2 iclg δ ho 269 kj and δs o 113 jk calculate the

2 ICl(g)
Δ H
o
= -26.9 kJ and Δ S
o
= 11.3 J/K.
Calculate the equilibrium constant K for this reaction at 298 K.
A. 2.3 x 10
3
B. 7.2 x 10
8
C . 2.0 x 10
5
D. 5.2 x 10
17
E. 5.0 x 10
-6
Δ G
o
= Δ H
o
- T Δ S
o
Δ G
o
= -26900 – (298 x 11.3) = -3.027 x 10
4
J
Δ G
o
= -RTln(K)
-3.027 x 10
4
= -8.314 x 298 x ln(K)
X = 2.023 x 10
5
= K

Chem 162-2012 Exam III review 87
Chem 162-2011 Exam III + Answers
Chapter 17 - Thermodynamics
Free Energy And Equilibria Calculations
25*. Calculate Δ G for the following reaction at 1200 K and 3.29 x 10
-3
atm of Cl
2
(g).
2AgCl(s) → 2Ag(s) + Cl
2
(g) Δ G
o
= 220 kJ
A . 163 kJ
B. 190 kJ
C. 181 kJ
D. 220 kJ
E. 277 kJ
Δ G = Δ G
o
+ RT ln(Q)
Q = P
Cl2
Δ G = 220000 + (8.314 x 1200 x (ln(3.29x10
-3
))) = 1.63 x 10
5
J = 1.63 x 10
2
kJ

Chem 162-2012 Exam III review 88
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Chem 162-2011 Final exam
Chapter 17 – Thermodynamics
Free energy calculations
The enthalpy change of a particular reaction is 15 kJ/mol at 27 °C. This
reaction will be spontaneous if the entropy change is:
A . > 50J mol
-1
K
-1
B. < -50J mol
-1
K
-l
C. 15 J mol
-l
K
-l
D. -50 J mol
-l
K
-l
E. -15 J mol
-1
K
-1
Δ G = Δ H – T Δ S
First determine the value of Δ S when the reaction is at equilibrium. Then
determine the value of Δ S when the reaction is spontaneous.
0 = 15000 – (300 x Δ S)
Δ S = 50J mol
-1
K
-1
when the reaction is at equilibrium.
If Δ S is greater than 50J mol
-1
K
-1
, then Δ G would
be negative, meaning that the
reaction would be spontaneous.

Chem 162-2012 Exam III review 89
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Chem 162-2011 Exam III + Answers
Chapter 17 - Thermodynamics
Free Energy And Equilibria Concepts
13. If the numerical value of the equilibrium constant increases as the
temperature increases then:
A. Δ S° < 0
B. Δ H° < 0
C. Δ G° < 0
D . Δ H° > 0
E. Δ S° > 0
LeChatelier’s principle states that for an endothermic reaction, if the
temperature increases then the reaction goes to the right and K increases. (This
is supported by the van’t Hoff equation.) Hence, Δ H° > 0.
A + B + heat
←
→
C + D

Chem 162-2012 Exam III review 90
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Chem 162-2011 Final exam
Chapter 17 – Thermodynamics
Free energy and equilibria concepts
Which one of the following statements is true about the equilibrium
constant for a reaction if ∆ G ˚ for the reaction has a negative value?
A. K= 0
B. K = 1
C. K = -1
D. K < 1
E . K > 1
∆ G ˚ = 0 K = 1
∆ G ˚ > 0 K < 1
∆ G ˚ < 0 K > 1
Note that the superscript zero means that everything is being measured
under standard state conditions, which means that all reactants and
products are at 1M or 1atm.
If ∆ G ˚ for the reaction is zero, then the reaction is at equilibrium, and K
= 1 (i.e., everything is at 1 atm or 1 molar).
If ∆ G ˚ for the reaction is positive, then the reaction is not spontaneous
(or spontaneous to the left), and K<1.
If ∆ G ˚ for the reaction is negative, then the reaction is spontaneous (or
spontaneous to the right), and K > 1.
E

Chem 162-2012 Exam III review 91
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Chapter 17 - Thermodynamics
Free Energy And Equilibria Calculations
9 . Consider the reaction:
Fe
2
O
3
(s) + 3H
2
(g) → 2Fe(s) + 3H
2
O(g)
Given: Δ H° = 100 kJ and Δ S° = 138 J/K, at what temperature would the equilibrium constant K = 1?
A. 465 K
B. 685 K
C
. 725 K
D. 575 K
E. 845 K
Δ G
o
= Δ H
o
- T Δ S
o
Δ G
o
= -RT ln(K)
Δ H
o
- T Δ S
o
= -RTln(K)
Rearranging terms:
T = - Δ H
o
/(- Δ S
o
+ Rln(K))
T = -100000/(-138 + (8.314 x ln(1)))
T = 725K
This also could have been solved in a simpler manner.