To accompany Inquiry into
Chemistry
PowerPoint Presentation
prepared by Robert Schultz
robert.schultz@ei.educ.ab.ca
Chemistry 30 – Unit 1
Thermochemical Changes
Preparation Info
• Systems: Open, closed, and isolated definitions
• First Law of Thermodynamics – Total energy
of the universe is constant (energy can’t be
created or destroyed)
• Second Law of Thermodynamics – In the
absence of energy input, a system becomes
more disordered (its entropy increases)
Preparation
• Meaning?
• A system at lower temperature will be
more ordered as the particles have less
average kinetic energy
• Two systems in thermal contact will
transfer energy such that the more
ordered (cooler) one gains energy and
becomes more disordered
• Consequence: heat always flows from
hotter systems to cooler ones
Preparation
Important Definitions:
• Thermal Energy: the total kinetic
energy of all particles of a system
• Temperature: a measure of the average
kinetic energy of the particles of a
system
• Heat: a transfer of thermal energy
between 2 systems
Chapter 9, Section 9.1
Questions:
• Which has more thermal energy, a hot
cup of coffee or an iceberg?
• Which has a larger average thermal
energy, a hot cup of coffee or an
iceberg?
• If an iceberg and a hot cup of coffee
come into contact, in which direction
will heat flow?
Preparation
• Heat energy transferred will be related
to the temperature change of the system
• It takes different amounts of heat energy
to change the temperature of
1 g of a substance by 1°C
• This number is called the specific heat
capacity, c, and is measured in units of:
J
g C
Preparation
• Water has a c value of
4.19 J g C
• This means that it takes 4.19 J of heat to
raise the temperature of 1 g of water by
1°C
• Water has a very large c compared to most
other common substances
Preparation
• To determine the amount of heat transferred
the formula used is
Q  mc t
• Despite what your text says on page 337, I
would always take ∆t as positive
• If heat is absorbed, temperature of surroundings
will decrease; if heat is released temperature of
surroundings will increase
• Examples: Practice Problems 1 and 4, page 337
Preparation
• Practice Problem 1, page 337
Q  mc t  0.100 kg  2.44 kJ kg C  25C  6.1kJ
• Since 1 J is such a small amount of
heat energy I start my questions in
kJ as shown above
• If necessary I move into MJ or GJ
Preparation
• Practice Problem 4, page 337
Q  mc t
c
4.937 kJ
Q

 0.790 kJ kg C  0.790 J g C
m t 0.25000 kg  25.0 C
• Putting kilo top and bottom cancels
out and c stays the same
• The substance is granite
• Worksheet: WS 43 (Nelson) then BLM
9.1.1 (back only)
Chapter 9, Section 9.1
• Energy changes in chemical reactions
crucial to life
• Not just in photosynthesis, fuels, and
batteries, but in the very way that your
body metabolizes food and makes the
energy available for life processes
• Thermodynamics: the study of energy
and energy changes
Chapter 9, Section 9.1
• Recall the first law of thermodynamics:
∆Euniverse= 0
• If a system loses energy, the
surroundings gain energy (get warmer)
• If a system gains energy, the
surroundings lose energy (get cooler)
∆Esystem = - ∆Esurroundings
Chapter 9, Section 9.1
• Energy types:
• Kinetic energy, Ek, energy of motion of
particles of a system
• Temperature is a measure of the
average Ek of the particles of a system
• Potential energy, Ep, stored energy,
usually in chemical bonds
Chapter 9, Section 9.1
• Transfer of Ek: heat flows from hotter objects
to cooler ones (Preparation section of notes)
• Breaking bonds always requires energy
(endothermic); forming bonds always releases
energy (exothermic)
• Chemical reaction:
breaking bonds + energy1
forming bonds + energy2
input
output
• If energy1 > energy2, reaction is endothermic
• If reverse is true, it is exothermic
• Worksheet BLM 9.1.3
Chapter 9, Section 9.1
• New term: enthalpy (not entropy)
• Enthalpy (change), ∆H: the difference
in potential energy between reactants
and products, measured at constant
pressure – measured in kJ (or MJ, etc)
• Molar Enthalpy (change), ∆rH: the
enthalpy change for 1 mole of a
specified substance – measured in
kJ/mol (or MJ/mol etc)
• In common usage the word change gets
left out
Chapter 9, Section 9.1
• Negative ∆H’s are exothermic (think
lose heat) and temperature of
surroundings increases
• Positive ∆H’s are endothermic (think
gain heat) and temperature of the
surroundings decreases
• Note: this increase → negative, and
decrease → positive is a stumbling
block for many students
Chapter 9, Section 9.1
• Chemical reactions can be written using
∆H notation:
C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l) ∆H=-2802.5 kJ
value for the reaction as written
4 NO(g) + 6 H2O(g)
4 NH3(g) + 5 O2(g)
∆H=+906 kJ
• They can also be written with the heat as
a term in the equation:
C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l) + 2802.5 kJ
4 NO(g) + 6 H2O(g) + 906 kJ
Do ∆H Worksheet!
4 NH3(g) + 5 O2(g)
Chapter 9, Section 9.1
• Potential energy diagrams for the same
2 reactions are shown below:
H (kJ)
C6H12O6(s) + 6 O2(g)
reactants
∆H = -2802.5 kJ
H (kJ)
6 CO2(g) + 6 H2O(l)
products
4 NO(g) + 6 H2O(g)
products
∆H = +906 kJ
4 NH3(g) + 5 O2(g)
reactants
Chapter 9, Section 9.2
• Recalling that breaking bonds always
endothermic and forming new bonds is
always exothermic, more complete Ep
diagrams might be shown as follows:
Endothermic
Exothermic
intermediate
products
ΔH
reactants
Ep (kJ)
Ep (kJ)
intermediate
reactants
products
ΔH
Chapter 9, Section 9.1
• Alternate forms of potential energy diagram (from
Chemistry 30 Diploma Exam Bulletin)
Chapter 9, Section 9.1
• Example: Practice Problem 3, page 346
a) C(s) + 2 H2(g)
CH4(g) + 74.6 kJ
b) C(s) + 2 H2(g)
CH4(g)
∆H = -74.6 kJ
H (kJ)
c)
C(s) + 2 H2(g)
reactants
∆H = -74.6 kJ
CH4(g)
products
Do Ep diagrams for formation of Cr2O3(s), simple decomp* of AgI(s),
and formation of SO2(g)
Chapter 9, Section 9.2
Formation of Cr2O3(s)
Ep (kJ)
2 Cr(s) + 3/2 O2(g)
ΔH=ˉ1139.7 kJ
Cr2O3(s)
reaction coordinate
formation of SO2(g)
simple decomposition of AgI(s)
ΔH=+61.8 kJ
AgI(s)
reaction coordinate
Ep (kJ)
Ep (kJ)
Ag(s) + ½ I2(s)
1/8 S8(s) + O2(g)
ΔH=ˉ296.8 kJ
SO2(g)
reaction coordinate
Chapter 9, Section 9.1
• Molar enthalpy of combustion: the
enthalpy change for the complete
combustion of 1 mol of a substance
• Complete combustions of fossil fuels
always yields CO2(g) and H2O
• Open systems – constant pressure – gases
escape – H2O(g)
• Isolated systems – H2O(l)
• Human body – cellular respiration - H2O(l)
Chapter 9, Section 9.1
• Table of Molar Enthalpies of Combustions of
alkanes, page 347
• Practice Problem 5b, page 347 (open system)
C4H10(g) + 13/2 O2(g)
OR:
2 C4H10(g) + 13 O2(g)
4 CO2(g) + 5 H2O(g) ∆H = -2657.3 kJ
note change in units!
8 CO2(g) + 10 H2O(g) ∆H = -5314.6 kJ
• In thermodynamics it is acceptable to write
equations with fractional coefficients – don’t
do this elsewhere
• Try question 5a, page 347
Chapter 9, Section 9.1
• Question 5a page 347
C5H12(l) + 8 O2(g)
5 CO2(g) + 6 H2O(g)
∆H = -3244.8 kJ
• Note that the value of ∆H varies directly as the
number of moles of reacting substances
H  n r H
• This formula gets used to calculate enthalpy
changes for ∆Ep like phase changes, chemical
reactions, and nuclear reactions
Chapter 9, Section 9.1
• Example Practice Problem 3a, page 349
r HC5H12  3244.8 kJ mol
Note: from table, page
347 - comment
H  n r H
Find H for 56.78 g of pentane
56.78 g
H  n r H 
 3244.8 kJ mol   2553 kJ
72.17 g / mol

mol of pentane

Chapter 9, Section 9.1
• Example Practice Problem 6, page 349
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g) ΔH = -906 kJ
• molar enthalpy change for?
906 kJ
• a) ammonia
H 
r
• b) oxygen
 227 kJ mol
4 mol
NH3
906 kJ
r HO2 
 181 kJ mol
5 mol
• c) nitrogen monoxide
906 kJ
r HNO 
 227 kJ mol
4 mol
• d) water
r HH2O 
906 kJ
 151 kJ mol
6 mol
Chapter 9, Section 9.1
• Do Worksheet BLM 9.1.6
Chapter 9, Section 9.2
• Finding the value of energy changes
experimentally: calorimetry
• Device: calorimeter
• The following diagrams show the principle
behind calorimetry – note arrow directions
Chapter 9, Section 9.2
• A simple calorimeter like the one you will
use
2 nested styrofoam
cups containing a
measured volume of
water
sitting in a beaker so
that it doesn’t fall over
3rd styrofoam cup
inverted on top with
hole for thermometer
(stirrer)
Chapter 9, Section 9.2
• Assumptions in styrofoam cup
calorimetry:
• Amount of energy transferred to cups
and thermometer is small and can be
ignored
• The system is isolated
• The solution produced has the same
density and specific heat capacity as
water
• The process occurs at constant pressure
Chapter 9, Section 9.2
• The enthalpy change of a chemical reaction =
energy lost or gained, and is indicated by the
symbol ΔH
• Energy gained or lost by the water causes a
temperature change and is indicated by the
symbol Q
• In an ideal calorimeter ΔH = Q
• But recall:
H  n r H
• Therefore
and
Q  mc t
n r H  mc t
system
calorimeter “water”
calorimetry
equation
Chapter 9, Section 9.2
• I will redo the example on page 354
using this formula
mc t mc t
r H 

n
cv
limiting reagent, if not stated,
or substance question asks
about
• remember m c Δt is for the “water” and
n (c v) for the CuSO4(aq)
2  0.05000 kg   4.19 kJ kg C   24.60  21.40  C

H
 89.4 kJ
r
0.300 mol L  0.05000 L
Since the temperature has gone up
the process is exothermic
Correct answer:
 89.4 kJ mol
mol
Chapter 9, Section 9.2
• Practice Problem 9, page 355
• Note that question asks for molar
enthalpy of reaction for sodium
• n will be moles of sodium (question asks)
n  r H  m c t
m c t 0.175 kg  4.19 kJ kg C   25.70 19.30  C
r H 

 2.9  10 2 kJ mol
0.37 g
n
22.99 g mol
• Since temperature increases, answer
is correctly expressed as
2.9  102 kJ mol or  0.29 MJ mol
Do Practice Problems 7, 10, 12, page 355
Chapter 9, Section 9.2
• Investigation 9.A page 356 (goes with
the questions you’ve been doing)
• Molar enthalpy of combustion:
Investigation 9.B, page 357
Chapter 9, Section 9.2
• Bomb Calorimetry: a bomb calorimeter is
used to make accurate and precise
measurements
Chapter 9, Section 9.2
• Reaction takes place inside an inner
container called the “bomb” that
contains pure oxygen
• Chemicals are electrically ignited and
heat is released to or absorbed from
calorimeter water
• Calorimeter materials: stirrer,
thermometer, containers are not
ignored
• With calorimeter filled to a set level
with water, all of their heat capacities
are combined as shown:
Chapter 9, Section 9.2
n r H  mH2OcH2O t  mther cther t  mstir cstir t  mcontainsccontains t


n r H  mH2OcH2O  mther cther  mstir cstir  mcontains ccontains t
n r H  C t
bomb calorimeter equation
Heat capacity of calorimeter
• Note that C contains the mass and specific
heat capacity of each component of the
calorimeter
• How do you know when to use
n r H  C t
versus
n r H  mc t ?
Chapter 9, Section 9.2
• Look for:
- words “bomb calorimeter”
- no mention of the mass or volume of water
- words “heat capacity” rather than “specific heat
capacity”
- units J/°C rather than J/g°C
• Question 2, Worksheet 46
n  r H  C t
C t 40.00 kJ C  3.54C
r H 

 286 kJ mol
1.00 g
n
2.02 g mol
• Since temperature increases, answer is
-286 kJ/mol
• Do rest of Worksheet 46
Chapter 9, Section 9.2
• More practice with
• WS 9.1.5
Q  mc t
Chapter 9, Section 9.2
• Review: page 366-7 good questions: 1,
3, 4 (no actual calculation needed), 5c
(data page 347), 6a (data page 347), 8,
10, 13, 15, 16, 17, 18, 19, 21
Chapter 9, Section 9.2