MULTIPLE CHOICE QUESTIONS: only one choice is correct
1.
A reaction will always be spontaneous when:
a)
b)
c)
d)
2.
In the reaction: Ti(s) + O2 (g)  TiO2 (s) entropy is expected to:
a)
b)
c)
d)
3.
G = 0
Stotal > 0
Stotal < 0
G < 0
If the potential of a redox reaction is +1.8 V,
a)
b)
c)
d)
5.
increase
decrease
remain unchanged
too little info to tell
If Q = Keq for a system then
a)
b)
c)
d)
4.
Ssys is positive
Ssurr is positive
both (a) and (b)
none of the above
Grxn < 0
Stotal > 0
the reaction cell is galvanic
all of the above
When a strip of zinc is placed in a solution of Cu(2+) ion, a reaction immediately occurs
and the voltage is initially +1.1 V. If the voltage is measured every hour, its value will be
observed to:
a)
b)
c)
d)
be the same as the initial value of +1.1 V
decrease from the initial value to -1.1 V
decrease from the initial value to 0.0 V
none of the above
SHORT ANSWER QUESTIONS:
6. What is the second law of thermodynamics?
The entropy of the universe is increasing.
Only spontaneous reactions cause total entropy to increase.
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7. Which of the four metals—Li, Na, Cs, Pt—is expected to react most violently with protons
(H+) in when the metal is dropped into water? Use the reduction potentials (below) to explain
your answer.
E° Li+/Li0 = -3.05 V because the reverse rxn Li Li+, has the largest +E value.
E° Na+/Na0 = -2.71 V
E° Cs+/Cs° = -2.92 V
E° Pt2+/Pt° = +1.20 V
8. Balance the following redox reaction:
2H+ + Cu (s) + NO3-  Cu(2+) + NO2- + H2O
9. The reaction of HCl gas and ammonia gas forms solid ammonium chloride that appears as a
white cloud, just as you saw in class. How does this one observation allow us to conclude that
the reaction must be exothermic?
The reaction is spontaneous so G is negative.
Two gases forma a solid so S is negative.
The only way G can be negative is if H is a more negative number than –TS, as described by
G = H–TS.
PROBLEMS
10.For this reaction on nickel: [Ni(H2O)6]2+ + 6 H3N  [Ni(NH3)6]2+ + 6 H2O Keq = 2.6 x 108
a) (a) What is the free energy change for this reaction at equilibrium? G = 0 at equilibrium
(b) What is the standard free energy for this reaction?
Go = -RTlnK, so Go = -2.48 kJ/mol ln (2.6 x 108); Go = -48 kJ/mol
(c) When the concentrations of all species are at the standard concentration of 1.00 M in water,
what is the free energy for the reaction?
That is the condition for standard free energy, Go = -48 kJ/mol
(d) When the concentrations of all species are 0.00100 M, what is the free energy for the
reaction?
G = Go + RTlnQ, G = -48 kJ/mol + 2.48 kJ/mol ln ( .001/[.001x(.001)6]) = +54 kJ/mol
(e) using your results from parts (c) and (d), under which conditions—(c) vs (d)— is the product
complex [Ni(NH3)6]2+ more stable?
G is more negative in (b) under standard conditions so the reaction is more favored to go
forward and the products are more stable.
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11. Combine any two of the redox half reactions below to (a) write a spontaneous, balanced
reaction, then (b) calculate the standard free energy change for your reaction.
i. NO3—  HNO2
E +0.93 V
2ii. Fe(3+)  [FeO4]
E +2.20 V
iii. S(s)  H2S
E +0.14 V
iv. IO3  IE +1.09 V
12. Use the data in the table below to answer the following questions.
Hºf, kJ/mol
Sºf, J/mol K

Gºf, kJ/mol
C2H2
+209
CaC2
CaO
CO(g)
C(graphite)
-59
-635
-110
+70
+38
+198
+5.6
-64
-603
-137
(a). Explain why acetylene, C2H2, is thermodynamically unstable.
Since it has a positive fee energy of formation +209 kJ/mol, it does not spontaneously form
from the elements. HOWEVER, the reverse reaction, acetylene decomposing to C and H2, IS a
spontaneous reaction with w free energy of -209. It is thermodynamically unstable because it
spontaneously should decompose to the elements.
(b). Explain why acetylene, C2H2, can exist even though it is thermodynamically unstable.
There must be a very high activation barrier to the decomposition making it a slow reaction
kinetically.
(c) Write and balance the equation that describes the formation of calcium carbide, CaC2, from
the reaction of calcium oxide, CaO, and carbon (graphite). Next, calculate the free energy of this
reaction to determine whether this process is thermodynamically favored, or not.
CaO + 3C (graphite)  CaC2 + CO(g)
Gºrxn = -64 +(-137)-(-603) = +402 kJ/mol, so this reaction is not spontaneous and is not
thermodynamically favored.
(d) How does changing the temperature, T, of the reaction in (c) affect its spontaneity? If there is
a change in spontaneity, calculate the temperature at which that change occurs.
Calculate enthalpy and entropy to see how change in temperature affects free energy:
Hºrxn = -59 + (-110) –(-635) = +466 kJ/mol
Sºrxn = 70 + (198) –(38 + 3(5.6)) = +213 J/mol K
Since entropy change is positive, increasing T will make TSºrxn a larger negative number that
can compensate for the positive enthalpy. The change in spontaneity will happen at equilibrium
when Gº = 0 then Hºrxn = TSºrxn or +466 kJ/mol = T (+213 J/mol K); T = 2188 K
=
3
Extra Credit
Use the standard reduction potentials for lead (below) to calculate the Ksp for lead sulfate.
Pb2+/Pb Eo = -0.130 V
Pb[SO4](s)/Pb Eo = -0.356 V
Ksp for lead sulfate refers to the reaction: Pb[SO4](s)  Pb2+ + SO42Use the redox couples to create that equation by adding tem appropriately:
Pb[SO4](s) + 2 e-  Pb + SO42-
Eo = -0.356 V
Pb
 Pb2+ + 2 eEo = -0.130 V
Pb[SO4](s)  Pb2+ + SO42Eo = -0.226 V not spontaneous, OK since
lead sulfate is very insoluble, reaction does NOT go forward.
Now convert E to a Keq:
Gº = -n F E = -RT lnKeq
n F E /RT = lnKeq
2 x -0.226 / 0.0257 = ln Keq
Keq = Ksp = 2.3 x 10-8
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