Honors Chemistry
Unit 5
Matter & Energy
 Label a heating / cooling curve
o Solid, liquid, gas
o Evaporation, condensation, freezing, melting
o Enthalpy of fusion, enthalpy of vaporization
o Specific heat
o Boiling point, melting point
 Separate mixtures based on physical properties.
o boiling point (distillation), magnetism, density etc.
 Evaluate energy changes of matter (Specific Heat & Calorimetry)
o Calculation of specific heat (q = mCΔT)
o Calorimetry of various systems involving exothermic and
endothermic heat exchange.
o The calculation of energy released from a food substance using
calorimetry.
1
We are looking for:
1a. Identification of all phase changes and energy change values
1b. Evaporation, condensation, freezing, melting
1c. Enthalpy of fusion, enthalpy of vaporization
1d. Specific heat
1e. Boiling point, melting point
1f. Solid, liquid, gas
2a. Physical properties such as boiling point, magnetism, density etc.
2b. Use boiling point in the distillation process to separate mixture of liquids and identify the liquids present.
3a. Calculations of energy released/gained using specific heat (q = mCΔT)
3b. Calorimetry of various systems involving exothermic and endothermic heat exchange.
3c. The calculation of energy released from a food substance using calorimetry.
Matter
Objectives:


Distinguish between a mixture and a pure substance
Describe the states of matter in terms of particles
Building Blocks of Matter
 Atom – smallest unit of an element that maintains the
properties of that element. Neutral charge
 Element – a pure substance made of only one kind of atom
 Compound – a substance that is made of the atoms from two or more elements that
are chemically combined
Classification of Matter
Pure substance – composed of one kind of atom or molecule


has a fixed composition
has the same physical and chemical properties throughout
Mixture – a blend of two or more kinds of matter
 Can be physically separated.
 Methods include: centrifugation, distillation, filtration, and magnets.
o Homogeneous – uniform in composition throughout
 Solution - one or more substances (the solutes) dissolved in another substance (the
solvent)
 Colloid - Particles suspended in a liquid that do not settle out
2
o Heterogeneous – NOT uniform throughout and will show Tyndall Effect
(scattering of light due to particles in a solution)
 Suspension - Particles are suspended in a fluid that will settle out
States of Matter:
1.
Solid – has definite volume and definite shape.
(same shape regardless of its container)
2. Liquid - has definite volume but an indefinite shape (assumes the
shape of its container)
3. Gas – has neither a definite volume nor definite shape. (expands to fill
any container)
4. Plasma – high temperature state in which atoms lose their electrons.
3
Heating and Cooling Curve Definitions
Specific Heat –
Solid –
LiquidGas –
Plasma –
Heating Curve Enthalpy of Fusion/ Molar heat of fusion–
Melting–
Melting Point –
Enthalpy of Vaporization/Molar heat of Vaporization–
Evaporation–
Boiling Point –
Sublimation-
Cooling Curve –
Condensation –
Condensation Point –
Freezing –
Freezing Point –
Deposition -
4
The graph below shows the relationship between heat (energy) added, in calories (cal), and temperature
for 1 g of water. A student applied heat to 1 g of ice that had been cooled to -40⁰C and measured the rise
in temperature.
Read and fill-in the notes below and on the following pages and label the steps/regions A, B, C, D, E on the
graph.
Step A:Solid Water (Ice) Rises in Temperature (Keep in mind the graph is for water!)

If the __________________ is not at 0oC, it will rise as heat is ____________to get there. (Kinetic
energy is _________________)

Each gram of water requires a constant amount of energy to increase 1o = specific heat

IMPORTANT – the ice has not________________ yet!
Step B: Solid Water (Ice) Melts
 By ______________energy the ice begins to _____________.

Temperature does not ___________ as more energy is being ______________ (Kinetic energy is
_____________________ but potential energy is ____________)

Each mole of water requires a given amount of energy to melt = molar heat of fusion (∆ Hfus) in
kJ /mole.
5

Energy is overcoming water molecules attraction for each other so it can be converted from a solid to
liquid.

How many calories of energy did it take to completely change the 1 gram of solid water (ice) at 0⁰C to
liquid water?________________________
Step C: Liquid Water Rises in Temperature

Now the ice is completely _________ and the water temperature begins to _________________ as
heat is ________________. (specific heat)

Kinetic energy is ______________________.

The water has not started to____________ yet.

How many calories of energy did it take to make the 1 gram of liquid water to change temperature
from 0⁰C to 100⁰C (just beginning to boil)?____________
Step D: Liquid Water Boils

As we __________ energy the temperature does not change.

Each mole of water will require a constant amount of energy to boil = molar heat of vaporization
(∆Hvap) KJ/mole.

The energy is being used to overcome water's attraction to each other to convert the liquid to a gas
(kinetic energy _________________ but potential energy is _________________).
How many calories of energy did it take to make the 1 gram of liquid water to completely turn to
steam once it hit 100⁰C?________________________

Step E: Steam Rises in Temperature

Temperature ___________ again when all water is turned to steam

Each gram of water requires a constant amount of energy to rise 1o = specific heat.
6
Specific Heat Capacity “C”
The amount of energy required to be absorbed to warm 1 gram of a substance by 1 oC (or 1
K) or the amount of energy required to be released to cool 1 gram of a substance by 1 oC
(or 1 K).
-orHow easily things warm up & cool down.
Energy Calculations Involving Specific Heat:
q = mC∆T
where:
q = Heat Energy
+ q means heat/energy is being absorbed (endothermic process)
- q means heat/energy is being released (exothermic process).
m = mass in grams
c = specific heat capacity (also “s”)
∆T = change in temperature (temperature final – temperature initial)
Energy Units:
Heat energy (q) is in joules(J), kilojoules (kJ) or calories (cal).
1 calorie = 4.184 joules
Mass (m) is in grams or kilograms
Specific heat capacity, c, is in J/g oC or kJ/kgoC
Water (L) = 4.184 J/goC
Water (s) = 2.03 J/goC
Water (g) = 2.0 J/goC
Temperature , T, is usually in oC (temperature can be in K)
7
Metals have low specific heat values
Aluminum
0.900 J/goC
Iron
0.450 J/goC
Gold
0.126 J/goC
Doesn’t take much heat to heat them up and they don’t hold the heat well!!! (better conductors of
heat/energy)
Water and organic materials hold heat much better – have higher specific heats also takes more energy to
heat them up. (better insulators of heat/energy)
Water = 4.184 J/goC
Wood = 1.76 J/goC
Specific heats and molar heat capacities for various substances at 20⁰ C
Substance
J/goC
cal/g K or Btu/lb F
Molar C (J/mol K)
Aluminum
0.900
0.215
24.3
Bismuth
0.123
0.0294
25.7
Copper
0.386
0.0923
24.5
Brass
0.380
0.092
...
Gold
0.126
0.0301
25.6
Lead
0.128
0.0305
26.4
Silver
0.233
0.0558
24.9
Tungsten
0.134
0.0321
24.8
Zinc
0.387
0.0925
25.2
Mercury
0.140
0.033
28.3
Alcohol(ethyl)
2.4
0.58
111
Solid water, Ice (-10 C)
2.05
0.49
36.9
Granite
.790
0.19
...
Glass
.84
0.20
...
8
Specific Heat Problems
Complete the following on a separate sheet of paper
1. How much heat energy does a copper sample absorb if its specific heat is 0.386 J/g oC, its mass is 12.5 g and
it is heated from 25.0 oC to 40.0 oC?
2. How much heat energy is released by 10.0 g of gold, when it is cooled from 35.0 oC to 25.0 oC? The specific
heat of gold is 0.129 J/g oC.
3. A 4.00 kg sample of iron was heated from 0.0 oC to 20.0 oC. It absorbed 35.2 kJ of energy as heat. What is
the specific heat of this piece of iron?
4. 42.6 J of energy is needed to heat 2.00 grams of carbon from 50.0 oC to what final temperature? The specific
heat of carbon is 0.790 J/g oC.
Specific Heat Problems2
Complete the following on a separate sheet of paper
1. What amount of heat is required to raise the temperature of 85.9 g of water by 7.0C?
2. When 1045 joules are absorbed by a certain mass of water, the temperature of the water increases
from 45.0 ºC to 50.0 ºC. What is the mass of the water sample?
3. How many joules are required to heat 38.0 grams of gold from 60.0 ºC to 260.0 ºC? The specific heat of
gold is 0.126 J/(g·ºC).
4. Iron has a specific heat of 0.450 J/(g·ºC). If 1400. joules are absorbed by a chunk of iron that weighs 40.0
grams, how much does the temperature of the iron increase?
Specific Heat Problems3
Complete the following on a separate sheet of paper
1. What is the specific heat value of a sample of unknown material, if it weighs 36.359 grams and 59.912 J of
heat raise its temperature 152.0 oC?
2. What would be the final temperature of a 73.174 g sample of cobalt with an initial temperature of 102.0 oC,
after it loses 800 J? (The specific heat of cobalt is 0.4210 J/goC)
3. What mass of iron would release 0.1854 kJ when its temperature changed from 1550.0 oC to 75.0 oC? (The
specific heat of iron is 0.450 J/g oC)
4. The specific heat of mercury is 0.0335 cal/g oC. If 152.00 g of mercury at 75.0 oC are cooled to 23.5 oC, what
is the value of q in Joules?
5. Kelly has 2.00 kg of water at 80.0 oC and wants it to cool to 45 oC. If the water releases 20.9 kJ of energy
every minute, how long will it take to cool?
9
Calorimetry
From the point of view of the system
Endothermic
Exothermic
Feels cold
Feels hot
Surroundings lose heat (energy)
Surroundings gain heat (energy)
System gains energy
System loses energy
(+) Energy term
Energy is absorbed
(-) Energy term
Energy is released
Measured in Joules
Measured in Joules
To convert between Joules and Calories:
1 calorie = 4.184 Joules
10
Calorimeter
Q water = -Q system
Mass
H2O
x CH2O x ∆TH2O = Mass
sys
x Csys x ∆Tsys
Mass
H2O
x CH2O x ∆TH2O = Mass
sys
x Csys x ∆Tsys
11
Calorimetry Lab
Name:________________________________________
Class Period:_______
Purpose: Using the Law of Conservation of Energy, energy transfer, and a calorimeter, you will determine the specific heat
value for a given solid sample.
Materials: coffee cup calorimeter
400 mL beaker
water
Procedure:
thermometer
ring stand
balance
solid sample
2 metal rings
crucible tongs
Bunsen burner
wire gauze
1) Set up a boiling water bath:
a. Set up a double ring stand with the wire gauze to support the beaker.
b. Fill the 400 mL beaker about half full with water and place it on the wire gauze/double ring setup.
c. Light the Bunsen burner to heat the water to boiling.
2) Determine the mass of your solid sample.
3) Carefully place the solid sample into the boiling water bath.
4) Add about 100 mL of room temperature water from the sink to the calorimeter and determine the mass of the
water used. This is best done by placing the empty calorimeter on the balance and taring/rezero the balance then
adding the water.
5) Determine the initial temperature of the water in the calorimeter to the nearest 0.1⁰C .
6) After the solid has been in the boiling water bath for at least 5 minutes, remove it and carefully place it into the
calorimeter and quickly place the lid on the calorimeter.
7) Gently swirl the calorimeter with the thermometer in it for several minutes while monitoring the temperature of
the water inside. Record the highest temperature reached to the nearest 0.1⁰C. *Do not let the thermometer
rest on the metal or sides of the calorimeter.*
8) Remove the solid and place it back into the boiling water bath to heat it up for the next trial. Repeat steps 4-7 for
two more trials.
Data Table:
Trial 1
Trial 2
Trial 3
Mass of solid (g)
Mass of water in the
calorimeter (g)
Initial temperature of the
solid (⁰ C)
Initial temperature of the
water in the calorimeter
(⁰ C)
Final temperature of the
water in the calorimeter
and the solid (⁰ C)
Change in temperature for
the solid (⁰ C)
Change in temperature for
the water in the
calorimeter (⁰ C)
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Analysis: Calculate the specific heat of the solid sample for each trial. (show your work!)
Trial One:
Trial Two:
Trial Three:
Average specific heat of the solid: _____________________
You the chart of specific heat values in your packet to try to identify your solid._______________________
Explain why the specific heat value you determined in the lab is lower than the real value for the solid:
13
Calorimetry Problems
Complete the following on a separate sheet of paper
1) A 2.8 kg sample of metal with a specific heat of 0.43 kJ/kg°C is heated to 100.0°C and then placed in a 50.0g sample
of water at 30.0°C. What is the final temperature of the water and the metal?
2) The specific heat of mercury is 0.033 cal/g°C. If 152g of mercury at 75.0°C is placed in 145g of water at 23.5°C,
what will be the final temperature of the water?
3) A 37.7 g piece of metal is heated to 100.0C and placed into 75.0 g of water in a coffee-cup calorimeter.
Initially, the temperature of the water in the calorimeter was 23.1C. After the metal was added to the
water the temperature of the water increased until its temperature and the temperature of the metal
were 30.6C.
a. What is the specific heat of the metal?
b. What kind of metal was added to the water in the calorimeter?
4) A 440.00 g sample of mercury (specific heat = 0.140 J/goC, initial temperature of 22.00oC) is placed into
134.00 g of water (initial temperature of 35.00oC). Find the final temperature of the system.
5) Abbey is testing her baby’s bath water and finds that it is too cool, so she adds some hot water from kettle on the
stove. If Abbey adds 2.00 kg of water at 80.0°C to 20.0 kg of water at 27.0°C, what is the final temperature of the
bath water?
6) Jason is emptying the dishwasher. He removes a 0.200 kg glass that has a temperature of 30.0°C. Into the glass,
he pours 0.100 kg of diet soda (mostly water) which comes out of the refrigerator with a temperature of 5.00°C.
Assuming no external heat loss, what will be the final equilibrium temperature of the glass of diet soda (no ice was
added)? (c for glass =0.84 J/g°C).
More Calorimetry Problems
Complete the following on a separate sheet of paper
1) 45.3 g of a shiny metal, with a specific heat of 0.561 cal/g ⁰C, is placed into a water bath that has a temperature of
99.7 ⁰C. It is then placed into a calorimeter that has 54.7 mL of water. If the water and the metal end up with a
temperature of 17.2 ⁰C, what was the initial temperature of the water in the calorimeter?
2) A metal with a mass of 97.4 g is heated to a temperature of 81.4 ⁰C. It is then placed into a calorimeter containing
0.246 kg of benzene, which has a specific heat of 1.74 J/g ⁰C. The temperature of the benzene rises from 15.5 ⁰C
to 32.5 ⁰C. What is the specific heat of the metal in calories?
3) A metal with a specific heat of 0.126 cal/g ⁰C is placed into a water bath with a temperature of 94.5 ⁰C. The metal
is then placed into a calorimeter containing 86.5 g of acetic acid at a temperature of 20.6 ⁰C. The acetic acid and
metal have a final temperature of 35.5 ⁰C. The acetic acid has a specific heat of 2.05 J/g ⁰C. What is the mass of
the metal?
4) A metal with a specific heat of 2.03 J/g ⁰C and a mass of 68.5 g is placed into a hot water bath with a temperature
of 74.5 ⁰C. The metal is then placed into a calorimeter containing acetic acid at a temperature of 14.5 ⁰C. The final
temperature of the acetic acid and metal is 45.5 ⁰C. The density of acetic acid is 1.04 g/mL and a specific heat of
0.49 cal/g ⁰C. What is the volume of acetic acid in the calorimeter?
14
Distillation
To separate a mixture of liquids, the liquid can be heated
to force components, which have different boiling points,
into the gas phase. The gas is then condensed back into
liquid form and collected.
impure liquid
(mixture)
Distilled liquid
(distillate)
Separation of Unknown with
Fractional Distillation
120
Temperature (⁰C)
100
80
60
40
20
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Volume (mL)
Material with lower boiling point = __________________
Material with higher boiling point = _________________
What material could still be present? ________________
Solvent Boiling Point °C
Acetone = 56.5
Methanol = 64.7
Hexane = 68.8
Ethyl Acetate = 77
2-Methyl-2-propanol = 82.2
Water = 100.0
Toluene = 110.6
1-Butanol = 117.2
15
Solvent Boiling Point °C
Acetone = 56.5
Methanol = 64.7
Hexane = 68.8
Ethyl Acetate = 77
2-Methyl-2-propanol = 82.2
Water = 100.0
Toluene = 110.6
1-Butanol = 117.2
Based on the above graph, how many different components are in this mixture?? ___________
What is the lowest distillation temperature on this graph? ____________________
What material boils at this temperature? ______________________
What is temperature when the next component begins to distill? _____________________
Identify this material bases on the given table _________________________________
What temperature does the last component boil? _______________________
This component ‘s chemical name is _____________________________________
16
Enthalpy and Energy Diagrams
Objectives:



Define enthalpy
Draw and label an energy diagram
Explain the concept of activation energy and activated complex
So far we’ve talked a lot about heat.



Heat is the change in energy due to a temperature difference.
Heat can be used to measure the energy in something
Heat moves from areas of hot to areas of cold.
A change in energy can also be described as a change in enthalpy when the system is at constant
pressure (open to the atmosphere = constant pressure)
Enthalpy: a measure of the energy content of a substance. This includes kinetic and potential. (It also
includes pressure-volume energy but we will ignore this in general chemistry).
Since we will always be dealing with constant pressure systems in chemistry (until we get to the gas laws
unit), we can say
q = H
Just like heat



H > 0 = Endothermic
H < 0 = Exothermic
Units: Joules, J
A visual for the change in energy of a reaction:
Activated
Complex
Reactants
Energy
Products
Reaction Coordinate
The diagram above shows that



Reactants are at a higher energy state than the products
The reaction is exothermic
An in-between compound (activated complex), is formed
Reaction Coordinate: a step in the process. Reactants have to “morph” into the products.
17
Activation Energy
Energy
H < 0
Reaction Coordinate
The diagram above shows that



The change in energy between reactants and products is the change in enthalpy
The reaction is exothermic
The reaction requires some energy before it will start, Activation Energy: Ea
Example of Activation Energy:
1.
2.
3.
4.
Candles don’t burn until you ignite them
Matches don’t light until you use some of your own energy as you strike them
The food didn’t burn until it was lit
The Nitrogen triiodide (remember explosion video) doesn’t explode until it was tickled
Analogy: Old Love, New Love
Reactants are like you and an old boyfriend/girlfriend…content until you break up. You are in a state of
agitation…activated complex…until you meet and bond with your new boy/girl and you are
happy…at a low energy state….a new product has been formed.
If you are happier than in previous relationship = exothermic
Happier in previous relationship = endothermic
What would an endothermic reaction look like graphically?
Activation
Energy
Energy
H
Reactants
>
Reaction Coordinate
0
18
Name:__________________________________
1. From the diagram below match the letter to the appropriate term.
activation energy of forward reaction _____
products _____
activation energy of reverse reaction _____
reactants _____
activated complex _____
ΔH of reaction _____
Enthalpy (H) of products (energy) _____
Enthalpy (H) of reactants (energy) _____
Enthalpy (H) of activated complex (energy) _____
Referring to the diagram answer the following questions:
2. What is the ΔH of reaction of the forward reaction?
3. Is this reaction endothermic or exothermic?
4. What is the ΔH of the reverse reaction?
5. Is this reaction endothermic or exothermic?
6. What is the enthalpy of the activated complex?
19
20
HEAT OF SOLUTION
Reminder – Goggles must be worn at all times in the lab
PRE-LAB DISCUSSION:
When salts are dissolved in water, there is often a temperature change associated with the process. Some salts
dissolve, releasing heat in the process. Others dissolve while absorbing heat. As you may remember, processes
that proceed with a release of heat energy are called “exothermic” processes, while those that absorb heat are
called “endothermic” processes.
Because energy is released to the surroundings in an exothermic process, the heat of solution would be given a
negative value because the energy of the system is decreasing.
NaOH(s)  Na+(aq) + OH-(aq)
ΔHsolution = - 44.51 kJ/mol
Because energy is absorbed from the surroundings in endothermic processes, the heat of solution would be given
a positive value because the energy of the system is increasing.
KNO3(s)  K+(aq) + NO3-(aq)
ΔHsolution = + 34.89 kJ/mol
In this experiment you will start with a known mass of calcium chloride, ammonium chloride, and a known volume
of water. You will determine the magnitude of the temperature change associated with the dissolving process, and
use the masses of the solute and solvent, the temperature change (called ΔT), and the known heat capacity
(specific heat) of water, 4.18 J/(g⋅ °C), to calculate the heat of solution for the 2 different compounds.
PURPOSE:
To apply the concepts of specific heat and temperature change in the experimental determination of the heat of
solution of two soluble salts, calcium chloride and ammonium chloride.
PROCEDURE:
1. Obtain a Styrofoam cup “calorimeter” and add to it 40.0 mL of distilled water.
2. Secure a thermometer to stand up in the calorimeter, using your ring stand (your instructor will show you
how.)
3. Weigh out close to 1.00 grams of calcium chloride. Record the exact mass used.
4. Record the initial temperature of the water in the calorimeter.
5. Add the calcium chloride to the water in the calorimeter, all at once, and begin stirring the solution to dissolve
the salt as rapidly as possible. Stir by swirling the cup with your hand. DO NOT use a glass stir rod, as it will
effect your results. Watch out for the fragile thermometer! Record the lowest temperature achieved during the
dissolving of the salt.
6. Cleanup: Rinse the solution down the sink with LOTS of water. Rinse the Styrofoam cup, and return it to the
side counter. DO NOT throw the Styrofoam cup away – we re-use them. Rinse the thermometer and return it
to your lab drawer.
7. Repeat the procedure for ammonium chloride.
RESULTS
Observations and Data:
1. Mass calcium chloride used ________________g
2. Volume of water used ________________mL
3. Initial temperature (T1) ________________°C
4. Final (lowest) temperature (T 2) ________________°C
21
1. Mass ammonium chloride used ________________g
2. Volume of water used ________________mL
3. Initial temperature (T1) ________________°C
4. Final (lowest) temperature (T 2) ________________°C
Calculations: Show your work on a separate sheet of paper!
1. Calculate the amount of energy absorbed as the calcium chloride and ammonium chloride dissolved. In order
to do this, we must assume that the solution has the same specific heat (cp) as pure water, and that the
density of the water was 1.00 gram/mL. Express your final answer in kilojoules, kJ!
q = cp x m x ΔT
2. Calculate the number of moles of each salt used. (Moles are a unit of measure). Divide the number of grams
you used for each salt by the molar mass of each salt. The molar mass of calcium chloride is 110.98g/mol
and the molar mass of ammonium chloride is 53.49g/mol.
3. Divide the heat absorbed (in kJ) by the moles of each salt dissolved. This is the heat of solution, expressed in
kJ/mol of solute.
4. Using the known value of the heat of solution of calcium chloride and the heat of solution of ammonium
chloride, calculate the absolute error in your experimental result.
5. Calculate your percent error for each.
22
Flaming Bugles Lab:
Problem: How can you determine the amount of heat energy (calories) released per gram when
burning a Bugle snack?
Background:
What do you already know about energy transfer?
You and your partner need to design a lab to determine the energy in a single Bugle snack.
You must complete a formal lab report and include all the parts of the scientific method. Use the
following headings in your report:
Problem:
Background information:
Hypothesis:
Experiment:



Materials
Procedure
Data collection
Data analysis & conclusion
23