Law - Energy of the universe is constant.
Teacher Tip:
“Energy can be converted from one form to another but cannot be created or destroyed.”
2nd Law – Entropy of universe increases.
PY
“The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium
process.”
3rd Law – At absolute zero, the entropy of a perfect crystal is 0.
C
O
“The entropy of a perfect crystalline substance is zero at the absolute zero of temperature (T = 0
K= -273.15 oC).”
A. SPONTANEOUS PROCESSES: Characteristics of Spontaneous Processes
D
EP
ED
What can you say about the pictures shown? Compare each pair and tell which one is more spontaneous than
the other?
Uphill and Downhill Skiing. (Image obtained from http://static.guim.co.uk/sys-images/
Guardian/Pix/audio/video/2013/1/22/1358854831497/Vertical-skiing-an-uphill-012.jpg) and
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• Practice problems should be incorporated
in the delivery of the instruction to provide
immediate feedback on students’
learning.
• Provide the problems to students in work
sheets.
PY
C
O
Rock rolled uphill and downhill. (Image obtained from http://millionairecorner.com/LibRepository/
5589d308-f280-44fa-90b8-6ec4f5dff07a.jpg and http://s3.amazonaws.com/
thumbnails.illustrationsource.com/huge.103.519587.JPG)
ED
What is a spontaneous process? Give examples of spontaneous processes.
Expected answers:
D
EP
A spontaneous process is a physical or chemical change that occurs by itself. These processes occur
without requiring an outside force and continue until equilibrium is reached.
Heat flows from a hotter object to a colder one.
An iron object rusts in moist air.
Sugar dissolves in a cup of coffee.
How does spontaneity apply to chemical reactions?
In a chemical reaction, ΔHreaction = Hproducts - Hreactants. If it is exothermic, then ΔHreaction = (-). To get a
negative ΔHreaction , the Hproducts must be lower than the Hreactants.
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A spontaneous process is one that takes place without energy from an external source. For a
chemical reaction to be spontaneous, it should proceed as written (from left to right), without an
input of energy.
PY
Examples of Reactions
Combustion of methane
+
2O2
—>
6CO2
+
2H2O
Acid-base neutralization reaction
OH- (aq)
—>
H2O (l)
ΔHo = - 56.2 kJ/mol
ED
H+(aq) +
ΔHo = - 890.4 kJ/mol
C
O
CH4
* Both of these reactions are very exothermic and are not reversible.
H2O (s)
—>
D
EP
Solid to liquid phase transition of water
H2O (l)
ΔHo = 6.01kJ/mol
Dissolution of ammonium nitrate in water
NH4NO3 (s)
—>
NH4+ (aq)
+
NO3- (aq)
ΔHo = 6.01 kJ/mol
* Ice melting above 0oC and ammonium nitrate dissolving in water are both spontaneous process yet
endothermic.
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•
From the examples given, both
endothermic and exothermic reactions
can be spontaneous.
•
The energy change in the system cannot
solely tell whether a chemical reaction
will occur spontaneously. To make this
kind of prediction, another
thermodynamic quantity is needed:
ENTROPY.
B. ENTROPY
What is entropy? How does it relate to spontaneity of a process?
The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.
C
O
•
PY
Entropy, S , is a thermodynamic quantity that is a measure of how spread out or dispersed the energy of
a system is among the different possible ways that system can contain energy. It is a quantity that is
generally used to describe the course of a process, that is, whether it is a spontaneous process and has a
probability of occurring in a defined direction, or a non-spontaneous process and will not proceed in the
defined direction, but in the reverse direction.
How do entropy changes occur?
Melting
Order
Solid
D
EP
Process
ED
Most processes are accompanied by entropy change. The following are processes that lead to an
increase in entropy of the system.
—>
Disorder
Teacher tip
—>
Liquid
•
Vaporization
Liquid
—>
Vapor
Dissolving
Solute
—>
Solution
Heating
System at T1
—>
System at T2 (T2 > T1)
The spreading out of more concentrated molecules and the spreading out of more concentrated energy
are changes from more order to more random. The changes that occur are the ones that lead to an
increasing randomness of the universe. Entropy is sometimes referred as the measure of randomness
and disorder.
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Entropy change examples:
➡ Molecules of gas at high pressure
always spread to lower pressure
regions.
➡ Gas in balloon spreads out into room
and deflates but never see balloon
spontaneous fill with air.
➡ Heat always goes from high
temperature into cooler regions.
➡ Hot coffee in a room gets cooler and
the heat spreads out into the room,
but never sees a cold cup of coffee
spontaneously warm up.
PY
C
O
<— heat <— (direction of heat flow)
http://www.physchem.co.za/Heat/Effects.htm
ED
At high enough temperature, the spontaneous change is from Solid Liquid Gas; gas is more random
than liquid and liquid is more random than solid. There is an increase in entropy (S) of the system by
going from solid to liquid to gas.
D
EP
C. Entropy and the Second Law of Thermodynamics
Focus Question:
Will water freeze at room temperature? Why?
State the Second Law of Thermodynamics. What does it say about the entropy of spontaneous processes and
processes at equilibrium?
The Second Law of Thermodynamics deals with entropy. It tells whether a process or chemical reaction
can occur. The connection between entropy and the spontaneity of a reaction is expressed by the
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second law of thermodynamics:
It states that:
•
It is possible for either ΔSsys or ΔSsur to be
negative, as long as their sum is greater
than zero.
•
At a stable equilibrium, ΔSuniv = 0, in this
case, ΔSsys or ΔSsur must be equal in
magnitude but opposite in sign.
PY
“The entropy of the universe increases in a spontaneous process and remains unchanged in
an equilibrium process.”
ΔSuniv
=
ΔSsys
+
ΔSsur
>
0
ΔSuniv
=
ΔSsys
+
ΔSsur
=
0
ΔSuniv
=
ΔSsys
+
ΔSsur
<
0
C
O
Because the universe is made up of the system and the surroundings, the entropy change in the universe
(ΔSuniv) for any process is the sum of the entropy changes in the system (ΔSsys) and in the surroundings
(ΔSsur).
Process is spontaneous.
Process tends not to occur; equilibrium
is attained.
Reverse process occurs spontaneously.
ED
Calculating Entropy Changes in the System: Standard Entropy of Reaction, ΔS°rxn
D
EP
How is entropy change determined? What data are needed for the calculation?
Suppose that the system is represented by the following reaction:
!
!
!
!
!
!
aA###### +######bB!
—>!
cC!
+!
dD!
As in the case for the enthalpy of a reaction, the standard entropy of reaction ΔS° rxn is given by the
difference in standard entropies between the products and the reactants.
!
!
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ΔS°
ΣnS° (products)
−
ΣmS° (reactants)
Where m and n are the stoichiometric coefficients in the reaction.
ΔS° rxn
=
[cSo (C)
+
d So (D)]
[aSo (A)
‒
+
bSo (B)]
PY
•
=
C
O
The standard entropy values of compounds have been measured in J/K mol. To calculate the ΔS° rxn
(which is the ΔSsys), the values may be found in the Thermodynamic Data Table. Thermodynamic tables
have absolute entropy of substances at 25°C and 1atm.
For convenience, ΔS° will be used instead of ΔS° rxn in the proceeding discussion.
•
Remember, the greater the value of ΔS then the greater is the increase in the randomness of the
system.!
ED
•
Sample Problem 1:
D
EP
From the standard entropy values in the Thermodynamic Data table, calculate ΔS° for the following reaction.
H2(g)
+
I2(s)
—>
From the table, S°(J/K·mol):
130.6
+
I2(s)
—>
116.7
•
Step 2: Using the equation for the standard entropy of reaction
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For problems such as this, it is more
convenient to write the thermodynamic
data below each formula.
Tracking the units:
•
The units (mol) for the number of moles
of reacting substances and (J/ K mol) for
the S° results to the unit (J/K) for the ΔS°
= ((mol) (J/ K mol) = J/K.
•
We expect a positive value for ΔS for the
system because the change involved 1
mole gas + 1 mole solid 2 moles gas
should increase randomness (entropy).
•
Gas is more random than solid. In
general, there is more randomness in a
gas or if the temperature is higher or if a
much larger molecule rather than smaller.
2HI(g)
206.3
Thermodynamic Data Tables are available
in chemistry books and will be used as
sources of data.
Here is another source:
http://bilbo.chm.uri.edu/CHM112/tables/
thermtable.htm
2HI(g)
Step 1. Write the standard entropy below each formula.
H2(g)
•
ΔS°
=!!
ΣnS° (products)
−
ΣmS° (reactants)
=
[(2) So HI] – [(1) So H2 + (1) So I2]
=
[ (2) (206.3) ] – [ (1) (130.6) + (1) (116.7) ]
=
[ 412.6 ] – [ 247.3 ]
ΔS° =
+165.3 J/K
PY
Step 3: Substitute the entropy values.
C
O
What general rules apply to predicting whether an entropy change is negative or positive?
General rules for predicting entropy change of the system:
1. If the reaction produces more gas molecules than it consumes, ΔS° is positive.
2. If the total number of gas molecules diminishes, ΔS° is negative.!
ED
3. If there is no net change in the total number of gas molecules, ΔS° may be positive or negative, but will
be relatively small numerically. !
Practice Problem 1:
O2(g) —> 2O(g)
D
EP
Predict whether the entropy change of the system in each of the following is positive or negative.
N2 (g, 10 atm) —> N2 (g, 1atm).
6CO2(g) + 6H2O(g) —> C6H12O6(g) + 6O2(g).
2 H2 (g) + O2 (g) —> 2 H2O (l)
NH4Cl (s) —> NH3(g) + HCl (g)
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Expected answer:
Positive
Increase in number of gas molecules
N2 (g, 10 atm) —> N2 (g, 1atm).
Positive
Decrease in pressure of the system
will increase entropy
Negative
Decrease in number of gas
particles
Negative
Net decease in number of molecules and
gases are converted to solids
Positive
A solid is converted to two gaseous products.
PY
O2(g) —> 2O(g)
C
O
6CO2(g) + 6H2O(g) —> C6H12O6(g) + 6O2(g).
2 H2 (g) + O2 (g) —> 2 H2O (l)
Practice Problem 2:
1. Determine S for the reaction:
Given:
S°(J/K·mol):
+
D
EP
SO3(g)
ED
NH4Cl (s) —> NH3(g) + HCl (g)
256.2
H2O(l)
—>
69.9
H2SO4(l)
156.9
2. Calculate S for the reaction
SO2(s)
Given:
S°(J/K·mol):
248.5
+
NO2(g) —> SO3(g)
240.5
256.2
+
NO(g)
210.6
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3. Calculate S at 25C for the reduction of given these absolute entropies:
2PbO(s)
Given:
S°(J/K·mol):
+
69.54
C(s)
—>
5.7
2Pb(s)
+
CO2(g)
64.89
PY
Expected answers:
213.6
1. -169.2 J/K·mol
2. -22.2 J/K·mol
Calculating the Entropy Changes in the Surroundings, ΔSsur
C
O
3. +198.8 J/K·mol
How is entropy change in the surrounding determined? What conditions are involved in the calculations?
ED
How can the spontaneity of a reaction be predicted using entropy change in the surrounding, ΔSsur?
D
EP
For constant-pressure process the heat change is equal to the enthalpy change of the system ΔHsys,. Then
the change in entropy of the surroundings ΔSsurr is proportional to the ΔHsys.
ΔSsurr
—>
− ΔHsys
The minus sign is used because, if the process is exothermic, ΔHsys is negative and the ΔSsurr is a positive
quantity, indicating an increase in entropy.
•
For an endothermic process, ΔHsys is positive and the negative sign ensures that the entropy of the
surroundings ΔHsurr decreases.
The change in entropy for a given amount of heat absorbed also depends on the temperature.
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Emphasize the importance of the positive
and negative signs in the processes and
their significance.
•
If the temperature of the surroundings is high, the molecules are already quite energetic. Therefore, the
absorption of heat from an exothermic process in the system will have relatively little impact on the
motion of the molecules and the resulting increase in entropy of the surroundings will be small.
•
However, if the temperature of the surroundings is low, then the addition of the same amount of heat will
cause a more drastic increase in molecular motion and hence a larger increase in entropy.
−ΔHsys
PY
=
ΔSsurr
C
O
T
Sample Problem 2:
Applying the procedure for calculating the ΔSsys and ΔSsurr to the synthesis of ammonia: Is the reaction
spontaneous at 25C?
Calculating the ΔSsur
=
D
EP
ΔSsurr
=
ΔSsurr
ΔH°rxn = - 92.6 kJ/mol
ED
N2 (g) + 3 H2 (g) —> 2 NH3 (g)
=
−ΔHsys
T
− (− 92.6 kJ/mol)(1 kJ/1000 J)
298 K
311 J/K ·mol
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Calculating the ΔSsys
N2 (g) + 3 H2 (g) —> 2 NH3 (g)
ΔS°
131
=
ΣnS° (products)
=
[(2) So NH3] – [(1) So N2 + (3) So H2]
=
[ (2) (193) ] – [ (1) (192) + (3) (131) ]
=
−199 J/K· mol
Determining Spontaneity of Reaction Using ΔSuniv
ΣmS° (reactants)
=
ΔSsys
ΔSsurr
=
−199 J/K· mol + 311 J/K ·mol
=
112 J/K ·mol
D
EP
ΔSuniv
+
−
ED
ΔSuniv
193
C
O
ΔS°
192
PY
From the table, S°(J/K· mol):
* Because the ΔSuniv is positive, we predict that the reaction is spontaneous at 25C.
The!Third!Law!of!Thermodynamics!and!Absolute!Entropy!
What is absolute entropy? How does it relate to the third law of thermodynamics?
The Third Law of Thermodynamics states:
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“The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.”
At absolute zero (T = 0 K= -273.15 oC), the entropy of a perfect crystal is 0.
•
As the temperature increases, the freedom of motion increases. The entropy of any substance at a
temperature above 0 K is greater than zero.
•
A solid like glass with imperfections built into it will not have entropy = 0 even at 0 K because it is not
a perfect crystal and still has some randomness left in it.
PY
•
C
O
At absolute zero, all atomic motion stops. Atoms and molecules are no longer vibrating around or
moving past each other. However, it becomes harder and harder to lower the temperature as you get
closer and closer to absolute zero. But even though scientists are not able to attain the T = 0 K, it is still a
useful reference point.
Absolute Entropies
D
EP
ED
The reference point for entropy of a substance is entropy at 0 K. Entropy increases as temperature
increases creating more randomness. The lowest entropy, therefore, occurs at this temperature (S = 0 J/
K mol at T = 0 K).
Image obtained from http://
wine1.sb.fsu.edu/chm1046/notes/
Thermody/MolBasis/MolBasis.htm
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Thermodynamic tables may have absolute entropy of substances at 25°C and 1atm. But the reference
point is S= 0 J/K for 1 mol of substance at 0 K.
If the ΔHf° for an element is 0 at the standard state of 25°C and 1 atm, the S° for an element is not
equal to 0 at 25°C and 1 atm. (The symbol ° is used to indicate the standard state of 25oC and 1
atm.)
•
If the ΔHf° for a compound is based on its formation from its elements, the S° for a compound is not
based on its formation from its elements. Entropy will be 0 only at -273 oC not at 25oC. If the
entropy of a pure element is not zero at standard state, it cannot be assumed as zero; its value has to
be determined. The absolute entropies could be used to calculate ΔS entropy change for a reaction.
C
O
PY
•
D. Gibbs Free Energy, G
ED
What is Gibbs free energy? How does it relate to the spontaneity of a reaction?
D
EP
Another thermodynamic function is used in order to express the spontaneity of a reaction more directly.
This is called Gibbs free energy, G. The use of G predicts changes that are focused on the system.
Gibbs free energy is defined as:
G = H – TS
•
All the quantities in the equation pertain to the system; the temperature T is the temperature of the
system.
•
G has units of energy; both H and TS are in energy units.
•
H, S and G are all state functions.
If the entropy of the universe increases then the ΔG of the system will decrease. The direction of
spontaneous change is negative ΔG for system. The ΔG tells us if a change can occur for a chemical
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reaction.
If ΔSuniv is (+) for universe then ΔG for system is ( - ). The ΔG for the system is a convenient way to
predict a change.
ΔG
ΔSuniv
Reaction
+
-
Increase
0
0
Stay the same
-
+
Decrease
PY
ΔSuniv
Spontaneous, will go
C
O
No change at
equilibrium
Not spontaneous, will
not go, the reverse will
Determining spontaneity of a reaction using ΔG
ED
How can the spontaneity of a reaction be predicted using Gibbs free energy change, ΔG?
D
EP
The change in free energy (ΔG) of a system for a constant-temperature process is
ΔG = ΔH –TΔS
In this context, free energy is the energy available to do work. If a particular reaction is accompanied by
a release of usable energy (ΔG is negative), the reaction is spontaneous.
•
Unless stated otherwise, the ΔH, ΔS and ΔG refer to the system at 25oC. Most of our calculations are for
values of ΔG at 25oC (298 K).
Summary of conditions for spontaneity and equilibrium at constant temperature and pressure in terms of ΔG
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•
If the ΔG = more negative and ΔSuniv =
more positive (increase, more random)
•
If ΔG of universe is more negative then
the change will occur until equilibrium is
reached.
!
<
0
ΔG!
!
>
0
ΔG!
!
=
0
The reaction is spontaneous in the forward
direction.
The reaction is nonspontaneous. The reaction
is spontaneous in the opposite direction.
The system is at equilibrium. There is no net
change.
Methods to determine change in Gibbs Free Energy (ΔG)
•
Determining ΔG using ΔH and ΔS data. !
Sample Problem 3:
ED
ΔG = ΔH –TΔS
C
O
How is Gibbs free energy change determined when ΔH and ΔS are available?
PY
ΔG!
The old camera flash bulb used Mg metal sealed in a bulb with oxygen. The reaction is:
S° J/K mol:
ΔHfo kJ/mol:
Calculating the ΔS°
+
½ O2
—>
D
EP
Mg
ΔS°
MgO
32.7
205.0
26.9
0
0
-601.2
=
ΣnS° (products)
−
ΣmS° (reactants)
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Tracking down units:
ΔG = Δ H ‒ T Δ S
= (kJ) (K) (J/K)(kJ/1000J)
= kJ or kJ/mol in
Thermodynamic tables
=
[(1) So MgO] – [(1) So Mg + (½) So O2]
=
[(1) (26.9)] – [(1)(32.7) + (½) (205.0)]
ΔS°
=
-108.3 J/K mol
ΔH°
=
Σ nΔHf° (products)
=
[(1) ΔHfo MgO] – [(1) ΔHfo Mg + (½) ΔHfo O2]
=
[(1) (-601.2)] – [(1) (0) + (½) (0)]
ΔH°
=
-601.2kJ
ΔG
=
ΔH
=
601.2kJ - (298K) (-108.3J/K) ( 1kJ/1000J)
=
-601.2 + 32.3kJ
=
-568.9kJ
ED
TΔS
D
EP
ΔG
Σ mΔHf° (reactants)
C
O
Calculating the ΔG
-
–
PY
Calculating the ΔH°
Since ΔG is negative the reaction will form MgO. Looking up ΔGf (MgO) = -569 kJ based on
ΔG = [ΔGf (MgO) ] – [ΔGf (Mg) + ½ ΔGf (O2)] this reaction occurs rapidly once initiated.
Practice Problem 4:
Ozone (O3) in the atmosphere can react with nitric oxide (NO):
O3(g) + NO(g) —> NO2(g) + O2(g).
431
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Calculate the G for this reaction at 25C.
(H = -199 kJ/mol, S = -4.1 J/K·mol)
Expected answer:
•
PY
-198 kJ/mol
Calculating ΔG from Standard Free Energies!
C
O
How is Gibbs free energy change determined if the standard free energy of formation ΔGf° for the reactants and
products are available?
If the data for ΔG of formation ΔGf° of the reactants and products are available, the following equation is
used.
Σ nΔGf° (products)
−
Σ mΔGf° (reactants)
Where m and n are the stoichiometric coefficients in the reaction and ΔGf° is the standard free
energy of formation
D
EP
•
=
ED
ΔG°
ΔGf° is the standard free energy of formation at 25 °C and 1atm for 1 mol of compound formed from
its elements. The ΔGf° can be used to get the ΔG of a reaction just like using ΔHf° to get ΔH for
reaction. The standard free energy ΔG° of element in the standard state is 0.
Sample Problem 4:
Will this reaction occur at 298 K ?
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Cu (g)
ΔGf° kJ/mol:
—>
CuO(s)
0
-127
ΔGf° for Cu and O2 are 0 since they are pure elements
=!
Σ nΔGf° (products)
−
Σ mΔGf° (reactants)
=
[(1) Gf° CuO] – [(1) Gf° Cu + (½) Gf° O2]
=
[(1) (-127) ] - [ (1)( 0) + (½) (0) ]
=!
- 127 kJ/mol
PY
ΔG°
½ O2 (g)
0
Use the Thermodynamic table to find ΔGf°
ΔG°
+
Factors affecting the sign of ΔG
ED
C
O
Negative ΔG means that the reaction will occur but because it is very slow at room temperature, it can
take years for a penny to get CuO coating and turn brown. The reaction can be sped up by raising the
temperature.
What!combinations of enthalpy change (ΔH) and entropy change (ΔS) can!happen?!
D
EP
In order to predict the sign of ΔG according to the equation: ΔG = ΔH- TΔS, we need to know both ΔH
and ΔS. Temperature may also influence the direction of the spontaneous reaction. The four possible
combinations are shown below.
ΔH
TΔS
ΔG
Effects
‒
+
‒
Reaction proceeds spontaneously at
all temperatures
+
‒
+
Example: 2 H2O2 (aq) —> 2 H2O (l) + O2 (g)
Reaction is spontaneous in the reverse
direction at all temperatures
Example: 3 O2 (g) —>
2 O3 (g)
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+
+
?
Reaction proceeds spontaneously at high
temperatures. At low temperatures, the
reverse reaction becomes spontaneous.
Example: 2 HgO (s) —> 2 Hg (l) + H2O (g)
?
‒
Reaction proceeds spontaneously at low
temperatures. At high temperatures, the
reverse reaction becomes spontaneous.
PY
‒
Example: NH3 (g) + HCl (g) —> NH4Cl (s)
C
O
Sample Problem Application:
Should you invest in an engine that is said to burn air at room temperature? You are told that a special
chamber allows O2 to combine with N2 to form NO2 (nitrogen dioxide) using reaction
+
½ N2(g)
+
Evaluate using thermodynamics
ΔHfo kJ/mol:
191.5
—>
O2 (g)
—>
0
NO2(g)
NO2(g)
205.0
240.5
0
+34
D
EP
S° J/K mol:
O2 (g)
ED
½ N2(g)
Calculate enthalpy change
ΔH°
ΔH°
=
Σ nΔHf° (products) – Σ mΔHf° (reactants)
=
[(1 ) ΔHfo NO2] – [(½)ΔHfo N2 +(1) ΔHfo O2]
=
[(1)(+34)] – [((½) (0) + (1)(0)]
=
+34 kJ
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* No heat given off only taken in.
Calculate entropy change
ΣnS° (products)
−
ΣmS° (reactants)
=
[(1 ) So NO2] – [(½) So N2 +(1) So O2]
=
[(1) (240.5) ] – [(½) (191.5) + (1) (205.0)]
=
240.5 -
=
-60.3 J/K mol
300.8
Calculate Gibbs Free Energy change from above
ΔG
= !
ΔH- TΔS
=
+34kJ – (298K) (-60.3J/molK) (1kJ/1000J)
=
34 kJ + 18.0 kJ
=
+52 kJ
ED
ΔG
PY
ΔS°
=
C
O
ΔS°
D
EP
* The reaction will NOT occur and NO heat is given off. Neither ΔH which is + or ΔS which is − favors
reaction.
Effect of Temperature Effect on Chemical Reactions
How does temperature affect the direction of a chemical reaction?
The ΔG can be found for different temperature values.
ΔG
=
ΔH
-
TΔS
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The values for ΔS° from S° and ΔH° from ΔHf° can be derived from the Thermodynamic Data Table for both
reactants and products at 25 oC and 1 atm. ΔH and ΔS will vary with temperature but the variation is
frequently small. In the calculations, it will be assumed that there is no variation in ΔH or ΔS but only in T
and ΔG.!
How can we determine the temperature that will cause a reaction to occur?
PY
Determining temperature when a reaction can occur
C
O
Calcium oxide (CaO), also called quicklime is prepared by decomposing limestone (CaCO3) in a kiln at high
temperature and the reaction proceeds as:
CaCO3 (s) ⇋ CaO (s) + CO2 (g)
D
EP
Sample Problem 5:
ED
It is a reversible reaction, where CaO readily combines with CO2 to for CaCO3. To promote the formation of
CaO, CO2 is constantly removed from the kiln to shift the equilibrium from left to right.
The important determination for the decomposition process to occur is the temperature that will promote
the forward reaction. Here we can use the thermodynamics data for the standard states of the reactants and
products at 25 °C.
CaCO3 (s)
⇋
CaO (s)
+
CO2 (g)
S° J/K mol:
92.9
39.8
213.6
ΔHfo kJ/mol:
-1206.9
-635.6
-393.5
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First we apply the equation for the enthalpy change in the reaction
!
!
=
Σ nΔHf° (products)
=
[(1 ) ΔHfo CaO + (1) ΔHfo!CO2] – [(1) ΔHfo!CaCO3]!
!
=
[(1)( -635.6) + (1)( -33.59) ] – [(1)( -1206.9)]
ΔH°
=
177.8 kJ/mol
–
Σ mΔHf° (reactants)
PY
ΔH°
Then we apply the equation for the entropy change in the reaction
ΔS°
=!!
ΣnS° (products)
−
ΣmS° (reactants)
=
[(1 ) So CaO + (1) So CO2] – [(1) So CaCO3]
=
[(1)( 39.8) + (1)( 213.6)] – [(1)(92.9)]
=!
160.5 J/K mol
C
O
ΔS°
Calculating Gibbs Free Energy change from the data obtained above
ΔG
= !
ΔH
-
TΔS
=
177.8 kJ/mol – (298K) (160.5 J/K mol) (1kJ/1000J)
=
130.0 kJ/mol
ED
ΔG
D
EP
Because the ΔG° is a large positive value, we conclude that the reaction is not favored for the product
formation at 25 °C (298 K). In order to make the ΔG° negative, we need to find the temperature at
which ΔG° is zero (at equilibrium).
ΔG° = ΔH° - T ΔS° = 0
0 = ΔH° - T ΔS°!
!
!
!
!
!
!
!T
=
ΔH°
ΔS°
=
(177.8 kJ/mol)(1000 J/ 1kJ)
160.5 J/K·mol
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T
=
1108 K or 835 °C
For example, at 840 °C (1113 K):
ΔG
=
ΔH
–
=
177.8 kJ/mol – (1113 K) (160.5 J/K·mol) (1kJ/1000J)
=
0.8 kJ/mol
Sample Problem 6:
ED
Given: ΔH°= +177kJ and ΔS°= +285J/K for the reaction
TΔS
C
O
ΔG
PY
At a temperature higher than 835 °C, ΔG° is now negative, indicating that the reaction now favors the
formation of CaO and CO2.
NH4Cl(s) —> NH3 (g) + HCl (g)
D
EP
Find the ΔG° at 25!°C and ΔG at 500 °C.
At T= 25 °C, then T=298 K
ΔG
=
ΔH
=
177kJ –
ΔG at 298 K =
–
TΔS
298 (285J/K)( 1kJ/1000J)
+ 92kJ
* The reaction will NOT occur at 25 °C
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At T= 500 °C, then T=773 K
ΔG
ΔH
–
=
177kJ –
=
- 43kJ
TΔS
773(285J/K)( 1kJ/1000J)
PY
ΔG at 773 K
=
* The reaction will occur at 500 °C and NH4Cl will decompose to NH3 and HCl. At high enough
temperature, the entropy change will dominate the enthalpy change term.
ΔH
‒
T
(‒)
‒
(+)
ΔS
for the system
(+)
If the surrounding is
more random ΔH sur =+
ED
ΔG =
C
O
Sign considerations
D
EP
Sample Problem 7:
If the system is more
random ΔS sys = +
At what temp will a reaction occur? Given the following reaction and data:
N2(g)
Δ S° J/K mol:
ΔHfo kJ/mol:
+
O2 (g)
—>
2 NO (g)
192
205.0
211
0
0
90
Calculate enthalpy change
ΔH°
=
Σ nΔHf° (products)
=
[(2 ) Hfo NO] – [(1)Hfo N2 +(1) Hfo O2]
–
Σ mΔHf° (reactants)
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ΔH°
=
[(2)(+90)] – [((1) (0) + (1)(0)]
=
+180 kJ
Calculate entropy change
ΔS°
=
ΣnS° (products)
–
ΣmS° (reactants)
=
[(2 ) So NO] – [(1) So N2 +(1) So O2]
=
[(2) (211) ] – [(1) (192) + (1) (205.0)]
=
422 - 397
=
+ 25 J/K mol
C
O
ΔS°
PY
No heat given off only taken in.
When ΔG = ΔH - T ΔS and to reach equilibrium where ΔG = 0
ED
0 = +180 kJ – [ (T) (+ 25 J/K) (kJ/1000J) ]
Assuming no change in ΔH and ΔS, we solve for T
T
=
[ 180kJ ]
T
D
EP
[0.025kJ/K]
=
7200 K
So if T > 7200 K then ΔS dominates and reaction will occur. In a lightning bolt very high temperatures
are created and nitrogen oxide can be formed in the atmosphere.
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•
Remember ΔH and ΔS are different at
high temperature but we are assuming
no change for this calculation.
Sample Problem Application:
Find ΔH, ΔS, and ΔG. Is this reaction exothermic and is it spontaneous?
+
½ O2 (g)
—>
NO2 (g)
Given
NO (g)
½ O2 (g)
211
205.0
ΔHfo kJ/mol:
+ 90
0
For the reaction
NO (g)
+
Calculate entropy change
ΔS°
—>
240.5
+ 34
NO2 (g)
=
ΣnS° (products)
=
[ (1) (240.5) ] – [ (1) (211) + (1/2) (205)]
=
−
ΣmS° (reactants)
D
EP
ΔS°
½ O2 (g)
NO2 (g)
C
O
Δ S° J/K mol:
—>
ED
Solution:
+
PY
NO (g)
-73 J/K
Calculate enthalpy change
ΔH°
=
=
=
ΔH°
=
Σ nΔHf° (products) – Σ mΔHf° (reactants)
[(1) (34) ] – [ (1) 90 + (1/2) (0)]
[(1)(+34)] – [((½) (0) + (1)(0)]
- 56 kJ
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Calculate Gibbs Free Energy change from above
ΔG
=
ΔH
-–
TΔS
=
-56kJ - (298K) (-73J/K)( 1kJ/1000J)
=
-56kJ + 21.8
=
- 34.2kJ
PY
ΔG
Findings:
The reaction is exothermic at a ΔH = ( – )
It becomes more ordered (less gas) at a ΔS = (+)
C
O
It is spontaneous at 298 K at a ΔG = (–)
Phase Transitions
ED
How is entropy change determined in phase transitions?
At the temperature at which a phase transition occurs (melting or boiling point), the system is at
equilibrium (ΔG = 0).
=
ΔH
—
T ΔS
0
=
ΔH
—
T ΔS
ΔS
=
ΔH
D
EP
ΔG
T
Sample!Problem!8:!
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Consider the ice-water equilibrium:
ΔHfus = 6.01 kJ/mol = 6010 J/mol
T = MPH2O = 0 °C = 273 K
ΔSice —> water =
ΔH
=
PY
T
6010 J/mol
273 K
22.0 J/K·mol
C
O
ΔSice —> water =
* When 1 mole of ice melts at 0 °C, there is an increase in entropy of 22.0 J/K·mol.
ΔSwater —> ice
=
- ΔH
T
=
ED
For the water-ice transition, the decrease in entropy is given by
− 6010 J/mol
273 K
Practice Problem 5:
=
− 22.0 J/K·mol
D
EP
ΔSwater —> ice
HI has a normal boiling point of -35.4 C, and its Hvap is 21.16 kJ/mol. Calculate the molar entropy of
vaporization (Svap).
Expected Answer:
89.0 J/K·mol
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Free energy and Chemical Equilibrium: Relating ΔG° to the Equilibrium Constant
How is free energy change in non-standard states (ΔG) related to the standard free energy change (ΔG°)?
ΔG
ΔG°
+
RT ln Q
Here Q is the thermodynamic form of the reaction quotient.
C
O
•
=
PY
The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is
related to the standard free energy change, ΔG°, by the following equation.
ΔG represents an instantaneous change in free energy at some point in the reaction approaching
equilibrium. At equilibrium, ΔG=0 and the reaction quotient Q becomes the equilibrium constant K.
•
ED
How does the standard free energy change (ΔG°) relate to the equilibrium constant (K)?
At equilibrium, ΔG=0 and the reaction quotient Q becomes the equilibrium constant K.
=
ΔG°
D
EP
0
+
RT ln K
This result easily rearranges to give the basic equation relating the standard free-energy change to the
equilibrium constant.
ΔG°
=
−RT ln K
•
When K > 1 , the ln K is positive and ΔG° is negative.
•
When K < 1 , the ln K is negative and ΔG° is positive.
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Sample Problem 9:
Find the value for the equilibrium constant, K, at 25 °C (298 K) for the following reaction. The standard free
energy change, ΔG°, at 25 °C equals –13.6 kJ.
PY
2NH3 (g) + CO2 (g) ⇋ NH2CONH2 (aq) + H2O(l)
Rearrange the equation
=
−RT ln K
ln K
=
ΔG°! !
!
!
− RT
!
!
!
!
!
!
Substituting numerical values into the equation,
=
D
EP
!
ED
to give equation
C
O
ΔG°
13.6 x 103 J
− (8.31 J/(mol K)( 298 K
ln K
=
5.49
K
=
e5.49
K
=
2.42×102
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Practice Problem 6:
Determine the equilibrium constant K at 25C for the reaction N2(g) + 3H2(g) ⇋ 2NH3(g)
(ΔG° NH3(g) = -16.6 kJ/mol)
PY
Expected answer:
6.60 x 105
C
O
PRACTICE (15 MINS)
Iron can be extracted from haematite, Fe2O3, using either C or CO2 as the reducing agent.
Fe2O3 (s)
Fe (s)
C (s)
CO (g)
CO2 (g)
Fe2O3(s) + 3 C(s) → 2 Fe(s) + 3 CO(g)
ΔH = +492.7 kJ /mol
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
ΔH = - 24.8 kJ/mol
S° (J/ K·mol)
D
EP
Substance
ED
The reactions are shown below.
87.4
27.3
5.7
197.6
213.6
Use available data given to:
1. Calculate the minimum temperature at which reduction with carbon is feasible
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Teacher tip
•
If pressed for time, the additional
practice may be given as homework,
allowing extra working time for the
students.
•
Assign appropriate ratings to the
students’ answers. Partial points may be
given.
2. Prove by calculation that reduction with carbon monoxide is feasible at all temperatures.
Expected Answer:
ΔS
=
+542.9 J/K·mol
T
=
908 K
PY
For reduction with carbon (C):
For reduction with carbon monoxide (CO):
=
+15.2 J K/mol
T
=
–1630 K.
C
O
ΔS
Since the lowest temperature possible is 0 K, the reaction is feasible at all temperatures.
ED
ENRICHMENT (60 MINS)
Perform Experiment on Enthalpy, Free Energy and Entropy.
EVALUATION (10 MINS)
1. A spontaneous process
D
EP
Written Homework. Give an example for each of the following and provide explanation based from what
you learned about thermodynamics.
Teacher tip
•
Provide a format for the students to
follow, if necessary. The assignment can
be handwritten or encoded, depending
on the convenience of the students.
•
The student’s work will be rated whether
it covers the required elements; it
presents information accurately; used
information creatively; and is evidence
based.
•
A sample rubric is provided and should be
modified according to suitability to the
teacher’s criteria and weight of each
criterion.
2. A process that would violate the first law of thermodynamics
3. A process that would violate the second law of thermodynamics
4. An irreversible process
5. An equilibrium process
Cite your references properly using APA style.
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PY
C
O
ED
D
EP
(Rubrics obtained from https://teaching.berkeley.edu/sites/teaching.berkeley.edu/
files/Rubric%20for%20Evaluating%20Written%20Assignments%20.pdf
* Assign appropriate weight for the different criteria in the rubric.
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Chemistry 2
90 MINS
Chemical Thermodynamics:
Enthalpy, Free Energy, and
Entropy
LESSON OUTLINE
Communicating learning objectives
5
Motivation
Questioning engagement
5
PY
Introduction
C
O
Instruction
Hands-on activity
Content Standard
The learners demonstrate an understanding of spontaneous change, entropy, and free Practice
Post-lab discussion
energy.
Enrichment
Analysis of reactions
Performance Standards
The learners shall prepare a poster on a specific application of one of the following: acid- Evaluation
Assessment of lab reports
base equilibrium or electrochemistry
Materials
Collect and organize data needed to determine minimum entropy change required for a
Calorimeter (two styrofoam cups with a cap, one cup nestled inside
the other cup. The outer cup serves as insulation); thermometer;
reaction
D
EP
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
ED
Learning Competencies
Predict the spontaneity of a process based on entropy.
(STEM_GC11CT-IVa-b-140)
Use Gibbs’ free energy to determine the direction of a reaction.
(STEM_GC11CT-IVa-b-143)!
45
15
10
10
graduated cylinder; balance; solid samples (NaNO3, NH4Cl,
NH4NO3, NaOH, CaCl2, MgSO4, Na2CO3, Na2CO3 etc.); distilled
water; activity/worksheets; calculator
Resources
(1) Chang, R. (2007) Chemistry, 9th Ed. McGraw-Hill, Inc., USA.
(2) ChemLab – Chemistry 3/5 – Hot and Cold Reactions – Introduction.
(2016). Dartmouth.edu. Retrieved from https://www.dartmouth.edu/
~chemlab/chem3-5/calor2/full_text/intro.html
•
determine enthalpy change and temperature of a chemical reaction;
(3) Whitten, K.W., et al (2007) Chemistry, 8th Ed. Thomson-Brooks/Cole,
USA.
•
use standard Gibbs free energy of formation to determine change in free energy in a
reaction;
(4) Zumdahl, S.S. and Zumdahl, S.A. (2000) Chemistry, 5 ed., Houghton
Mifflin.
•
predict whether a reaction is spontaneous using Gibbs free energy change; and
•
use data from experiment and calculations to determine entropy change for a
spontaneous reaction.
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INTRODUCTION (5 MINS)
1. Communicate learning competencies and objectives to the learners using any of the suggested protocols
(Verbatim, Own words, Read-aloud).
a. Determine enthalpy change and temperature of a chemical reaction;
c. Predict when a reaction is spontaneous using Gibbs free energy change;
PY
b. Use standard Gibbs free energy of formation to determine change in free energy in a reaction;
d. Use data from experiment and calculations to determine entropy change for a spontaneous reaction
C
O
2. Present relevant vocabulary the students should know that will be used in the lesson.
Endothermic process
Exothermic process
Process that gives off heat to the surroundings
Enthalpy of reaction, ΔHrxn
ED
Process that absorbs heat from the surroundings
Spontaneous process
D
EP
The difference between the enthalpies of the products and the enthalpies of reactants
A physical or chemical change that occurs by itself. A process that takes place without energy from an external
source
Entropy, S
A thermodynamic quantity that is a measure of how spread out or dispersed the energy of a system is among
the different possible ways that system can contain energy
450
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Standard entropy, S°
The absolute entropy of a substance at 1 atm and 25 °C
Gibbs Free Energy, G
PY
Energy available to do useful work. Used to express the spontaneity of a reaction more directly
Standard Free Energy, ΔGrxn°
C
O
The free energy change for reaction when it occurs under standard state conditions, when reactants in their
standard states are converted into products in their standard states
State functions
Properties that can be expressed as (final – initial) states
ED
3. Connect the lesson with prerequisite knowledge
Calculate the mass needed to prepare 10.0 mL of 1.0 M of solution.
Teacher tip
Calculate MM: NH4NO3 (in g/mol)
N
=
2 x 14.01 = 28.02
H
=
4 x 1.008 = 4.03
O
=
3 x 16.00 = 48.00
80.05
For example: To prepare 10.0 mL of 1.0 M NH4NO3
Lsoln =
=
Msoln =
D
EP
MM NH4NO3 = 80.05 g/mol
(10.0 mL)(1L/1000 mL)
0.01 L
Track Units
1.0 M (mol/L)
gsolute
g solute = Msoln
= mol/L
=
Msoln
MM solute
Lsoln
=
1.0 M
80.05 g/mol
0.01 L
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g solute = g
MMsolute
Lsoln
g/mol
L
gsolute
=
0.8005 g
Calculate the number of moles of a substance n.
Number of moles n
=
gsolute
=
PY
MM solute
0.8005 g
Number of moles n
=
0.01 mol
C
O
80.05 g/mol
Determine heat change q of a substance with increase in temperature:
q = m S Δt
ED
q = C Δt
Determine the molar heat of solution ΔHsoln in a dissolution process.
qsoln
n
D
EP
ΔHsoln =
where qsoln = Csoln Δt = msoln ssoln ( Δt)
Calculate the change in Gibbs free energy, G° using the thermodynamic data for the solids used.
For example, for NH4NO3(s), dissolution reaction is:
NH4NO3(s)
G° kJ/mol
- 184.02
—>
NH4+ (aq)
+
NO3- (aq)
- 79.31
- 108.74
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Teacher tip
•
Thermodynamic Data Tables are available
in chemistry books and will be used as
sources of data.
•
Here is another source:
http://bilbo.chm.uri.edu/CHM112/
tables/thermtable.htm
ΔG°
=
Σm ΔGf° (products) − Σn ΔGf° (reactants)
=
[(1) Gf° NH4+ + (1) Gf° NO3-] – [(1) Gf° NH4NO3]
=
[ (1)( - 79.31) + (1) (- 108.74) ] - [(1) - 184.02) ]
=
(- 188.05) – (- 184.020)
=
- 4.03 kJ/mol
PY
ΔG°
ΔG
=
ΔH - TΔS
ΔS
=
ΔH −Δ G
Focus question:
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MOTIVATION (5 MINS)
ED
T
C
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Calculate the change in entropy of reaction S using the relationship:
1. What happens to the temperature of the solution for exothermic and endothermic reactions?
2. In this experiment, how can you tell whether a reaction is spontaneous or not?
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INSTRUCTION (45 MINS)
Teacher Tip:
•
If availability of materials and time
permit, learners can work on several solid
samples.
•
Two solid samples may be used
(endothermic and exothermic), with one
group working on one of the solids. Two
groups can share data.
A. Pre-lab Discussion
Check if learners are ready for the experiment. Make sure that they have read through the purpose,
procedure, the data table, and understand what needs to be recorded during the lab exercise.
PY
Remind learners of the proper handling of substances and apparatuses they will be using. It is best if
they use apron or lab gown, safety gloves, masks, and goggles. The solids to be used may come in powder
form and dangerous when inhaled.
C
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1. Tell the learners to work in groups of four members. One of them acts as the recorder of data. Assign
two groups to work cooperatively in sharing data. Each group works with a solid; one group with an
exothermic solid and the other with an endothermic solid.
2. Give each learner a data sheet for the results of the experiment.
3. Check the availability of the materials for the activity.
4. Make sure that learners record their data properly and accurately.
ED
5. Allow them to compare results with the results of the bigger group.
Procedure:
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B. Laboratory Proper: Learners perform the ENTHALPY, FREE ENERGY AND ENTROPY
1. Obtain an improvised calorimeter and thermometer. For your improvised calorimeter, get two styrofoam
cups. Let one cup nest inside the other cup. The outer cup serves as insulation.
2. Punch a hole on the lid for the thermometer.
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-
Exothermic substances:
NaOH, CaCl2, Na2CO3, NaC2H3O2
-
Endothermic substances:
NH4NO3, NH4Cl, MgSO4
Teacher Tip:
• A member of the group should be
assigned to record data.
ED
C
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PY
• Calculations should be a collaborative
work for all the members.
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Image obtained from https://www.dartmouth.edu/~chemlab/chem3-5/calor2/overview/
techniques.html
3. Weigh and record the mass of the calorimeter mcal.
4. Place 10.0 mL of distilled water in the calorimeter. Weigh again to get the mass of calorimeter and water
mcal+water. Subtract the mass of the calorimeter to determine the mass of the water mwater.
5. Measure the temperature of the water ti.
6. Calculate the mass of solid msolute needed to prepare 10.0 mL of a 1.00 M solution of the solid to be
used.
7. Weigh the sample, and record the mass in the data table.
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• Remind the learners of the need for
correct signs (+/-) of quantities.
8. Add the solid to the water, and place the lid on the calorimeter. Stir gently, and record the temperature
tf, when the entire solid has dissolved.
PY
9. Calculate the heat change involved in the reaction using the change in temperature Δt. The specific heat
of the aqueous solution is usually close to that of pure water (4.18 J/goC). Assume that the specific heats
of solution are the same as water (ssoln = 4.18 J/goC):
qrxn = (msoln) (ssoln )( Δt)
C
O
10. Calculate the H for the reaction using the heat of reaction qrxn, and the number of moles of the solid n
used.
Note: The heat capacity of the calorimeter will not be included in the calculation.
ΔHsoln
=
qrxn
ED
n
11. Calculate the change in Gibbs free energy ΔG°, using the thermodynamic data for the solids used.
ΔG° = Σm ΔGf° (products) − Σn ΔGf° (reactants)
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12. From the data collected and recorded, calculate the entropy change involved for the spontaneous
reaction.
13. Repeat the procedure two more times. Average the data of your trials.
14. Ask learners to show their solutions and calculations.
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Trial 1
Mass of calorimeter,
Trial 2
Trial 2
mcal
Mass of calorimeter + water,
mcal+water
Mass of water (mcal+water − mcal),
mwater
msolute
Mass of solution,
msoln
Moles of solute,
nsolute
Initial temperature,
ti
Final temperature,
tf
Δt
Heat change in the reaction,
qrxn
Heat of reaction (solution), J
ΔHsoln, J
ED
Temperature change (tf - ti ),
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Mass of solid,
PY
Solute used, formula
Gibbs free energy,
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ΔHsoln, kJ/
mole
ΔG°
Entropy change of reaction,
ΔS
Average Value for ΔS
•
Notes for the teacher:
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•
In this experiment, the heat capacity of the calorimeter is not included in the calculations. Only the heat
of solution is considered.
qsoln = Csoln Δt = msoln ssoln ( Δt)
One typically determines the heat capacity of the aqueous solution (Csoln) from the mass of the solution
(msoln), and the specific heat capacity of the solution (ssoln).
•
The mass of the solution is the sum of the masses of the water and solid substance originally placed in
the calorimeter.
•
The specific heat of the aqueous solution is usually close to that of pure water (4.18 J/goC). Assume that
the specific heats of solution are the same as water (ssoln = 4.18 J/goC):
qrxn = (msoln) (ssoln )( Δt)
Similarly one can report a specific heat of solution, which is the heat of a solution per gram of solute.
More commonly though, the molar heat of solution (ΔHsoln) or the heat of solution (qrxn) per mole of
solute (n), is reported.
ED
•
C
O
PY
•
qrxn
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ΔHsoln =
n
•
To calculate the change in Gibbs Free Energy, use the thermodynamic data for the solids used.
PRACTICE (15 MINS)
Teacher tip
•
1. Have the leaners complete the data table for their activity.
2. Have them compare the results between an endothermic and exothermic reaction.
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Learners submit their laboratory reports
with the questions answered and
calculations of data shown.
3. Ask the following questions:
a. Write a balanced equation for the reaction the group studied (including the heat of solution).
b. Was the reaction spontaneous? How do you know this?
c. From the temperature change of your trials, what must be the sign for H?
d. From question 3, what must be true about the sign for S? Why?
Many learners believe that a reaction must be exothermic to be spontaneous. Comment on this in
terms of this experiment.
C
O
f.
PY
e. What are the units for entropy, S?
ENRICHMENT (10 MINS)
Thermodynamic Analysis of Reactions
Steel is made by the high temperature reaction of iron oxide (Fe2O3) with coke (a form of carbon) to produce
metallic iron and CO2:
ED
Fe2O3 (s) + C (s) Fe (s) + CO2 (g)
This same reaction can NOT be done with alumina (Al2O3) and carbon to make metallic Al and CO2.
Why not?
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Al2O3 (s) + C (s) Al (s) + CO2 (g)
Use thermodynamic reasons to explain this. Further information from outside sources on the properties of the
substances involved.
Teacher tips
•
A laboratory report sheet should include
the necessary answers to data collection,
calculations, post-lab questions and
enrichment task.
•
Learners’ answers will be rated based on
the correct use of data and accuracy and
application of concepts.
EVALUATION (10 MINS)
Assessment of learners’ submitted laboratory reports.
The learners’ laboratory reports include the questions answered and calculations of data as shown in the
Enrichment task.
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Chemistry 2
120 MINS
Chemical Equilibrium
(STEM_GC11CE-IVb-e-146)
Explain the significance of the value of the equilibrium constant.
(STEM_GC11CEIVb-e-147)!
15
60
10
30
(2) Chang, R. Chemistry.2007. 9th ed. New York: McGraw-Hill Companies,
Inc.
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Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
5
(1) Brown, TL et. al. Chemistry The Central Science. 2009. 11th ed. New
Jersey: Pearson Prentice Hall
ED
Write expressions for the reaction quotient/equilibrium constants.
C
O
PY
LESSON OUTLINE
Content Standards
The learners demonstrate an understanding of chemical equilibrium and Le Chatelier’s Introduction Communicating learning objectives
Principle
Motivation
Conductivity test
Performance Standard
Instruction
Class discussion
The learners shall prepare prepare a poster in a specific application of either Acid-base
equilibrium or Electrochemistry. Include in the poster the concepts, principles, and
Enrichment
Reflection questioning / journal
chemical reactions involved, and diagrams of processes and other relevant materials.
Evaluation
Poster/Brochure making
Learning Competencies
Describe reversible reactions. (STEM_GC11CE-IVb-e-144)
Materials
Explain chemical equilibrium in terms of the reaction rates of the forward and the reverse Laptop/computer; projector/TV; Tarpapel
reaction. (STEM_GC11CE-IVb-e-145)
Resources
•
give examples of reversible and irreversible processes;
•
explain the equilibrium condition in terms of reaction rates of forward and backward
reactions and concentrations of reactants and products;
•
write the mass action expression for a given balanced chemical equation for
homogeneous and heterogeneous equilibria; and
•
predict the direction in which a reaction at equilibrium will shift given the values of
the reaction quotients and the equilibrium constant.
460
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INTRODUCTION (5 MINS)
Teacher Tip
•
The teacher may prepare several “blue
bottles” so that the students may do the
actual shaking of the bottle.
•
Remind the students to not shake too
much to avoid spillage of the solution.
1. Recall to the students the questions being addressed by the previously discussed aspects of chemical
reactions and that of the current topic:
Chemical thermodynamics answers the question “Why do some reactions proceed spontaneously while
some or non-spontaneous?”
•
Chemical kinetics answers the question “How fast a chemical reaction proceeds?”
•
Chemical equilibrium, on the other hand, basically answers the question “How far do reactions
proceed?”
PY
•
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2. Relay the learning competencies to the students.
MOTIVATION (15 MINS)
Two possible demonstration activities to illustrate reversible processes may be done:
ED
1. The “blue-bottle” experiment
A “blue bottle” is a rubber/cork-stoppered test-tube half-filled with a colorless liquid that turns blue
when shaken. Upon standing of the blue solution undisturbed, it becomes colorless again. The process
can be repeated several times.
•
Procedure for the preparation of the solution (to be done before the class):
•
Dissolve 1 g NaOH and 1.67 g glucose in 50 mL distilled water. To this solution, add a previously
prepared solution of 0.008 g methylene blue in 8 mL ethanol. The blue solution will turn colorless after
about a minute.
•
The stopper may be secured with a parafilm or masking tape.
•
The results are best observed when the solution is freshly prepared.
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•
2. The “camote-tops extract” indicator experiment
•
Camote tops extract changes color depending on the acidity or basicity of the solution. It turns purple or
red when it is acidic and yellow when basic. The idea of a reversible process may be illustrated by
461
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observing color changes upon adding an acid or a base to the camote tops extract.
Camote tops extract is prepared by boiling red camote tops in water. The resulting mixture is filtered and
the supernatant is cooled to room temperature before it can be used.
•
The demonstration is done by placing about 20 mL of the camote tops extract in a beaker. To this
solution, samples of acidic or basic solutions are added alternately to demonstrate reversible color
changes. You may use vinegar or calamansi extract for the acidic solutions. For the basic solutions, you
may use household bleach, “liquid sosa” or soap solution.
PY
•
C
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The teacher may wrap up the motivation part by correlating the observations from the demonstration
activity with the idea of reversibility. For the blue bottle experiment, the forward process occurs when the bottle
was shaken and the solution inside turns blue. After standing undisturbed, the solution turns colorless again
which indicates that the solution returned to its initial state, hence the occurrence of the reverse process. For the
camote tops extract indicator experiment, the reversibility of the color changes of the indicator if a base and an
acid are added alternately to the extract.
ED
The teacher may also cite examples of irreversible processes such as the burning of a matchstick. The
said reaction will just be proceeding to one direction as it is impossible to regenerate the same matchstick.
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INSTRUCTION (60 MINS)
A. The Concept of Equilibrium
The teacher may begin the delivery of the lesson by stating that many chemical reactions do not proceed
to just one direction or proceed essentially to completion. These are called reversible reactions.
What happens in a reversible reaction?
In reversible reactions, the reactants are not completely converted into products and some of the
products may be converted back into reactants.
462
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How do we distinguish an irreversible reaction from a reversible reaction in a chemical equation?
Teacher tip
Unlike in irreversible reactions where a single headed arrow is used (), reversible reactions use a double
headed arrow ()!to indicate that the forward and backward reactions are occurring simultaneously. In
general terms, a reversible reactions may be represented as follows:
•
Q: What can you say about the number
of cars inside San Francisco if the rate at
which cars enter the city is also the same
as the rate at which the cars leave the
city through the Golden Gate Bridge?
PY
aA + bB —> !cC + dD
where the lower case letters represent the stoichiometric coefficients of the reactants and products.
C
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What is chemical equilibrium?
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ED
To describe the state of equilibrium, consider Figure 1. The figure shows a time-lapsed photograph of
the traffic entering and leaving the City of San Francisco in California, USA.
Figure 1. The Golden Gate Bridge. (Image obtained from https://upload.wikimedia.org/
wikipedia/commons/thumb/d/da/
Golden_Gate_Bridge_at_Night_Long_Exposure_7105222661.jpg/800pxGolden_Gate_Bridge_at_Night_Long_Exposure_7105222661.jpg
463
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The teacher may first present Figure 1 to
the class. He/She then may ask the
students:
A: The number of cars inside the city is
constant.
•
The teacher then relates the Golden
Gate Bridge analogy to chemical
equilibrium. He/She presents Figure 2 to
the class and ask the students to
interpret the graph in terms of what
happens to the rate of the forward and
backward reactions as the reaction
progresses.
Assuming the rate (say number of cars per hour) at which cars enter the city is the same as the rate at
which the cars leave the city, then the two opposing processes are in balance. This also means that there
is a constant number of cars inside the city.
Figure 2.
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ED
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PY
A state of balance is also referred to as a state of equilibrium. In a reversible reaction, when the reactants
start to form the products, the products would then start to reform the reactants. The two opposing
processes happen at different rates but a certain point in the reaction will be reached where the rates of
the forward and backward reactions are the same (marked by the broken line in Figure 2). This is the
state of chemical equilibrium.
Changes in the rate of the forward and backward reactions in a reversible reaction.
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In a state of chemical equilibrium, since the rate of product formation is equal to the rate of the
reformation of the reactants, then the concentrations of the reactants and products remain becomes
constant (Figure 3).
Figure 3.
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ED
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PY
!
Changes in the amount of reactants and products in a reversible reaction.
The state of chemical equilibrium is a highly dynamic state. This means that though there are no change
in the composition of the reaction mixture and no visible changes taking place, the particles are
continuously reacting. Also, a system at chemical equilibrium can be easily disturbed by changes in the
reaction conditions.
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•
The teacher presents Figure 3 at this
point. He/She then asks the students to
interpret the graph in terms of what
happens to the amount of reactants and
products as the reaction progresses.
•
The teacher may complete the analogy
of the Golden Gate Bridge to the state of
chemical equilibrium by consolidating the
students’ answers with the explanation
provided on this guide.
B. The Law of Mass Action Expression/Equilibrium Constant Expression
•
The teacher should emphasize to the
students the importance of writing a
balanced chemical equation since the law
of mass action expression is very much
dependent to this.
•
An unbalanced chemical equation may
be given so that the students will first
have to check on it. This way, the
importance of having a balanced
chemical equation in writing equilibrium
constant expressions will be emphasized.
The relationship between the concentrations of the reactants and products may be expressed using the
law of mass action expression/equilibrium constant expression. For the general equilibrium reaction:
!
cC + dD
PY
aA + bB
the law of mass action expression is written as
=
[C]c[D]d
[A]a[B]b
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Keq
ED
where the [ ] is the concentration expressed in molarity and Keq is the equilibrium constant. If molar
concentrations are used, Keq may also be referred to as Kc The law of mass action is basically the ratio of
the concentrations of the products raised to their respective stoichiometric coefficients to that of the
reactants.
Example:
2NO2(g), the law of mass action expression is written as
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For the reaction: N2O4(g)
Kc = [NO]2
[N2O4]
Other examples:
Balanced Chemical Equation
1. 2 O3(g) !
!
Equilibrium Constant Expression
3 O2(g)
Kc = [O2]3
[O3]2
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2. 2 NO(g) + Cl2(g)
2 NOCl(g)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Kc = [NOCl]2
[NO] [Cl2]
2
3. HF(aq) + C2O42–(aq)
2 F–(aq) + H2C2O4(aq)
Kc = [F–]2[H2C2O4]
PY
[HF][C2O42–]
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Equilibrium constants for homogeneous gaseous equilibria may also be expressed in terms of partial
pressures. The expression is written in much the same way as described, only that the partial pressure is
raised to the coefficient instead of the molar concentration.
Example:
! 3 O2(g)
Kp = (P O2)3
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1. 2 O3(g) !
Equilibrium Constant Expression
ED
Balanced Chemical Equation
2 NOCl(g)
2. 2 NO(g) + Cl2(g) !!!!!!!!!!!!!!!!!!!!!!
(P O3)2
Kp = (P NOCl)2
(PNO)2 (PCl )
2
For equilibrium reactions where the reactants and products are in different phases (heterogeneous
equilibria), pure solids and pure liquids are excluded in writing equilibrium constant expressions. For
example, consider the reaction:
PbCl2(s)
!!!!!!!
Pb2+(aq) + 2 Cl–(aq)
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•
The equilibrium constant expression for the reaction is:
Kc = [Pb2+][Cl-]2
Why do we omit pure solids and pure liquids in the equilibrium constant expression?
PY
When the mass of a certain pure solid substance is doubled, its volume is also doubled. Therefore, when
the mass and volume is related to get the concentration, a constant value is obtained. Only reactants
and products whose concentration varies during a chemical reaction are included in the expression.
C
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Here are other examples:
Balanced Chemical Equation
CO2(g) + H2(g) !!
!
Equilibrium Constant Expression
CO(g) + H2O(l)
Kc =
[CO]
! Sn(s) + 2 CO2(g)
Kc =
[CO2]2
[CO]2
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SnO2(s) + 2 CO(g) !
ED
[CO2][H2]
C. The Equilibrium Constant, K
The equilibrium constant, K, is the numerical value that is obtained when equilibrium concentrations are
substituted to the equilibrium constant expression. The value of K may vary from very large to very small
values. This value provides an idea of the relative concentrations of the reactants and products in an
equilibrium mixture.
How can the value of the equilibrium constant be used to determine the relative composition of
the reaction mixture at equilibrium?
468
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Consider the reaction for the production of phosgene (COCl2), a toxic gas used in the manufacture of
certain polymers and insecticides, at 100°C:
CO(g) + Cl2(g) !!!!!!!!!!!!!!!! COCl2(g)
[CoCl2]
= 4.56 x 109
PY
Kc =
[CO][Cl2]
C
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Q: If a certain fraction equates to a very large value, then which has a larger value between the
numerator and the denominator?
A: The numerator should have the larger value.
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ED
The large value of 4.56 × 109 suggests that the concentration of COCl2 must be very large as compared
to the individual concentrations of CO and Cl2. The value suggests that at equilibrium, the reactions
mixture contains more of the product COCl2 than the reactants CO and Cl2. This is experimentally
verified. In other words, the equilibrium lies to the right or towards the product side based from the
chemical equation given.
Q: Gaseous hydrogen iodide is placed in a closed container at 425°C, where it partially decomposes to
H2(g) + I2(g). At equilibrium, it is found that [HI] = 3.53 × 10–3 M, [H2]
hydrogen and iodine: 2 HI(g)
= 4.79 × 10–4 M and [I2] =4.79 × 10–4 M. What is the value of Kc at this temperature?
A:
Kc =
[H2][I2]
[HI]
=
(4.79 x 10-4)(4.79 x 10-4)
= 0.0184
(3.53 x 10-3)2
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•
This brochure/poster making can be tied
up with the Arts class or Computer class
of your students.
•
The teacher should emphasize to the
students that when calculating the value
of the equilibrium constant, the
concentrations to be used must be
equilibrium concentrations.
•
Figure 4 may be presented to the class
upon discussion of the two examples so
that the students can have a visual
representation of the relationship of the
K value with the composition of the
equilibrium mixture.
In the second example, the value of Kc is small (< 1). For this to happen, the value of the denominator
must be larger compared to the numerator. In the case of the given reaction, the equilibrium
concentration of HI is higher than the equilibrium concentrations of the decomposition products. This
means that the equilibrium lies on the left or on the reactant side.
PY
In general,
If K >> 1 (large K value), the equilibrium lies to the right and the products predominate in the equilibrium
mixture.
C
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If K << 1 (small K value), the equilibrium lies to the left and the reactants predominate in the equilibrium
mixture.
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ED
This is summarized in Figure 4.
Figure 4. The relationship of K value and the composition of the equilibrium mixture.
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•
D. The Reaction Quotient, Q
•
Q: What if the concentrations of reactant and products given are not equilibrium concentrations?
PY
A: If the concentrations given are not equilibrium concentrations, we can calculate for the reaction
quotient.
What is a reaction quotient, Q
C
O
It is the value obtained when product and reactant concentrations or partial pressures at any point of the
reaction is plugged in the equilibrium constant expression. It is calculated in the same way as K. Thus for
a general equilibrium reaction
aA + bB !!!!!!!!!!!!!!!!!!!!!!!!!!!
ED
The reaction quotient can be expressed as:
cC + dD
Qc = [C]c[D]d
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[B]b[A]a
or if in terms of partial pressures,
Qc = (Pc)c(PD)d
(PA)a(PB)b
What is the significance of the reaction quotient?
The reaction quotient may be used to determine if a particular reaction is at equilibrium, and if not, in
which direction the reaction will proceed to attain the equilibrium.
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•
The teacher should emphasize that the
values that must be plugged in the
equilibrium constant expression should
be molar concentrations. The given
moles of the reaction components must
therefore be divided first with the volume
of the container in liters.
•
Let the students analyze what should
happen to the concentrations of the
products to the reactants in order for the
ratio to decrease from 0.500 to 0.150.
Consider the following example:
PY
The Kc value for the reaction N2(g) + 3
H2(g) 2 NH3(g) at 472°C is 0.105. Suppose a mixture of 2.00
mol of H2, 1.00 mol of N2 and 2.00 mol of NH3 is placed on a sealed 1.00-L container.
Q: How would we know if the reaction is already at equilibrium?
Qc
=
[NH3]2
=
[N2][H2]
C
O
A: We can calculate for the reaction quotient and compare it to the reported equilibrium constant value.
If the two values are equal, then the mixture is already in equilibrium.
(2.00)2
=
0.500
(1.00)(2.00)3
ED
Since the Qc ≠ Kc, then the given mixture is not in equilibrium.
In what direction will the reaction proceed in order to attain the equilibrium?
In general:
D
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To attain equilibrium, the quotient must decrease to 0.150. This will only happen if the concentration of
NH3 will decrease and the concentrations of N2 and H2 will increase. Thus, the reaction must proceed in
the backward direction until equilibrium is attained.
If Q = K , then the system is already at equilibrium;
If Q > K, the products dominate the reaction mixture so the products must react to form the reactants;
reaction proceeds in the backward direction until equilibrium is attained;
If Q < K, the reactants dominate the reaction mixture so the reactants must react to form the products;
reaction proceeds in the forward direction until equilibrium is attained.
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]
This is summarized in Figure 5.
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ED
C
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PY
•
Figure 5. Predicting the direction of equilibrium shift given the values of K and Q.
ENRICHMENT (20 MINS)
Ask the following to the students:
1. Think of at least two examples each of a reversible and an irreversible process that is evident in everyday
situations.
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2. Research for at least two important applications of chemical equilibrium. Briefly describe the significance of
chemical equilibrium in these applications.
•
Teacher Tip
•
For question no. 1, the students may
work in pairs. They may write their
answers to their notebooks.
•
Question no. 2 may be given as an
assignment individually or by group/pair.
They can write a short narrative about
this.
•
The teacher may select only a few
representative examples from the given
pool.
Possible answers:
✓ Equilibrium in various industrial processes particularly in manufacturing
PY
✓ Equilibrium in biochemical systems (eg. physiological buffers, oxygen transport through hemoglobin,
etc.)
EVALUATION (30 MINS)
N2O(g) + NO2(g)
2. CH4(g) + 2 H2S(g)
3. Ni(CO)4(g)
4. HF(aq)
CS2(g) + 4 H2(g)
Ni(s) + 4 CO(g)
H+(aq) + F–(aq)
5. 2 Ag(s) + Zn2+(aq)
2 Ag+(aq) + Zn(s)
6. 2 C2H4(g) + 2 H2O(g)
8. 4 HCl(aq) + O2(g)
2 C2H6(g) + O2(g)
CH4(g)
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7. C(s) + 2 H2(g)
ED
1. 3 NO(g)
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A. Write the equilibrium constant expression, Kc, for the following reactions. Indicate also if the equilibrium
is homogeneous or heterogeneous.
2 H2O(l) + 2 Cl2(g)
Answers
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•
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B. For the following reactions at equilibrium, identify which between the reactants and products is
dominant.
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C. At 450 °C, the Kp for the reaction N2(g) + 3 H2(g) 2NH3(g) is 4.5 × 10–5. For each mixture listed, indicate
whether the mixture is at equilibrium at 450 °C. If it is not at equilibrium, indicate the direction (toward
product or toward reactants) in which the mixture must shift to achieve equilibrium.
1. 98 atm NH3, 45 atm N2, 55 atm H2
2. 57 atm NH3, 143 atm N2, 79.6 atm H2
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3. 13 atm NH3, 172 atm N2, 82 atm H2
1 (NOT VISIBLE)
2 (NEEDS
IMPROVEMENT)
3 (MEETS
EXPECTATIONS)
475
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4 (EXCEEDS
EXPECTATIONS)
Chemistry 2
180 MINS
Chemical Equilibrium
Content Standard
The learners demonstrate an understanding of chemical equilibrium and Le Chatelier’s Introduction
Principle
LESSON OUTLINE
15
Instruction
Class discussion
105
Evaluation
Poster/Brochure making
60
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Communicating learning objectives and
Review
Performance Standard
The learners shall prepare prepare a poster in a specific application of either Acid-base
equilibrium or Electrochemistry. Include in the poster the concepts, principles, and
chemical reactions involved, and diagrams of processes and other relevant materials.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
perform calculations on the following:
(1) Brown, TL et. al. Chemistry The Central Science. 2009. 11th ed. New
Jersey: Pearson Prentice Hall
(2) Chang, R. Chemistry.2007. 9th ed. New York: McGraw-Hill Companies,
Inc.
a. determination of K when all equilibrium concentrations or pressures are known;
b. determination of K from initial and equilibrium concentrations or partial pressures;
and
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Resources
ED
Perform calculations involving equilibrium of gaseous reactions.
(STEM_GC11CE-IVb-e-152)!
Scientific calculator (for students)
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Learning Competencies
Calculate equilibrium constant and the pressure or concentration of reactants or products
in an equilibrium mixture. (STEM_GC11CE-IVb-e-148)
Materials
c. determination of equilibrium concentrations/partial pressures from initial
concentrations/partial pressures.
476
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INTRODUCTION (5 MINS)
The teacher may first recall how to write equilibrium constant expressions by giving some examples as
equilibrium calculations would basically require such knowledge. Examples are available in the abovementioned references.
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INSTRUCTION (105 MINS)
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Chemical equilibrium calculations usually involve calculating equilibrium constants given equilibrium or initial
concentrations or partial pressures or vice versa.
A. Calculating K when all equilibrium concentrations/partial pressures are known
Determining equilibrium constants when equilibrium concentrations or partial pressures are known involves
straightforward substitution to the equilibrium constant expression.
Example A.1:
ED
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 °C. The
equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2 and 0.166 atm
NH3. From these data, calculate the equilibrium constant Kp for the reaction:
Given:
Balanced chemical equation
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N2(g) + 3 H2(g)
2 NH3(g).
Equilibrium partial pressures: 7.38 atm H2, 2.46 atm N2 and 0.166 atm NH3
Strategy:
Using the balanced chemical equation, write the equilibrium constant expression, Kp, then substitute the given
equilibrium partial pressures to it.
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Solution:
Example A.2:
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An aqueous solution of acetic acid is found to have the following concentrations at 25 °C: [CH3COOH] =
1.65 × 10–2 M; [H3O+] = 5.44 × 10–4 M; and [CH3COO–] = 5.44 × 10–4 M. Calculate the equilibrium
constant Kc for the ionization of acetic acid at 25 °C. The reaction is:
Given:
Balanced chemical equation
H3O+(aq) + CH3COO–(aq)
ED
CH3COOH(aq) + H2O(l)
Equilibrium concentrations: [CH3COOH] = 1.65 × 10–2 M; [H3O+] = 5.44 × 10–4 M; and
Strategy:
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[CH3COO–] = 5.44 × 10–4 M
Using the balanced chemical equation, write the equilibrium constant expression, Kc, then substitute the given
equilibrium concentrations to it.
Solution:
478
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Other sample problems:
A3. Gaseous hydrogen iodide is placed in a closed container at 425 °C, where it partially decomposes to
hydrogen and iodine: 2 HI(g)
H2(g) + I2(g). At equilibrium it is found that [HI] = 3.53 × 10–3 M, [H2] = 4.79 ×
10–4 M and [I2] = 4.79 × 10–4 M. What is the value of Kc at this temperature?
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Answer: Kc = 0.0184
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A4. Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:
CH3OH(g). An equilibrium mixture in a 2.00-L vessel is found to contain 0.0406 mol
CO(g) + 2 H2(g)
CH3OH, 0.170 mol CO and 0.302 mol H2 at 500 K. Calculate Kc at this temperature.
Answer: Kc = 10.5
Example B1:
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B. Calculating K from initial and equilibrium concentrations/partial pressures
In most cases, what is known to the experimenter is the equilibrium constant at a certain temperature and the
initial concentrations or partial pressures of the species present. This means that equilibrium quantities must be
determined before calculating for K. This can be done by treating the change as a variable where the
stoichiometric coefficients from the balanced equation can be used to denote the relationship between the
changes in the concentration/partial pressure of the reactants and products.
A closed system initially containing 1.000 × 10–3 M H2 and 2.000 × 10–3 M I2 at 448 °C is allowed to reach
equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 × 10–3 M. Calculate Kc
at 448 °C for the reaction taking place which is
H2(g) + I2(g)
2 HI(g)
479
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Given:
balanced chemical equation
initial concentrations: 1.000 × 10–3 M H2 and 2.000 × 10–3 M I2
equilibrium concentration: 1.87 × 10–3 M HI
Strategy:
Solution:
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Let x = amount of H2 that changes to attain equilibrium
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Using the equilibrium table, tabulate the given initial and equilibrium concentration. Denote the changes in
concentration using a variable as guided by stoichiometric coefficients in the balanced chemical equation. Solve
for the unknown to calculate for the equilibrium concentrations of H2 and I2 and then solve for Kc.
Example B2:
Sulfur trioxide decomposes at high temperature in a sealed container according to the reaction
2 SO3(g)
2 SO2(g) + O2(g)
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Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3
partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.
Given:
Balanced chemical equation
PY
Initial partial pressure for SO3(g) = 0.500 atm
Equilibrium partial pressure for SO3(g) = 0.200 atm
Strategy:
Solution:
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Let x = amount of SO3 that changes to attain equilibrium
C
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Using the equilibrium table, tabulate the given initial and equilibrium partial pressures. Denote the changes in
partial pressure using a variable as guided by stoichiometric coefficients in the balanced chemical equation.
Solve for the unknown to calculate for the equilibrium partial pressures of SO2 and O2 and then solve for Kp.
481
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Other sample problems:
B3. A mixture of 0.100 mol of NO, 0.050 mol of H2 and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K.
N2(g) + 2 H2O(g). At equilibrium [NO] = 0.062
The following equilibrium is established: 2 NO(g) + 2 H2(g)
M. Calculate the equilibrium concentrations of H2, N2 and H2O and Kc.
PY
Answers: [H2]eq = 0.012 M, [N2]eq = 0.019 M, [H2O]eq = 0.138 M, Kc = 653.7
B4. A mixture of 1.374 g of H2 (MM = 2.016 g/mol) and 70.31 g of Br2 (MM = 70.9 g/mol) is heated in a 2.00-L
2 HBr(g). At equilibrium, the vessel is
vessel at 700 K. These substances react as follows: H2(g) + Br2(g)
found to contain 0.566 g of H2. Calculate the equilibrium concentrations of H2, Br2 and HBr and Kc.
C
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Answers: [H2]eq = 0.140 M, [Br2]eq = 0.295 M, [HBr]eq = 0.4016 M, Kc = 3.91
B3. A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel at
CO(g) + H2O(g).
226 °C. The following equilibrium is established: CO2(g) + H2(g)
a. Calculate the initial partial pressures of CO2, H2 and H2O.
b. At equilibrium, . Calculate the equilibrium partial pressures of CO2, H2 and CO.
ED
c. Calculate Kp for the reaction.
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Answers:
B4. A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25 °C, and the following equilibrium is
2NO2(g). After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.
achieved: N2O4(g)
a. What is the equilibrium partial pressure of N2O4?
b. Calculate the value of Kp for the reaction.
482
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Answer:
•
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B5. At 900 K, the following reaction has Kp = 0.345: 2 SO2(g) + O2(g)
2 SO3(g). In an equilibrium mixture,
the partial pressure of SO2 and O2 are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial
pressure of SO3 in the mixture?
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Answer:
Example C1
ED
C. Calculating equilibrium concentrations/partial pressures from initial concentrations/partial pressures and
K values
Given:
Balanced chemical equation
Kp = 7.0 (at 400 K);
Solution:
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For the equilibrium Br2(g) + Cl2(g)
2 BrCl(g), the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is
charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium, what is
the equilibrium partial pressure of BrCl?
Let x = amount of BrCl that changes to attain equilibrium
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C
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Example C2:
For the reaction I2(g) + Br2(g)
2 IBr(g), Kc = 280 at 150 °C. Suppose that 0.500 mol IBr in a 2.00-L flask is
allowed to reach equilibrium at 150 °C. What are the equilibrium concentrations of IBr, I2 and Br2?
Given:
Balanced chemical equation
Kc = 280 (at 150 °C)
[IBr]i =
0.500 mol = 0.250 M
2.00 L
484
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Solution:
Other sample problems:
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ED
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Let x = amount of BrCl that changes to attain equilibrium
C3. When 1.50 mol CO2 and 1.50 mol H2 are placed in a 3.00-L container at 395 °C, the following reaction
occurs: CO2(g) + H2(g) CO(g) + H2O(g). If Kc = 0.802, what are the concentrations of each substance in the
equilibrium mixture?
Answers: [CO2]eq = [H2] = 0.264 M ; [CO]eq = [H2O]eq = 0.236 M
485
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4. A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25 °C, and the following equilibrium is
2NO2(g). After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.
achieved: N2O4(g)
•
a. What is the equilibrium partial pressure of N2O4?
b. Calculate the value of Kp for the reaction.
PY
Answer:
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2 SO3(g). In an equilibrium mixture,
5. At 900 K, the following reaction has Kp = 0.345: 2 SO2(g) + O2(g)
the partial pressure of SO2 and O2 are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial
pressure of SO3 in the mixture?
Answer:
ED
N2(g) + O2(g) has a Kc value of 2400 at 2000 K. If 0.850 M each of N2 and O2 are
6. The reaction 2 NO(g)
initially present in a 3.00-L vessel, calculate the equilibrium concentrations of NO, N2, and O2.
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Answer:
1 (NOT VISIBLE)
2 (NEEDS
IMPROVEMENT)
3 (MEETS
EXPECTATIONS)
487
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4 (EXCEEDS
EXPECTATIONS)
PY
C4. Chloromethane, CH3Cl, which has been used as a refrigerant and a local anesthetic, can be made from the
following reaction: CH3OH(g) + HCl(g) CH3Cl(g) + H2O(g); Kp = 5.9 × 103. If enough methanol and hydrogen
monochloride are added to a container at 120 °C to yield an initial pressure of 0.75 atm for each, what will be
the equilibrium pressures of all the reactants and products?
EVALUATION (60 MINS)
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1. The equilibrium 2 NO(g) + Cl2(g)
2 NOCl(g) is established at 500 K. An equilibrium mixture of the three
gases has partial pressures of 0.095 atm, 0.171 atm and 0.28 atm for NO, Cl2 and NOCl, respectively.
Calculate the Kp for this reaction at 500 K.
Answer: Kp = 51
ED
2. Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g) + Cl2(g)
PCl5(g). A gas vessel is charged with a mixture of PCl3 and Cl2 which is allowed to equilibrate at 450 K.
At equilibrium, the partial pressures of the three gases are What is the value of Kp at this temperature?
Answer: Kp = 66.8
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3. A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel at
CO(g) + H2O(g).
226 °C. The following equilibrium is established: CO2(g) + H2(g)
a. Calculate the initial partial pressures of CO2, H2 and H2O.
b. At equilibrium, . Calculate the equilibrium partial pressures of CO2, H2 and CO.
c. Calculate Kp for the reaction.
Answers:
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•
Chemistry 2
180 MINS
Chemical Equilibrium: Le
Chatelier’s Principle
LESSON OUTLINE
Pre-lab discussion
15
Content Standard
Instruction
The learners demonstrate an understanding of chemical equilibrium and Le Chatelier’s
Enrichment
Principle.
Experiment proper
135
Question & answer
5
Evaluation
Complete the table
25
Materials
C
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Performance Standards
The learners prepare a poster in a specific application of one of the following:
PY
Introduction
a. Acid-base equilibrium
none
b. Electrochemistry
Resources
(1) Brown, TL et. al. Chemistry The Central Science. 2009. 11th ed. New
Jersey: Pearson Prentice Hall
Learning Competencies
State the Le Chatelier’s principle and apply it qualitatively to describe the effect of
changes in pressure, concentration and temperature on a system at equilibrium.
(STEM_GC11CE-IVb-e-149)
(3) Slowinski, EJ et al. Chemical Principles in the Laboratory. 1985. 4th ed.
Philadelphia: Saunders College Publishing
ED
Include in the poster the concepts, principles, and chemical reactions involved, and
diagrams of processes and other relevant materials.
(2) Chang, R. Chemistry.2007. 9th ed. New York: McGraw-Hill Companies,
Inc.
Describe the behavior of reversible reactions (STEM_GC11CE-!Vb-e-150)
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Describe the behavior of a reaction mixture when the following takes place:
•
change in concentration of reactants or products
•
change in temperature (STEM_GC11CE-IVb-e-151)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
predict the effects of changes in concentration, pressure and temperature on a
system in chemical equilibrium;
•
explain the effects of the abovementioned factors in terms of the Le Chatelier’s
principle; and
•
discuss some practical applications of the Le Chatelier’s principle.
488
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INTRODUCTION (15 MINS)
•
Teacher Tip
Recall to the students the characteristics of a system at chemical equilibrium:
➡ It is a state of “BALANCE” – the rate of product formation is equal to the rate of reactant reformation.
PY
➡ It is a “DYNAMIC SITUATION” – the forward and the reverse processes continue to take place even
though it appears to have stopped because there is no change in the relative concentrations of the
reactants and products.
➡ It is mathematically described by the LAW OF MASS ACTION.
However, systems at chemical equilibrium may be disturbed by the changes in various experimental
conditions. The effect of any change in these conditions to a system at chemical equilibrium can be
described by the LE CHATELIER’S PRINCIPLE.
•
Le Chatelier’s Principle states that if a stress (changes in reaction conditions) is applied to a system in
equilibrium, then the systems adjusts in order to reduce the effect of the stress applied.
•
The stress that may affect a system at chemical equilibrium include changes in the concentration of either
products or reactants, changes in temperature and changes in pressure for gaseous equilibria. In this
laboratory experiment, the effects of the changes in the first two factors will be investigated.
INSTRUCTION (135 MINS)
Experiment Proper (90 mins)
ED
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•
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A. Effect of the changes in concentration of reactants or products
1. Put 5 mL of 0.02 M KSCN and 5 mL of distilled water into a test tube. Add 2 to 4 drops of 0.02 M Fe(NO3)3
solution in to the test tube and gently shake. Note the observations. The color of the solution that was
formed is due to FeSCN2+ complex. The reaction involved is:
Fe3+(aq) + SCN– (aq) !!!!
FeSCN2+(aq)
2. Divide the prepared solution in to four test tubes labeled 1 to 4. Use the test tube no. 1 as the reference.
Add a small crystal of KSCN to test tube no. 2. Note your observations.
3. To test tube no. 3, add 1 drop of 0.02 M Fe(NO3)3. Note your observations.
4. To test tube no. 4, add 2 drops of 0.02 M NaH2PO4. Note your observations. The reaction involved in this
part is:
489
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At this point, the teacher will just give the
statement of the Le Chatelier’s principle and
enumerate the changes in reaction conditions
that may affect a system at chemical
equilibrium.
Fe3+(aq) + H2PO4–(aq)
[Fe(H2PO4)]2+(aq)
Expected results:
Addition of Fe(NO3)3 solution to KSCN solution produces a deep-red colored solution. The color is
due the formation of the FeSCN2+ complex.
•
Upon addition of KSCN crystal to the FeSCN2+ solution on test tube 2 and Fe(NO3)3 solution to test
tube 3, the deep-red color intensifies.
•
Upon addition of NaHPO4 solution to test tube 4, the deep-red color of the solution is lost
PY
•
[Co(H2O)6]2+(aq) + 4Cl–(aq) !!!
(pink)
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B. Effect of changes in temperature
1. Place about 30 mL of 10% CoCl2 solution into a beaker. Add dropwise concentrated HCl until the color of
the solution changes from pink to violet/lilac. Divide the resulting solution equally into three test tubes. Use
one test tube as the reference. The chemical reaction involved is:
[CoCl4]2+(aq) + 6H2O(aq)
blue)
ED
2. Heat one test tube using a hot water bath (not boiling) while swirl the other tube in an ice bath for 5 minutes.
Observe what happens.
Expected results:
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3. Transfer the tube from the hot water bath to the ice bath and vice versa. Stand for 5 minutes. Note your
observations
•
The CoCl2 solution is pink due to the complex formed by Co2+ with H2O which is [Co(H2O)6]2+. Upon
addition of HCl, the solution changes from pink to purple due to the gradual formation of the
[CoCl4]2+ complex. If excess acid was added, the solution turns dark blue. This can be corrected by
adding distilled water dropwise until the desired purple color is achieved.
•
Upon submerging the tube in a hot water bath, the solution should turn dark blue indicating the
formation of [CoCl4]2+.
•
Upon submerging the tube in an ice bath, the solution should turn pink, indicating that the formation
of [Co(H2O)6]2+ is favored.
490
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Post-Lab Discussion (45 MINS)
For each part of the experiment, the student should be able identify the reactions involved and explain the
effect of each factor as one which can affect the rate of reactions (either the forward or the backward reaction).
Teacher tip
A. Effect of changing the concentration of reactants or products
The net ionic equation involved is
Fe3+(aq) + SCN– (aq) !!!!
•
C
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Upon addition of KSCN crystal, the concentration of SCN– increases. There is an increase in the
frequency of collisions of SCN– with Fe3+ forming more FeSCN2+ as indicated by the increase in intensity
of the deep red color. The equilibrium shifts forward toward the direction of the formation of the
product.
The same thing is observed when Fe(NO3)3 is added. Since the concentration of Fe3+ is increased,
greater frequency of collisions with SCN– occurs forming more FeSCN2+ causing the increased intensity
of the deep red color. The equilibrium also shifts forward toward the direction of the formation of the
product.
ED
•
FeSCN2+(aq) (deep red color)
Addition of NaH2PO4 caused the loss of the deep red color of the solution due to the backward shift of
the equilibrium that consumes the FeSCN2+ complex to reform the reactants. The added H2PO4– reacts
with Fe3+ to form a colorless complex [Fe(H2PO4)]2+ according to the reaction:
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•
PY
•
Fe3+(aq) + H2PO4–(aq)
[Fe(H2PO4)]2+(aq) (colorless)
Since Fe3+ concentration was decreased, the equilibrium has to shift backward to replenish the lost Fe3+
in order to establish a new state of equilibrium.
•
Generalization: Increasing the concentration of a substance in an equilibrium mixture displaces the
equilibrium in the direction which consumes some of the added material. Conversely, decreasing the
491
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•
Go over the procedure of the experiment
before letting the students perform it.
•
It’s a must that the students and the
teacher wear the proper laboratory attire
(lab gown and safety goggle)
•
The experiment must be done inside a
fumehood. If there’s no fumehood, make
sure that the laboratory is well ventilated.
•
Concentrated HCl is corrosive. Students
should be strongly advised to handle it
carefully.
•
Wastes generated must be placed in a
waste bottle labeled inorganic wastes and
not on the sink.
concentration of a substance favors the reaction which produces it.
In the context of the Le Chatelier’s principle, the stress referred here is the change in concentration.
When the concentration of either a reactant or a product is increased, the equilibrium shifts into the
direction that would consume that added component. If the concentration is decreased, then the
equilibrium shifts into the direction that replenishes the lost component.
PY
•
B. Effect of changing the temperature
The net ionic equation involved is
[Co(H2O)6]2+(aq) + 4Cl–(aq) !!!
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•
[CoCl4]2+(aq) + 6H2O(aq)
(pink)
(blue)
Upon submerging the tube containing the solution into a hot water bath, the solutions turns blue and
pink when placed in an ice bath
•
From these observations, it can be concluded that the forward reaction is endothermic and is therefore
favored upon application of heat.
•
Generalization: When the temperature is increased, the reaction which consumes the applied heat is
favored i.e. the reaction which is endothermic. When the temperature is decreased, the reaction which
produces heat is favored i.e. the reaction which is exothermic.
•
Another way of interpreting the results is to treat heat as either a reactant or product. When heat is
added, equilibrium shifts to the reaction that consumes it. The same generalization on the effect of
changes in concentration is applied.
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ED
•
The following video clip may be used to summarize the discussions on the effects of changing the
concentration and temperature of a system at chemical equilibrium:
492
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Video URL: https://youtu.be/crr5ZMHCJ_Y
C. Discussion on the Effect of Changes in Pressure (or Volume) on Gaseous Equilibria
For gaseous equilibria, changes in pressure due to changes in the volume of the container affect
chemical equilibrium. When the pressure is increased (by decreasing the volume of the container), the
system adjusts by reducing the total pressure exerted by the gas particles present. This can be done by
shifting the equilibrium towards the formation of a lesser number of gas particles which may be
predicted using a balanced chemical equation.
•
For example, consider the gaseous equilibrium:
C
O
2 NO2(g) (brown)
PY
•
N2O4(g) (colorless)
D.
ED
When the volume of the container is decreased, the total pressure inside is increased and the
equilibrium has to shift to the direction that has a lesser number of gaseous particles in order to relieve
the pressure, i.e. towards the formation of N2O4. According to the balanced equation, there are two
moles of N2O on the reactant side and 1 mole of N2O4 on the product side.
When the volume of the container is increased, then the total pressure is decreased. When this happens,
the equilibrium shifts to the direction that would produce more gas particles, i.e. towards the formation
of N2O.
•
The total pressure of the gaseous reaction may also be increased by adding an inert gas which is not
involved in the equilibrium reaction. For example, addition of neon gas on an equilibrium mixture of N2O
and N2O4, will be able to change the total pressure but not the partial pressures of the gases that are
involved in the equilibrium process. This will therefore not affect the value of the equilibrium constant
and will not cause a shift of the equilibrium.
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•
Practical Application of the Le Chatelier’s Principle
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•
Before discussing this part, the teacher
may first recall the definition of pressure
as a measure of the frequency of collision
of gas particles against the walls of its
container. Also, it is worth mentioning
that the pressure may be altered by
changing the volume of a container
(decrease in volume increases the
pressure)
Practitioners of chemistry in the industry find Le Chatelier’s principle of great importance especially in the
manufacturing of many products. For a more efficient production process, they formulate ways by which
product yield can be maximized and waste generation can be minimized. This can be done by
examining the effects of changing the reaction conditions such as temperature, pressure and
concentrations of the substances involved on the yield of the process.
•
For example, when Fritz Haber developed the process of producing ammonia, NH3, from N2 and H2,
according to the reaction:
N2(g) + 3H2(g)
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•
2NH3(g)
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he tried to vary the pressure and temperature conditions to determine what conditions will maximize the
yield of ammonia (Figure 1)
Figure 1. Varying conditions of temperature and pressure in the production of ammonia. (Image obtained
from http://www.bbc.co.uk/staticarchive/df8a4f732521999a55ac8841453ed3e2bad4e156.gif)
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•
The teacher may first present the figure
and can then ask the student to interpret
the graph in terms of the effect of
varying pressure and temperature
conditions to the yield of ammonia.
From the plot, it can be concluded that the yield of ammonia can be maximized with increasing pressure
and decreasing temperature. Both conditions shift the equilibrium towards the direction of producing
more ammonia.
•
Another way by which the yield of ammonia can be maximized is to continuously remove the ammonia
product from the equilibrium mixture. This can be done by condensing the gas in to a liquid form and
withdrawing it from the reaction. In this way, the equilibrium will shift to the right favoring the formation
of more ammonia. (Figure 2)
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•
Figure 2. Diagram of a Haber process reactor. (Image obtained from Image URL: http://
images.flatworldknowledge.com/averillfwk/averillfwk-fig15_015.jpg)
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ENRICHMENT (5 MINS)
Teacher tip
This part should be given as an assignment.
The teacher may ask the students to look for other specific industrial processes that are made more efficient
using the Le Chetalier’s principle. Their research outputs should include the following:
1. Significance of the process
2. Schematic diagram of the process
PY
3. The equilibrium reaction involved in the process
4. Explanation on how Le Chatelier’s principle is applied to make the process more efficient.
EVALUATION (25 MINS)
Complete the following table.
2 HCl(g)
Fe3O4(s) + 4 H2(g) + heat
3 Fe(s) + 4 H2O(g)
2 NO(g) + O2(g)
2 NO2(g) + heat
H2CO3(aq)
+ H2O(l)
Stress applied
CO2(aq)
Effect
Equilibrium Shift
(encircle your answer)
(encircle your answer)
Increasing the temperature
(increase, decrease, no change) in No Shift
the volume of water vapor
collected
Addition of helium
(increase, decrease, no change) in No Shift
the number of moles of NO
Removal of CO2
(increase, decrease, no change) in No Shift
the amount of H2CO3
Decreasing the pressure
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H2(g) + Cl2(g)
Keq
expression
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Equilibrium reaction
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They may present their outputs through a powerpoint presentation or any other creative means.
(increase, decrease, no change) in Forward
the number of moles of HCl(g)
Backward
No Shift
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4 NH3(g) + 5 O2(g)4
NO(g) + 6 H2O(g) + heat
Increasing the temperature
(increase, decrease, no change)
in the amount of O2
Forward
Backward
Increasing the volume of the
container
Answers:
H2(g) + Cl2(g)
Keq
expression
2 HCl(g)
Decreasing the pressure
!
3 Fe(s) + 4 H2O(g)
2 NO(g) + O2(g)
2 NO2(g) + heat
H2CO3(aq)
+ H2O(l)
Increasing the temperature
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Fe3O4(s) + 4 H2(g) + heat
Stress applied
!
!
CO2(aq)
No Shift
(increase, decrease, no change)
in the amount of NH3
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Equilibrium reaction
Forward
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Removing O2
(increase, decrease, no change)
in the amount of H2O
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No Shift
Addition of helium
Backward
Forward
Backward
No Shift
Effect
Equilibrium Shift
(encircle your answer)
(encircle your answer)
(increase, decrease, no change) Forward
in the number of moles of HCl(g) Backward
No Shift
(increase, decrease, no change)
in the volume of water vapor
collected
Forward
(increase, decrease, no change)
in the number of moles of NO
Forward
Backward
No Shift
Backward
No Shift
Removal of CO2
(increase, decrease, no change)
in the amount of H2CO3
!
Forward
Backward
No Shift
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Increasing the temperature
!
(increase, decrease, no change)
in the amount of O2
Forward
Backward
No Shift
2 (NEEDS
IMPROVEMENT)
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1 (NOT VISIBLE)
(increase, decrease, no change)
in the amount of NH3
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Removing O2
(increase, decrease, no change)
in the amount of H2O
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Increasing the volume of the
container
3 (MEETS
EXPECTATIONS)
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4 NH3(g) + 5 O2(g)4
NO(g) + 6 H2O(g) + heat
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Forward
Backward
No Shift
Forward
Backward
No Shift
4 (EXCEEDS
EXPECTATIONS)
Chemistry 2
150 MINS
Acid -Base Equilibria and Salt
Equilibria: Bronsted Acids &
Bases and Acid-Base Property of
Water
Introduction
Communicating learning objectives
5
Motivation
Illustration
5
Instruction
Analogy, video, & class discussion
80
Content Standards
The learners demonstrate an understanding of
Practice
Guided exercises
20
Enrichment
Reflection questioning / journal
20
Poster/Brochure making
20
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LESSON OUTLINE
1. acid-base equilibrium and its applications to the pH of solutions and the use of buffer Evaluation
solutions; and
Materials
2. solubility equilibrium and its applications.
ED
Performance Standard
The learners shall prepare a poster on a specific application of one of the following: acidbase equilibrium or electrochemistry
Laptop/computer; projector/TV; Tarpaper
Resources
(1) Hill, John W., and Kolb, Doris K. (1998). Chemistry for Changing Times
(8th ed.).
(2) Snyder, Carl H. (1995). The Extraordinary Chemistry of Ordinary Things
(2nd ed.).
Learning Competencies
Define Bronsted acids and bases. (STEM_GC11ABIVf-g-153)
(3) Masterton, William L. and Hurley, Cecile N. Hurley. (2009). Chemistry
Principles and Reactions (6th ed.).
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*Include in the poster the concepts, principles, and chemical reactions involved, and
diagrams of processes and other relevant materials
Discuss the acid-base property of water. (STEM_GC11ABIVf-g-154)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
explain the acid-base property of water;
•
write the equilibrium constant expression for the auto-ionization of water; and
•
describe how Bronsted acids and bases act in a chemical reaction.
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Teacher Tip
•
Depending on the learners’ response,
•
Ask them to give more examples of
tools/objects that can be used for more
than one purpose. For example, a knife
can be used to cut things and also to
scale fish.
•
Some students may be accustomed in
using the term amphoteric. Give the
following explanation: The term
amphoteric is a general term for
substances that can react both as an acid
and a base. On the other hand,
amphiprotic (protic refers to hydrogen
ion) is a more specific term used to
describe a substance which can both
donate and accept hydrogen ions
(protons)..
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INTRODUCTION (5 MINS)
1. Communicate the learning objectives by presenting the essential questions below:
a. How can a substance, like water, show its amphiprotic nature?
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b. What information can be derived from the auto-ionization of water?
MOTIVATION (5 MINS)
1. Engage the learners by showing the illustration below. Ask them, “What information can you get from
this illustration?”
ED
Note: All amphiprotic substances are
also amphoteric, but not all amphoteric
substances are amphiprotic.
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2. A hammer can be used to put and also remove nails, depending on the need. Similarly, amphiprotic
substances can act both as an acid and as a base.!
If learners still have difficulty grasping
this concept, give more examples like the
case of aluminium oxide. It is amphoteric,
but not amphiprotic.
•
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Post the questions on the board to guide
learners throughout the lesson.
Teacher tip
•
Post the illustrations side by side on the
board.
•
Writing, analyzing and interpreting
chemical equations should be constantly
practiced in class.
•
Note: Protons do not exist as an
identifiable species in water. All protons
that might form in water bond firmly to
the electron pair of another water
molecule to produce the hydronium ion
(H+ = H3).
INSTRUCTION (80 MINS)
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Ask the learners to examine the equation below:
How can this equation be related to the hammer previously discussed?”
ED
Water, like the hammer, can serve two purposes: as a proton donor and as proton acceptor.
Add that water molecules exhibit their amphiprotic property even in trace amounts.
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Ask a learner to write the chemical reaction for the auto-ionisation of water on the board. The auto-ionization of
water molecules follows the reaction below:
H2O
H+
+
OH-
“From the equation, what are the products of the auto-ionization of water?”
“How to simplify this equation further?”
The products of the ionization of water molecules include a hydrogen ion and a hydronium ion. The reaction
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can also be written as:
Teacher tip
2H2O
H3O+
+
•
Give the learners time to analyze the
information before giving the correct
expression on the board.
•
Be alert on their possible difficulties or
misconceptions about chemical
equilibrium expression.
•
Understanding the acid-base property of
water will be applied and is very
important in the next topic: pH.
•
Supplementary videos can be used to
enrich this lesson: A) L29-5
Autoionization, Autodissociation of
amphoteric water - pH of pure water is 7
at 25°C (2:46mins) at https://
www.youtube.com/watch?
v=DpDewqtha8o, B) Autoionization in
Liquid Water (3:24mins) at https://
www.youtube.com/watch?
v=zeFSzt5x9uo.
•
The first video reiterates the key points
covered in the preceding discussion,
while the second video shows autoionization of water at the molecular level.
Video B can help your students visualize
the reaction better.
•
You may cut the lesson up to this point.
OH-
1. At this point, ask the learners to write the equilibrium constant expression (K reaction).
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2. Recall the following points in writing equilibrium constant expression:
a. Solutes enter as their molarity, [ ].
The equilibrium constant expression can be written as: K
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b. Terms for pure liquids need not appear in the expression. Its concentration is the same for all dilutes.
Experimentally, the hydronium ion and the hydroxide ion is present at almost exactly
0.0000001 molar at 25.
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“Why do you think these two ions are present in equal amounts?”
They must be equal since ionization of water molecule produces equal number of the two ions.
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3. Ask learners to compute the value of K At 250C, Kw.
(Answer Practice Problem 1 and 2)
4. The reaction discussed above can be regarded as a Bronsted acid-base reaction. To give an idea of
Bronsted acids and bases, watch the video The Bronsted and Lowry Definition of Acids and Bases (2
minutes) on this link https://www.youtube.com/ watch?v=-Yv6LOUK7_8.
Pre-viewing questions:
a. What are the limitations of Arrhenius definition of acids and bases?
b. What similarity does Arrhenius have with Bronsted-Lowry definition?!
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c. How does it improve Arrhenius’ definition?
•
If video viewing in the classroom is not
possible, give the link to the learners for
viewing. They can also use their gadgets
to view the video in the classroom,
provided there is internet connection.
Worst scenario, just give the questions
below in advance as an assignment.
•
If the video link is no longer active by the
time of using, the teacher can choose
other available videos which features
Bronsted definition of acids and bases,
and why it replaces Arrhenius’s definition.
The video replacement should not be
more than 5 minutes long. Remember
that properties of acids and bases were
already taken in Grade 7.
•
On the board, clearly mark the direction
of transfer of proton from the acid to the
base.
Post-viewing questions:
a. What is Bronsted and Lowry’s definition of acids and bases?
b. How does it impact chemistry?
To illustrate, examine the equation:
+
H2O —>
H3O+
+
NO3-
C
O
HNO3
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c. Based on Bronsted’s definition, explain the relation between H+, acids and base.
“Which reactant loses a proton? Which reactant gains a proton?”
ED
Nitric acid is the Bronsted acid because it loses one proton. Water is the Bronsted base because it
gains one proton
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5. Examine the resulting species in the equation. After an acid has lost its proton, the resulting species is
capable of acting as a base. The same applies to Bronsted base. The stronger the acid, the weaker the
conjugate base. The weaker the acid, the stronger the conjugate base.
Acids and bases in the Bronsted model therefore exist as conjugate pairs whose formulas are related by the
gain or loss of a hydrogen ion
“How to identify conjugate pairs?”
An easy way to identify the conjugate base is it differs from the acid by one proton.
(Answer Practice number 3 and 4)
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PRACTICE (20 MINS)
2. In the equation depicting the autoionization of water,
H2O
H+
OH-
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The reaction proceeds far to the _________. Explain.
+
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1. Given the amphiprotic nature of water, why is water neutral?
3. Explain the principle, STRONG ACID + STRONG BASE & WEAK ACID + WEAK BASE in terms of the
tendencies of substances to donate and accept proton.
ED
4. Box the conjugate base and circle the conjugate acid in the following equations:
+
H2O
—>
ClO2-
+
H3O+
OCl-
+
H2O
—>
HOCl
+
OH-
HCl-
+
H2PO4-
—>
Cl-
+
H3PO4
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HClO2
ENRICHMENT (20 MINS)
Ask learners to write a journal entry about the ideas below. They may also take this opportunity to raise
questions or clarifications they may have about the lesson.
1. “An acid cannot exist in the absence of a base.” Is this statement true or false? Explain.
2. “There is nothing in the Universe but alkali and acid, from which Nature composes all things.”-Otto
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Teacher Tip:
•
The teacher may require the learners to
write at least one question related to the
lesson which they want to be answered.
•
After going through the learners’ journal
entries, the teacher may select a few
entries for sharing in class. Depending
on the answers of learners, the teacher
might need to allot more minutes on this
part for the processing of their ideas.
Tachenius (1671)
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EVALUATION (20 MINS)
Performance Task
1. Make a poster or a brochure showing the importance of Bronsted acid-base definition and acid-base
property of water.
Chemistry
Organization
Creativity
3 (MEETS
EXPECTATIONS)
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Use the rubric below to evaluate learners’ output:
2 (NEEDS
IMPROVEMENT)
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1 (NOT VISIBLE)
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2. Draw your poster/brochure in a letter size bond paper. Specifically use crayons colouring material.
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4 (EXCEEDS
EXPECTATIONS)
Chemistry 2
150 MINS
Buffer Solutions
Content Standard
The learners demonstrate an understanding of Acid-base equilibrium and its application Introduction
to the pH of buffer solutions
Motivation
Performance Standard
Instruction
The learners shall prepare a poster on a specific application of pH.
10
Autobiography
10
Answering drills
30
Practice
Guided exericse
45
Enrichment
pH analysis of biological systems
25
Evaluation
Poster/brochure making
30
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Calculate pH from the concentration of hydrogen ion or hydroxide ions in aqueous
solutions (STEM_GC11AB-IVf-g-156).
Communicating learning objectives
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Learning Competency
Define pH (STEM_GC11AB-IVf-g-155).
LESSON OUTLINE
Materials
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Laptop/computer; projector/TV; Tarpaper
Resources
define pH;
•
relate pH and hydronium ion concentration;
•
explain the importance of determining pH of solution;
•
calculate pH and pOH in aqueous solutions; and
•
appreciate the importance of the knowledge of pH on different natural systems.
(1) Chemistry for Changing Times.8th Ed.John W. Hill and Doris K. Kolb.
1998
(2) The Extraordinary Chemistry of Ordinary Things.2nd Ed. Carl H.
Snyder. 1995
(3) Chemistry Principles and Reactions. 6th Ed.William L. Masterton and
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•
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INTRODUCTION (10 MINS)
Teacher Tip
Communicate the learning objectives:
•
Post the objectives on the board and
refer to them frequently during the
discussion. Put a check-mark beside
each objective to give the students an
idea of the class' progress.
•
Write on the board the molar
concentration of both H+ and OH-:
1. Define pH,
2. Relate pH and hydronium ion concentration,
4. Calculate pH and pOH of aqueous solutions, and
5. Appreciate the importance of the knowledge of pH on different natural systems.
1. “What are the products of the auto-ionization of water?”
If [H+] > 0.0000001 M < [OH-] 1x10-7
the solution is acidic
If [OH-] > 0.0000001 M <or [OH-]
1x10-7 the solution is basic
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Recall the key points discussed last meeting about the acid-base property of water:
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3. Explain the importance of determining pH of solution,
2. “Which chemical species is responsible for the acid property of water? Base property of water?”
3. “What is the molar concentration of these two species at 250C?”
ED
Ask the students, "Have anyone encountered in your readings a concentration expressed as 0.0000001 M or
1x10-7 M?"
This expression is very rarely heard. Ask a follow up question, "Why?"
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Some students might say that the number is very small. Or some might have observed that the concentration of
hydronium ion in solutions is simply stated in terms of “pH".
MOTIVATION (10 MINS)
•
To make it more catchy, you can call this
portion of your lesson as the "scientist of
the day", “who's who" or "the man
behind the science" segment. You can
call it anyway you like.
•
The presentation of the autobiography of
Soren Sorensen should only be brief and
not exceed the allotted time for this
section (10mins).
At this point, introduce how Soren Sorensen introduced the pH scale as a method of specifying the acidity of a
solution.
Visit the links http://www.humantouchofchemistry.com/soren-sorensen-introduces-the-ph-scale.htm and http://
www.chemheritage.org/discover/online-resources/chemistry-in-history/themes/electrochemistry/sorensen.aspx
for the biography and works of Soren Sorensen.
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Questions about the presented autobiography:
1. What moves Soren Sorensen to devise the pH scale?
2. What two methods of measuring acidity did he propose?
INSTRUCTION (30 MINS)
PY
As devised by Sorensen, pH stands for the “power of hydrogen.”
[H+]=[H3O+]= 10-pH
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pH=-log10[H+]=-log10[H3O+]
The pH of a solution is a measure of the solution’s acidity. Do the sample problem below.
Step 1. Press the (-) button on your calculator.
Step 3. Enter 0.0003.
The answer is 3.52.
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Step 2. Find and press the Log button.
ED
1. Calculate the pH of a solution with a hydrogen ion concentration of 0.0003 M.
Note: The +/- button on your calculator can be pressed either before the log button or after entering the
solution concentration.
2. What is the hydronium ion concentration of a solution which has a pH of 4?
Step 1. Look for the 10X function in your calculator. Normally, 10x is the
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Teacher tip
•
Ask your students to bring individual
scientific calculators. Students should be
familiar with the functions of their
calculators.
second function in the log key. Press shift + log.
Step 2. Find and press the (-) button.
Step 3. Enter 4.
•
PY
The answer is 0.0001.
With your class, complete a table showing the relationship between pH and the concentration of the
hydronium ions. Let your students compute for the pH and later decipher the trend/relation.
•
Column 1 of the table shows the concentration, Column 2 is for the pH.
Concentration, M
pH
0
1x10
1x10-3
1x10-4
1x10-5
1x10-6
1x10-7
1x10-8
1x10-9
1x10-10
1x10-11
1x10-12
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1x10-2
ED
1x10-1
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•
1x10-13
1x10-14
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Hint for Practice problem no. 4, the
students should compute first the
ammonia/ammonium ratio.
“From the completed table, what relationship of pH and hydronium ion concentration can be established?”
The following answers could be deduced from the students:
•
Help your students arrive at the
conclusions on the left. Give your
students time to keenly observe the
concentration and pH table.
•
Recall their activity during Grade 7
science on determining pH using natural
indicators. If they have missed a lab
activity on pH determination using
natural indicators, include this short
activity in the lesson:
•
Cut several (10-15) purplish camote
leaves into small pieces. Boil with
enough water until the solution is deep
purple. Let it cool.
•
Engage the students by observing the
camote top solution in a cup. Ask the
students to describe the characteristics
of the camote top solution.
•
Ask them to infer what will happen if you
add another liquid into it (dilute
ammonia). After they have given their
answers, ask a student to pour the dilute
ammonia into the deep purple camote
top solution until it turns changes color.
•
Give the students time to process their
observations. Then, ask another students
to blow the solution using a straw until it
turns back into deep purple color. Ask
another students to add another liquid
(vinegar) to change the colour of the
solution back.
•
Reveal the identity and the acidity of the
unknown liquids.
1. The base 10 and the negative sign were removed from the concentration expression to give the pH of the
solution.
PY
2. The pH goes down when the acidity goes up.
3. A decrease in pH by one unit represents a tenfold increase in acidity. A two unit decrease in pH, increases
the concentration by a factor of 100.
pOH=-log10[OH-]
ED
[OH-]= 10-pOH
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"For basic solutions, how can you make a similar scale for hydroxide ions?"
Because [H+][OH-]=1x10-14 at 25 0C, it follows that pH + pOH = 14.
“How can you use the pH scale in differentiating acidic, neutral and basic solutions?”
D
EP
If pH > 7, the solution is acidic
If pH = 7, the solution is neutral
If pH < 7, the solution is basic
•
pH can also be determined using pH meter and acid-base indicators. Common indicators such as
phenolphthalein, methyl red, and bromothymol blue are used to indicate pH ranges of about 8 to 10, 4.5 to
6, and 6 to 7.5, accordingly. On these ranges, phenolphthalein changes from colorless to pink, methyl red
from red to yellow, and bromothymol blue from yellow to blue. For universal indicators, however, the pH
range is much broader and the number of color changes is much greater. Usually, universal pH indicators are
in the paper strip form.
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PRACTICE/DRILL (45 MINS)
ENRICHMENT (25 MINS)
"Why is it important to take note of pH of biological systems?"
pH is important in everyday life. Show your class the pictures (or similar) below:
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ED
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(Images obtained from commons.wikimedia.org)
PY
See attached drill
Human Survival
Human body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH.
Large amount of acids are being released in the stomach every time you eat. It helps in the digestion of food
without harming the stomach. Tooth decay starts when the pH of the mouth is lower than 5.5. Tooth enamel,
made up of calcium phosphate, is corroded when the pH in the mouth is below 5.5.
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•
The drill can be done with a partner.
•
The students should be able to finish the
drill within the lab period.
•
Whenever students ask for clarifications
or hints about an item in the drill,
address the question or answer to the
whole the class. Try to create an
atmosphere of collaboration during the
answering of the drill.
PY
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Plants and Animal Survival
When pH of rain water is less than 5.6, it is called acid rain. When acid rain flows into the rivers, it lowers the pH
of the river water. Aquatic life cannot survive in acidic water.
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Acid rain has indirect effects on plants. It weakens the trees by damaging their leaves, limiting the nutrients
available to them, or poisoning them with toxic substances slowly released from the soil. Trees affected by acid
rain do not grow as quickly as usual, their leaves and needles turn brown and fall off when they should be green
and healthy.
Soil Acidity
Plants require a specific pH range for their healthy growth. Soil pH should be maintained at >5.5. Acidic soil
prevents or limits root development. Plants grown in acidic soil cannot absorb water and nutrients, are stunted,
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•
You may extend your discussion on the
following topics: a) pollutants which
cause acid rain, b) fertiliser and soil
acidity, c) how antacids work, and d) oral
hygiene and tooth decay.
•
Due to limited time, you may ask your
students to submit a reflection paper on
any one of the topics: a) how they can
participate in minimizing pollution
specifically the ones which cause acid
rain, b) how agriculturist mend soil
acidity, and c) the role of science in being
healthy.
and exhibit nutrient deficiency symptoms (especially those for phosphorus).
EVALUATION (30 MINS)
PY
Performance Task
Make a poster or a brochure showing the importance of pH scale. Draw your poster/brochure in a letter size
bond paper. Specifically use crayons colouring material.
1-3
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You may use the rubric below in evaluating student output:
4-6
7-8
9-10
Incorrect or little
concept
Some concepts are
accurate, included
some details
Organization
Unclear
Not organised
enough
Well oraganized,
clearly presented
Flows nicely
Simple design, but
not engaging
Attractive and
invites the reader
Outstanding visual
appeal
Little to no layout
or design
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Creativity
Substantial concept Many concepts with
and details
clear descriptions
and examples
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Chemistry
Content
Easy to follow even
without assistance
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Drill on pH Calculation
Name(s):__________________________________________________________ Date:______________________________________
Calculate pH from the following hydrogen ion concentration (M). Identify each as an acidic pH or a basic pH.
Hydrogen Ion Concentration
pH
1. 0.0015
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2. 5.0x10-9
3. 1.0
4. 1.0x10-12
-9
8. 5.63 × 10
-6
9. 3.67 × 10
-4
10. 3.25 × 10
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-8
7. 1.00 × 10
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5. 0.0001
-4
6. 1.00 × 10
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I.
II. Show complete solution for the following problems. Box your final answer.
1. If the H+ concentration is 0.00001 M, what is the OH- concentration?
2. If the H+ concentration is 0.00005 M, what is the pH?
3. If the pH is 4.5, what is the pOH?
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Basic/Acidic
4. What is the pH of a solution with a pOH of 1.36?
5. What is the pOH of a solution with a [OH−] of 2.97 X 10−10
6. An apple juice was tested to have a hydrogen ion concentration of 0.0003. Is the juice acidic or basic?
8. If a soil has a pH of 4.7, what is the H+ concentration of the soil solution?
PY
7. You test some ammonia and determine the hydrogen ion concentration to be 1.3 × 10–9. Compare its acidity with the apple juice (no. 6)
9. Hair stylists recommend slightly acidic and near neutral shampoo for smoother hair. You find 5 brands that you like. The first has a pH of 3.6, the
second of 13, the third of 8.2, the fourth of 6.8 and the fifth of 9.7. Which one should you buy? Explain why it is better.
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10. 0.00026 moles of acetic acid is added to 2.5 L of water. What is the pH of the solution?
III. Briefly answer the questions below.
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1. Arrange the following liquids in order of increasing acidity.
Wine, cow’s milk, tomato juice, sea water, beer, lemon juice, drinking water, vinegar
2. Based on the auto-ionization of water, draw a graph showing the relationship of [OH-] and [H+]. Explain the relationship.
3. 5.00 g NaOH and 5.00 g Ba(OH)2 were used to prepare 1L solution each. Which solution will be more basic? (MM NaOH=40, Ba(OH)2=171.3)
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4. 4.00 g KOH was added to the less basic solution in number 3. What will be the resulting pH of the solution? (MM KOH=56)
5. You decide to test the pH of your brand new swimming pool. The instruction manual advises to keep it between 7.2-7.6. Shockingly, you found out
that the pH of your pool is 8.3! What kind of chemical should you add?
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Answer Key
4. pOH=9.5
I.
pH
6. pOH=9.52
1. 2.8
Acidic
2. 8.3
Basic
8. (pH=8.8) Ammonia is basic.
3. 0
Acidic
9. [H+]=2.0 x 10-5
4. 12
Basic
5. 4
Acidic
10. The fourth brand that has a pH of 6.8. It is close to being neutral
(pH=7), but is below pH 7 making it slightly acidic.
6. 4.00
Acidic
7. 7.10
Basic
8. 8.25
Basic
9. 5.44
Acidic
PY
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11. Msoltn=1.04x10-4 mol/L, pH=3.99
III.
1. sea water, drinking water, cow’s milk, beer, tomato juice, wine, lemon
juice
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Acidic
7. (pH=3.52) The juice is acidic.
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10. 3.49
II.
5. pH=12.64
Basic/Acidic
3. NaOH solution is more basic. The [OH-] in the NaOH solution is
0.125 (pOH=0.90). The [OH-] in the Ba(OH)2 solution is 0.058
(pOH=0.12).
4. KOH contributes 0.0714 moles OH-. [OH-]=(0.058mol+0.0714mol)/
1L= 0.1294 M
2. pH= 5, pOH=9, [OH-]=1x10-9
3. pH=4.3
2. (x-axis [H+], y-axis [OH-]) The graph is similar to the V vs. P graph for
gases. It shows inverse relationship between the two species.
pOH=0.89, pH=13.11
5. Add an acidic solution to the pool.
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Chemistry 2
120 MINS
Strength of Acids and Bases
Content Standards
The learners demonstrate an understanding of the acid-base equilibria in solutions of Introduction
weak acids and weak bases.
Motivation
Performance Standard
Instruction
The learners shall prepare a a poster showing the relationship between acid/base
strength and ionisation constant.
Practice
3
Conductivity test
12
Class discussion
45
Guided exercises
20
Enrichment
Reflection questioning / journal
20
Evaluation
Poster/Brochure making
20
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Communicating learning objectives
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Learning Competencies
Determine the relative strength of an acid or a base from the value of the ionisation
constant of a weak acid or base. (STEM_GC11ABIVf-g-157)
LESSON OUTLINE
Determine the pH of a solution of weak acid or weak base. (STEM_GC11ABIVf-g-158)
Materials
Laptop/computer; projector/TV; Tarpapel
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Resources
explain the acid-base property of water;
•
write the equilibrium constant expression for the auto-ionization of water; and
•
describe how Bronsted acids and bases act in a chemical reaction.
(1) Chemistry for Changing Times.8th Ed.John W. Hill and Doris K. Kolb.
1998
(2) Chemistry. 4th Ed. John McMurry and Robert Fay. 2004
(3) Chemistry Principles and Reactions. 6th Ed.William L. Masterton and
Cecile N. Hurley. 2009
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•
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INTRODUCTION (3 MINS)
Teacher Tip
•
Elaborate on objective letter A. Ask
students their idea of a “weak” acid or
base. Some students might think that
weak acids/bases are safer or less
corrosive than stronger acids/bases.
•
Elaborate on objective letter D. Ask the
students the possible changes or
additions in the steps in pH
determination for weak acids and bases.
•
Post the questions on the board to guide
the students throughout the lesson.
•
You can ask help from your physics
laboratory instructor or technology and
livelihood education instructor if they
have available pre-constructed
conductivity tester (ei a circuit).
•
If the chemicals suggested in this guide
are not available in your laboratory, you
can try other combinations of 2 strong
acid and base and 2 weak acid and base.
•
Just like in any science demonstration,
try and practice the procedure before
doing it in class.
1. Communicate the learning objectives by presenting the essential questions below:
a. What is acid/base strength?
c. How do acid dissociation constant relate to acid/base strength?
MOTIVATION (12 MINS)
Do a quick demonstration of acid and base conductivity.
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d. How can you compute for the pH of a solution of weak acid/base?
PY
b. How can you determine acid/base strength?
ED
Prepare a conductivity tester, pure water and solutions of hydrochloric acid, acetic acid, sodium hydroxide and
ammonia.
It is expected that hydrochloric acid and sodium hydroxide will produce bright light in the tester. Water will not
light the build, while ammonia and acetic acid will only show dim to very dim light.
During the demonstration:
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Strong acids and bases conducts electricity due to large number of ions it contains. Substance which conducts
electricity weakly are weak acids and bases. They contain less ions and mostly molecules.
1. Ask the students what they expect will happen before dipping the electrodes in the tests materials.
2. Immediately after they observed the light in the tester, ask them if the substance is a strong or weak acid/
base.
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INSTRUCTION (45 MINS)
Teacher tip
•
Here are some points that might be
raised during the discussion:
a. Even for the strongest acids, there
will always be few HA molecules left
in the solution.
b. The process of proton transfer still
happens at equilibrium, however at
equal rates.
•
List of strong acids and strong bases are
provided in reference books and
textbooks. It will also be beneficial if
students can memorise the list, it is not
that long, anyway.
Ask the students to recall the definition of a Bronsted acid and base.
PY
Bronsted acid is a substance that loses a proton, H+, and a Bronsted base gains a proton.
From this definition, build the concept of a acid/base strength by asking similar questions:
What characterizes a strong acid/base?
•
What species should be formed from the strong acid/base?
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•
Acid strength refers to its ability to lose a proton. A strong acid is one that completely ionizes (dissociates) in a
solution yielding a mole of hydronium ion. A strong base always gets the proton.
As seen in the previous activity, strong acids/bases produce ions.
How does it happen?
ED
•
The proton is highlighted in red in the equation HA + H2O. After mixing, the proton can go to either the H2O
molecule or stick with A-. There are two possibilities for this reaction:
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1. If H2O is a strong base it will get the proton and the reaction will yield high concentration of H3O+
and A-.
2. If A- has a stronger attraction to the proton (low tendency to lose a proton), we say that it is a weak
acid, then the reaction will yield low concentration of H3O+ and A-.
At this point, try if your students can write the equation and predict the direction of equilibrium for the two
cases described above.
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•
Reiterate that the water being the
solvent is not shown in the expression.
Before proceeding further, ask your students to try writing the equilibrium constant for the reactions:
•
Scaffolding of concepts learned will help
students achieve higher competencies.
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PY
The acid/base strength can also be described quantitatively. The acid dissociation constant measures the
strength of an acid and the base dissociation constant for the strength of a base.
Also, ask the students to describe the reaction depicted by each equation.
For the first reaction, the acid HA dissociates to form hydronium ion and an anion. The acid donates it proton to
water.
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The equilibrium constants should be written as:
ED
For the second reaction, the base B accepts a proton to form the conjugate base and hydroxide ion.
Ask the students what the subscripts a and b mean in the expression above.
They will easily observe that a and b stands for acid and base equilibrium constant, respectively.
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Teacher tip
In what cases will the acid dissociation constant be grater that 1? Less than 1?
•
This important generalization should be
posted on the board and written in the
notes of the students.
•
Point out that a new set of steps should
be used for computing pH of weak acids
and weak bases. The change is due to
the incomplete dissociation weak acids
and weak bases at equilibrium in water.
•
•
Valuing
Although the computation is more
complicated than the pH computation for
a strong acid or base solution, it is all
worth it because most biochemically
important acids and bases are
considered weak.
Examining the equilibrium expressions, Ka will be greater than 1 if:
a) the reaction involves the dissociation of a strong acid,
c) the concentration of HA is low.
PY
b) the concentration of hydronium ion and anion is high, and
Otherwise, the value of Ka will be less than 1. For small Ka, the undissociated acid is favoured.
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In what cases will the base dissociation constant be grater that 1? Less than 1?
Examining the expression as we did with Ka, Kb will be greater than 1 if:
a)
the reaction involves a strong base,
c) the concentration of unionized B is low.
ED
b) the concentration of hydroxide ion and cation is high, and
Otherwise, the value of Kb will be less than 1. For small Ka, the unionized base is favoured.
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Answer Practice no. 1 for a quick understanding check.
Given Ka and [HA], equilibrium concentrations and pH can be easily computed. The problem statement we will
discuss is:
Determine the pH of a _____ M solution of _____ acid, Ka = _____.
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Follow the steps below:
Determine the concentration of the solution (if not given) and the Ka (from standard Ka list)
b)
Write the balanced dissociation equation for the acid dissolving in water.
c)
Write the equilibrium expression for the acid.
d)
Make an ICE (initial concentration, change in concentration and concentration at equilibrium) table.
e)
Using the Ka expression in C, solve for [H3O+]
f)
Calculate the pH of the solution.
PY
a)
Guide your students in solving the practice problem 2 and 4.
ED
PRACTICE (20 MINS)
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The steps enumerated above also applies to weak bases.
1. As acid strength increases, indicate if the following will be higher or lower:
a. [H3O+]
c. [HA]
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b. pH
Answer: a. Higher, b. Lower, c. Lower
2. What is the pH of a 0.50 M acetic acid? (Ka=1.8 x 10-5)
Solution:
a.
The concentration of the acetic acid solution is 0.50 M and the Ka of the acid is 1.8 x 10-5.
b.
The balanced equation is:
c.
The Ka expression is:
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•
The ICE table is an aid for students in
keeping track of the concentrations of
the different species involved in the
reaction. Construct the ICE based on the
balanced equation.
•
Recognizing the nature of the acid/base
as strong or weak is essential before
proceeding to the pH computation.
The ICE table for the reaction is:
•
e.
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PY
d.
Rearranging the Ka expression to solve for the concentration of H3O+ (H+):
1.8 x 10-5 = [x][x]
ED
0.5
x2 = 1.8 x 10-5(0.5)
x = 0.003
f.
Calculate the pH of the solution.
3.
Calculate the pH of 0.050 M HF.
Answer: pH=2.40
4.
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-log(0.003) = pH = 2.5
Calculate the pH of 6x10-3 M NH3 solution. (pKb=4.74)
Solution:
a)
Determine the dissociation constant of ammonia.
pKb=-logKb=4.74
Kb=10-4.74
Kb=1.8 x 10-5
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Notice that the expression 0.5-x was
estimated as solely 0.5. Since we are
dealing with weak acids/bases, we can
assume that the change in concentration,
x, is negligible compared to the initial
concentration of the acid/base.
Note: This assumption is valid only if x is less
than 5% of the initial concentration of the
acid/base. For Item no. 2,
0.003 x100% = 0.6%
0.5
•
To illustrate this, do Practice Problem no.
3. In this problem, concentration x is
more than 5% of the initial concentration
of the acid. Thus, the quadratic equation
should be solved without
approximations.
•
b)
The balanced equation is:
c)
The equilibrium expresion is:
3.29 x 10-4 x100% = 0.0658 (less than 5%)
0.5
The ICE table for the reaction is:
e)
Calculate x, [OH-].
ED
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d)
PY
•
1.8 x 10-5
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=
[x][x]
0.5
x2 = 1.8 x 10-5(6x10-3)
x = 3.29 x 10-4
f)
The value of x is negligible compared
with the initial concentration of ammonia.
Calculate the pH of the solution.
-log(3.29 x 10-4)
=
14-3.48 =
pOH = 3.48
10.52
=
pH
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Thus, [OH-] can be computed with
approximation.
ENRICHMENT (20 MINS)
1.
Teacher tip
Watch a video showing the dissociation of strong and weak acids at the molecular level. Follow the link
https://www.youtube.com/watch?v=rKqYE5sZi1s&app=desktop.
Do the terms "strong" and "weak" tell something about corrosiveness or being caustic? Explain by citing
hydrofluoric acid as an example.
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2.
PY
After viewing the video, ask the students to draw molecules of a strong acid, a weak acid, a strong base
and a weak base in an aqueous solution. Students should be able to explain their choice of acids,
illustrations. Assume 3 formula units or molecules for each substance. Their outputs may be drawn in
their journal.
Answer to Enrichment questions:
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ED
1. Strong acids and strong bases form more hydronium ions and hydroxide ions respectively. The molecular
representation of the students can be patterned to the picture below. The hydronium ion should be
replaced with an appropriate drawing showing an oxygen atom and a hydrogen atom (hydroxide ion).
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•
If the video is not available, they can still
do with the activity through research and
library work. The corrosiveness of HF is
shown in a portion in the movie Saw 6.
(If the copy of the movie is available, you
can show the class the portion where HF
is used, but not the whole movie.)
If you feel that your students can do
better with computer lay outing, you can
allow them to lay out the brochure/
poster using a computer. You best know
the capabilities and resources of your
students.
•
This brochure/poster making can be tied
up with the Arts class or Computer class
of your students.
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PY
2. The terms "strong" and "weak" do not relate to how corrosive or caustic a subtance is. The term “weak”
simply means that a particular acid/base doesn’t ionize completely in water. It doesn’t tell us the extent of
ionization, or concentration of the ions in the solution. For example, corrosiveness is a function of the properties
of that acid, as well as its concentration. We cannot conclude that a strong acid is more dangerous than a weak
acid. Hydroflouric acid is a weak acid, but it is extremely dangerous. As shown in the video and in other
references, hydroflouric acid is capable of eating through glass and flesh.
•
EVALUATION (20 MINS)
Performance Task
1. Make a poster or a brochure showing the different strengths of acids and bases. Draw your poster/brochure
in a letter size bond paper. Specifically use crayons colouring material.
Chemistry
Organization
Creativity
2 (NEEDS
IMPROVEMENT)
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1 (NOT VISIBLE)
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2. You may use the rubric below in evaluating student output:
3 (MEETS
EXPECTATIONS)
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4 (EXCEEDS
EXPECTATIONS)
Chemistry 2
90 MINS
Buffer Solutions: Henderson
Hasselbalch Equation
LESSON OUTLINE
Introduction
Communicating learning objectives
10
Guided practice
30
Problem solving
10
Enrichment
Poster making
30
Evaluation
Problem solving
10
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Performance Standard
The learners shall prepare a poster showing the usefulness of Henderson-Hasselbalch
equation in pH calculations
ED
Learning Competency
Calculate the pH of a buffer solution using the Henderson-Hasselbalch equation
(STEM_GC11AB-IVf-g-161)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
PY
Content Standards
Instruction
The learners demonstrate an understanding of Acid-base equilibrium and its application
Practice
to the pH of buffer solutions
identify an appropriate buffer system for a specific pH; and
•
calculate the pH of a buffer solution using the Henderson-Hasselbalch equation.
Metacards/Tarpapel
Resources
(1) Chemistry. 4th Ed. John McMurry and Robert Fay. 2004
(2) Chemistry Principles and Reactions. 6th Ed.William L. Masterton and
Cecile N. Hurley. 2009
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Materials
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Teacher Tip
•
Expected student’s answer:
1.1 Buffers are widely use in
maintaining pH during pharmaceutical
preparation and in food preparation.
1.2 Its large equilibrium constant
results to near complete consumption
of added OH- and H+ into buffer
solutions.
1.3 Buffered solutions are more
resistant to pH changes compared to
unbuffered solutions.
•
The equation on the Introduction
Delivery part is derived from the
dissociation of a weak acid (HB) in water.
INTRODUCTION (10 MINS)
PY
Recall the concepts learned from the previous lesson on the common ion effect and buffer solutions:
1. What is the importance of buffer systems learned in food industries and in medicine?
2. What is the implication of the large equilibrium constants for buffer systems with added small amounts of
acids/bases?
Present the learning objective by posing the questions below:
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3. How would you compare a buffered solution with an unbuffered solution?
1. How can you choose an appropriate buffer system for a desired pH?
2. How can you calculate the pH of a buffer solution?
INSTRUCTION (30 MINS)
ED
3. What is the Henderson-Hasselbalch equation?
[H+]=Ka x [HB]
[B-]
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Suppose a certain laboratory procedure requires a buffer solution at pH 4, 7 and 10, respectively. How should
you prepare the buffers? Examine the equation below to know the answer.
Taking the log to determine pH and multiplying by -1:
pH=pKa + log [B-]
[HB]
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HB(aq)
H+(aq) + B-(aq)
Ka=[H+][B-]
[HB]
If there is time, you can ask your
students to derive the equation.
This is the Henderson-Hasselbalch equation.
On what factors does [H+] depend on?
PY
Clearly, the pH of a buffer solution depends on Ka and on the ratio of the concentration of the weak acid and its
conjugate base.
Teacher tip
•
Expected student’s answer:
Choose lactic acid-lactate ion pair for
pH 4.
Choose dihydrogen phosphatephosphate ion pair for pH 7.
Choose hydrogen carbonatecarbonate ion pair for pH 10.
•
The students can easily remember the
buffer systems for each desired pH using
a mnemonics (for example LACDHAH).
•
From the equations, the students should
recognize the use of HendersonHasselbalch equation in determining pH
of buffer solutions and in determining the
amounts of weak acids and its conjugate
base to be used in preparing the buffer.
•
Your students might ask: “Is pH of buffer
solutions affected by volume of the
solution?”
ED
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When the amounts of the weak acid and its conjugate base is equal, pH will greatly depend on Ka. Therefore,
choose a conjugate weak acid-weak base pair in which the value of Ka of the weak acid is close to the desired
pH. Examine the table below.
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Which buffer system should you use if the desired pH is 4? 7? 10?
As a rule, the pKa of the weak acid should be within ±1 unit front eh desired pH.
No, pH of buffers depends only on
the pKa and the ratio of the weak
acid and its conjugate base.
At this point, answer practice no. 1.
How can we use the Henderson-Hasselbalch equation in preparing buffer solutions?
[H+]=Ka x [HB]
[B-]
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•
The equation is straight forward. Be
careful in committing a common mistake
of inverting the base/acid ratio.
pH=pKa + log [B-]
[HB]
PY
In buffer solutions, we assume that the concentration of both HB and B- is not changed at equilibrium. Also,
since the two species are present in the same solution, the ratio of their concentration is also their mole ratio.
The Henderson-Hasselbalch equation can be re-written as:
[H+]=Ka x nHB
nB-
Let your students answer Practice item no. 4, on their own.
PRACTICE (10 MINS)
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Do a guided practice for Practice items no. 2 and 3.
ED
1. What buffer system should you use if the desired pH is 6? desired pH is 9?
Answer: Carbonic acid-hydrogen carbonate ion pair and ammonium ion-ammonia pair, respectively.
2. How would your prepare a an ammonium-ammonia buffer solution at pH 9?
pH=pKa + log [NH3]
[NH4+]
log [NH3] = pH-pKa
[NH4+]
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Solution: First, derive an expression shoring the relative amounts of ammonium and ammonia.
[NH3] = antilog (pH-pKa)
[NH4+]
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•
Hint for Practice problem no. 4, the
students should compute first the
ammonia/ammonium ratio.
•
The following generalizations might help
students in analyzing the correctness of
their answers:
•
If the concentration of the weak acid and
its conjugate base is equal, pH is equal to
pKa.
•
If the concentration of the weak acid is
higher, pH will be less than the pKa.
•
If the concentration of the conjugate
base is higher, pH will be greater than
the pKa.
[NH3] = antilog (9-9.25)
[NH4+]
[NH3] = antilog (-0.25)
[NH4+]
=
10-0.25 = 0.56
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3. What is the pH of the solution containing 0.20 M NH3 and 0.15 M NH4Cl?
Answer: 9.37
pH=9.37 + log 0.20
0.15
pH=9.37
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Equation: pH=pKa + log [B-]
[HB]
ED
Solution:
Given: [NH3]=0.20 (conjugate base)
[NH4Cl]= 0.15 (weak acid)
pKa=9.25 (forNH4+, from the table)
PY
Your solution must contain 0.56 mole ammonia for every mole of ammonium ion. If the concentration of
the reactants is the same, then 100mL ammonium chloride must be mixed with 56 mL ammonia to achieve the
computed ratio.
•
Comments on Enrichment activity no. 2
Buffer solutions ca also be prepared
from the partial neutralization of a
weak acid or a weak base.
•
Hints for Evaluation:
a. Use the equation below for
Evaluation no. 1
[B-] = antilog (pH-pKa)
[HB]
b. Calculate first the moles of each
species. Don’t forget to convert mL to
L. Then substitute the values into the
Henderson-Hasselbalch equation.
c. Determine first the moles of acid.
Rearrange the equation
[H+]=Ka x nHB
nBto:
nB- =Ka x nHB
[H+]
4. Why is ammonium-ammonia pair a poor choice for a buffer at pH 7?
Since the pH of the buffer solution is
9, [H+] is equal to 1.0x10-9.
Answer: The computed ratio of ammonium-ammonia for pH 7 (0.0056) is very small and very far from
1. For example, given the same concentration of ammonium salt and ammonia, 100mL of ammonium salt
solution will contain only 0.56mL ammonia. Addition of small amount of acids or base will result to large change
The pH is lower than the pKa of
ammonium (9.25), therefore, the
concentration of the weak acid is
higher than its conjugate base.
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in pH.
ENRICHMENT (30 MINS)
1. Make a poster showing the concepts and principles of the Henderson-Hasselbalch equation. Draw the
poster in a bond paper.
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EVALUATION (10 MINS)
PY
2. Research on other ways of preparing buffer solution.
1. Calculate the ratio of [CH3COOH]/NaCH3COO] to give a buffer solution at pH 5.
Answer: 1.82
Answer:3.65
ED
2. What is the pH of the solution of 35 mL 0.20 M formic acid and 55mL 0.10 M sodium formate in 110 mL
water? pKa formic acid=3.75
3. How many gram of ammonia should be added to 500mL of 1M ammonium chloride to give a pH 9? Ka
ammonium=5.6x10-10
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Answer: 0.28 moles ammonia x 17g/mol ammonia = 4.76 g ammonia
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Chemistry 2
60 MINS
Solubility Product Constant
LESSON OUTLINE
Performance Standard
The learners shall prepare a poster on the solubility product constant in our life.
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
•
explain the meaning of solubility product constant;
• write the solubility product constant expression;
Motivation
Video Showing
12
Instruction
Lecture
20
Practice
Exercises
10
Enrichment
Research work
5
Evaluation
Poster/Brochure making
10
Materials
Tarpapel; computer/laptop; projector
Resources
(1) Chemistry. 4th Ed. John McMurry and Robert Fay. 2004.
apply the solubility product constant to predict the solubility of salts.
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•
3
ED
• compute for Ksp using the concentration of ions; and
Communicating learning objectives
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Learning Competencies
Explain and apply the solubility product constant to predict the solubility of salts
(STEM_GC11AB- IVf-g-164)
Introduction
PY
Content Standards
The learners demonstrate an understanding of acid-base equilibrium
533
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INTRODUCTION (3 MINS)
Teacher Tip
Post the learning objectives on the board. Read the objectives aloud, then ask the students to write the learning
objectives in their notebooks word-per-word.
b. Write the solubility product constant formula, and
c. Apply the solubility product constant to predict the solubility of salts.
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MOTIVATION (12 MINS)
PY
a. Explain the meaning of solubility product constant.
Watch a short animation showing ions in solution. See the link https://www.youtube.com/watch?v=EBfGcTAJF4o
Answering the following questions after the viewing:
2. What happens to the ions after dissociation?
3. What is solubility limit?
ED
1. How does chemical equilibrium apply to dissolution of salt in water?
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In a saturated solution, any added salt doesn't seem to be dissolved anymore. An equilibrium between the ion
and the salt is reached.
INSTRUCTION (20 MINS)
Solids dissociating into ions has limits to their solubilities. These sparingly soluble ionic compounds are
Let us now describe the equilibrium of a saturated solution of CaF.
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•
The lesson on Solution was presented
earlier in the semester. A brief recall will
help students connect their learnings.
•
If the link is not available, you can choose
any other video showing solution
formation at the molecular level. A
drawing can also be used to show the
separation of ions of salt.
Teacher tip
•
The equilibrium expression for solubility
product constant is straightforward. Give
your students time be familiar with the
format.
•
Expected student answer:
In writing the solubility product expression, follow the format below:
MmXx(s)
n+ m
PY
mMn+(aq) + xXy-(aq)
y- x
Ksp=[M ] [X ]
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For CaF2, the equilibrium constant expression for Ksp is: Ksp=[Ca2+][F-]2.
At this point, answer practice no. 1.
Ask the students to operationally define Ksp.
ED
Discuss the following with your students:
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What does Ksp say about the solubility of ionic compounds.
What information can we get from the solubility product
constant?
Consider the dissolution of stannous hydroxide: Sn(OH)2 Sn2+ + 2 OH¯ and its Ksp expression:
Ksp = [Sn2+] [OH¯]2.
The ratio of Sn2+ and OH¯ is one-to-two. For every Sn2+ dissolves, we get twice that amount of OH¯.
535
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Solubility product constant (Ksp) is the
product of the dissolved ion
concentrations raised to the power of
their stoichiometric coefficients.
•
When the concentration of ions is high,
the salt has high solubility, and Ksp is also
high.
•
Emphasize that Ksp values are based on
saturated solutions where equilibrium has
been reached. Unsaturated solutions
don't have Ksp values because they are
not at equilibrium.
•
The ratio of the ions in the Ksp
expression will be used in the
computation of solubility of ions.
Representing it mathematically, 'x' Sn2+ gives '2x' OH¯.
We can use Ksp to compute for solubility of ions, and vice versa.
At this point, answer practice no. 2 and 3.
PY
Ask your students to answer no. 4 and 5, on their own.
PRACTICE (10 MINS)
1. Write the equilibrium expression for Ksp of the following:
a. GaBr3
b. PbI2
e. Co(OH)2
f.
BaCO3
g. Mg(OH)2
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d. Ca3(PO4)2
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c. Al2O3
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After no. 4 and 5, proceed to practice no. 6 and 7. Students should be able to answer practice no. 8 and
9 on their own.
Hint for the answers: Write first the reaction describing the dissolution of the ionic compound. The subscript of
each atom in the formula will be exponent of the concentration of the ions. The charges of each atom can be
determined by criss-crossing the subscripts of each atom in the formula.
Example:
a. Ksp=[Ga3+][Br-]3
2. Calculate Ksp of saturated solution of CuCO3 containing [Cu+2]=[CO3-2]= 1.58x10-5
536
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Answer:
Teacher Tip
Step 1: Write the reaction describing the dissolution. CuCO3 (s)
—>
Cu
+2
(aq) + CO3-2 (aq)
•
Make sure to use balanced equation in
Step 1. The concentration unit derived
fromt the Ksp expression is molarity.
•
Depending on the available time, you
may give some of the practice items as
an assignment or quiz.
Step 2: Determine the concentration of each ion. [Cu+2]=[CO3-2]= 1.58x10-5
Step 3: Write solubity product constant expression. Ksp = [Cu+2][CO3-2]
Ksp= (1.58x10-5)2
Ksp = 2.5 x10-10
3. Compute the Ksp of Ag2S if a saturated solution contains [S-2] = 2.92 x10-17.
—>
2 Ag+ (aq) + S-2 (aq)
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Step 1: Write the reaction describing the dissolution. Ag2S (s)
PY
Step 4: Substitute the values.
Step 2: Determine the concentration of each ion. The ratio of Ag+ to S-2 is 2:1. Therefore,
[Ag+1] = 2(2.92 x10-17)
[S-2] = 2.92 x10-17
Step 4: Substitute the values.
Ksp= (5.84x10-17)2 (2.92x10-17)
Ksp = 9.96 x10-50
ED
Step 3: Write solubity product constant expression. Ksp =[Ag+1]2[S-2]
Answer: Ksp=1.1 x 10-10
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4. Solid magnesium fluoride dissolves and forms a saturated solution in water at 25oC. The concentration of
[Mg2+] = 3.0 x 10-4 M. Calculate the Ksp value.
Pb2+ (aq) + SO42- (aq). The measured Pb2+ in the saturated
5. Consider the reaction: PbSO4 (s) —>
solution is 2.0 x 10-5 M. Find the value of Ksp.
Answer: Ksp=4.0 x 10-10
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6. The Ksp of silver bromide is 3.2 10-13. Calculate the molar solubility of the ions present in the resulting
solution.
Ag+ (aq) + Br- (aq)
Step 1: Write the reaction describing the dissolution. AgBr (s) —>
PY
Step 2: Determine the ratio of the concentration of the ions in the right side. The ratio of Ag+ to Br- is
1:1. (x:x)
Step 3: Write the solubility product expression. Ksp=[Ag+][Br-]
•
In practice problem no. 7, take note that
the number 2 should appear twice in the
Ksp expression. One number 2 is from
the exponent in the Ksp expression and
the other number 2 from the ratio of the
two ions from the balanced chemical
equation.
•
The concentration of the ions can also be
computed from a given mass of salt and
volume of solution.
Step 4: Compute for the concentration (x) of each ion using the solubility product expression.
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3.2 x10-13=[x][x]=x2
√3.2 x10-13=√x2
5.66 x10-7 = x
ED
7. The Ksp of Sn(OH)2 is 5.45 x 10¯27. Calculate the solubility (in g/L) of the ions present in the resulting
solution.
Step 1: Write the reaction describing the dissolution. Sn(OH)2(s)
—>
Sn+2 (aq) + OH- (aq)
Step 2: Determine the ratio of the concentration of the ions in the right side. The ratio of Sn+2 to
OH- is 2:1. (2x:x)
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Step 3: Write the solubility product expression. Ksp=[Sn+2]2[OH-]
Step 4: Compute for the concentration (x) of each ion using the solubility product expression.
5.45 x 10¯27=[2x]2[x]=4x3
Solving for x, we get x=1.1 x 10-9 M
Step 5: Convert mol/L to g/L. The molar mass of the ionic compound is 152.71 g/mol.
1.1 x 10-9 M = 1.67 x 10-7 g/L
538
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8. Calculate the solubility of AlPO4. (Ksp = 9.8 x 10-21)
•
Answer: 9.9 x 10-11 M = 1.21x10-8 g/L
The topic in the enrichment activity can
be used in the evaluation part.
9. Calculate the solubility of CdCO3 in g/L. (Ksp = 1.0 x 10-12) Answer: 1.0 x 10-6 M
PY
ENRICHMENT (5 MINS)
Ask the students to research on how solubility equilibria works on the following situations.
1. Formation of salt lakes
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2. Formation of tooth decay
3. Formation of kidney stones
EVALUATION (10 MINS)
ED
Comment on enrichment activity:
The systems in the enrichment activity are affected by precipitation or dissolution of ionic compounds.
Make a brochure about the application of solubility product constant in biological and environmental systems.
4-6
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1-3
7-8
9-10
Chemistry
Content
Incorrect or little concept
Some concepts are accurate,
included some details
Substantial concept and
details
Many concepts with clear descriptions
and examples
Organization
Unclear
Not organised enough
Well oraganized, clearly
presented
Flows nicely to assist readers even
without assistance
Creativity
Little to no layout or
design
Simple design, but not
engaging
Attractive and invites the
reader
Outstanding visual appeal
539
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Chemistry 2
120 MINS
Buffer Solutions and Solutions
Equilibria
LESSON OUTLINE
Introduction
3
Content Standards
Motivation
Review of laboratory techniques
The learners demonstrate an understanding of acid-base equilibrium and its application
Instruction
Lab activity
to the pH of solutions and the use of buffer solutions
Enrichment
Buffer in different biological systems
Performance Standard
The learners shall perform a laboratory activity on preparing and testing buffer solutions
Evaluation
Laboratory report
Learning Competencies
Materials
Determine the pH of solutions of a weak acid at different concentrations and in the
See laboratory protocol for the list
presence of its salt (STEM_GC11AB-IVf-g-167)
Resources
Determine the behaviour of the pH of buffered solutions upon addition of a small
12
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
prepare buffer solutions;
•
measure pH of different buffer solutions;
•
determine the buffering capacity of different solutions; and
•
appreacite the importance of the different buffers in natural systems.
5
10
(2) Boundless. “The Common Ion Effect.” Boundless Chemistry. Boundless,
21 Jul. 2015. Retrieved 23 Oct. 2015 from https://www.boundless.com/
chemistry/textbooks/boundless-chemistry-textbook/acid-baseequilibria-16/homogeneous-versus-heterogeneous-solutionequilibria-116/the-common-ion-effect-473-3639/
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•
30
(1) Waterman, Thompson: Small-Scale Chemistry Lab Manual, AddisonWesley Publishing Company, 1993
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amount of acid and base (STEM_GC11AB-IVf-g-168)
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Communicating learning objectives
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INTRODUCTION (3 MINS)
In this laboratory activity, you should be able to:
1. Prepare buffer solutions,
2. Demonstrate the effect of concentration of the weak acid and the concentration of its salt on pH of buffer
solution, and
PY
3. Observe resistance of the buffer solutions when small amounts of acid or base is added,
Pre-laboratory Preparation
Safety Precaution:
Use goggles or safety glasses.
•
Use small scale pipettes or syringes in dispensing the liquids.
•
Read labels of reagent containers before dispensing.
•
Label your solutions.
Laboratory Proper
Materials:
•
Sodium hydroxide (NaOH)
•
Hydrochloric acid (HCl)
•
Solutions
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INSTRUCTION (20 MINS)
ED
•
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MOTIVATION (12 MINS)
1 M acetic acid (CH3COOH)
1 M sodium acetate (CH3COONa)
1 M phosphoric acid (H3PO4)
1 M sodium dihydrogen phosphate (NaH2PO4)
545
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Teacher Tip
•
You can use other buffer systems and
other concentrations, depending on what
is available in your laboratory.
•
The use of indicators will better aid visual
students by observing the colour change
accompanying the change in pH.
•
If you’re using a pH meter, you might
need to increase the amount of liquids to
cover the bulb of the pH meter. Adjust
the amount accordingly.
Clear cups
Teacher tip
•
Small-scale reaction surface
•
•
pH meter/paper
Label the cups properly. Sixteen
unlabelled cups will surely cause
confusion.
•
Universal indicator or natural indicators
•
Micro-pipettes/Syringes
•
For the buffer preparations Part A and B,
the resulting buffer solutions should total
to 16.
•
If your class has limited resources
(materials, time and space), you can
perform the buffer preparation and the
buffer capacity test for each pair of weak
acid/base and its salt simultaneously.
•
You can also divide the class into three
groups and assign each group a pair of
weak acid/base and its salt to work on.
•
The number of drops of acid/base added
before a pH change is a crude measure
of buffer capacity.
•
For the clean up part, use paper towels
in cleaning the reaction surface.
Thoroughly wash the cups with water.
Everyone should wash hands thoroughly
with soap and water.
PY
•
Procedure:
ED
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Part A: Buffer Solutions with Varying Concentrations of the Weak Acid
Buffer preparation
1. Place 4 cups in a clean surface. In each cup, add 10 drops of sodium acetate. Then, add 3 drops of water
in the first cup and 3 drops of acetic acid in the remaining 3 cups.
2. Measure and record the pH of each solution.
3. Repeat steps 1 and 2, twice. First repetition, use 7 drops of water in the first cup and acetic acid in the
remaining three. Second repetition, use use 12 drops of water in the first cup and acetic acid in the
remaining three.
4. Repeat steps 1 to 3, replacing acetic acid with phosphoric acid and the sodium acetate with sodium
dihydrogen phosphate.
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Part B: Buffer Solutions with Varying Concentrations of Salt
1. Place 4 cups in a clean surface. In each cup, add 10 drops of acetic acid. Then, add 10 drops of water in
the first cup and 3 drops of sodium acetate in the remaining 3 cups.
2. Measure and record the pH of each solution.
3. Repeat steps 1 and 2, twice. First repetition, use 7 drops of water in the first cup and acetic acid in the
remaining three. Second repetition, use use 12 drops of water in the first cup and acetic acid in the
remaining three.
4. Repeat steps 1 to 3, replacing acetic acid with phosphoric acid and the sodium acetate with sodium
dihydrogen phosphate.
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PY
Buffer capacity (For Part A and B)
1. For the buffer capacity test, use the last three cups in each buffer preparations.
2. Add a drop of the universal indicator in each cup.
3. In the first cup, count the number of drops of NaOH to a distinct color change.
4. In the second cup, count the number of drops of HCl to a distinct color change. The third cup will serve as
the control for colour comparison.
5. Do the same for the other set of buffer solutions.
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Note: If the indicator is not available, dip the pH meter electrode as you drop the acid/base to monitor pH
change until 1 unit. Make sure to clean the electrode after every test.
Data:
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Questions for Analysis
ED
Record your data in a similar table.
1. What is the effect of concentration of the weak acid on pH of the buffer solution?
2. What is the effect of concentration of the salt on pH of the buffer solution?
3. Explain your observation in buffer preparation Part B using the principle of the Common Ion Effect and Le
Chatelier’s principle.
4. What happens to pH of buffer solutions when small amounts of acid and base is added?
5. Which mixture in each set of buffer solutions has the highest buffer capacity?
6. From the activity, operationally define buffer capacity.
Answer to Analysis Questions
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1. The higher concentration of the weak acid decreases the pH of the buffer solution.
2. The higher concentration of the salt increases the pH of the buffer solution.
3. According to Le Chatelier's principle, altering the equilibrium by addition of a common ion shifts the
reaction to favor the deionized acid. Additional ion prevents the ionisation of the acid, placing the
equilibrium to the left. In the case of the buffers in the activity, the hydrogen ion concentration decreases,
and the resulting solution is less acidic (higher pH) than a solution containing the pure weak acid.
4. Very small to no change in pH is observed when small amounts of acid and base is added.
5. Buffer capacity is highest for buffer solution where the ratio of acetic acid:acetate ion and phosphoric
acid:dihydrogen phosphate ion is 1:1; that is, when pH = pKa.
6. Buffer capacity measures how effective a buffer is in resisting changes in pH.
ENRICHMENT (5 MINS)
Research on health conditions which result from sudden drop or increase in pH.
Answer:
ED
Health conditions related to pH and buffer are respiratory acidosis, metabolic acidosis,
respiratory alkalosis and metabolic alkalosis
Visit the website http://scifun.chem.wisc.edu/chemweek/biobuff/biobuffers.html for more information.
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EVALUATION (10 MINS)
Remind students to submit a laboratory report that contains relevant data and results to further understand and
comprehend how the experiment connects with the concepts learned in class.
548
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Chemistry 2
60 MINS
Electrochemistry: OxidationReduction Reactions
LESSON OUTLINE
Communicating learning objectives
5
Content Standard
Motivation
The learners learners demonstrate an understanding of redox reactions as applied to
Instruction
galvanic and electrolytic cells.
Enrichment
Performance Standard
The learners
Evaluation
Demonstration of a Redox Reaction
5
Discussion
30
Balancing using half-reaction method
10
Quiz on balancing redox reactions
10
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Introduction
•
prepare a poster on a specific application of electrochemistry,
•
include in the poster the concepts, principles and chemical reactions involved, and
diagrams of processes and other relevant materials.
Balance redox reactions using the change in oxidation number method.
(STEM_GC11AB-lvf-g-170)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
recognize redox reactions;
•
identify reducing and oxidizing agents in a given redox reaction;
•
balance redox equations using change in oxidation method;
•
write the oxidation and reduction half reactions for a given redox reaction;
•
Resources
(2) Chang, R. (2007). Chemistry (9th ed). New York: McGraw-Hill.
(3) Kotz, J.C,, Treichel, P.M. and Townsend, J.R. (2009). Chemistry and
Chemical Reactivity. (7th ed. Pp. 141-149). Canada: Thomson Brooks/
Cole.
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•
and
Periodic table
(1) Brown, T. L. et al. (2009). Chemistry: The Central Science (11th ed.).
Pearson Education Inc.
ED
Learning Competencies
Define oxidation-reduction reactions. (STEM_GC11AB-lvf-g-169)
Materials
balance a redox reaction using the half-reaction method.
549
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INTRODUCTION (5 MINS)
Teacher Tip
PY
1. Ask the students to name some types of reactions they have encountered in the course of their lessons.
Among these would be combustion reactions, neutralization reactions, double displacement reactions,
and precipitation reactions.
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2. Introduce oxidation-reduction reactions as a large group of reactions that would include many of the
reactions the students have worked on in previous lessons, but were not expressedly identified as
oxidation-reduction reactions. Examples of these are burning of fuels, which are actually combustion
reactions, rusting of metals, reactions for the manufacture of many important chemicals, and those
involved in the metabolic processes of organisms.
•
The students have been introduced to
different types of reactions in their lesson
on Stoichiometry in CHEMISTRY 1.
•
Classifications of reactions are not
mutually exclusive, and a reaction can
belong to two or more classes. However,
all chemical reactions can be divided into
two big groups: redox reactions and nonredox reactions.
•
You may ask students to read aloud the
objectives of the lesson.
3. Inform the students that oxidation-reduction reactions are given the shortcut name of “redox” reactions,
and they would meet this “nickname” often.
recognize redox reactions, and
b.
balance equations of redox reactions.
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a.
ED
4. Let the students know that in this lesson, they will learn how to
MOTIVATION (5 MINS)!
Teacher Tip
Carry out a demonstration of an oxidation-reduction reaction between aluminum and copper sulfate. Do the
final step in the set-up in front of the class before doing the introduction.
What you need:
•
An empty aluminum can of pop soda (Coke or Pepsi or other drinks), on which you have etched using a
knife point a thin circular line around the can about 2inches from the bottom. Make sure that the etched
line has exposed the aluminum metal and removed the wax and paint layer, but not too deep to cut
550
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While the demonstrated reaction appears to
be some kind of magic trick, many redox
reactions can be less and some even more
spectacular. All combustion reactions, and
many that are involved in explosions are
redox. On the other hand, rusting of iron,
another redox reaction, is so ordinary that it
does not attract too much attention.
through the metal.
•
A beaker containing a 0.5 M copper sulfate solution that would cover up to the etched line on the soda
can when this is immersed in the solution.
PY
Show the students the materials for the set-up and place the soda can into the solution of CuSO4. You may
have to place a small weight on the can to keep it from floating and ensure that the bottom part where the
etched line is located is in constant contact with the solution.
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You may now proceed with the introduction to the lesson and return to the set-up after the introduction.
What should happen in this demonstration?
The soda can should be cut in two, at the location of the etched line. The exposed aluminum would react
with Cu2+: Al(s) + CuSO4(aq) —> Al2(SO4)3(aq) + Cu(s)
Lecture/Discussion
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INSTRUCTION (30 MINS)
ED
This reaction is a redox reaction.
A. Recognizing Redox Reactions
1. What are redox reactions?
Teacher Tip
Historically, the term “oxidation” referred to reactions of substances with oxygen, while “reduction” involved
removal of oxygen. Many known redox reactions today do not involve reactions with oxygen.
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•
Each of the two half-processes cannot
happen independently without the other
half-process. Hence, when one reads
phrases like “the oxidation of iron”, there
is emphasis placed on what is happening
to the iron metal, but there is always an
accompanying reduction process.
PY
Oxidation-reduction reactions are those that involve a movement of electron or electrons from one
particle to another. Movement of electrons can be a complete transfer, such as in the formation of some ions,
or a partial transfer due to rearrangements in the formation of new covalent bonds. When electrons transfer,
there should be atoms that would give away electrons, and atoms that would accept the electrons. Redox
reactions are therefore made up of two half-processes that occur together: the losing of electrons or oxidation,
and the gaining of electrons or reduction.
Examples:
2Mg(s) + O2(g) —> 2MgO(s)
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Example 1: Consider the reaction between Mg metal and oxygen gas. The equation for the reaction is given
below.
ED
The product of the reaction of the two elements, Mg metal and O2 gas, is a white solid, magnesium oxide,
MgO. MgO is an ionic compound, and is made up of Mg2+ and O2- ions.
Ask students to describe how the ions Mg2+ and O2- are formed from neutral atoms.
The +2 charge means that the Mg atom lost two electrons.
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Where did the electrons lost by Mg go?
O2-, on the other hand, is formed when an O atom gains 2 electrons.
Where did the electrons gained by O come from?
In the reaction between Mg and O2, the electrons lost by Mg were gained by O.
Electrons from Mg transferred to O.
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•
To capture the attention of the students,
you may light a piece of Mg ribbon (do
this outdoors or in a well-ventilated area).
Alternatively, you may use a video clip or
at least a picture showing a burning Mg
ribbon.
A short video showing this
reaction is available in Youtube or at this
site:
http://www.sciencephoto.com/media/
234533/view
•
Ask the students to write the equation
for the reaction on the board. They
should have learned how to do this in
Chem 1.
•
The students should recognize that MgO
is an ionic compound (combination of a
metal and a nonmetal), and is made up
of ions, which are formed in accordance
with the Octet Rule. Now, you do not
stop at ion formation, but continue on to
ask where the electrons lost go or those
gained are coming from.
The burning of Mg is a reaction that involved a transfer of electrons between Mg and O. It is a
redox reaction.
When Mg lost electrons, it was OXIDIZED. Loss of electrons is OXIDATION.
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When O gained electrons, it was REDUCED. Gain of electrons is REDUCTION.
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Figure: Representation of Oxidation and Reduction. (Image source: http://
classes.midlandstech.edu/carterp/courses/bio225/chap05/Slide11.GIF)
Suggested Activity 1: Ask your students to make a similar description of what happens when sodium
metal, Na, and chlorine gas, Cl2, react to form table salt, NaCl. Which atom was oxidized? Which atom
was reduced?
Example 2. The reaction of hydrogen gas, H2, and fluorine gas F2 yields hydrogen fluoride, HF, a covalent
compound. No ions were formed, but the reaction is a redox reaction. Was there electron transfer?
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In H2, the electrons in each of the hydrogen atoms are evenly distributed between the atoms since the
two atoms attract the bonding electrons equally. That is also true in F2.
•
However, the H – F bond is a polar bond, with the partially negative end of the dipole at the fluorine end
of the bond. This is because F is very electronegative, and has the ability to pull the bonding electrons
towards itself, and away from H, as shown in the illustration below.
•
Ask the class to recall what makes a bond
polar. A covalent bond formed by atoms
of different electronegativities is a polar
bond. The electronegativity of H is 2.2,
while that of F is 4.0, the highest
electronegativity value among the
elements.
•
Since O is more electronegative than N,
the partially negative end of the N–O
bond is at oxygen. Hence, the electron
transfer must be from N to O: N is
oxidized while O is reduced.
PY
•
ED
C
O
Figure 2: Polarity in an HF molecule
(Image source:
www.chemwiki.ucdavis.edu)
In the H – F bond, the bonding electrons are found closer to F, making it appear that the electrons being
shared have moved closer to F than to H. It is as if H “lost” its electron and was “gained”, although just
partially, by F.
•
The formation of HF from H2 and F2 is a redox reaction.
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•
H was OXIDIZED; F was REDUCED.
Suggested Activity 2: Ask the students to describe the redox reaction in the formation of NO from N2
and O2. Which is oxidized? Which is reduced?
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2. Bookkeeping of electrons: Oxidation Numbers
A simple way of determining if a reaction is redox or not is by assigning oxidation numbers to the elements
involved in the reaction. The oxidation number of an atom may be its actual or apparent charge in the
substance.
Assigning oxidation numbers is a like a bookkeeping technique that allows
A more detailed list of rules to follow is
available in General Chemistry books.
For example, elements of the main block,
especially the metallic elements, form
ions consistent with the Octet Rule. In
their compounds, these elements will
have oxidation states equal to the charge
of their ions. For example, K is a Group
1 element and forms a +1 ion in its
compounds, such as in KCl, and K2SO4.
ON of each K in these compounds is +1.
•
For ionic compounds with a polyatomic
ion, it may be useful to the student to
separate the ions and determine the ON
based on the charge of the ion. For
example, in (NH4)2S, split the compound
into the ions. NH4+ and S2-.
Then
determine the ON of the atoms in the
ions. The net charge of NH4+ is +1, and
each H has ON of +1. Th ON of N is -3.
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a. counting of the number of electrons lost /gained by a reactant, and
•
b. identifying the atom oxidized and the atom reduced.
C
O
There are some easy rules to follow in assigning oxidation numbers (ON).
1. A metal or nonmetal in the free state, that is, occurring as an element is assigned an
oxidation number of 0.
E.g., all atoms in the following have ON of 0: Zn, H2, P4
2. A monoatomic ion has an oxidation number equal to its charge.
E.g.,
ON of Ca2+
-
is +2, ON of Br is -1.
ED
3. In its compounds, a hydrogen atom is usually assigned an ON of +1.
E.g., all H in H2O, HF, NH3, CH3COOH, H2SO4 are all +1.
4. In its compounds, an oxygen atom is assigned an ON of -2.
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E.g. all O atoms in the oxygen-containing examples given for previous rule is -2.
5. In all its compounds, F has an ON of -1.
E.g., HF, NF3, Cl2F2, ClF3.
6. The sum of the ON of all atoms in a polyatomic group is equal to the net charge of the
group.
E.g. In H2O, each H has an ON of +1 and O has -2. Total ON is 2(+1) + -2 = 0
In H2SO4, each H is +1, each O is -2, and the ON of S is 6.
0 = 2(+1) + x + 4(-2)
[ x = ON of S]
x = +6
In PO43-, the ON of P is calculated as follows:
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Net charge (or total ON) = -3
-3 = y + 4(-2)
•
The students can draw lines to connect
an element in the reactant and in the
product side to find out if the element
changed in oxidation number.
•
Alert the students that they should find
atoms that increase in ON and atoms
that decrease in ON. If they see only one
type of change – either an increase or a
decrease in ON, but not both, they
should go back and check the assigned
ON for each atom in the equation.
•
What if no atom changed in oxidation
number? Then the reaction may not be
redox!
•
It is important to remember that an
electron is negatively-charged. The loss
of an electron from an atom results in the
ON becoming more positive e.g. from +2
to +3, (or less negative, from -2 to 0). On
the other hand,
a gain of electrons
results in the oxidation number of an
atom to be more negative, e.g. from 0
to -3 (or less positive, e.g. +4 to +2).
[ y = ON of P]
y = +5
Suggested Activity 3: Using the above rules, assign oxidation numbers for all elements:
PY
1. HCOOH (Answer: H = +1; O = -2; C = +2 )
2. Ba(OH)2 ( Answer: H = +1; O = -2; Ba = +2 )
3. (NH4)2S ( Answer: S = -2; H = +1; N = -3)
C
O
4. Na2Cr2O7 (Answer: Na = +1, O = -2; Cr = +6 )
Suggested Activity 4: Assign oxidation numbers to all atoms in the following equation:
HNO3 + SO2 —> H2SO4 + NO2
+1 +5 -2
HNO3
+4 -2
+1 +6 -2
ED
( Answer:
+
SO2 —>
H2SO4
+
+4 -2
NO2 )
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3. Is the reaction redox?
Use the equation in Activity 4 above to show if a reaction is a redox reaction. Ask the students the following
questions about the equation:
a. Are there atoms that changed in oxidation numbers from the reactant side to the product side? Which
are these?
The elements that changed in ON are:
N from +5 to +4
S from +4 to +6
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The presence of elements that change in oxidation number in the equation is indication that
the reaction is indeed redox. However, there should always be one that will show an increase
in ON (or the ON becoming more positive) and one that will decrease in ON (or the ON
becoming less positive).
PY
b. For the elements that changed in ON, which lost electrons? Which gained electrons?
S was oxidized (change in ON from +4 to +6).
N was reduced (change in ON from +5 to +4).
C
O
( H and O did not change in ON, and were neither reduced or oxidized)
Are there reactions that are not redox? An example of a nonredox reaction is the neutralization re a c t i o n
between HCl and NaOH. No change in ON can be seen from reactant side to product side for all the elements
involved. Let your students assign ONs to the atoms in the equation to confirm that no atom changed in ON.
H2O + NaCl
ED
HCl + NaOH —>
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Suggested Activity 5: Assign oxidation numbers to each of the atoms in the equation and determine if the
reaction is redox:
1. Fe2O3 (l) + CO (g) —> Fe (l) + CO2 (g)
(redox; Fe and C changed ON)
2. Na2CO3 (aq) + 2HClO4 (aq) —> CO2 (g) + H2O (l) + 2NaClO4 (aq) (not redox)
3. Pb(NO3)2 (aq) + 2KI(aq) —> PbI2(s) + 2KNO3 (aq) (not redox)
4. 2S2O32- (aq) + I2 (aq) —> S4O62- (aq) + 2I- (aq)
(redox; S and I changed ON)
Even without having to go through the details above to show change in oxidation number and electron
transfer, some redox reactions are easy to recognize.
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•
In ionic equations, monoatomic ions
appear as if they are uncombined, but
they are unable to exist isolated
or
independently, unlike the elements in
their standard states.
If in the reaction, a reactant or product is an element (neutral independent form consisting of only one
element), and the same element is in combined or in ion form on the other side of the equation, the reaction is
definitely redox.
Here are some examples to illustrate this:
Al + H2SO4 —> Al2(SO4)3 + H2
(Al is uncombined or in element state in the reactant
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side.)
SO2 + O2 —> SO3
( Fe is obtained as the element in the product side.)
C
O
FeO + CO —> Fe + CO2
( Oxygen is in element form in the reactant side.)
B. The key reactants in a redox reaction
•
Some confusion arise here for many
students because there appears to be an
interchange of terms. Why will the one
oxidized be the reducing agent?
•
When we refer to soap as a cleaning
agent, it is not the soap that gets
cleaned, but it is the one that causes the
cleaning. In a similar manner, a reducing
agent is the substance that causes the
reduction of another substance. To
enable students to remember this,
introduce the mnemonic LEORA – Lose
Electrons – Oxidation – Reducing Agent
(the reactant that loses electrons
undergoes oxidation and is the reducing
agent). The students just have to
remember one of the half-reaction and
they would know the other half – Gain
Electrons – Reduction – Oxidizing Agent.
The main reactants in a redox reaction are the oxidizing agent and the reducing agent.
ED
In the example reaction of the burning of Mg, Mg lost electrons. We say Mg was oxidized. On the other hand,
O gained electrons and was reduced.
—> 2MgO(s)
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2Mg(s) + O2(g)
The reactant that carries the atom or element oxidized is called REDUCING AGENT (RA).The reactant that
carries the atom or element reduced is called OXIDIZING AGENT (OA). In this case, Mg is the reducing agent.
While it is said that O is reduced, it is O2, and not just O, that is the oxidizing agent.
The reducing agent and the oxidizing agent are reactants, not just the atoms oxidized or reduced. It is also
important to remember that RA and OA cannot be any of the products of the reaction.
Consider another example. the ON of each element has been placed above their respective symbols.
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+3 -2
+2 -2
0
+4 -2
C
Atom reduced:
Fe
Reducing agent:
CO
Oxidizing agent:
Fe2O3
C
O
Atom oxidized:
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Fe2O3 (l) + CO(g) —> Fe(l) + CO2 (g)
Suggested Activity 6: Identify the RA and OA in the following reactions:
1. SiO2(s) + C(s)
Si(s) + CO(g)
S4O62- (aq) + I- (aq)
(RA – S2O32-; OA – I2)
ED
2. S2O32- (aq) + I2 (aq)
(RA – C; OA – SiO2)
C. Balancing Redox Equations by the Change in Oxidation Number Method
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Redox equations are balanced based on the same conservation principles used in balancing any chemical
equation. In previous lessons, the students have balanced many redox equations without being told that those
were redox reactions. However, some redox reactions may not be easy to balance, and the additional
knowledge that the reaction involves electron transfer is a useful one. In redox reactions, reduction of an atom
cannot happen without another being oxidized. More importantly, the total number of electrons lost by the
reducing agent is equal to the number of electrons gained by the oxidizing agent.
Example 1: In the burning of magnesium,
0
0
Mg + O2 —>
•
+2 -2
MgO
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You can read the equation and the
processes described at the molar level.
Mg lost 2 electrons per Mg atom. Each atom of oxygen gains 2 electrons, and since oxygen occurs as
O2 molecules, each O2 molecule will gain a total of 4 electrons Therefore, 2 Mg atoms have to be oxidized to
reduce a molecule of O2. The total number of electrons transferred is four (4). The balanced equation for this
reaction is therefore
2Mg + O2 —>
2MgO.
+2 -2
0
+4 -2
Fe2O3 (l) +
CO (g) —> Fe (l) + CO2 (g)
C
O
+3 -2
PY
Example 2. Balance the following equation using the conservation of electrons transferred. The oxidation
numbers of the elements have been determined earlier in this lesson.
Oxidation: C +2 —> +4
2 electrons lost / C atom
2 electrons lost/CO
Reduction: Fe +3 —> 0
3 electrons gained / Fe atom
6 electrons gained/ Fe2O3
ED
To make the number of electrons lost equal to number of electrons gained, 3 CO must be oxidized to
reduce one Fe2O3. Place a coefficient of 3 for CO.
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1Fe2O3 (l) + 3CO (g) —> Fe (l) + CO2 (g)
The balancing process can now be finished.
Fe2O3 (l) + 3CO (g) —> 2Fe (l) + 3CO2 (g)
The equation is now balanced.
This method that we used leading to balancing of the equation is called the change in oxidation number
method.
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•
Suggested Activity 7: Balance the following redox equations using the change in oxidation number method:
1. SiO2(s) + C(s) —> Si(s) + CO(g)
2. S2O32- (aq) + I2 (aq) —> S4O62- (aq) + I- (aq)
PY
Example 3. You may wish to further impress the importance of conserving the number of electrons lost and
gained with the following equation:
Al + Cu2+ —> Al3+ + Cu
C
O
Ask the students if the equation is balanced. Some will say yes. It does look balanced, but it is not. Let
the students determine the number of electrons lost and gained, and balance the
equation using this
conservation principle.
ED
The balanced equation is
2Al + 3Cu2+ —> 2Al3+ + 3Cu
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ENRICHMENT (10 MINS)
Balancing Redox Equations by the Half-Reaction Method
Some redox equations are given in net ionic forms and at times, there are oxygen or hydrogen atoms on one
side of the equation but none on the other side. These are not easy to balance by inspection or by the change
in oxidation number method but can be balanced by the half-reaction method or the ion-electron method.
This method makes use of another conservation principle, that of balancing charges: the sum of the charges of
all substances on the reactant side should be equal to the sum of the charges of all substances on the product
side.
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•
•
In this method, there is no mention of
oxidation numbers. Neither is there
mention of atoms oxidized or reduced,
but we can identify the reduction and
oxidation processes.
The introduction of half-reactions here is
a good preparation for the next lesson,
which is on galvanic cells or batteries,
where electrode reactions are presented
as half-reactions.
Try this method in balancing the equation for the reaction between MnO2 and HCl. The reaction occurs in acid
condition. The equation to be balanced is given in a net ionic form.
MnO2(s) + Cl-(aq) —> Mn2+(aq) + Cl2(g)
PY
In this method, a redox reaction is seen as a pair of half reactions that occur simultaneously: the oxidation
and reduction half reactions. The half-reactions are balanced separately, and then added to each other to arrive
at the balanced equation.
Half-reaction 1:
MnO2(s) —> Mn2+(aq)
Half-reaction 2:
Cl-(aq) —> Cl2(g)
ED
For each half-reaction,
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2. Balance elements other than O and H.
Half-reaction 1
C
O
1. Split the equation into half-reactions. It is not necessary to assign oxidation numbers, nor to identify which is
the oxidation half-reaction or the reduction half-reaction at this point.
Half-reaction 2
MnO2(s) —> Mn2+(aq)
2Cl-(aq) —> Cl2(g)
(there is 1 Mn on both sides, so no change
made)
(a coefficient of 2 is placed for Cl- since there
are 2 Cl atoms in Cl2)
3. Balance the O atoms by adding the appropriate number of H2O molecules.
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•
•
MnO2(s) —> Mn2+(aq) + 2H2O
2Cl-(aq) —> Cl2(g)
PY
(add 2 molecules of H2O to the product side to (no change made since there are no O atoms
on either side of the equation)
balance the 2 O atoms in the reactant side)
4. Balance the H atoms by adding the appropriate number of H+ .
2Cl-(aq) —> Cl2(g)
C
O
4H+ + MnO2(s) —> Mn2+(aq) + 2H2O
(add 4 H+ to the reactant side to balance the H (no change made since there are no H atoms
on either side of the equation)
atoms at the product side)
2e- + 4H+ + MnO2(s) —> Mn2+(aq) + 2H2O
ED
5. Balance charges on both sides by adding electrons to the more positive side.
2Cl-(aq) —> Cl2(g) + 2e-
Sum of charges on reactant side: -2
Sum of charges on reactant side: +4
Sum of charges on product side: 0
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Sum of charges on product side: +2
(add 2 electrons to the reactant side to make
the sum of charges on this side equal to that
in product side)
(add 2e- to the product side to make charges
on this side equal to the reactant side)
The half reaction representing oxidation and that representing reduction can be identified at this point
by the position of the electrons used to balance charges. It is
In Half-reaction 1, the electrons are in the reactant side. This means electrons have to be added or
gained by the reactant for it to be transformed into the product. This half-reaction is the reduction half reaction
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(RHR), and MnO2 is the oxidizing agent.
•
PY
In Half-reaction 2, the electrons are in the product side. This means that for the reactant to be
transformed into the product, it has to give up or lose electrons. This half-reaction is the oxidation half-reaction
(OHR), and Cl- is the reducing agent.
6. Make the number of electrons lost equal to the number of electrons gained by multiplying the half reaction
with the appropriate factor.
Mn2+(aq) + 2H2O
2Cl-(aq)
The number of electrons gained and lost are equal.
Cl2(g) + 2e-
C
O
2e- + 4H+ + MnO2(s)
ED
7. Add the two half-reactions. Simplify the equation by removing appropriate numbers of substances that
appear on both sides. These would be electrons and probably H2O molecules.
The balanced equation is
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2e- + 4H+ + MnO2(s) + 2Cl-(aq) —> Mn2+(aq) + 2H2O + Cl2(g) + 2e-
4H+ + MnO2(s) + 2Cl-(aq) —> Mn2+(aq) + 2H2O + Cl2(g)
If the redox reaction occurs in basic or alkaline conditions, the half reactions are balanced as in
acid conditions but an additional step to convert to basic condition is done before the balanced
half-reactions are added to form the whole equation. An example is worked on below.
HS-(aq) + ClO3- (aq) —> S(s) + Cl-(aq)
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•
Step
Half-reaction 1
Half-reaction 2
ClO3- (aq) —> Cl-(aq)
2
HS-(aq) —> S(s)
ClO3- (aq) —> Cl-(aq)
3
HS-(aq) —> S(s)
ClO3- (aq) —> Cl-(aq) + 3H2O
4
HS-(aq) —> S(s) + H+
6H+ + ClO3- (aq) —> Cl-(aq) + 3H2O
5
HS-(aq) —> S(s) + H+ + 2e-
6e- + 6H+ + ClO3- (aq) —> Cl-(aq) + 3H2O
PY
1 HS-(aq) —> S(s)
3HS-(aq) —> 3S(s) + 3H+ + 6e-
C
O
6 3 [HS-(aq) —> S(s) + H+ + 2e-
6e- + 6H+ + ClO3- (aq) —> Cl-(aq) + 3H2O
7 3OH- + 3HS-(aq) —> 3S(s) + 3H+ + 6e- +
6OH- + 6e- + 6H+ + ClO3-(aq) —> Cl-(aq) + 3H2O +
6OH-
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3OH-
ED
Step 7: For each half-reaction, add as many OH- as there are H+ on both sides of the half-reactions
Step 8: Combine H+ and OH- to form water, H2O. Simplify the half-reactions by cancelling similar
substances.
8 3OH- + 3HS-(aq) —> 3S(s) + 3H2O + 6e-
6e- + 6H2O + ClO3- (aq) —> Cl-(aq) + 3H2O + 6OH-
6e- + 3H2O + ClO3- (aq) —> Cl-(aq) + 6OH-
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Step 9. Add the half-reactions and simplify the equation if necessary.
•
3OH- + 3HS-(aq) —> 3S(s) + 3H2O + 6e6e- + 3H2O + ClO3- (aq) —> Cl-(aq) + 6OH-
C
O
PY
3HS-(aq) + ClO3- (aq) —> 3S(s) + Cl-(aq) + 3OH-(aq)
EVALUATION (10 MINS)
1. Balance the following redox reactions:
a. CH4 + NO2 —> N2 + CO2 + H2O (oxidation number method)
b. Zn + Cr2O72- —> Zn2+ + Cr3+ (half-reaction method, in acidic medium)
ED
2. When silver metal, Ag, is exposed to hydrogen sulfide gas, H2S, it tarnishes.
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Is the reaction of Ag with H2S a redox reaction? How did you arrive at your answer? If your answer is yes,
identify the RA and OA.
Evaluation
1
(Not Visible)
2
3
4
(Needs Improvement)
(Meets Expectations)
(Exceeds Expectations)
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Chemistry 2
60 MINS
Electrochemistry: Redox
Reactions
LESSON OUTLINE
Communicating learning objectives
5
Illustration
5
Analogy, video, & class discussion
40
Guided exercises
10
C
O
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Introduction
Content Standard
The learners demonstrate an understanding of redox reactions as applied to galvanic and Motivation
electrolytic cells.
Instruction
Performance Standard
Practice
Learning Competencies
Materials
Define oxidation-reduction reactions. (STEM_GC11AB-IVf-g-169)
Balance redox reactions using the change in oxidation number and half-reaction
methods. (STEM_GC11AB-IVf-g-171)
Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
classify reactions as redox or non-redox;
•
identify reducing and oxidizing agents in a given redox reaction;
•
balance redox equations using change in oxidation method;
•
write the oxidation and reduction half-reactions for a given redox reaction; and
•
balance a redox reaction using the half-reaction method.
Resources
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ED
•
Periodic table of elements
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INTRODUCTION (5 MINS)
Teacher Tip
Learners would take cue that some redox
reactions produce energy.
•
Some students may be accustomed in
using the term amphoteric. Give the
following explanation: The term
amphoteric is a general term for
substances that can react both as an acid
and a base. On the other hand,
amphiprotic (protic refers to hydrogen
ion) is a more specific term used to
describe a substance which can both
donate and accept hydrogen ions
(protons)..
PY
The lesson is about oxidation-reduction reactions, a very important type of chemical reaction. A large extent
of producing many substances important to industry or energy in various forms rely on this type of reaction.
These reactions are given the shortcut name “redox” reactions. Ask learners what familiar processes they
think involve “redox” reactions, i.e., burning of wood, combustion of fuel gases, etc.
•
MOTIVATION (5 MINS)
C
O
1. Brownout! You need a flashlight. Do you know how a flashlight works?
Note: All amphiprotic substances are
also amphoteric, but not all amphoteric
substances are amphiprotic.
ED
2. Group the class into three or four groups and let each describe what makes the flashlight bulb light when
the switch is turned on. This will be discussed again after lesson on galvanic cells (acknowledge that a
chemical reaction is involved in the production of electricity that makes the bulb light).
Lecture/Discussion
•
What is a redox reaction?
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INSTRUCTION (40 MINS)
Light a piece of Mg ribbon, or show a picture of a burning Mg ribbon. Ask the learners to
write the equation for the reaction on the board:
Mg(s) + O2(g) —> MgO(g)
The product of the reaction of the two elements, Mg and O2, is MgO.
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If learners still have difficulty grasping
this concept, give more examples like the
case of aluminium oxide. It is amphoteric,
but not amphiprotic.
•
Post the questions on the board to guide
learners throughout the lesson.
What kind of compound is MgO?
Teacher tip
Post the illustrations side by side on the
board.
It is an ionic compound.
What ions compose MgO?
PY
Mg2+ and O2-.
Consider Mg2+. What does the charge it carries means?
It has lost electrons, two in fact.
C
O
Where did the electrons go?
They were accepted by O to form O2-. This reaction involved a loss of electrons in Mg,
and a gain of electrons in O. In other words, a transfer of electrons occurred between Mg and O.
ED
The reaction of Mg and O2 is an example of a redox reaction. Oxidation-reduction reactions involve a transfer of
electrons.
•
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Oxidation is the loss of electrons; reduction is gain of electrons. A gain of electrons cannot happen if no loss of
electrons happen simultaneously.
Key reactants in a redox reaction
In the reaction, Mg lost electrons. We say Mg was oxidized. O gained electrons. It was
reduced.
Image obtained from http://classes.midlandstech.edu/carterp/courses/bio225/chap05/Slide11.GIF
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C
O
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Teacher tip
•
Give the learners time to analyze the
information before giving the correct
expression on the board.
•
Be alert on their possible difficulties or
misconceptions about chemical
equilibrium expression.
•
Understanding the acid-base property of
water will be applied and is very
important in the next topic: pH.
•
Supplementary videos can be used to
enrich this lesson: A) L29-5
Autoionization, Autodissociation of
amphoteric water - pH of pure water is 7
at 25°C (2:46mins) at https://
www.youtube.com/watch?
v=DpDewqtha8o, B) Autoionization in
Liquid Water (3:24mins) at https://
www.youtube.com/watch?
v=zeFSzt5x9uo.
•
The first video reiterates the key points
covered in the preceding discussion,
while the second video shows autoionization of water at the molecular level.
Video B can help your students visualize
the reaction better.
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You may cut the lesson up to this point
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Illustration of reduction and oxidation. Image obtained from http://classes.midlandstech.edu/
carterp/courses/bio225/chap05/Slide11.GIF
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Activity:
Can the learners identify the atom oxidized and the atom reduced in these redox equations?
1. Na(s) + Cl2(g) —> NaCl(s)
2. Al(s) + Cu2+(aq) —> Al3+(aq) + Cu(s)
3. H2(g) +
F2(g) —> HF(g)
If they can, identify these atoms and show how they arrived at their answer. If not, state the difficulty
encountered.
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(the presence of atoms in elemental form on one side of the equation and in combination with other atoms on
the other side already indicates that the reaction is redox.)
•
If video viewing in the classroom is not
possible, give the link to the learners for
viewing. They can also use their gadgets
to view the video in the classroom,
provided there is internet connection.
Worst scenario, just give the questions
below in advance as an assignment.
•
If the video link is no longer active by the
time of using, the teacher can choose
other available videos which features
Bronsted definition of acids and bases,
and why it replaces Arrhenius’s definition.
The video replacement should not be
more than 5 minutes long. Remember
that properties of acids and bases were
already taken in Grade 7.
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On the board, clearly mark the direction
of transfer of proton from the acid to the
base.
Reducing and oxidizing agents
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Activity: Identify Reducing Agents and Oxidizing Agents
In the first example, Mg + O2 —> MgO,
the oxidation of Mg can be represented by the equation:
Mg —> Mg2+
O2 —> 2 O2-
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The reduction of O can be represented by:
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Half-reactions
These representations are called half-reactions: oxidation half-reaction, and reduction half-reaction.
Activity:
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A redox reaction is a pair of half-reactions that occur simultaneously.
Split the equations in Activity 1 and identify which is the oxidation half-reaction, and reduction halfreaction respectively.
(Again, learners might encounter some difficulty in the second or third equation).
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Oxidation Numbers: a system of electron bookkeeping
Discuss the following:
Rules
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Examples
PRACTICE (10 MINS)
Ask the learners:
1. Is the reaction of Ag with H2S a redox reaction?
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When silver metal Ag, is exposed to hydrogen sulfide gas H2S, it tarnishes.
2. How did you arrive at your answer? If you answereds yes, identify the reducing and oxidizing agent.
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Balancing redox reactions
Identify if the following equations are balanced. If not, balance them.
a. 2Na + Cl2 —> 2NaCl
b. Fe2O3 + CO —> 2Fe + CO2
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c. Al + Cu2+ —> Al3+ + Cu
d. Zn + 2HNO3 —> Zn(NO3)2 + NO2 + H2O
Sample response:
Only the first one is balanced.
Many redox reactions can be easily balanced by inspection. But one important feature distinct to balanced
redox reactions is that the number of electrons lost is equal to the number of electrons gained.
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Teacher Tip:
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The teacher may require the learners to
write at least one question related to the
lesson which they want to be answered.
•
After going through the learners’ journal
entries, the teacher may select a few
entries for sharing in class. Depending
on the answers of learners, the teacher
might need to allot more minutes on this
part for the processing of their ideas.
1. Balance redox reactions using the change in oxidation number method.
2. Consider the second equation above:
Fe2O3 + CO —> Fe + CO2
3. Illustrate each step in the method using the process below
A. Assign ON to each element in the equation.
B. Identify the element that changed its ON.
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C. Identify the element oxidized (reduced). Determine the number of electrons lost (gained) /
atom; per formula unit. (Why?)
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E. Balance remaining substances/atoms.
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D. Make the number of electrons lost and number of electrons gained equal by adjusting the
coefficients of the oxidizing and reducing agents.
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Chemistry 2
120 MINS
Electrochemistry: Corrosion
LESSON OUTLINE
Communicating learning objective
2
Short observation activity
18
Class discussion
40
Poster/Brochure making
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Content Standard
The learners demonstrate an understanding of redox reactions as applied to galvanic and Introduction
electrolytic cells.
Motivation
Performance Standards
Instruction
The learners prepare a poster in a specific application of one of the following:
a. Acid-base equilibrium
Evaluation
b. Electrochemistry
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Materials
Include in the poster the concepts, principles, and chemical reactions involved, and
diagrams of processes and other relevant materials.
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Learning Competency
Apply electrochemical principles to explain corrosion. (STEM_GC11AB-IVf-g-181)
(1) Brown, TL et. al. Chemistry The Central Science. 2009. 11th ed. New
Jersey: Pearson Prentice Hall
(2) Chang, R. Chemistry.2007. 9th ed. New York: McGraw-Hill Companies,
Inc.
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Specific Learning Outcomes
At the end of the lesson, the learners will be able to:
Resources
explain corrosion in terms of the electrochemical reactions involved; and
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discuss the economic impacts of corrosion and some measures by which metals can
be protected from corrosion
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INTRODUCTION (2 MINS)
Teacher Tip
The teacher shall communicate the objectives of the lesson to the class.
Just let the students do the qualitative
description using their own words. Do not
introduce yet technical terms such as
oxidation
MOTIVATION (18 MINS)
The teacher will be asking the students to roam around the vicinity of the classroom or certain areas around
the school where they can find metal objects (regardless if it is still on it’s lustrous/untarnished/uncorroded
state or already rusting/tarnishing/corroding. Students will be asked to describe the physical appearance of
the metal object.
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➡ Possible answers:
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✓ For relatively “new” metal objects/fixtures, appearance may be shiny/lustrous when light strikes its
surface and hard
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✓ For relatively “old” metal objects/fixtures, some parts may appear to have brown substance forming
in its surface which is more commonly known as rust. The surface may not anymore appear to be the
same as the other parts of the object that are not yet rusting. In some cases, the metal is already flaky
and can easily disintegrate.
✓ If the metal is coated with paint, corroded parts, if any, usually appear at portions where the metal is
already stripped of paint.
The teacher then asks the students to report on their observations.
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The teacher then gives this question to the students: “Can we restore the original appearance of the metal
by just washing its surface with soap?”
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➡ Answer: No. The formation of “rust” is actually a chemical reaction that consumes the metal. Washing
with water will just remove the deposits on the surface but the metal will not appear exactly as the
original.
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The teacher then proceeds in discussing the lesson.
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INSTRUCTION (40 MINS)
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Corrosion of Metals
Corrosion is a general term used to refer to deterioration of metals through an electrochemical process. There
are many examples of metal corrosion such as the tarnish in silver, green patina in copper and brass and the
most common which is the rust in iron. (Figure 1)
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Figure 1. Examples of metal corrosion. (a) tarnish in silver; (b) green patina in brass/copper; (c) rust in iron.
(Images obtained from https://www.wholeheartedmen.com/wp-content/uploads/2014/09/silver.jpg, http://
i757.photobucket.com/albums/xx218/itsnotworkitsgardening/July%202011/IMGP2769.jpg, http://
s.hswstatic.com/gif/rusty-nail-tetanus-1.jpg)
Silverwares tend to form a layer of silver sulfide, Ag2S when it comes into contact with foodstuffs over time. This
is referred to as the tarnish in silver. Silver tends to be oxidized to Ag+. It’s negative oxidation potential suggests
that the process takes place slowly.
Ag(s) → Ag+(aq) + e– (E°oxd’n = –0.80 V)
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Copper is also prone to corrosion. Upon atmospheric exposure, copper forms a layer of green patina which is
basically copper (II) carbonate, CuCO3. This is a result of the oxidation of Cu metal into Cu2+ which also occurs in
a relatively slow pace as suggested by the negative oxidation potential:
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Cu(s) → Cu2+(aq) + 2e– (E°oxd’n = –0.34 V)
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The patina layer that forms on the surface of the copper metal protects the metal underneath from further
corrosion. The Statue of Liberty in New York, USA, is made from copper. Prior to its restoration in 1986, the
statue appears green because of the green patina layer. Likewise, the second tallest statue of National Hero
Jose Rizal in Calamba City, which was made from bronze (an alloy of primarily made up of copper), is already
showing evidences of corrosion (Figure 2)
Figure 2. The Statue of Liberty and Dr. Jose Rizal with layers of patina. (Images obtained from http://
www.kidport.com/reflib/socialstudies/landmarks/images/StatueLiberty.jpg, http://
polymu.smugmug.com/Other/Portfolio/i-PgTQrGV/0/L/Largest%20Rizal%2001-L.jpg)
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Rusting of Iron
Perhaps the most familiar example of corrosion is the formation of rust in iron. The reaction requires the
presence of water and oxygen. Rusting of iron involves a series of redox reactions that occur at different portions
of the same iron sample. (Figure 3)
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!
Figure 3. Electrochemical processes involved in rust formation. (Image obtained from Chang, R.
Chemistry.2007. 9th ed. New York: McGraw-Hill Companies, Inc.)
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The anode reaction occurs in one region of the metal where the oxidation of elemental iron occurs:
Fe(s) → Fe2+(aq) + 2e– (E°oxd’n = +0.44 V)
O2(g) + 4H+(aq) + 4e– → 2H2O(l) (E°red’n = +1.23 V)
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On another region of the metal, the electrons given up at the anode are used to reduce atmospheric oxygen to
water. This region serves as the cathode.
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This results into an overall redox reaction that is spontaneous as evident in the positive value of the overall cell
potential:
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l)
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E°cell = E°oxd’n + E°red’n
= 1.23 V + 0.44 V
= 1.67 V
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The protons (H+) are supplied partially by the reaction of atmospheric carbon dioxide with water to produce
carbonic acid, H2CO3.
Another reaction takes place at the anode where Fe2+ is further oxidized into Fe3+ by oxygen:
4Fe2+(aq) + O2(g) + (4+2x)H2O(l) → 2Fe2O3·xH2O(s) + 8H+(aq)
The iron (III) oxide with varying amount of water associated with it is the rust that deposits at the surface of the
iron.
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Since the process involves migration of ions and electrons, it is greatly accelerated in the presence of salts. This
is why rusting occurs more rapidly if iron is exposed to saltwater such as what happens in ships.
Figure 4. A rusted ship. (Image obtained from http://hd-covers.com/wp-content/uploads/
Rusted-Ship.png
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Unlike in patina layers in copper, rust do not protect the iron underneath because the latter is porous.
ENRICHMENT (60 MINS)
Ask the students to do a short library research activity on the following:
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1. Economic impacts of corrosion
2. Measures to prevent corrosion of metals (much better if they will be able to look for the electrochemical
reactions involved and the use of the activity series of metals in explaining the mechanism of the preventive
measure)
The outputs of their research will be reported in class.
A summary video caps the lesson on corrosion:
Link: https://youtu.be/jQoE_9x37mQ
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Chapter 8: Electrochemistry
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Biographical Notes
MA. CORAZON B. BARRAMEDA
Writer
Dr. Myrna S. Rodriguez is Assistant Professor 7 at the
Institute of Chemistry in the University of the Philippines Los
Banos, Laguna. She finished her bachelor’s and master’s degrees
in Agricultural Chemistry at the UP Los Banos. She received her
doctorate in Chemistry and Biochemistry at LaTrobe University in
Victoria, Australia. For many years now, Dr. Rodriguez has been
working with different agencies, such as the Department of
Education, UP Open University, Network for Inter-Asian
Chemistry Educators, and various Local Government Units, in
upgrading science literacy in the country. She has served as the
President of the Philippine Association of Chemistry Teachers for
seven years, and is a member of other organizations such as the
Kapisanan ng mga Kimiko sa Pilipinas – Southern Tagalog
Chapter.
Professor Ma. Corazon B. Barrameda is an Associate
Professor at the Bicol University College of Science. She finished
her bachelor’s degree in Chemistry at University of Nueva
Caceres and accomplished her master’s degree in Chemistry
Education at Bicol University Graduate School. Prof. Barrameda
also served as a Laboratory Facilities Coordinator for Natural
Science Laboratories under Bicol University College of Science.
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MYRNA S. RODRIGUEZ, PH.D.
Team Leader
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Dr. Rodriguez’s published scholarly works include
researches such as 6-Phosphogluconate dehydratase from
Zymomonas mobilis: An iron-sulfur manganese enzyme, policy
papers, and educational manuals for students and faculty of UP
Los Banos. She is a recipient of numerous distinctions in
Chemistry including the recent Achievement Award for Chemical
Education, Tertiary Level, given by the Philippine Federation of
Chemistry Societies (PFCS) on April 2016.
SHIRLEY R. JUSAYAN, PH.D.
Writer
Dr. Shirley R. Jusayan is an Associate Professor V in
Chemistry & ADS at Western Visayas State University, La Paz,
Iloilo City for almost ten years now. She finished her bachelor’s
degree in Chemistry at the University of Iloilo and accomplished
her doctorate degree in Educational Management at Western
Visayas State University. Before serving as a professor, Dr.
Jusayan has served as a secondary teacher in Chemistry for 25
years. She has been a regular resource speaker and consultant in
the field of Math and Science Education since 2008. Dr. Jusayan
is recipient of various distinctions, including the Metrobank
Foundation Inc: Outstanding Teacher for Secondary - National
Level, awarded in 2001.
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APRHODITE M. MACALE
Writer
Ms. Aprhodite is currently an Assistant Professor IV at UP
Rural High School. She accomplished her bachelor’s degree in
Chemistry for Teachers and master’s degree in Chemistry at
Philippine Normal University. Presently, she is pursuing a
doctorate degree in Educator, Major in Chemistry, at UP Open
University. Aprhodite has experience as a paper presenter, guest
lecturer, and resource speaker. She has also co-authored journals
abstracts in the field of chemistry.
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Dr. Sabularse is a retired professor of Chemistry at UP Los
Baños where she has taught Chemistry for 29 years. She finished
her doctorate degree in Food Science at the Louisiana State
University, USA. She earned her master’s degree in Food Science
at UPLB and her bachelor’s degree in Food Technology at UP
Diliman. Dr. Sabularse has been actively involved in numerous
research projects related to improvements in the field of food
technology, processing, and development. Dr. Sabularse has
made significant contributions in numerous publications,
conferences, thesis researches, and training workshops inside
and outside the country for the last 40 years. Dr. Sabularse has
been awarded a number of grants which involved researches in
the field of Chemistry and Food Technology.
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VERONICA C. SABULARSE, PH.D.
Writer
JOSEPH CARMELO K. SAN PASCUAL
Writer
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Joseph Carmelo K. San Pascual is an Assistant Professor I
in Chemistry at the UP Los Baños. He earned both his bachelor’s
degree in Chemistry and master’s degree in Biochemistry at
UPLB. Joseph has been teaching chemistry for almost 8 years at
UPLB. Mr. San Pascual has been a part of numerous seminars,
workshops, and training conventions since 2008. He is also
actively involved in facilitating teacher training and workshops
and being a resource speaker as well in such activities.
MARIA CRISTINA D. PADOLINA, PH.D.
Technical Editor
Maria Cristina Damasco-Padolina is the current and
seventh President and Chief Academic Officer of Centro Escolar
University in Manila, Philippines. She held various positions at the
University of the Philippines, the most notable of which are her
appointments as Director for Instruction of the University of the
Philippines Los Baños (1980-1984) and as the first Chancellor of
the UP Open University (March 1995 - February 2001).
Afterwards, she was appointed one of the Commissioners of the
Commission on Higher Education. She earned her doctorate
degree in Inorganic Chemistry from the University of Texas at
Austin, her Master's Degree in Chemistry from the Ateneo de
Manila University, and her Bachelor's Degree in Chemical
Engineering from the University of the Philippines, Diliman. As a
commissioner of CHED, Dr. Padolina focused on the
improvement and enhancement of teacher education &
information technology education.
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Juan Miguel M. Razon
Illustrator
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Kevin Mark R. Gomez is a junior Visa Officer at the
Embassy of Italy in Manila and an instructor at the De La Salle
University. He earned his Master’s degree in International
Relations (with distinction) at the St. Petersburg State University,
St. Petersburg, Russian Federation and he received his bachelor’s
degree in Public Administration at the University of the Philippines
Diliman, Quezon City. Throughout his college years, he served as
a Writer for UP Diliman’s official publication, the Philippine
Collegian. Mr Gomez has worked as Language Editor at the
Department of Agrarian Reform, Research Associate of IBON
International and National Chairperson and Consul-general of the
UP SOLIDARIDAD.
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KEVIN MARK R. GOMEZ
Copyreader
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Mr. Juan Miguel M. Razon graduated with a degree in
Bachelor of Science in Management and Bachelor of Science in
Information Technology Entrepreneurship, Minor in Literature
from the Ateneo de Manila University. He worked at IBM
Philippines and contributed in the ideation and implementation of
the intranet-based “knowledge hub” for the employees of IBM.
He also served as the Finance Commissioner of the Ateneo
Commission on Elections and the Vice President for Public
Relations for Ateneo Kaingin. He intends to pursue a long-term
career in business intelligence, corporate finance, and graphic
design.
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Chemistry 2 - Colored Pages
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Lesson 1: Intermolecular Forces and Liquids and Solids
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Lesson 2: Physical Properties of Solutions
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Lesson 3: Thermochemistry
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Lesson 4: Chemical Kinetics
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Lesson 5: Chemical Thermodynamics
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Lesson 6: Chemical Equilibrium
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Chapter 7: Acid-Base Equilibria and Salt Solution Equilibria
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