CHAPTER 6
Thermochemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
6.1. Substitute into the formula Ek = ½mv2 using SI units:
Ek = ½ 9.11 10−31 kg (5.0 106 m/s)2 = 1.13 10−17 = 1.1 10−17 J
1 cal
1.13 10−17 J = 2.72 10−18 = 2.7 10−18 cal
4.184 J
6.2. The change in the internal energy of the system, U, equals q + w. Relative to the system, q is
negative because the system gives off heat and w is positive because the weights do work on the
system by compressing the gas. The value of q is straightforward, i.e., −1.50 J. The value of w is
equal to the change in the gravitational potential energy of the weights, mgh. Thus,
9.81 m
U = q + w = (−1.50 J) + (2.20 kg 0.250 m) = −1.50 J + 5.396 J = +3.90 J
s2
6.3. Heat is evolved; therefore, the reaction is exothermic. The value of q is −1170 kJ.
6.4 The change in volume, V, can be calculated by taking the difference in the volume of gas before
and after the reaction, using the ideal gas equation in each case:
n final RT ninitial RT (n final ninitial ) RT
V V final Vinitial
P P P
L•atm
(1 mol 3 mol)(0.08206 )(350 K)
V = mol K 57.44 L
1.00 atm
Because the change is from two moles of gas to one mole of gas, this represents a compression
being performed on the system. In other words, work is done on the system so should be positive.
This pressure-volume work is calculated as follows:
w = −P V = −(1.00 atm) (−57.44 L) = −(1.01 105 Pa) (−57.44 10-3 m3)
w = + 5801 J = + 5.80 kJ
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 6: Thermochemistry 201
By the first law of thermodynamics, U = q + w. We are given q directly in the problem
statement, i.e., q = −885.5 kJ. Combining q and w, U can now be calculated.
U = q + w = −885.5 kJ + 5.80 kJ = −879.70 = −879.7 kJ
6.5. The balanced thermochemical equation is
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g); H = −1049 kJ
6.6. a. N2H4(l) + 1/2N2O4(l) 3/2N2(g) + 2H2O(g); H = −524.5 kJ
b. 4H2O(g) + 3N2(g) 2N2H4(l) + N2O4(l); H = 1049 kJ
6.7. The reaction is
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g); H = −1049 kJ
1 mol N 2 H 4 1 mol N 2 O4 1049 kJ
10.0 g N2H4 = −163.7 = −164 kJ
32.05 g 2 mol N 2 H 4 1 mol N 2 O4
6.8. Substitute into the equation q = s m t to obtain the heat transferred. The temperature change
is
t = tf − ti = 100.0°C − 20.0°C = 80.0°C
Therefore,
q = s m t = 0.449 J/(g°C) 5.00 g 80.0°C = 1.796 102 = 1.80 102 J
6.9. The total mass of the solution is obtained by adding the volumes together and by using the
density of water (1.000 g/mL). This gives 33 + 42 = 75 mL, or 75 g. The heat absorbed by the
solution is
q = s m t = 4.184 J/(g°C) 75 g (31.8°C − 25.0°C) = 2133.8 J
The heat released by the reaction, qrxn, is equal to the negative of this value, or −2133.8 J. To
obtain the enthalpy change for the reaction, you need to calculate the moles of HCl that reacted.
This is
Mol HCl = 1.20 mol/L 0.033 L = 0.0396 mol
The enthalpy change for the reaction can now be calculated.
2133.8 J
H = = −53884 J/mol = −54 kJ/mol
0.0396 mol
Expressing this result as a thermochemical equation, you have
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l); H = −54 kJ
6.10. Use Hess's law to find H for 4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s) from the following data
for equations 1 and 2:
2Al(s) + 3/2O2(g) Al2O3(s); H = −1676 kJ (1)
Mn(s) + O2(g) MnO2(s); H = −520 kJ (2)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 202 Chapter 6: Thermochemistry
If you take equation 1 and multiply it by 2, you obtain
4Al(s) + 3O2(g) 2Al2O3(s); H = 2 (−1676 kJ) = −3352 kJ
Since the desired reaction has three MnO2 on the left side, reverse equation 2 and multiply it by 3.
The result is
3MnO2(s) 3Mn(s) + 3O2(g) H = −3 (−520 kJ) = 1560 kJ
If you add the two equations and corresponding enthalpy changes, you obtain the enthalpy change
of the desired equation.
4Al(s) + 3O2(g) 2Al2O3(s)
H = −3352 kJ
3MnO2(s) 3Mn(s) + 3O2(g)
H = 1560 kJ
4Al(s) +3MnO2(s) 2Al2O3(s) + 3Mn(s)
H = −1792 kJ
6.11. The vaporization process, with the Hf values given below the substances, is
The calculation is
H vap
o
= n H of (products) − m H of (reactants)
= (1 mol) H of [H2O(g)] − (1mol) H of [H2O(l)]
= (−241.8 kJ) − (−285.8 kJ) = 44.0 kJ
6.12. The reaction, with the Hf values given below the substances, is
3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)
33.10 -285.8 -207.4 90.29 (kJ)
The calculation is
H rxn
o
= n H of (products) − m H of (reactants)
H rxn
o
= [(2 mol) H of (HNO3) + (1 mol) H of (NO)] − [(3 mol) H of (NO2) + (1 mol) H of (H2O)
H rxn
o
= [2(−207.4) + (90.29)] kJ − [3(33.10) + (−285.8)] kJ = −138.01 = −138.0 kJ
6.13. The net chemical reaction, with the Hf values given below the substances, is
2 NH4+(aq) + 2OH-(aq) 2NH3(g) + 2H2O(l)
2(-132.5) 2(-230.0) 2(-45.90) 2(-285.8) (kJ)
The calculation is
H rxn
o
= [2 H of (NH3) + 2 H of (H2O)] − [2 H of (NH4+) + 2 H of (OH−)]
= [2(−45.90) + 2(−285.8)] − [2(−132.5) + 2(−230.0)] = 61.60 = 61.6 kJ
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 6: Thermochemistry 203
■ ANSWERS TO CONCEPT CHECKS
6.1. The photovoltaic cells collect the sun’s energy, converting it to electrical energy. This electrical
energy is stored in the battery as chemical energy, which is later changed back to electrical
energy that runs a motor. As the motor rotates, it changes the electrical energy to kinetic energy
(energy of motion) of the motor, then of water, which in turn is changed to potential energy
(energy of position) of water as the water moves upward in the gravitational field of earth.
6.2. The change in the internal energy of the system, U, equals q + w. Relative to the system, q is
negative because the system evolves (gives off) heat . The value of q is −89.1 kJ. Because the
weights do work on the system by compressing the gas, w is positive relative to the system. The
value of w is +2.5 kJ Thus,
U = q + w = (−89.1 kJ) + (+2.5 kJ) = −86.6 kJ (Selection d is the correct answer.)
6.3. a. This reaction is the one shown in the problem, and it has a positive H, so the reaction is
endothermic.
b. This reaction is simply twice reaction a, so it is also endothermic.
c. This reaction is the reverse of reaction a, so it is exothermic.
d. This reaction is twice reaction c, so it is more exothermic than reaction c. Thus, reaction d
is the most exothermic reaction.
6.4. You can think of the sublimation of ice as taking place in two stages. First, the solid melts to
liquid; then the liquid vaporizes. The first process has an enthalpy Hfus. The second process has
an enthalpy Hvap. Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum
of these two enthalpies:
Hsub = Hfus + Hvap
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
6.1. Energy is the potential or capacity to move matter. Kinetic energy is the energy associated with
an object by virtue of its motion. Potential energy is the energy an object has by virtue of its
position in a field of force. Internal energy is the sum of the kinetic and potential energies of the
particles making up a substance.
6.2. In terms of SI base units, a joule is kgm2/s2.
6.3. Originally, a calorie was defined as the amount of energy required to raise the temperature of one
gram of water by one degree Celsius. At present, the calorie is defined as 4.184 J.
6.4. At either of the two highest points above the earth in a pendulum's cycle, the energy of the
pendulum is all potential energy and is equal to the product mgh (m = mass of pendulum,
g = constant acceleration of gravity, and h = height of pendulum). As the pendulum moves
downward, its potential energy decreases from mgh to near zero, depending on how close it
comes to the earth's surface. During the downward motion, its potential energy is converted to
kinetic energy. When it reaches the lowest point (middle) of its cycle, the pendulum has its
maximum kinetic energy and minimum potential energy. As it rises above the lowest point, its
kinetic energy begins to be converted to potential energy. When it reaches the other high point in
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Thermochemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
6.1. Substitute into the formula Ek = ½mv2 using SI units:
Ek = ½ 9.11 10−31 kg (5.0 106 m/s)2 = 1.13 10−17 = 1.1 10−17 J
1 cal
1.13 10−17 J = 2.72 10−18 = 2.7 10−18 cal
4.184 J
6.2. The change in the internal energy of the system, U, equals q + w. Relative to the system, q is
negative because the system gives off heat and w is positive because the weights do work on the
system by compressing the gas. The value of q is straightforward, i.e., −1.50 J. The value of w is
equal to the change in the gravitational potential energy of the weights, mgh. Thus,
9.81 m
U = q + w = (−1.50 J) + (2.20 kg 0.250 m) = −1.50 J + 5.396 J = +3.90 J
s2
6.3. Heat is evolved; therefore, the reaction is exothermic. The value of q is −1170 kJ.
6.4 The change in volume, V, can be calculated by taking the difference in the volume of gas before
and after the reaction, using the ideal gas equation in each case:
n final RT ninitial RT (n final ninitial ) RT
V V final Vinitial
P P P
L•atm
(1 mol 3 mol)(0.08206 )(350 K)
V = mol K 57.44 L
1.00 atm
Because the change is from two moles of gas to one mole of gas, this represents a compression
being performed on the system. In other words, work is done on the system so should be positive.
This pressure-volume work is calculated as follows:
w = −P V = −(1.00 atm) (−57.44 L) = −(1.01 105 Pa) (−57.44 10-3 m3)
w = + 5801 J = + 5.80 kJ
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 6: Thermochemistry 201
By the first law of thermodynamics, U = q + w. We are given q directly in the problem
statement, i.e., q = −885.5 kJ. Combining q and w, U can now be calculated.
U = q + w = −885.5 kJ + 5.80 kJ = −879.70 = −879.7 kJ
6.5. The balanced thermochemical equation is
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g); H = −1049 kJ
6.6. a. N2H4(l) + 1/2N2O4(l) 3/2N2(g) + 2H2O(g); H = −524.5 kJ
b. 4H2O(g) + 3N2(g) 2N2H4(l) + N2O4(l); H = 1049 kJ
6.7. The reaction is
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g); H = −1049 kJ
1 mol N 2 H 4 1 mol N 2 O4 1049 kJ
10.0 g N2H4 = −163.7 = −164 kJ
32.05 g 2 mol N 2 H 4 1 mol N 2 O4
6.8. Substitute into the equation q = s m t to obtain the heat transferred. The temperature change
is
t = tf − ti = 100.0°C − 20.0°C = 80.0°C
Therefore,
q = s m t = 0.449 J/(g°C) 5.00 g 80.0°C = 1.796 102 = 1.80 102 J
6.9. The total mass of the solution is obtained by adding the volumes together and by using the
density of water (1.000 g/mL). This gives 33 + 42 = 75 mL, or 75 g. The heat absorbed by the
solution is
q = s m t = 4.184 J/(g°C) 75 g (31.8°C − 25.0°C) = 2133.8 J
The heat released by the reaction, qrxn, is equal to the negative of this value, or −2133.8 J. To
obtain the enthalpy change for the reaction, you need to calculate the moles of HCl that reacted.
This is
Mol HCl = 1.20 mol/L 0.033 L = 0.0396 mol
The enthalpy change for the reaction can now be calculated.
2133.8 J
H = = −53884 J/mol = −54 kJ/mol
0.0396 mol
Expressing this result as a thermochemical equation, you have
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l); H = −54 kJ
6.10. Use Hess's law to find H for 4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s) from the following data
for equations 1 and 2:
2Al(s) + 3/2O2(g) Al2O3(s); H = −1676 kJ (1)
Mn(s) + O2(g) MnO2(s); H = −520 kJ (2)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 202 Chapter 6: Thermochemistry
If you take equation 1 and multiply it by 2, you obtain
4Al(s) + 3O2(g) 2Al2O3(s); H = 2 (−1676 kJ) = −3352 kJ
Since the desired reaction has three MnO2 on the left side, reverse equation 2 and multiply it by 3.
The result is
3MnO2(s) 3Mn(s) + 3O2(g) H = −3 (−520 kJ) = 1560 kJ
If you add the two equations and corresponding enthalpy changes, you obtain the enthalpy change
of the desired equation.
4Al(s) + 3O2(g) 2Al2O3(s)
H = −3352 kJ
3MnO2(s) 3Mn(s) + 3O2(g)
H = 1560 kJ
4Al(s) +3MnO2(s) 2Al2O3(s) + 3Mn(s)
H = −1792 kJ
6.11. The vaporization process, with the Hf values given below the substances, is
The calculation is
H vap
o
= n H of (products) − m H of (reactants)
= (1 mol) H of [H2O(g)] − (1mol) H of [H2O(l)]
= (−241.8 kJ) − (−285.8 kJ) = 44.0 kJ
6.12. The reaction, with the Hf values given below the substances, is
3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)
33.10 -285.8 -207.4 90.29 (kJ)
The calculation is
H rxn
o
= n H of (products) − m H of (reactants)
H rxn
o
= [(2 mol) H of (HNO3) + (1 mol) H of (NO)] − [(3 mol) H of (NO2) + (1 mol) H of (H2O)
H rxn
o
= [2(−207.4) + (90.29)] kJ − [3(33.10) + (−285.8)] kJ = −138.01 = −138.0 kJ
6.13. The net chemical reaction, with the Hf values given below the substances, is
2 NH4+(aq) + 2OH-(aq) 2NH3(g) + 2H2O(l)
2(-132.5) 2(-230.0) 2(-45.90) 2(-285.8) (kJ)
The calculation is
H rxn
o
= [2 H of (NH3) + 2 H of (H2O)] − [2 H of (NH4+) + 2 H of (OH−)]
= [2(−45.90) + 2(−285.8)] − [2(−132.5) + 2(−230.0)] = 61.60 = 61.6 kJ
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 6: Thermochemistry 203
■ ANSWERS TO CONCEPT CHECKS
6.1. The photovoltaic cells collect the sun’s energy, converting it to electrical energy. This electrical
energy is stored in the battery as chemical energy, which is later changed back to electrical
energy that runs a motor. As the motor rotates, it changes the electrical energy to kinetic energy
(energy of motion) of the motor, then of water, which in turn is changed to potential energy
(energy of position) of water as the water moves upward in the gravitational field of earth.
6.2. The change in the internal energy of the system, U, equals q + w. Relative to the system, q is
negative because the system evolves (gives off) heat . The value of q is −89.1 kJ. Because the
weights do work on the system by compressing the gas, w is positive relative to the system. The
value of w is +2.5 kJ Thus,
U = q + w = (−89.1 kJ) + (+2.5 kJ) = −86.6 kJ (Selection d is the correct answer.)
6.3. a. This reaction is the one shown in the problem, and it has a positive H, so the reaction is
endothermic.
b. This reaction is simply twice reaction a, so it is also endothermic.
c. This reaction is the reverse of reaction a, so it is exothermic.
d. This reaction is twice reaction c, so it is more exothermic than reaction c. Thus, reaction d
is the most exothermic reaction.
6.4. You can think of the sublimation of ice as taking place in two stages. First, the solid melts to
liquid; then the liquid vaporizes. The first process has an enthalpy Hfus. The second process has
an enthalpy Hvap. Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum
of these two enthalpies:
Hsub = Hfus + Hvap
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
6.1. Energy is the potential or capacity to move matter. Kinetic energy is the energy associated with
an object by virtue of its motion. Potential energy is the energy an object has by virtue of its
position in a field of force. Internal energy is the sum of the kinetic and potential energies of the
particles making up a substance.
6.2. In terms of SI base units, a joule is kgm2/s2.
6.3. Originally, a calorie was defined as the amount of energy required to raise the temperature of one
gram of water by one degree Celsius. At present, the calorie is defined as 4.184 J.
6.4. At either of the two highest points above the earth in a pendulum's cycle, the energy of the
pendulum is all potential energy and is equal to the product mgh (m = mass of pendulum,
g = constant acceleration of gravity, and h = height of pendulum). As the pendulum moves
downward, its potential energy decreases from mgh to near zero, depending on how close it
comes to the earth's surface. During the downward motion, its potential energy is converted to
kinetic energy. When it reaches the lowest point (middle) of its cycle, the pendulum has its
maximum kinetic energy and minimum potential energy. As it rises above the lowest point, its
kinetic energy begins to be converted to potential energy. When it reaches the other high point in
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
