South Pasadena • AP Chemistry
Name
6 ▪ Thermodynamics
Period
6.3
PROBLEMS
1. We can summarize the relationship between
thermodynamic and equilibrium quantities.
a. Table 1: ∆G gives the direction of the reaction.
When
Direction of spontaneous reaction Q vs. K
∆G < 0
Forward
Q<K
∆G = 0
At Equilibrium
Q=K
∆G > 0
Reverse
Q>K
b. Table 2: ∆G° gives the magnitude of Keq.
When
Magnitude of K
∆G° < 0
K>1
∆G° = 0
K=1
∆G° > 0
K<1
3. One of the reactions that destroys ozone in the
upper atmosphere is
NO (g) + O3 (g)  NO2 (g) + O2 (g)
a. Using the Table of Thermodynamic Values,
calculate the value of ∆G°rxn at 25°C. Is this
process spontaneous? Why?
∆G° = [1(+34) + 1(0)] – [1(90) + 1(143)]
= −199 kJ/mol
The process is spontaneous because
∆G° < 0.
b. Calculate the value of the equilibrium
constant, Kp, at this temperature.
−∆G°rxn/R·T
−(−199000/(8.314)(298))
=e
= 7.63 × 1034
b. Calculate the values of ∆G°rxn and Kp at 25°C.
∆G° = ∆H° − T·∆S°
= (−92 kJ/mol)–(298 K)(−0.199 kJ/mol·K)
= −32.7 kJ/mol
−∆G°rxn/R·T
Keq = e
−(−32700/(8.314)(298))
= e13.2 = 5.39 × 105
c. Calculate the values of ∆G°rxn and Kp at
227°C.
∆G° = ∆H° − T·∆S°
= (−92 kJ/mol)–(500 K)(−0.199 kJ/mol·K)
= +7.5 kJ/mol
−∆G°rxn/R·T
Keq = e
=e
NON-STANDARD STATES
Keq = e
2. Consider the following process:
N2 (g) + 3 H2 (g)  2 NH3 (g)
a. Using the Table of Thermodynamic Values,
calculate the values of ∆H°rxn and ∆S°.
∆H° = [2(−46)] – [1(0) + 3(0)] = −92 kJ/mol
∆S° = [2(193)] – [1(192) + 3(131)]
= −199 J/mol·K
=e
–
Date
−(+7500/(8.314)(500))
= e−1.804 = 0.165
= e80.3
c. Calculate the value of ∆G at the following
conditions at 25°C:
PNO = 1.00 × 10−6 atm
PO3 = 2.00 × 10−6 atm
PNO2 = 1.00 × 10−7 atm
PO2 = 1.00 × 10−3 atm
(PNO2)(PO2) (1.00 × 10−7)(1.00 × 10−3)
Q=
=
(PNO)(PO3) (1.00 × 10−6)(2.00 × 10−6)
= 50
∆Grxn = ∆G°rxn + R·T·ln Q
8.314
= (−199) + 
 1000  (298) ln(50)
= −199 + 9.7 = −189 kJ/mol
AP Chemistry 2011B #3
Answer the following questions about glucose, C6H12O6, an important biochemical energy source.
(a) Write the empirical formula of glucose.
CH2O
In many organisms, glucose is oxidized to carbon dioxide and water, as represented by the following equation.
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
A 2.50 g sample of glucose and an excess of O2(g) were placed in a calorimeter. After the reaction was initiated
and proceeded to completion, the total heat released by the reaction was calculated to be 39.0 kJ.
(b) Calculate the value of ΔH°, in kJ mol−1, for the combustion of glucose.
qrxn
−39.0 kJ
180.16 g C6H12O6 1 mol C6H12O6 = −2810 kJ/mol
∆H° =
=
nrxn 2.50 g C6H12O6  1 mol C6H12O6   1 mol rxn 
(c) When oxygen is not available, glucose can be oxidized by fermentation. In that process, ethanol and carbon
dioxide are produced, as represented by the following equation.
C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g)
ΔH° = −68.0 kJ mol−1 at 298 K
The value of the equilibrium constant, Kp , for the reaction at 298 K is 8.9 × 1039.
(i) Calculate the value of the standard free-energy change, ΔG°, for the reaction at 298 K. Include units with
your answer.
8.314 J/mol·K
39
∆G°rxn = −R·T·ln Keq = − 
 1000 J/kJ  (298 K) ln (8.9 × 10 ) = −228 kJ/mol
(ii) Calculate the value of the standard entropy change, ΔS°, in J K−1 mol−1, for the reaction at 298 K.
∆G° = ∆H° − T·∆S°
−288 kJ/mol = (−68.0 kJ/mol) – (298 K)(∆S°)
∆S° = +0.537 kJ/mol·K = +537 J/mol·K
(iii) Indicate whether the equilibrium constant for the fermentation reaction increases, decreases, or remains
the same if the temperature is increased. Justify your answer.
Because ∆H° < 0, at higher temperatures, reaction shifts to the left so Keq decreases.
(d) Using your answer for part (b) and the information provided in part (c), calculate the value of ΔH° for the
following reaction.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
1
1
C2H5OH(l) + CO2(g) → C6H12O6(s)
∆H° = − (−68.0 kJ/mol) = +34.0 kJ/mol
2
2
1
1
C H O (s) + 3 O2(g) → 3 CO2(g) + 3 H2O(l)
∆H° = + (−2810.0 kJ/mol) = −1405 kJ/mol
2 6 12 6
2
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
∆H° = +34.0 kJ/mol + (−1405 kJ/mol) = −1371 kJ/mol
AP Chemistry 2009 #5
Reaction
Equation
∆H°298
∆S°298
∆G°298
X
C (s) + H2O (g)  CO (g) + H2 (g)
+131 kJ mol−1
+134 J mol−1 K−1
+91 kJ mol−1
Y
CO2 (g) + H2 (g)  CO (g) + H2O (g)
+41 kJ mol−1
+42 J mol−1 K−1
+29 kJ mol−1
Z
2 CO (g)  C (s) + CO2 (g)
?
?
?
Answer the following questions using the information related to reactions X, Y, and Z in the table above.
(a) For reaction X, write the expression for the equilibrium constant, Kp.
(PCO)(PH2)
Kp =
(PH2O)
(b) For reaction X, will the equilibrium constant, Kp, increase, decrease, or remain the same if the temperature
rises above 298 K ? Justify your answer.
Because ∆H° > 0, at higher temperatures, the reaction will shift to the right so Kp increases.
(c) For reaction Y at 298 K, is the value of Kp greater than 1, less than 1, or equal to 1? Justify your answer.
Because ∆G° > 0, Kp is less than 1.
(d) For reaction Y at 298 K, which is larger: the total bond energy of the reactants or the total bond energy of the
products? Explain.
Because ∆H° > 0, the energy required to break bonds in the reactants is greater than the energy
released to form bonds in the products, so the total bond energy of reactants is larger than that of
products.
(e) Is the following statement true or false? Justify your answer.
“On the basis of the data in the table, it can be predicted that reaction Y will occur more rapidly than reaction
X will occur.”
The statement is false because thermodynamic data predicts whether the reaction is favorable, but
not the rate that the reaction occurs.
(f) Consider reaction Z at 298 K.
(i) Is ΔS° for the reaction positive, negative, or zero? Justify your answer.
ΔS° is negative because the reaction produces a solid from a gas.
(ii) Determine the value of ΔH° for the reaction.
CO (g) + H2 (g)  C (s) + H2O (g)
CO (g) + H2O (g)  CO2 (g) + H2 (g)
2 CO (g)  C (s) + CO2 (g)
∆H° = −(+131 kJ/mol) = −131 kJ/mol
∆H° = −(+41 kJ/mol) = −41 kJ/mol
∆H° = (−131) + (−41) = −172 kJ/mol
(iii) A sealed glass reaction vessel contains only CO(g) and a small amount of C(s). If a reaction occurs and
the temperature is held constant at 298 K, will the pressure in the reaction vessel increase, decrease, or
remain the same over time? Explain.
The pressure in the reaction vessel will decrease, because the reaction proceeds from 2 moles of
gas in the reactants to 1 mole of gas in the products.
AP Chemistry 2001 #2
2 NO(g) + O2(g)  2 NO2(g)
ΔH° = –114.1 kJ, ΔS° = –146.5 J K–1
The reaction represented above is one that contributes significantly to the formation of photochemical smog.
(a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).
1 mol NO  1 mol rxn −114.1 kJ
q = 73.1 g NO 
30.01 g NO 2 mol NO  mol rxn  = −139 kJ/mol
(b) For the reaction at 25°C, the value of the standard free-energy change, ΔG°, is –70.4 kJ.
i. Calculate the value of the equilibrium constant, Keq, for the reaction at 25°C.
−∆G°rxn/R·T
Keq = e
=e
−(−70400/(8.314)(298))
= e+28.415 = 2.2 × 1012
ii. Indicate whether the value of ΔG° would become more negative, less negative, or remain unchanged as
the temperature is increased. Justify your answer.
Since ∆G° = ∆H° − T·∆S°, as the temperature increases, ∆G° would be less negative because ∆S° < 0.
(c) Use the data in the table below to calculate the value of the standard molar entropy, S°, for O2(g) at 25°C.
Standard Molar Entropy, S°
(J K–1 mol–1)
NO(g)
NO2(g)
210.8
240.1
∆S°= [(2 mol NO2)(S°NO2)] – [(2 mol NO)(S°NO) + (1 mol O2)(S°O2)]
−146.5 = [(2)(240.1)] – [(2)(210.8) + (1)(S°O2)]
S°O2 = 205.1 J/mol·K
(d) Use the data in the table below to calculate the bond energy, in kJ mol–1, of the nitrogen-oxygen bond in NO2.
Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy).
Bond Energy (kJ mol–1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in NO2
?
Break 2 N−O bonds in NO = +2(607) = +1214
Break 1 O=O bond in O2 = +1(495) = + 495
Form 4 N−O bonds in NO2 = −4(x)
∆H° = (+1214) + (+495) + (−4x) = −114.1
x = 456 kJ/mol