Unit 13:
Thermochemistry
Chapter 17
By: Jennie Borders
Section 17.1 – The Flow of Energy
Energy is the capacity to do work or supply
heat.
 Energy has no mass or volume.
 Chemical potential energy is energy stored in
chemicals.
 The kinds of atoms and the arrangement of
the atoms in a substance determine the
amount of energy stored in the substance.

Heat
Heat is a form of energy that always flows from
a warmer object to a cooler object.
 Heat is represented by q.

Thermochemistry
Thermochemistry is the
study of the heat changes
that occur during chemical
reactions and physical
changes of state.
 The law of conservation of
energy states that in any
chemical or physical process,
energy is neither created nor
destroyed.

The Great Debate
1. Exothermic reactions lose heat.
2. Endothermic reactions absorb heat.
Exothermic and Endothermic
Thermochemistry is concerned with the flow of
heat between a chemical system (reaction) and
its surroundings.
 A system is the specific part of the universe on
which you focus your attention.
 The surroundings include everything outside the
system.
 The system and the surroundings constitute the
universe.

Exothermic and Endothermic
In thermochemical calculations
the direction of the heat flow is
given from the point of view of
the system.
 A process that absorbs heat from
the surroundings is called an
endothermic process.
 A process that loses heat to the
surroundings is called an
exothermic process.

Exothermic and Endothermic
Units of Heat
A calorie is the quantity of
heat that raises the
temperature of 1 gram of
pure water 1oC.
 A Calorie, or dietary Calorie,
is equal to 1000 calories.
 A Joule is the SI unit of heat
and energy.

1 Calorie = 1000 cal = 1 kcal
= 4184 J
1 cal = 4.184 J
Practice Problems
Make the following conversions.
1. 444 calories to Joules
444 cal x 4.184 J = 1857.7 J
1 cal
2.
850 Joules to calories
850 J x 1 cal = 203.2 cal
4.184 J
Heat Capacity
The heat capacity of an object is the amount of
heat it takes to change an object’s temperature by
exactly 1oC.
 The greater the mass of an object, the greater the
heat capacity.
 The heat capacity of an object also depends on its
chemical composition.

Specific Heat
The specific heat capacity
of a substance is the
amount of heat it takes to
raise the temperature of
1 gram of the substance
1oC.
 Specific heat is
represented by C.
 The units of specific heat
are J/goC.
 Water has a higher
specific heat than most
substances.

Heat
Heat = mass x specific heat x change in temp
q = m.C.DT
Mass is in grams
Specific heat is in J/goC
Change in temp is in oC
Sample Problem

The temperature of a 95.4g piece of
copper increases form 25oC to 48oC when
the copper absorbs 849J of heat. What is
the specific heat of copper?
q = m.c.DT
c=
q
m.DT
DT = 48oC – 25oC = 23oC
c=
849J
= 0.39 J/goC
95.4g.23oC
Practice Problems
1. When 435J of heat is added to 3.4g of
olive oil at 21oC, the temperature increases
to 85oC. What is the specific heat of the
olive oil?
q = m.c.DT
c=
c=
q
m.DT
DT = 85oC – 21oC = 64oC
435J
= 1.99 J/goC
3.4g.64oC
Practice Problems
2. How much heat is required to raise the
temperature of 250g of mercury 52oC?
(specific heat of mercury = 0.14 J/goC)
q = m.c.DT
q = 250g (0.14J/goC) (52oC) = 1820J
Section 17.1 Review
1.
2.
3.
4.
In what direction does heat flow between two
objects?
How do endothermic processes differ from
exothermic processes?
On what factors does the heat capacity of an
object depend?
How many kilojoules of heat are absorbed
when 1000g of water is heated from 18oC to
85oC? (specific heat of water = 4.184 J/goC)
q = 1000g (4.184J/goC) (67oC) = 280328J
280328J x 1kJ = 280.328kJ
1000 J
Section 17.1 Review
5.
Using calories, calculate how much heat 32.0g
of water absorbs when it is heated from 25oC
to 80oC. How many joules is this? (specific heat
of water = 4.184 J/goC)
q = 32g (4.184 J/goC) (55oC) = 7363.8J
7363.8J x 1 cal = 1759.9 cal
4.184 J
Section 17.2 – Measuring and
Expressing Enthalpy Changes
Calorimetry is the accurate and precise
measurement of heat change for chemical
and physical processes.
 Calorimeters are devices used to measure
the amount of heat absorbed or released
during chemical and physical processes.
 Enthalpy is the heat content of a system
at constant pressure.
 Enthalpy is represented by H.

Calorimeter
q = DH = m . C . DT
Heat Change Sign Convention
Direction of
Heat Flow
Sign
Reaction Type
Heat Flows Out
of the System
-DH
Exothermic
Heat Flows Into
the System
+DH
Endothermic
Thermochemical Equations
An equation that included the heat change is
a thermochemical equation.
 A heat of reaction is the heat change for the
equation exactly as written.
 Ex:
CaO(s) + H2O(l)  Ca(OH)2(s)
DH = -65.2 KJ
2NaHCO3(s)  Na2CO3(s) +H2O(g) +CO2(g)
DH = +129 KJ

Sample Problem

Calculate the amount of heat (in kJ)
required to decompose 2.24 moles of
NaHCO3.
2NaHCO3  Na2CO3 + H2O + CO2
DH = 129kJ
2.24 mol NaHCO3 x
129kJ
= 144.48kJ
2 mol NaHCO3
Practice Problems
1.
Calculate the amount of heat (in kJ)
absorbed when 5.66g of carbon disulfide
is formed.
C + 2S  CS2
DH = 89.3kJ
5.66g CS2 x 1 mol CS2 x 89.3kJ = 6.65kJ
76g CS2
1 mol CS2
Practice Problems
2. How many kilojoules of heat are
produced when 3.40 mole Fe2O3 reacts with
an excess of CO?
Fe2O3 + 3CO  2Fe + 3CO2
DH = -26.33 kJ
3.40 mol Fe2O3 x
-26.3kJ
= -89.42kJ
1 mol Fe2O3
Section 17.2 Review
1.
When 2 mol of solid magnesium
combines with 1 mol of oxygen gas, 2
mol of solid magnesium oxide is formed
and 1204kJ of heat is release. Write the
thermochemical equation for this
combustion reaction.
2Mg(s) + O2(g)  2MgO DH = -1204kJ
Section 17.2 Review
2. How much heat is released when 12.5g of
ethanol burns?
C2H5OH + 3O2  2CO2 + 3H2O DH = -1368kJ
12.5g x 1mol x -1368kJ = -371.74kJ
46g
1 mol
Section 17.3 – Heat in Changes of
State

The heat of combustion is
the heat of reaction for the
complete burning of one
mole of a substance.
DH (fusion and solidification)
The heat absorbed by one mole of a
substance melting from a solid to a liquid
at constant temperature is the molar heat
of fusion.
 The heat lost when one mole of a liquid
changes to a solid at a constant
temperature is the molar heat of
solidification.
DHfus = - DHsolid

DH (vaporization and condensation)
The heat absorbed by one mole of a
substance changing from a liquid to a
vapor is the molar heat of vaporization.
 The heat released by one mole of a
substance changing from a vapor to a
liquid is the molar heat of condensation.

DHvap = - DHcond
Practice Problems
1. What is the DHvap
of acetone?
29 kJ/mol
2. What is the DHcond
of water?
-41 kJ/mol
3. What is the DHfus
of rubbing
alcohol?
6 kJ/mol
4. What is the DHsolid
of diethyl ether?
-7 kJ/mol
DH (solution)
The heat change caused by dissolution of one
mole of a substance is the molar heat of
solution.
 Ex.
NaOH(s)  Na+(aq) + OH-(aq)

DHsoln = -445.1 KJ
Heating Curve for Water
Sample Problem

How much heat (in kJ) is absorbed when
24.8g H2O(l) at 100oC and 101.3kPa is
converted to steam at 100oC.
24.8g H2O x 1 mol H2O x 40.7kJ = 56.08kJ
18g H2O
1 mol
Practice Problems
1. How much heat is absorbed when 63.7g
H2O(l) at 100oC and 101.3kPa is converted to
steam at 100oC?
63.7g H2O x 1 mol H2O x
40.7kJ = 144.03kJ
18g H2O
1 mol H2O
Practice Problems
2. How many kilojoules of heat are absorbed
when 0.46g of chloroethane (C2H5Cl)
vaporizes at its boiling point? (The molar
heat of vaporization for chloroethane is 26.4
kJ/mol.)
0.46g C2H5Cl x 1 mol C2H5Cl x 26.4kJ = 0.19kJ
64g C2H5Cl
1 mol
Practice Problems
3. How much heat (in kJ) is released when
2.5 mol of NaOH is dissolved in water? (The
molar heat of solution is -445.1 kJ/mol.)
2.5 mol NaOH x -445.1kJ = -1112.75kJ
1 mol
Practice Problems
4. How many moles of NH4NO3 must be
dissolved in water so that 88kJ of heat is
absorbed from the water? (The molar heat
of solution 25.7 kJ/mol.)
88kJ x 1 mol = 3.42mol NH4NO3
25.7kJ
Heating Curve Problems
Sometimes a question might ask you
about a physical change that involves a
temperature change and a phase change.
 You will have to do these problems in
multiple steps.
 You have to work your way through the
heating curve.

Heating Curve
Sample Problem

Calculate the amount of heat required to
change 50g of water at 75oC to steam.
q = m.c.DT
q = 50g (4.184 J/goC) (25oC) = 5230J
50g H2O x 1 mol H2O x 40.7kJ = 113.06kJ
18g H2O
1 mol
113.06kJ = 113060J
113060J
+ 5230J
118290J
Practice Problems
1. Calculate the amount of heat needed to
change 175g of ice at -20oC to water at 15oC.
q = 175g (2.1 J/goC) (20oC) = 7350J
175g H2O x 1 mol H2O x 6.01kJ = 58.43kJ
18g H2O
1 mol
q = 175g (4.184 J/goC) (15oC) = 10983J
58.43kJ = 58430J
10983J
58430J
+7350J
76763J
Practice Problems
2. Calculate the amount of heat released when
100g of steam at 130oC is cooled to water at
60oC.
q = 100g (1.7 J/goC) (-30oC) = -5100J
100g H2O x 1 mol H2O x -40.7kJ = -226.11kJ
18g H2O
1 mol
q = 100g (4.184 J/goC) (-40oC) = -16736J
-226.11kJ = -226110J
-226110J
-16736J
-5100J
-247946J
Section 17.3 Review
1.
2.
How does the molar heat of fusion of a
substance compare to its molar heat of
solidification?
How does the molar heat of vaporization
of a substance compare to its molar heat
of condensation?
Section 17.3 Review
3. Identify each enthalpy change by name
and classify each change as exothermic
or endothermic.
a. 1 mol C3H8(l)  1 mol C3H8(g) Endothermic
Exothermic
b. 1 mol Hg(l)  1 mol Hg(s)
Exothermic
c. 1 mol NH3(g)  1 mol NH3(l)
d. 1 mol NaCl(s) + 3.88kJ/mol  1 mol
NaCl(aq)
Endothermic
e. 1 mol NaCl(s)  1 mol NaCl(l) Endothermic
Section 17.4 – Calculating Heats of
Reaction

Hess’ Law of heat summation states that if you
add two or more thermochemical equations to
give a final equation, then you can also add the
heat changes to give the final heat change.
Sample Problem

Calculate the enthalpy change for the
following reaction given the following
information:
PbCl2 + Cl2  PbCl4 DH = ?
Pb + 2Cl2  PbCl4
Pb + Cl2  PbCl2
DH = -329.2kJ
DH = -359.4kJ
1st reaction: keep the same
2nd reaction: flip the reaction (change sign)
DH = -329.2kJ
+DH = 359.4kJ
DH = 30.2kJ
Practice Problems
1.
Find the enthalpy change for the reaction
using the following information.
2P + 5Cl2  2PCl5
DH = ?
PCl5  PCl3 + Cl2
2P + 3Cl2  2PCl3
DH = 87.9kJ
DH = -574kJ
1st reaction: flip reaction (change sign) and x2
2nd reaction: keep the same
DH = -175.8kJ
+DH = -574kJ
DH = -749.8kJ
Practice Problems
2. Calculate the enthalpy change for the
reaction using the following information:
N2 + O2  2NO
DH = ?
4NH3 + 3O2  2N2 + 6H2O
4NH3 + 5O2  4NO + 6H2O
1st reaction: flip reaction (change sign) /2
2nd reaction: /2
DH = -1530kJ
DH = -1170kJ
DH = 765kJ
+DH = -585kJ
DH = 180kJ
Practice Problems
3. Calculate the enthalpy change for the
reaction using the following information.
C2H4 + H2  C2H6
DH = ?
2H2 + O2  2H2O
C2H4 + 3O2  2H2O + 2CO2
2C2H6 + 7O2  6H2O + 4CO2
DH = -572kJ
DH = -1401kJ
DH = -3100kJ
1st reaction: /2
DH = -286kJ
2nd reaction: keep the same
DH = -1401kJ
3rd reaction: flip reaction (change sign) /2 +DH = 1550kJ
DH = -137kJ
Standard Heat of Formation
The standard heat of formation of a compound
is the change in enthalpy that accompanies the
formation of one mole of the compound from its
element with all substances in their standard
states at 25oC.
 The DHfo of a free element in its standard state
is zero.

DHo = DHfo (products) – DHfo (reactants)
Sample Problem

Calculate the standard heat of formation
for the following reaction.
2CO(g) + O2(g)  2CO2(g)
CO2(g) = 2mol x -110.5 kJ/mol = -221kJ
O2(g) = 1 mol x 0 kJ/mol = 0kJ
CO2(g) = 2 mol x -393.5 kJ/mol = -787kJ
DHo = DHfo (products) – DHfo (reactants)
DHo = (-787kJ) – (-221kJ + 0kJ) = -566kJ
Practice Problems
1.
Calculate the standard heat of formation
for the following reaction.
CaCO3(s)  CaO(s) + CO2(g)
CaCO3(s) = 1 mol x -1207 kJ/mol = -1207kJ
CaO(s) = 1 mol x -635.1 kJ/mol = -635.1kJ
CO2(g) = 1 mol x -393.5 kJ/mol = -393.5kJ
DHo = DHfo (products) – DHfo (reactants)
DHo = (-635.1kJ + -393.5kJ) – (-1207kJ) = 178.4kJ
Practice Problems
2. Calculate the standard heat of formation
for the following reaction.
2NO(g) + O2(g)  2NO2(g)
NO(g) = 2 mol x 90.37 kJ/mol = 180.74kJ
O2(g) = 1 mol x 0 kJ/mol = 0kJ
NO2(g) = 2 mol x 33.85 kJ/mol = 67.7kJ
DHo = DHfo (products) – DHfo (reactants)
DHo = (67.7kJ) – (180.74kJ + 0kJ) = -113.04kJ
Section 17.4 Review
1.
Calculate the enthalpy change in kJ for
the following reaction.
2Al + Fe2O3  2Fe + Al2O3
Use the enthalpy changes for the combustion of
aluminum and iron:
2Al + 3/2O2  Al2O3 DH = -1669.8kJ
2Fe + 3/2O2  Fe2O3 DH = -824.2kJ
1st reaction: keep the same
DH = -1669.8kJ
2nd reaction: flip reaction (change sign) +DH = 824.2kJ
DH = -845.6kJ
Section 17.4 Review
2. What is the standard heat of reaction for
the decomposition of hydrogen peroxide?
2H2O2(l)  2H2O(l) + O2(g)
H2O2(l) = 2 mol x -187.8kJ/mol = -375.6kJ
H2O(l) = 2 mol x -285.8kJ/mol = -571.6kJ
O2(g) = 1 mol x 0kJ/mol = 0kJ
DHo = DHfo (products) – DHfo (reactants)
DHo = (-571.6kJ + 0kJ) – (-375.6kJ) = -196kJ
THE END