AP Chemistry Applications of Thermodynamics — Worked Answer Explanations

Unit 9 · 12 questions explained

Below is a complete answer key for our AP Chemistry Applications of Thermodynamics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Applications of Thermodynamics practice test and come back here to review, or head back to the Applications of Thermodynamics unit overview.

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Question 1 · Easy
Which of the following processes has a positive entropy change ( $Δ S > 0$ )?
- A
  $H_{2} O (l) \to H_{2} O (s)$ (freezing)
  Why not A: Freezing decreases disorder; $Δ S < 0$ .
- B
  $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
  Why not B: Four moles of gas → two moles; fewer particles means less disorder; $Δ S < 0$ .
- C
  $NaCl (s) \to N a^{+} (aq) + C l^{-} (aq)$ Correct
- D
  $C O_{2} (g) \to C O_{2} (s)$ (dry ice)
  Why not D: Gas to solid is a large decrease in disorder; $Δ S ≪ 0$ .
Explanation
When NaCl dissolves, one mole of ordered solid disperses into many ions in solution, greatly increasing the number of microstates. More particles spread through larger volume = increase in entropy ( $Δ S > 0$ ).

Key takeaway
Entropy increases: solid → liquid → gas; increasing moles of gas; dissolving solids; mixing; higher temperature.
Question 2 · Easy
For a spontaneous process at constant temperature and pressure, which condition must be met?
- A
  $Δ H < 0$ only
  Why not A: Some endothermic processes ( $Δ H > 0$ ) are spontaneous at high temperature if $T Δ S$ is large enough.
- B
  $Δ S > 0$ only
  Why not B: Some processes with $Δ S < 0$ are spontaneous at low temperature when $Δ H$ is sufficiently negative.
- C
  $Δ G = Δ H - T Δ S < 0$ Correct
- D
  $Δ H < 0$ and $Δ S < 0$ simultaneously
  Why not D: A reaction can be spontaneous even if $Δ S < 0$ (exothermic processes at low T), or if $Δ H > 0$ (endothermic processes at high T).
Explanation
Gibbs free energy: $Δ G = Δ H - T Δ S$ . A process is spontaneous (thermodynamically favorable) when $Δ G < 0$ . At equilibrium $Δ G = 0$ ; non-spontaneous in the forward direction when $Δ G > 0$ .

Key takeaway
$\Delta G = \Delta H - T\Delta S < 0$ is the criterion for spontaneity at constant $T$ and $P$.
Question 3 · Easy
In a galvanic (voltaic) cell, which statement correctly describes the anode and cathode?
- A
  Oxidation occurs at the cathode; reduction at the anode.
  Why not A: This is reversed; OX at ANOde = oxidation at anode.
- B
  Reduction occurs at the cathode; oxidation at the anode.Correct
- C
  Both electrodes undergo oxidation simultaneously.
  Why not C: In a galvanic cell, one electrode is oxidized and the other is reduced.
- D
  The cathode is the positive electrode in both galvanic and electrolytic cells.
  Why not D: In electrolytic cells, the cathode is the negative electrode; the positive/negative designation differs between cell types.
Explanation
Mnemonics: OIL RIG (Oxidation Is Loss; Reduction Is Gain) and AN OX, RED CAT. In a galvanic cell, the anode is negative (where oxidation occurs) and the cathode is positive (where reduction occurs). Electrons flow from anode to cathode through the external circuit.

Key takeaway
Anode: oxidation (negative in galvanic cell). Cathode: reduction (positive in galvanic cell). AN OX, RED CAT.
Question 4 · Easy
In an electrolytic cell, which statement correctly describes the process?
- A
  A spontaneous reaction drives electron flow through the external circuit.
  Why not A: That describes a galvanic cell; electrolytic cells use external electrical energy for non-spontaneous reactions.
- B
  External electrical energy forces a non-spontaneous redox reaction.Correct
- C
  The anode is negative; the cathode is positive.
  Why not C: In an electrolytic cell, the anode is positive and cathode is negative (opposite of galvanic cell's internal sign convention).
- D
  Both oxidation and reduction occur at the same electrode.
  Why not D: Oxidation occurs at the anode and reduction at the cathode in both cell types.
Explanation
Electrolytic cells use an external voltage source (e.g., battery) to supply electrical energy, forcing a non-spontaneous reaction ( $Δ G > 0$ , $E_{cell}^{\circ} < 0$ ). Examples: electrolysis of water, electroplating, aluminum smelting.

Key takeaway
Galvanic cell: spontaneous reaction → electricity. Electrolytic cell: electricity → non-spontaneous reaction.
Question 5 · Medium
Calculate the standard cell potential $E_{cell}^{\circ}$ for the reaction:

$Zn (s) + C u^{2 +} (aq) \to Z n^{2 +} (aq) + Cu (s)$

Given: $E_{red}^{\circ} (C u^{2 +} /Cu) = + 0.34 V$ ; $E_{red}^{\circ} (Z n^{2 +} /Zn) = - 0.76 V$
- A
  $- 1.10 V$
  Why not A: This is the negative; Zn is oxidized (anode), so subtract Zn's reduction potential.
- B
  $+ 1.10 V$ Correct
- C
  $+ 0.42 V$
  Why not C: Added or subtracted incorrectly; did not account for signs.
- D
  $+ 0.34 V$
  Why not D: Used only the copper half-cell potential.
Explanation
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{red}^{\circ} (C u^{2 +} /Cu) - E_{red}^{\circ} (Z n^{2 +} /Zn)$

$= (+ 0.34) - (- 0.76) = + 0.34 + 0.76 = + 1.10 V$

Positive $E_{cell}^{\circ}$ confirms the reaction is spontaneous ( $Δ G^{\circ} < 0$ ).

Key takeaway
$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ (using reduction potentials). Positive $E^\circ$ = spontaneous.
Question 6 · Medium
The relationship between standard cell potential and Gibbs free energy is $Δ G^{\circ} = - n F E_{cell}^{\circ}$ , where $F = 96, 485 C/mol$ . For a cell with $E^{\circ} = + 0.50 V$ and $n = 2$ , what is $Δ G^{\circ}$ ?
- A
  $+ 96.5 kJ/mol$
  Why not A: Positive $Δ G^{\circ}$ would indicate non-spontaneous; sign is wrong.
- B
  $- 96.5 kJ/mol$ Correct
- C
  $- 48.2 kJ/mol$
  Why not C: Used $n = 1$ instead of $n = 2$ .
- D
  $- 1.00 kJ/mol$
  Why not D: Did not multiply by Faraday's constant correctly.
Explanation
$Δ G^{\circ} = - n F E^{\circ} = - (2) (96, 485) (0.50) = - 96, 485 J/mol \approx - 96.5 kJ/mol$

Negative $Δ G^{\circ}$ confirms spontaneity, consistent with positive $E^{\circ}$ .

Key takeaway
$\Delta G^\circ = -nFE^\circ$. Positive $E^\circ$ ↔ Negative $\Delta G^\circ$ ↔ Spontaneous reaction.
Question 7 · Medium
The standard Gibbs free energy is related to the equilibrium constant by $Δ G^{\circ} = - R T ln K$ . If $Δ G^{\circ} = 0$ , what is $K$ ?
- A
  $K = 0$
  Why not A: $ln K = 0$ gives $K = 1$ , not $K = 0$ .
- B
  $K = 1$ Correct
- C
  $K = \infty$
  Why not C: $K \to \infty$ corresponds to $Δ G^{\circ} \to - \infty$ .
- D
  $K = e$
  Why not D: $K = e$ would require $Δ G^{\circ} = - R T$ , not $Δ G^{\circ} = 0$ .
Explanation
$Δ G^{\circ} = 0 \Rightarrow - R T ln K = 0 \Rightarrow ln K = 0 \Rightarrow K = e^{0} = 1$ . When $K = 1$ , products and reactants are in equal concentrations at equilibrium, and neither is thermodynamically favored.

Key takeaway
$\Delta G^\circ = -RT\ln K$: $\Delta G^\circ = 0 \to K = 1$; $\Delta G^\circ < 0 \to K > 1$ (products favored); $\Delta G^\circ > 0 \to K < 1$ (reactants favored).
Question 8 · Medium
Which of the following correctly ranks the entropy of three states of water at the same temperature?
- A
  $S (ice) > S (water) > S (steam)$
  Why not A: Entropy increases as disorder increases: solid < liquid < gas.
- B
  $S (steam) > S (water) > S (ice)$ Correct
- C
  $S (water) > S (steam) > S (ice)$
  Why not C: Liquid water is more ordered than steam — gas has the highest entropy.
- D
  All three have equal entropy at the same temperature.
  Why not D: Phase determines the degree of molecular disorder regardless of temperature.
Explanation
Third Law and qualitative entropy: solid is most ordered ( $S$ lowest), liquid intermediate, and gas most disordered ( $S$ highest). For water: $S^{\circ} (H_{2} O (s)) < S^{\circ} (H_{2} O (l)) < S^{\circ} (H_{2} O (g))$ .

Key takeaway
Entropy order by phase: solid < liquid < gas. Molecules in gas phase occupy far more microstates.
Question 9 · Medium
For the reaction $A \to B$ , $Δ H = - 50 kJ/mol$ and $Δ S = - 100 J/ (mol \cdot K)$ . Which statement is correct about spontaneity?
- A
  Always spontaneous because $Δ H < 0$ .
  Why not A: Spontaneity depends on both $Δ H$ and $T Δ S$ ; at high $T$ , $- T Δ S$ can dominate.
- B
  Never spontaneous because $Δ S < 0$ .
  Why not B: At low $T$ , the $T Δ S$ term is small and $Δ H < 0$ drives spontaneity.
- C
  Spontaneous at low temperature; non-spontaneous at high temperature.Correct
- D
  Non-spontaneous at all temperatures because both $Δ H < 0$ and $Δ S < 0$ .
  Why not D: $Δ H < 0$ and $Δ S < 0$ is the temperature-dependent case — low $T$ gives spontaneous.
Explanation
$Δ G = Δ H - T Δ S = - 50, 000 - T (- 100) = - 50, 000 + 100 T$

For low $T$ : $100 T < 50, 000 \Rightarrow Δ G < 0$ (spontaneous)

Crossover: $T = 500 K$ . For $T > 500 K$ : $Δ G > 0$ (non-spontaneous).

Key takeaway
$\Delta H < 0$, $\Delta S < 0$: spontaneous at low $T$, crossover at $T = \Delta H/\Delta S$.
Question 10 · Hard
For a reaction with $Δ H = + 120 kJ/mol$ and $Δ S = + 400 J/ (mol \cdot K)$ , at what temperature does the reaction become spontaneous?
- A
  Always spontaneous, regardless of temperature.
  Why not A: $Δ H > 0$ makes low-temperature spontaneity impossible; it becomes spontaneous only above a threshold.
- B
  Above $300 K$ Correct
- C
  Below $300 K$
  Why not C: At low $T$ , the $T Δ S$ term is small and $Δ H > 0$ dominates, giving $Δ G > 0$ .
- D
  Never spontaneous, because $Δ H > 0$ .
  Why not D: With $Δ S > 0$ , the $T Δ S$ term eventually overcomes the positive $Δ H$ at high enough $T$ .
Explanation
Set $Δ G = 0$ to find the crossover temperature:

$0 = Δ H - T Δ S$

$T = \frac{Δ H}{Δ S} = \frac{120 , 000 J/mol}{400 J/ ( mol \cdot K )} = 300 K$

For $T > 300 K$ : $T Δ S > Δ H$ , so $Δ G < 0$ (spontaneous).

Key takeaway
Spontaneity crossover temperature: $T = \Delta H / \Delta S$ (in Kelvin). Use consistent units (J, not kJ, for both).
Question 11 · Hard
How many grams of copper are deposited when a current of $2.00 A$ passes through a $CuS O_{4}$ solution for $1.00 hour$ ? ( $M (Cu) = 63.5 g/mol$ ; $F = 96, 485 C/mol$ ; Cu deposits as Cu $^{2 +} + 2 e^{-} \to Cu$ )
- A
  $2.37 g$ Correct
- B
  $4.74 g$
  Why not B: Used $n = 1$ instead of $n = 2$ electrons per Cu $^{2 +}$ .
- C
  $0.75 g$
  Why not C: Forgot to multiply by molar mass, or used 30 min instead of 60 min.
- D
  $9.48 g$
  Why not D: Used 4 hours or forgot to divide charge by 2 (two electrons per Cu).
Explanation
Charge: $q = I \times t = 2.00 A \times 3600 s = 7200 C$

Moles of electrons: $\frac{7200}{96 , 485} = 0.0746 mol$

Moles Cu: $\frac{0.0746}{2} = 0.0373 mol$ (2 electrons per Cu $^{2 +}$ )

Mass Cu: $0.0373 \times 63.5 = 2.37 g$

Key takeaway
Electrolysis: $q = It$; moles $e^-$ = $q/F$; moles product = moles $e^-$ / (electrons per ion).
Question 12 · Hard
In which of the following does the reaction become spontaneous as temperature increases from low to high?
- A
  $Δ H = - 80 kJ$ , $Δ S = - 200 J/K$
  Why not A: With $Δ H < 0$ and $Δ S < 0$ , the reaction is spontaneous at LOW temperature and becomes non-spontaneous at high temperature.
- B
  $Δ H = + 60 kJ$ , $Δ S = + 150 J/K$ Correct
- C
  $Δ H = - 60 kJ$ , $Δ S = + 150 J/K$
  Why not C: $Δ H < 0$ and $Δ S > 0$ means $Δ G < 0$ at ALL temperatures — always spontaneous.
- D
  $Δ H = + 80 kJ$ , $Δ S = - 200 J/K$
  Why not D: $Δ H > 0$ and $Δ S < 0$ means $Δ G > 0$ at ALL temperatures — never spontaneous.
Explanation
For $Δ H > 0$ and $Δ S > 0$ : $Δ G = Δ H - T Δ S$ . At low $T$ , $T Δ S$ is small and $Δ G > 0$ (non-spontaneous). At high $T$ , $T Δ S > Δ H$ and $Δ G < 0$ (spontaneous). The reaction becomes spontaneous above $T = Δ H /Δ S = 60, 000/150 = 400 K$ .

Key takeaway
$\Delta H > 0$, $\Delta S > 0$: non-spontaneous at low $T$; becomes spontaneous above $T = \Delta H / \Delta S$.