Name:
AP Chemistry
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Quiz 6.1 ANSWER KEY – ΔG and Spontaneity
1. Use the following data table to aid you with question a – d.
Enthalpy of Absolute
Combustion, H
Entropy, S
Substance
(kiloJoules/mol)
(Joules/mol-K)
C(s)
-393.5
5.740
H2(g)
-285.8
130.6
C2H5OH(l)
-1366.7
160.7
H2O(l)
-69.91
(a) Write a separate, balanced chemical equation for the combustion of each of the following: C (s), H2(g), and
C2H5OH(l). The only possible products are CO2 and/or H2O(l) (though both do not have to be used).
3 points possible; 1 point per equation
C (s) + O2 (g)  CO2 (g)
2H2 (g) + O2 (g)  2H2O (l)
C2H5OH + 3O2 (g)  2CO2 (g) + 3H2O (g)
(b) In principle, ethanol can be prepared by the following reaction:
2 C(s) + 2 H2(g) + H2O(l)  C2H5OH(l)
Calculate the standard enthalpy change, H, for the preparation of ethanol, as shown in the reaction above.
2 points possible. 1 for recognizing Hess’s Law and setup. 1 point for answer
C  O2 CO2
H = -393.5
2H 2  O2 2H 2O
H = -285.8
C2 H 5OH  3O2 2CO2  3H 2O H = -1366.7
Hess’s Law can be used to get the enthalpy of the reaction of interest:
2C  2O2 2CO2
H = 2(-393.5)
2H 2  O2 2H 2O
H = -285.8

2CO2  3H 2O C2 H 5OH  3O2
H = -1(-1366.7)
H = 2(-393.5) + (-285.8) + 1366.7 = 293.9 kJ/mol
(c) Calculate the standard entropy change, S, for the reaction given in part (b).
3 points possible.
1 point for correct equation, 1 point for correct answer, 1 point for units.
S   n(S products)   n(Sreactants)
S  [1(160.7)]  [2(5.740)  2(130.6) 1(69.91)]  181.89 J mol K
(d) Calculate the value of G, for the reaction given in part (b).
3 points possible. 1 point for correct equation, 1 point for correct answer/conversion, 1 point for units.
G o  H o  TS o
G o  (293.9 1000)  ((298)(181.89))  348,103  348,100 J mol

QUIZ