CHAPTER 6
Thermochemistry
CHAPTER TERMS AND DEFINITIONS
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred
terms are italicized in the text. Where a term does not fall directly under a text section heading,
additional information is given for you to locate it.
thermodynamics* science of the relationships between heat and other forms of energy (6.1,
introductory section)
thermochemistry* one area of thermodynamics; study of the quantity of heat absorbed or evolved
by chemical reactions (6.1, introductory section)
energy
potential or capacity to move matter (6.1)
kinetic energy (Ek)
joule (J)
watt*
energy associated with an object by virtue of its motion (6.1)
SI unit of energy, kg ∙ m2/s2 (6.1)
measure of quantity of energy used per unit time; 1 J/s (6.1)
calorie (cal) non-SI unit of energy commonly used by chemists; originally defined as the amount of
energy required to raise the temperature of one gram of water by one degree Celsius; 1 cal = 4.184 J
(exact definition) (6.1)
potential energy (Ep)
internal energy (U)
(6.1)
Etot*
(6.1)
an object’s energy because of its position in a field of force (6.1)
sum of kinetic and potential energies of the particles making up a substance
total energy of a substance; sum of the kinetic, potential, and internal energies of the substance
law of conservation of energy
energy may be converted from one form to another, but the total
quantity of energy remains constant (6.1)
thermodynamic system (or system)
substance or mixture of substances under study in which a
physical or chemical change occurs (6.2)
surroundings
everything in the vicinity of a thermodynamic system (6.2)
heat (q) energy that flows between system and surroundings because of a difference in temperature
between the thermodynamic system and its surroundings (6.2)
thermal equilibrium* state in which energy does not flow as heat between system and
surroundings; temperature equality (6.2)
heat of reaction value of q required, at a given temperature, to return a system to the given
temperature at the completion of the reaction (6.2)
exothermic process
(6.2)
chemical reaction or physical change in which heat is evolved (q is negative)
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128
Chapter 6: Thermochemistry
endothermic process
(6.2)
qp*
chemical reaction or physical change in which heat is absorbed (q is positive)
heat of reaction at constant pressure (6.3)
enthalpy (H) extensive property of a substance used to obtain the heat absorbed or evolved in a
chemical reaction (6.3)
state function
property of a system that depends only on its present state, which is determined by
variables such as temperature and pressure and is independent of any previous history of the system
(6.3)
enthalpy of reaction (ΔH) change in enthalpy for a reaction at a given temperature and pressure;
equals the heat of reaction at constant pressure (6.3)
enthalpy diagram*
pictorial representation of the enthalpy change for a reaction (6.3)
pressure–volume work* energy required by a system to change volume against the constant
pressure of the atmosphere (6.3)
thermochemical equation chemical equation for a reaction (including phase labels) in which the
equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written
directly after the equation (6.4)
heat capacity (C) quantity of heat needed to raise the temperature of the sample of substance one
degree Celsius (or one kelvin) (6.6)
specific heat capacity (specific heat)
quantity of heat required to raise the temperature of one gram
of a substance by one degree Celsius (or one kelvin) at constant pressure (6.6)
calorimeter
change (6.6)
device used to measure the heat absorbed or evolved during a physical or chemical
bomb calorimeter*
calorimeter used for reactions involving gases (6.6)
Hess’s law of heat summation
for a chemical equation that can be written as the sum of two or
more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the
individual steps (6.7)
standard state standard thermodynamic conditions chosen for substances when listing or comparing
thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C) (6.8)
standard enthalpy of reaction (ΔH°)* enthalpy change for a reaction in which reactants in their
standard states yield products in their standard states (6.8)
allotrope
one of two or more distinct forms of an element in the same physical state (6.8)
reference form stablest form (physical state and allotrope) of the element under standard
thermodynamic conditions (6.8)
standard enthalpy of formation (standard heat of formation) ( H fo )
enthalpy change for the
formation of one mole of a substance in its standard state from its elements in their reference forms and
in their standard states (6.8)
fuel*
(6.9)
any substance that is burned or similarly reacted to provide heat and other forms of energy
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Chapter 6: Thermochemistry
129
CHAPTER DIAGNOSTIC TEST
1.
How much heat is produced when 8.95 g C2H5OH is burned in a constant-pressure system? The
equation and enthalpy of reaction are
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
2.
ΔH = 1.367  103 kJ
Calculate the enthalpy change at 298 K for the reaction
Ni(s) + 4CO(g)  Ni(CO)4(g)
Heats of formation at 298 K are
H fo for CO = 110.5 kJ/mol
H fo for Ni(CO)4 = 605 kJ/mol
3.
Calculate ΔH° at 298 K for the reaction C (graphite) + CO2(g)  2CO(g)
using the following ΔH° data at 298 K:
H2(g) + CO(g)  C (graphite) + H2O(g)
FeO(s) + H2(g)  Fe(s) + H2O(g)
FeO(s) + CO(g)  Fe(s) + CO2(g)
ΔH° = 131.38 kJ
ΔH° = 24.69 kJ
ΔH° = 16.32 kJ
4.
What is the kinetic energy of an oxygen molecule traveling at a speed of 479 m/s in a tank at
21°C? (Hint: Use Avogadro’s number and the molar mass of O2 to get the actual mass of an
oxygen molecule.)
5.
For the reaction H2S(g) + 4H2O2(l)  H2SO4(l) + 4H2O(l), ΔH° is 1.186  103kJ. The enthalpy
change per mole of H2O2 is
6.
a.
1.301  102 kJ.
b.
4.742  103 kJ.
c.
2.965  102 kJ.
d.
4.741  103 kJ.
e.
none of the above.
Indicate whether each of the following statements is true or false. If a statement is false, change it
so that it is true.
a.
The enthalpy of reaction is independent of the exact state of the reactants or products.
True/False:________________________________________________________________
_________________________________________________________________________.
b.
If the enthalpy of reaction for N2(g) + O2(g)  2NO(g) is 180.5 kJ, then the enthalpy of
reaction for ½N2(g) + ½O2(g)  NO(g) is 90.25 kJ. True/False:
_________________________________________________________________________
_________________________________________________________________________.
c.
A calorimeter is a useful apparatus for determining heats of reaction. True/False:
_________________________________________________________________________
_________________________________________________________________________.
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130
Chapter 6: Thermochemistry
d.
For reactions involving gases and carried out at constant pressure, ΔH = qp. True/False:
_________________________________________________________________________
_________________________________________________________________________.
e.
Hess’s law permits the calculation of ΔH values for reactions from H fo values for
reactants and products and the calculation of ΔH values for hypothetical reactions.
True/False: _______________________________________________________________
_________________________________________________________________________.
f.
The H fo value for an element is always zero. True/False: __________________________
_________________________________________________________________________.
7.
Use heat of formation data in Appendix C in the text to calculate the enthalpy of the transition
from the liquid to the gaseous state for 1 mole of HCN. Report the answer in kilojoules and
kilocalories.
8.
Determine the enthalpy of ionization H Io for Cs(g) using all the thermodynamic data below. The
equation is Cs(g) 
 Cs+(g) + e(g).
F(g)+ e(g)  F(g)
F2(g)  2F(g)
Cs(s)  Cs(g)
Cs(s) + ½F2(g)  CsF(s)
CsF(s)  Cs+(g) + F(g)
9.
ΔH = 336 kJ
ΔH = 158 kJ
ΔH = 78 kJ
ΔH = 555 kJ
ΔH = 757 kJ
When 2.89 g N2H4(g) is combusted in a constant-pressure calorimeter containing exactly 1000 g
of water, a temperature increase of 6.68°C is observed. The heat capacity of the calorimeter is
1.00 kJ/°C, and the specific heat of water is 4.184J/(g∙°C). All products are gaseous. Determine
the enthalpy change per mole of N2H4 combusted.
10. A 17.9-g sample of an unknown metal was heated to 48.31°C. It was then added to 28.05 g of
water in an insulated cup. The water temperature rose from 21.04 to 23.98°C. What is the specific
heat of the metal?
11. Ethane gas, C2H6, burns in oxygen to form carbon dioxide gas, CO2, and gaseous water. For each
mole of ethane burned, 1.60 kJ of heat is evolved at constant pressure. Write the thermochemical
equation for this reaction, including labels for the states of all reactants and products.
ANSWERS TO CHAPTER DIAGNOSTIC TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
2.66  102 kJ (6.5, PS Sk. 4)
2.
163 kJ (6.8, PS Sk. 9)
3.
172.39 kJ (6.7, PS Sk. 7)
4.
6.10  1021 J/molecule (6.1, PS Sk. 1)
5.
e (6.4, PS Sk. 3)
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Chapter 6: Thermochemistry
131
6.
a.
False. The enthalpy of reaction depends on the exact state of the reactants or products. (6.4)
b.
True. (6.4, PS Sk. 3)
c.
True. (6.6)
d.
True. (6.3)
e.
True. (6.7)
f.
False. The H fo value for an element is always zero when the state of the element is the
form of the element that exists at standard conditions. (6.8)
7.
3.0  101 kJ; 7.2 kcal (6.8, PS Sk. 8)
8.
381 kJ (6.7, PS Sk. 7)
9.
383 kJ (6.6, PS Sk. 6)
10. 0.792 J/(g ∙ °C) (6.6, PS Sk. 5)
11. C2H6(g) +
7
O2(g)  2CO2(g) + 3H2O(g), ΔH = 1.60 kJ (6.4, PS Sk. 2)
2
SUMMARY OF CHAPTER TOPICS
Students find thermochemistry one of the more difficult topics in chemistry. This is said not to scare
you but to assure you that if you find yourself really scratching your head, you are not alone. It is also
said to let you know that this material is going to take a great deal of time and study to master. Your
text presents the subject quite well, but you probably will need to read it over several times before the
concepts begin to make sense. One of the biggest stumbling blocks for students is the arithmetic signs
(+ and ) that go with almost every term. If you memorize the sign conventions stressed in the text, you
will find things considerably easier. If you are stuck in your thinking or in an exercise or problem, go
back and review the sign conventions to see if your error is there.
6.1 Energy and Its Units
Learning Objectives

Define energy, kinetic energy, potential energy, and internal energy.

Define the SI unit of energy joule, as well as the common unit of energy calorie.

Calculate the kinetic energy of a moving object. (Example 6.1)

State the law of conservation of energy.
Problem-Solving Skill
1.
Calculating kinetic energy. Given the mass and speed of an object, calculate the kinetic energy
(Example 6.1).
Energy has many forms. Some of these are electromagnetic energy (such as light, heat, and x rays),
electrical energy, sound energy, gravitational potential energy, elastic potential energy, and chemical
potential energy. The forms of energy we will be concerned with in this course are electromagnetic
energy, electrical energy, and chemical potential energy.
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132
Chapter 6: Thermochemistry
Exercise 6.1
An electron, whose mass is 9.11  1031 kg, is accelerated by a positive charge to a speed of 5.0  106
m/s. What is the kinetic energy of the electron in joules? in calories?
Known:
Ek = ½mv2; 1 J = 1 kg ∙ m2/s2; 1 cal = 4.184 J
Solution:
Ek =
1J
1
 9.11  10–31 kg  (5.0  106 m/s)2 
2
1 kg  m 2 /s 2
= 1.1  10–17 J
Ek = 1.14  10–17 J 
1 cal
= 2.7  10–18 cal
4.184 J
6.2 Heat of Reaction
Heat is a difficult term to understand. The word heat makes a good verb but a poor noun. Heat is
energy in transit from a hotter object to a colder one. Objects do not possess heat. They possess energy
that can be transferred as heat. Once the energy arrives at its destination, it is absorbed and is no longer
called heat.
In an exothermic reaction, the reactants are always at a higher state of enthalpy than are the products.
Heat is given off, and the heat of reaction is negative (). In an endothermic reaction, it is just the
reverse. The reactants are at a lower state of enthalpy than are the products. Energy must be added to
the reactants to get the products. Thus the heat of the reaction is positive (+). The following diagrams
illustrate these concepts.
Exothermic Reaction
Endothermic Reaction
Learning Objectives

Define a thermodynamic system and its surroundings.

Define heat and heat of reaction.

Distinguish between an exothermic process and an endothermic process.
Exercise 6.2
Ammonia burns in the presence of a platinum catalyst to give nitric oxide, NO.
Pt
4NH3(g) + 5O2(g) 
 4NO(g) + 6H2O(l)
In an experiment, 4 mol NH3 is burned and evolves 1170 kJ of heat. Is the reaction endothermic or
exothermic? What is the value of q?
Wanted: whether reaction is endothermic or exothermic; value of q
Given:
1170 kJ of heat evolves.
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Chapter 6: Thermochemistry
Known:
133
definitions of two terms; sign is + if heat is absorbed.
Solution: Reaction is exothermic; q = 1170 kJ.
6.3 Enthalpy and Enthalpy Change
Learning Objectives

Define enthalpy and enthalpy of reaction.

Explain how the terms enthalpy of reaction and heat of reaction are related.

Explain how enthalpy and internal energy are related.
6.4 Thermochemical Equations
Learning Objectives

Define a thermochemical equation.

Write a thermochemical equation given pertinent information. (Example 6.2)

Learn the two rules for manipulating (reversing and multiplying) thermochemical equations.

Manipulate a thermochemical equation using these rules. (Example 6.3)
Problem-Solving Skills
2.
Writing thermochemical equations. Given a chemical equation, states of substances, and the
quantity of heat absorbed or evolved for molar amounts, write the thermochemical equation
(Example 6.2).
3.
Manipulating thermochemical equations. Given a thermochemical equation, write the
thermochemical equation for different multiples of the coefficients or for the reverse reaction
(Example 6.3).
Exercise 6.3
A propellant for rockets is obtained by mixing the liquids hydrazine, N2H4, and dinitrogen tetroxide,
N2O4. These compounds react to give gaseous nitrogen, N2, and water vapor, evolving 1049 kJ of heat
at constant pressure when 1mol N2O4 reacts. Write the thermochemical equation for this reaction.
Solution: The equation and heat of reaction are
2N2H4(l) + N2O4(l)  3N2(g) + 4H2O(g)
ΔH = 1049 kJ
Exercise 6.4
(a) Write the thermochemical equation for the reaction described in Exercise 6.3 for the case involving
1 mol N2H4. (b) Write the thermochemical equation for the reverse of the reaction described in Exercise
6.3.
a.
b.
Known:
Each coefficient would be divided by 2, as would ΔHrxn.
Solution:
N2H4(l) +
Solution:
3N2(g) + 4H2O(g)  2N2H4(l) + N2O4(l)
1
3
N2O4(l)  N2(g) + 2H2O(g)
2
2
Note that ΔH in this case is positive.
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ΔH = 5.245  102 kJ
ΔH = 1.049  103 kJ
134
Chapter 6: Thermochemistry
6.5 Applying Stoichiometry to Heats of Reaction
Learning Objective

Calculate the heat absorbed or evolved from a reaction given its enthalpy of reaction and
mass of a reactant or product. (Example 6.4)
Problem-Solving Skill
4.
Calculating the heat of reaction from the stoichiometry. Given the value of ΔH for a chemical
equation, calculate the heat of reaction for a given mass of reactant or product (Example 6.4).
Exercise 6.5
How much heat evolves when 10.0 g of hydrazine reacts according to the reaction described in Exercise
6.3?
Wanted: ΔH per 10.0 g hydrazine
Given:
10.0 g hydrazine; reaction from Exercise 6.3
Known:
ΔHrxn = 1.049  103 kJ per 2 mol hydrazine; formula = N2H4 = 32.0 g/mol.
Solution: Find the moles of N2H4 and then kilojoules:
10.0 g N2H4 
1 mol N 2 H 4
1.049  103 kJ

= 1.64  102 kJ
2 mol N 2 H 4
32.0 g N 2 H 4
A Chemist Looks at: Lucifers and Other Matches
Questions for Study
1.
What element must have been present in the mixture used in the first friction match?
2.
What was the purpose of the glue in the early white phosphorus match?
3.
Write the chemical reaction that occurs when the “strike anywhere” match ignites.
4.
How do safety matches differ from “strike anywhere” matches?
Answers to Questions for Study
1.
Sulfur, because the ignition produced SO2.
2.
The glue held the match mixture together and protected the white phosphorus from air.
3.
P4S3(s) + 8O2(g)  P4O10(s) + 3SO2(g)
4.
With the safety match, the striking surface contains the red phosphorus, and the match head
contains the oxidizing agent. In the “strike anywhere” match, the head contains both the
phosphorus and the oxidizing agent that ignite from the friction of rubbing the match head against
a surface.
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Chapter 6: Thermochemistry
135
6.6 Measuring Heats of Reaction
Learning Objectives

Define heat capacity and specific heat.

Relate the heat absorbed or evolved to the specific heat, mass, and temperature change.

Calculate using this relation between heat and specific heat. (Example 6.5)

Define calorimeter.

Calculate the enthalpy of reaction from calorimetric data (its temperature change and heat
capacity). (Example 6.6)
Problem-Solving Skills
5.
Relating heat and specific heat. Given any three of the quantities q, s, m, and Δt, calculate the
fourth one (Example 6.5).
6.
Calculating ΔH from calorimetric data. Given the amounts of reactants and the temperature
change of a calorimeter of specified heat capacity, calculate the heat of reaction (Example 6.6).
In calculations in this section, the temperature used is in °C. Note that whether you use the Celsius
or Kelvin scale, the number of units of temperature change is the same because both scales use the
same-sized unit.
When you calculate the temperature difference (Δt), be sure you always take tfinal  tinitial (tf  ti).
This way the sign of the heat or enthalpy always will come out correctly.
Exercise 6.6
Iron metal has a specific heat of 0.449 J/(g ∙ °C). How much heat is transferred to a 5.00-g piece of iron,
initially at 20.0°C, when it is placed in a pot of boiling water? Assume that the temperature of the water
is 100.0°C and that the water remains at this temperature, which is the final temperature of the iron.
Wanted: heat transferred (J)
Given:
5.00 g of iron; tinitial = 20.0°C; tfinal = 100.0°C; specific heat of iron is 0.449 J/(g ∙ °C).
Known:
Heat transferred = specific heat  mass  temperature change
Solution: Heat transferred =
0.449 J
 5.00 g  80.0°C = 1.80  102 J
g • C
Exercise 6.7
Suppose that 33 mL of 1.20 M HCl is added to 42 mL of a solution containing excess sodium
hydroxide, NaOH, in a coffee-cup calorimeter. The solution temperature, originally 25.0°C, rises to
31.8°C. Give the enthalpy change ΔH for the reaction
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Express the answer as a thermochemical equation. For simplicity, assume that the heat capacity and the
density of the final solution in the cup are those of water. (In more accurate work, these values must be
determined.) Also assume that the total volume of the solution equals the sum of the volumes of
HCl(aq) and NaOH(aq).
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136
Chapter 6: Thermochemistry
Wanted: ΔHrxn
Given:
33 mL 1.20 M HCl + 42 mL NaOH = 75 mL; ti = 25.0°C; tf = 31.8°C. Use density and
C of water for calorimeter (cal).
Known: Mass = density  volume; ΔHrxn = qrxn = CcalΔt; dwater = 1.0 g/mL; specific heat of
water = 4.184 J/(g ∙ °C), so use 4.18 J/(g ∙ °C) for solution.
Solution: First, find Ccal as follows, using C = specific heat  mass:
Mass soln = 75 mL 
C=
1.0 g
= 75 g
mL
4.184 J
 75 g = 3.14  102 J/°C
g • C
Then find ΔHrxn for 33 mL of 1.20 M HCl:
ΔHrxn = Ccal Δt = 3.14  102
J
 (31.8  25.0) °C = 2.14 kJ
C
To find ΔHrxn, determine the moles HCl in 33 mL of 1.20 M HCl:
0.033 L soln 
ΔHrxn =
1.20 mol HCl
= 0.0396 mol HCl
1 L soln
2.14 kJ
= 54 kJ/mol HCl
0.0396 mol HCl
The thermochemical equation is
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
ΔH = 54 kJ
6.7 Hess’s Law
Learning Objectives

State Hess’s law of heat summation.

Apply Hess’s law to obtain the enthalpy change for one reaction from the enthalpy changes
of a number of other reactions. (Example 6.7)
Problem-Solving Skill
7.
Applying Hess’s law. Given a set of reactions with enthalpy changes, calculate ΔH for a reaction
obtained from these other reactions by using Hess’s law (Example 6.7).
It is important to understand that the enthalpies (H) of the reactants or products describe the state of a
given reaction system, just as volumes (V) and temperatures (T) do. This means that the values of H
depend only on the state of the chemical system, not on the history of how it got to that state. Thus we
can use Hess’s law, combining as many reactions as needed, to come out with the specific ΔH of the
reaction wanted. The physical states of the chemical species in the reaction are important to the energy
of reaction. Be sure that you use the correct symbols and cancel them properly. Remember that the
value given for an enthalpy change (ΔH) of a reaction is specific for that written reaction: for the
physical states of each species and for the amounts of materials specified by the reaction.
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Chapter 6: Thermochemistry
137
Exercise 6.8
Manganese metal can be obtained by reaction of manganese dioxide with aluminum.
4Al(s) + 3MnO2(s)  2Al2O3(s) + 3Mn(s)
What is ΔH for this reaction? Use the following data:
2Al(s) + 3 O2(g)  Al2O3(s)
ΔH = 1676 kJ
Mn(s) + O2(g)  MnO2(s)
ΔH =  521 kJ
2
Wanted: ΔHrxn
Given:
reactions and enthalpies
Known: Use Hess’s law. Reversing an equation changes the sign of ΔH; if you multiply
coefficients by a factor, ΔHrxn must be multiplied by the same factor.
Solution: Write equation (1), multiplying coefficients and ΔH by 2; reverse equation (2),
changing the sign of ΔH, and multiply the coefficients and ΔH by 3. Then add the equations and
ΔH.
4Al(s) + 3O2(g)  2Al2O3(s)
3MnO2(s)  3Mn(s) + 3O2(g)
4Al(s) + 3MnO2(s)  2Al2O3(s) + 3Mn(s)
(1)
(2)
ΔH = 3352 kJ
ΔH = 1563 kJ
ΔH = 1789 kJ
6.8 Standard Enthalpies of Formation
Learning Objectives

Define standard state and reference form.

Define standard enthalpy of formation.

Calculate the heat of a phase transition using standard enthalpies of formation for the
different phases. (Example 6.8)

Calculate the heat (enthalpy) of reaction from the standard enthalpies of formation of the
substances in the reaction. (Example 6.9)
Problem-Solving Skills
8.
Calculating the heat of phase transition from standard enthalpies of formation. Given a table
of standard enthalpies of formation, calculate the heat of phase transition (Example 6.8).
9.
Calculating the enthalpy of reaction from standard enthalpies of formation. Given a table of
standard enthalpies of formation, calculate the enthalpy of reaction (Example 6.9).
The summarizing statement to remember for calculating enthalpy changes for reactions is “the enthalpy
change for a reaction at standard conditions is the sum of the enthalpies of formation of the products
minus the sum of the enthalpies of formation of the reactants.” It is written
o
ΔH = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
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138
Chapter 6: Thermochemistry
where the Greek letter Σ (capital sigma) indicates summation. Memorize this equation. Later, you will
see other summation statements in similar form. As your text cautions, pay particular attention to signs.
This will be your greatest source of error.
Exercise 6.9
o
Calculate the heat of vaporization Δ H vap
of water using standard enthalpies of formation (text Table
6.2).
Known:
ΔH° = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
Solution: Write the reaction. Then write the Δ H fo values under each formula and subtract as
indicated above.
H2O(l)  H2O(g)
285.8
241.8
ΔH° = Δ H
o
vap
(kJ)
= Δ H [H2O(g)]  Δ H fo [H2O(l)]
o
f
= 241.8 kJ – (285.8 kJ)
= 44.0 kJ per mole of water vaporized
Exercise 6.10
Calculate the enthalpy change for the following reaction:
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
Use standard enthalpies of formation.
Known:
ΔH° = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
Solution: Write Δ H fo under each species in the equation.
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
3  33.1 285.8
2  207.4
90.3
(kJ)
Then use the ΔH equation:
ΔH° = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
= [2(207.4) + 90.3] – [3(33.1) + (–285.8)] kJ
= –138 kJ
You have now seen two methods for calculating the enthalpy change of a reaction: (1) by using Hess’s
law and enthalpies of other reactions and (2) by using heats of formation of reactants and products.
Exercise 6.11
Calculate the standard enthalpy change for the reaction of an aqueous solution of barium hydroxide,
Ba(OH)2, with an aqueous solution of ammonium nitrate, NH4NO3, at 25°C. (Figure 6.1 illustrated this
reaction using solids instead of solutions.) The complete ionic equation is
Ba2+(aq) + 2OH–(aq) + 2NH4+(aq) + 2NO3–(aq)  2NH3(g) + 2H2O(l) + Ba2+(aq) + 2NO3–(aq)
Known:
ΔH° = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
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Chapter 6: Thermochemistry
139
Solution: Write Δ H fo under each species in the net ionic equation:
2OH–(aq) + 2NH4+(aq)  2NH3(g) + 2H2O(l)
2  230.0 2  132.5 2  45.9 2  285.8 (kJ)
Then use the ΔH° equation:
ΔH° = (2{Δ H fo [NH3(g)]} + {Δ H fo [H2O(l)]})
– (2{Δ H fo [OH–(aq)]} + {Δ H fo [NH4+(aq)]})
= [2(45.9) + 2(285.8)] – [2(230.0) + 2(132.5)] kJ
= [(663.4) – (725.0)] kJ = 61.6 kJ
6.9 Fuels—Foods, Commercial Fuels, and Rocket Fuels
Learning Objectives

Define fuel.

Describe the three needs of the body that are fulfilled by foods.

Give the approximate average values quoted (per gram) for the heat values (heats of
combustion) for fats and for carbohydrates.

List the three major fossil fuels.

Describe the processes of coal gasification and coal liquefaction.

Describe some fuel–oxidizer systems used in rockets.
ADDITIONAL PROBLEMS
1.
When butane gas, C4H10, burns in oxygen gas, O2, to form carbon dioxide gas, CO2, and gaseous
water, 2845 kJ of thermal energy is evolved per mole of butane burned.
a.
Write the thermochemical equation for the reaction using whole-number coefficients.
b.
Calculate the thermal energy evolved when 25.0 g of butane is burned.
2.
A 100.0-g sample of water at 25.30°C was placed in an insulated cup. Then 45.00 g of lead pellets
at 100.00°C was added. The final temperature of the water was 34.34°C. What is the specific heat
of lead?
3.
A 23.3-g sample of copper at 75.7°C is added to 100.0 mL of benzene at 20.0°C. Assuming that
no thermal energy is used in evaporation, calculate the final temperature of the benzene. [The
specific heat of copper is 0.389 J/(g ∙ °C), the specific heat of benzene is 1.70 J/(g ∙ °C), and the
density of benzene is 0.879 g/mL.] How would this value change if we took into account the
energy that actually would go toward the increased evaporation?
4.
A 0.875-g sample of anthracite coal was burned in a bomb calorimeter. The temperature rose from
22.50 to 23.80°C. The heat capacity of the calorimeter was found in another experiment to be 20.5
kJ/°C.
a.
What was the heat evolved by the reaction?
b.
What is the energy released on burning 1 metric ton (exactly 1000 kg) of this type of coal?
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140
5.
Chapter 6: Thermochemistry
POCl3 is formed according to the following reaction:
PCl5(g) + H2O(g)  POCl3(g) + 2HCl(g)
The enthalpy of reaction is 126 kJ per mole of POCl3 formed. Write the thermochemical
equation for the reverse reaction, doubling the coefficients.
6.
What is the ratio of speeds of H2 and N2 molecules at 125°C?
7.
Write the equation that represents the heat of formation of each of the following species. Include
states of matter.
8.
a.
Sb2O3(s)
b.
C6H12O6(s)
c.
N2H4(l)
d.
SF6(g)
e.
Ca(NO3)2(s)
Calculate ΔH° for the reaction
2NO(g) + 2CO(g)  N2(g) + 2CO2(g)
given that Δ H fo [NO(g)] = 90.3 kJ/mol, Δ H fo [CO(g)] = 110.5 kJ/mol, and Δ H fo [CO2(g)] =
393.5 kJ/mol.
9.
Using Δ H fo data given in text Table 6.2, calculate the amount of heat released from the
combustion of a mixture of 15.0 g C5H12(l), n-pentane, and 66.6 g O2(g).
Δ H fo [C5H12(l)] = 173.2 kJ/mol
10. Calculate the number of grams of natural gas at 25.0°C that are required to heat 1.00kg H2O from
25.0 to 100.0°C. Assume that the natural gas is 80.0% CH4 and 20.0% C2H6 by mass.
Δ H fo [CH4(g)] = 74.9 kJ/mol
Δ H fo [C2H6(g)] = 84.7 kJ/mol,
and the specific heat of liquid water is 4.184 J/(g ∙ °C).
ANSWERS TO ADDITIONAL PROBLEMS
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
a.
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
ΔH = 5.690  103 kJ (6.4, PS Sk. 2)
b.
2.
25.0 g C4H10 
1 mol C4 H10
2845 kJ

= 1.22  103 kJ (6.5, PS Sk. 4)
58.1 g C4 H10
mol C4 H10
ΔHlead = ΔHwater
Final temperature of both is the same.
–(45.0 g) (specific heat) (34.34  100.00)°C =
(100.0 g) [4.184 (J/g ∙ °C)] (34.34 – 25.30) °C
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 6: Thermochemistry
Specific heat =
3.
(100.0)(4.184 J)(9.04)
= 1.28 J/(g ∙ °C) (6.6, PS Sk. 5)
(45.0 g)(65.66 C)
ΔHcopper = ΔHbenzene
–(23.3 g) [0.389 J/(g ∙ °C)] (tf  75.7°C)
= (100.0 mL  0.879 g/mL) [1.70 J/(g ∙ °C)] (tf  20.0°C)
9.064tf + 686.1°C = 149.4tf  2989°C
tf = 23.2°C
The final temperature would be lower because not all the energy would go to raising the
temperature of the benzene. (6.6)
4.
a.
ΔHrxn = ΔHcal
ΔHcal = 20.5 kJ/°C  (23.80 – 22.50) °C = 26.6 kJ
ΔHrxn = 26.6 kJ
b.
106 g 
26.6 kJ
= 3.04  107 kJ (6.6, PS Sk. 6)
0.875 g
5.
4HCl(g) + 2POCl3(g)  2H2O(g) + 2PCl5(g)
6.
The kinetic energies of gases are the same at the same temperature, so
ΔH = 252 kJ (6.4, PS Sk. 3)
Ek (N2) = Ek (H2)
1
1
mN2 vN2 2 = mH2 vH2 2
2
2
mN2
mH2
=
vH2 2
vN 2 2
and since we are interested in the mass ratio, we can use molar masses. Thus
v
M m (N 2 )
= H2 =
M m (H 2 )
vN 2
28.0 g/mol
= 3.72 (6.1)
2.02 g/mol
7.
8.
3
O2(g)  Sb2O3(s)
2
a.
2Sb(s) +
b.
6C (graphite) + 6H2(g) + 3O2(g)  C6H12O6(s)
c.
N2(g) + 2H2(g)  N2H4(l)
d.
S(s)+ 3F2(g)  SF6(g)
e.
Ca(s) + N2(g) + 3O2(g)  Ca(NO3)2(s) (6.8)
ΔH° = ΣnΔ H fo (products) – ΣmΔ H fo (reactants)
= [0 + 2(393.5)]  [2(90.3) + 2(110.5)]
= 787.0  (180.6  221.0) = 746.6 kJ (6.8, PS Sk. 9)
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141
142
9.
Chapter 6: Thermochemistry
The balanced equation for the combustion of n-pentane is
C5H12(l) + 8O2(g)  5CO2(g) + 6H2O(l)
o
The calculation of Δ H rxn
is
ΔH° = 5(393.5 kJ) + 6(285.8 kJ)  (173.2 kJ)  8(0 kJ) = 3509.1 kJ
Find the limiting reagent:
15.0 g C5H12 
66.6 g O2 
1 mol C5 H12
5 mol CO2

= 1.040 mol CO2
1 mol C5 H12
72.1 g C5 H12
1 mol O 2
5 mol CO2

= 1.301 mol CO2
32.0 g O 2
8 mol O2
C5H12 is the limiting reagent. The amount of heat released in the combustion of 0.208 mol C5H12
is
15.0 g C5H12 
1 mol C5 H12
3509.1 kJ

= 7.30  102 kJ (6.8, PS Sk. 4, 9)
1 mol C5 H12
72.1 g C5 H12
10. Calculate the heats of combustion of CH4 and C2H6.
2O2(g) + CH4(g)  CO2(g) + 2H2O(l)
ΔH°= 393.5 kJ + 2(285.8 kJ)  (74.9 kJ) = 890.2 kJ
7 O2(g) + C2H6(g)  2CO2(g) + 3H2O(l)
2
ΔH° = 2(393.5 kJ) + 3(285.8 kJ)  (84.68 kJ) = 1559.7 kJ
The amount of heat required to raise the temperature of 1.00 kg of liquid water
(100.0  25.0)°C = 75.0°C is
75.0°C  1.00 kg  4.184 J/(g ∙ °C)  1000 g/kg  1 kJ/1000 J = 3.138  102 kJ
The heat released from the combustion of 1 g of natural gas (0.800 g CH4 and 0.200 g C2H6) is
1 mol CH 4
0.800 g CH 4
890.2 kJ


1 mol CH 4
16.0 g CH 4
1 g gas
+
1 mol C2 H6
0.200 g C2 H 6
1559.7 kJ


= 54.87 kJ/g gas
30.1 g C2 H6
1 mol C2 H6
1 g gas
The amount of natural gas needed to provide 313.8 kJ of heat is
313.8 kJ 
g gas
= 5.72 g gas (6.6, 6.8, PS Sk. 5, 6, 9)
54.87 kJ
CHAPTER POST-TEST
1.
Calculate the heat produced when 3.76 g CO at 298 K reacts with an excess of Ni to give Ni(CO)4
in a constant-pressure system.
Ni(s) + 4CO(g)  Ni(CO)4(g)
ΔH° = 163 kJ
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Chapter 6: Thermochemistry
2.
143
Calculate the enthalpy change at 298 K for the reaction
CaO(s) + 2HCl(g)  CaCl2(s) + H2O(l)
from the following heats of formation at 298 K:
CaO
HCl
3.
Δ H fo
635.1 kJ/mol
92.31 kJ/mol
CaCl2
H2O
Determine Δ H fo for CH4(g) from the following ΔH° data and corresponding reactions at 298 K:
C (graphite) + O2(g)  CO2(g)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H2(g) + 1 O2(g)  H2O(l)
2
4.
5.
Δ H fo
795.0 kJ/mol
258.8 kJ/mol
ΔH° = 393.55 kJ
ΔH° = 890.36 kJ
ΔH° = 285.85 kJ
If we consider a hot kitchen stove as a system, then the transfer of heat from the stove to the
kitchen is taken to be ___________________________.
(positive/negative)
Which of the following is false? For the false statement, change it so it is true.
a.
If qp for a reaction is negative, the reaction is endothermic. _________________________
_________________________________________________________________________
_________________________________________________________________________
b.
ΔH is the amount of heat released or absorbed by a system at constant pressure.
_________________________________________________________________________
c.
If heat is evolved in a reaction, the reaction is exothermic. __________________________
_________________________________________________________________________
d.
ΔH = H (products) – H (reactants). _____________________________________________
_________________________________________________________________________
e.
Enthalpy is a state function. __________________________________________________
_________________________________________________________________________
6.
The enthalpy change for a reaction system open to the atmosphere is
a.
dependent on the identity of the reactants and products, and/or
b.
zero, and/or
c.
negative, and/or
d.
qp, and/or
e.
positive.
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144
7.
Chapter 6: Thermochemistry
The following thermochemical equation for the decomposition of gaseous ammonia, NH3,
NH3(g)  1 N2(g) + 3 H2(g)
2
2
ΔH = 45.9 kJ
indicates that the formation of gaseous ammonia
8.
a.
evolves 45.9 kJ for each mole of ammonia formed, and/or
b.
evolves 23 kJ for each mole of nitrogen used, and/or
c.
absorbs 45.9 kJ for each mole of ammonia formed, and/or
d.
absorbs 23 kJ for each mole of nitrogen used, and/or
e.
is an exothermic process.
The following reaction is carried out in a bomb calorimeter:
CaO(s) + SO3(g)  CaSO4(s)
If 1.00 mol SO3 is reacted with an excess of CaO, 2.000  103 g of water increases in temperature
from 17.5 to 55.5°C. The calorimeter has a measured heat capacity of 0.126 kJ/°C. Calculate the
heat released in this reaction. [Specific heat of water is 4.184 J/(g ∙ °C).]
9.
What is the speed of a CCl4 molecule at 22°C when its kinetic energy is 6.13  10–21 J/molecule?
10. SiH4(g) reacts with gaseous oxygen to form SiO2(s) and liquid water. How much heat is liberated
when 5.75 g SiH4 burns in an excess of oxygen at 298 K?
Δ H fo [SiH4(g)] = +34.3 kJ/mol.
11. In a blast furnace, solid iron(III) oxide, Fe2O3, is reduced by gaseous carbon monoxide, CO, to
liquid iron and gaseous carbon dioxide, CO2. For each mole of iron oxide reduced, 27.6 kJ of heat
is evolved. Write the complete equation for this reaction, including the heat of reaction and labels
for the states of all reactants and products.
12. A 48.0-g sample of copper was heated to 98.5°C. It was then added to 105 g of water in an
insulated container. The temperature of the water rose from 25.6 to 28.8°C. What is the specific
heat of copper?
ANSWERS TO CHAPTER POST-TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
5.47 kJ (6.5, PS Sk. 4)
2.
261.1 kJ (6.8, PS Sk. 9)
3.
74.89 kJ/mol (6.7, 6.8, PS Sk. 7)
4.
negative (6.2)
5.
a.
If qp for a reaction is negative, the reaction is exothermic. (6.2)
6.
a and d (6.3)
7.
a and e (6.4, PS Sk. 3)
8.
323 kJ released (6.6, PS Sk. 6)
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Chapter 6: Thermochemistry
9.
219 m/s (6.1, PS Sk. 1)
10. 272 kJ (6.5, 6.8, PS Sk. 4, 9)
11. Fe2O3(s) + 3CO(g)  2Fe(l) + 3CO2(g)
12. 0.42 J/(g ∙ °C) (6.6, PS Sk. 5, 6)
Copyright © Houghton Mifflin Company. All rights reserved.
ΔH = 27.6 kJ (6.4, PS Sk. 2)
145