Unit 9 Practice: Hard
Applications of Thermodynamics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
A reaction has:
ΔH° = +85.0 kJ/mol ΔS° = +220 J/(mol·K)
Calculate ΔG° at 298 K and determine whether the reaction is favored.
Show answer and explanation
Answer: +19.4 kJ/mol; unfavored
Work:
Convert entropy:
220 J/(mol·K) = 0.220 kJ/(mol·K)
TΔS:
298(0.220) = 65.56 kJ/mol
ΔG°:
85.0 − 65.56
= +19.44 kJ/mol
Because ΔG° > 0, the reaction is unfavored under standard conditions at 298 K.
Question 2
Using the reaction from Question 1, approximately above what temperature will it become favored?
Show answer and explanation
Answer: 386 K
Work:
At crossover:
ΔG = 0
T = ΔH/ΔS
= 85,000 J/mol ÷ 220 J/(mol·K)
≈ 386 K
Because ΔH and ΔS are positive, the reaction becomes favored above this temperature.
Question 3
A reaction has:
ΔH° = −120 kJ/mol
ΔS° = −250 J/(mol·K)
Predict whether increasing temperature makes the reaction more or less favorable.
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Answer: Less favorable
Explanation:
Convert conceptually:
ΔG = ΔH − TΔS
Because ΔS is negative:
−TΔS becomes positive.
Increasing T makes this positive contribution larger, so ΔG becomes less negative and eventually may become positive.
Question 4
For the reaction in Question 3, estimate the crossover temperature.
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Answer: 480 K
Work:
T = ΔH/ΔS
Both values are negative, so the ratio is positive:
T = (−120,000)/(−250)
= 480 K
Below 480 K the reaction is favored; above 480 K it is unfavored.
Question 5
A student says, "If a reaction has ΔG < 0, it must occur rapidly." Explain the error.
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Answer: ΔG describes thermodynamic favorability, not reaction rate.
Explanation: A reaction can have ΔG < 0 but still have a large activation energy. If the activation barrier is high, the reaction may proceed very slowly.
Question 6
Explain why dissolving a salt can be endothermic but still thermodynamically favored.
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Answer: The increase in entropy can outweigh the positive enthalpy change.
Explanation:
If:
ΔH > 0
but:
ΔS > 0
then at a suitable temperature:
TΔS > ΔH
making:
ΔG = ΔH − TΔS < 0
Question 7
At 298 K, a reaction has:
ΔG° = −5.70 kJ/mol
Estimate K.
Use:
ΔG° = −RT ln K
R = 8.314 J/(mol·K)
Show answer and explanation
Answer: Approximately 10
Work:
Convert:
ΔG° = −5700 J/mol
−5700 = −(8.314)(298)lnK
5700 = 2477.6 lnK
lnK ≈ 2.30
K ≈ e²·³⁰
≈ 10.0
Question 8
A reaction has K = 1.0 × 10⁵ at 298 K. Is ΔG° positive or negative?
Show answer and explanation
Answer: Negative
Explanation:
K > 1 means lnK > 0.
Therefore:
ΔG° = −RT lnK
is negative.
Question 9
At equilibrium, why can ΔG equal zero even though ΔG° may not equal zero?
Show answer and explanation
Answer: Because ΔG depends on the actual reaction conditions through Q.
Explanation:
ΔG = ΔG° + RT lnQ
At equilibrium:
Q = K
and the two terms cancel, giving ΔG = 0.
ΔG° refers specifically to standard-state conditions.
Question 10
A reaction has:
ΔG° = +12 kJ/mol
but under current conditions Q is very small. Could the forward reaction still be favored?
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Answer: Yes
Explanation: The actual free energy is:
ΔG = ΔG° + RT lnQ
If Q is less than 1, lnQ is negative. A sufficiently small Q can make ΔG negative even when ΔG° is positive.
This is why standard favorability and actual favorability are not always identical.
Question 11
Calculate ΔG° for a galvanic cell with:
E°cell = 1.10 V
and:
n = 2
Use:
F = 96,485 C/mol e⁻
Show answer and explanation
Answer: −212 kJ/mol
Work:
ΔG° = −nFE°
= −(2)(96,485)(1.10)
= −212,267 J/mol
≈ −212 kJ/mol
Question 12
A cell has:
E°cell = −0.25 V
What can you conclude about ΔG° and K for the reaction as written?
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Answer:
ΔG° > 0 and K < 1
Explanation:
From:
ΔG° = −nFE°
a negative E° gives a positive ΔG°.
Positive ΔG° corresponds to K < 1.
Question 13
Given:
Ag⁺ + e⁻ → Ag E° = +0.80 V
Cu²⁺ + 2e⁻ → Cu E° = +0.34 V
Which species is reduced in a spontaneous galvanic cell containing these half-cells?
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Answer: Ag⁺
Explanation: Ag⁺ has the more positive reduction potential, so it is more strongly favored to undergo reduction.
Cu therefore undergoes oxidation.
Question 14
Calculate E°cell for the Ag/Cu cell from Question 13.
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Answer: +0.46 V
Work:
Cathode = Ag
Anode reduction potential = Cu
E°cell:
0.80 − 0.34
= +0.46 V
Question 15
Write the balanced overall reaction for the cell in Questions 13–14.
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Answer:
2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq)
Explanation:
Reduction:
2Ag⁺ + 2e⁻ → 2Ag
Oxidation:
Cu → Cu²⁺ + 2e⁻
Add and cancel electrons.
Question 16
For the reaction:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
write Q.
Show answer and explanation
Answer:
Q = [Zn²⁺]/[Cu²⁺]
Explanation: Pure solids Zn and Cu are excluded.
Question 17
For the Zn/Cu cell, E° = 1.10 V. At 25°C:
[Zn²⁺] = 1.0 M
[Cu²⁺] = 0.010 M
Calculate E using:
E = E° − (0.0592/n)logQ
Show answer and explanation
Answer: About 1.04 V
Work:
For the reaction:
Q = [Zn²⁺]/[Cu²⁺]
= 1.0/0.010
= 100
n = 2
E = 1.10 − (0.0592/2)log(100)
log(100) = 2
E = 1.10 − (0.0296)(2)
E = 1.10 − 0.0592
= 1.04 V
Question 18
Why does the voltage of a galvanic cell generally decrease as the cell operates?
Show answer and explanation
Answer: Reactants are consumed and products accumulate, causing Q to move toward K.
Explanation: As Q approaches K, the thermodynamic driving force decreases. From the Nernst equation, E decreases.
At equilibrium:
Q = K
and:
E = 0
Question 19
A current of 5.00 A passes through an electrolytic cell for 30.0 minutes. Calculate the moles of electrons transferred.
Use:
F = 96,485 C/mol e⁻
Show answer and explanation
Answer: 0.0933 mol e⁻
Work:
Convert time:
30.0 min × 60 = 1800 s
Charge:
q = It
= (5.00)(1800)
= 9000 C
Moles electrons:
9000/96,485
= 0.0933 mol e⁻
Question 20
The electrons from Question 19 are used to plate Cu according to:
Cu²⁺ + 2e⁻ → Cu
How many grams of Cu are deposited?
Molar mass Cu = 63.55 g/mol
Show answer and explanation
Answer: 2.96 g Cu
Work:
From Question 19:
0.0933 mol e⁻
Stoichiometry:
0.0933 mol e⁻ × (1 mol Cu / 2 mol e⁻)
= 0.04665 mol Cu
Mass:
0.04665 × 63.55
≈ 2.96 g Cu