← Back to practice questions

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)

Show answer and explanation

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)

Show answer and explanation

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)

Show answer and explanation

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.

Show answer and explanation

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?

Show answer and explanation

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?

Show answer and explanation

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?

Show answer and explanation

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

Show answer and explanation

Answer: −150 kJ/mol

Work:

ΔG°rxn = −600 − (−450)

= −150 kJ/mol

Question 9

A reaction has ΔG° < 0. Which statement is correct?

Show answer and explanation

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?

Show answer and explanation

Answer: Negative

Explanation: The system must move toward products to reach equilibrium.

Question 11

If Q > K, what is the sign of ΔG?

Show answer and explanation

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°.

Show answer and explanation

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)

Show answer and explanation

Answer: Zn

Explanation:

Zn loses electrons:

Zn → Zn²⁺ + 2e⁻

Loss of electrons is oxidation.

Question 14

In the same reaction, which substance is reduced?

Show answer and explanation

Answer: Cu²⁺

Explanation:

Cu²⁺ + 2e⁻ → Cu

Cu²⁺ gains electrons.

Question 15

In a Zn/Cu galvanic cell, which electrode is the anode?

Show answer and explanation

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?

Show answer and explanation

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?

Show answer and explanation

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

Show answer and explanation

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°?

Show answer and explanation

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.

Show answer and explanation

Answer: 200 C

Work:

q = It

q = (2.00 C/s)(100 s)

= 200 C