Unit 9 Practice: Medium
Applications of Thermodynamics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Predict the sign of ΔS for:
2SO₂(g) + O₂(g) → 2SO₃(g)
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Answer: Negative
Explanation: There are 3 mol of gas on the reactant side and 2 mol on the product side. Fewer gas particles generally means fewer possible arrangements.
Question 2
Calculate ΔS°rxn for:
A → B
Given:
S°A = 150 J/(mol·K) S°B = 210 J/(mol·K)
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Answer: +60 J/(mol·K)
Work:
ΔS° = products − reactants
= 210 − 150
= +60 J/(mol·K)
Question 3
Calculate ΔS°rxn for:
A + B → C
Given:
A = 100 J/(mol·K) B = 150 J/(mol·K) C = 180 J/(mol·K)
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Answer: −70 J/(mol·K)
Work:
ΔS° = 180 − (100 + 150)
= −70 J/(mol·K)
Question 4
A reaction has:
ΔH = −50.0 kJ/mol ΔS = +100 J/(mol·K) T = 298 K
Calculate ΔG.
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Answer: −79.8 kJ/mol
Work:
Convert entropy:
100 J/(mol·K) = 0.100 kJ/(mol·K)
TΔS:
298(0.100) = 29.8 kJ/mol
ΔG = −50.0 − 29.8
= −79.8 kJ/mol
Question 5
Is the reaction in Question 4 favored?
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Answer: Yes
Explanation: ΔG is negative.
Question 6
A reaction has:
ΔH = +40.0 kJ/mol ΔS = +150 J/(mol·K)
Above approximately what temperature does it become favored?
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Answer: 267 K
Work:
At crossover:
ΔG = 0
So:
T = ΔH/ΔS
Convert:
40.0 kJ = 40,000 J
T = 40,000/150
≈ 267 K
Because both ΔH and ΔS are positive, temperatures above this value favor the reaction.
Question 7
A reaction has ΔH < 0 and ΔS < 0. Is it more likely to be favored at high or low temperature?
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Answer: Low temperature
Explanation: The unfavorable +T|ΔS| contribution becomes larger as temperature increases.
Question 8
Calculate ΔG°rxn if:
ΣΔG°f(products) = −600 kJ/mol
ΣΔG°f(reactants) = −450 kJ/mol
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Answer: −150 kJ/mol
Work:
ΔG°rxn = −600 − (−450)
= −150 kJ/mol
Question 9
A reaction has ΔG° < 0. Which statement is correct?
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Answer: K > 1
Explanation: Negative standard Gibbs free energy corresponds to a product-favored equilibrium.
Question 10
If Q < K, what is the sign of ΔG for the forward reaction?
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Answer: Negative
Explanation: The system must move toward products to reach equilibrium.
Question 11
If Q > K, what is the sign of ΔG?
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Answer: Positive
Explanation: The forward direction is unfavored, and the reaction tends to shift toward reactants.
Question 12
Reaction A has ΔG° = +25 kJ. Reaction B has ΔG° = −60 kJ. They are properly coupled. Find the overall ΔG°.
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Answer: −35 kJ
Work:
+25 + (−60)
= −35 kJ
Explanation: The favorable reaction provides enough thermodynamic driving force to make the combined process favored.
Question 13
Identify the oxidized substance:
Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)
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Answer: Zn
Explanation:
Zn loses electrons:
Zn → Zn²⁺ + 2e⁻
Loss of electrons is oxidation.
Question 14
In the same reaction, which substance is reduced?
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Answer: Cu²⁺
Explanation:
Cu²⁺ + 2e⁻ → Cu
Cu²⁺ gains electrons.
Question 15
In a Zn/Cu galvanic cell, which electrode is the anode?
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Answer: Zn
Explanation: Zinc is oxidized, and oxidation always occurs at the anode.
Question 16
What happens to the mass of the Zn electrode as the cell operates?
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Answer: It decreases.
Explanation: Zn atoms leave the solid electrode and become Zn²⁺ ions.
Question 17
What happens to the mass of the Cu electrode?
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Answer: It increases.
Explanation: Cu²⁺ ions gain electrons and deposit as Cu metal.
Question 18
Calculate E°cell.
Given reduction potentials:
Cu²⁺ + 2e⁻ → Cu E° = +0.34 V
Zn²⁺ + 2e⁻ → Zn E° = −0.76 V
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Answer: +1.10 V
Work:
E°cell = E°cathode − E°anode
= 0.34 − (−0.76)
= 1.10 V
Question 19
If a half-reaction is multiplied by 3 to balance electrons, what happens to E°?
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Answer: Nothing
Explanation: Electrode potential is an intensive property and is not multiplied by stoichiometric coefficients.
Question 20
A current of 2.00 A flows for 100 s. Calculate the charge transferred.
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Answer: 200 C
Work:
q = It
q = (2.00 C/s)(100 s)
= 200 C