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AP Chemistry › Unit 9

AP Chemistry Unit 9: Thermodynamics and Electrochemistry

7 to 9% of the exam · 180 practice questions · 5 free-response questions

The final unit asks whether a reaction is thermodynamically favorable. You will predict entropy changes, calculate Gibbs free energy from enthalpy and entropy, and link free energy to K and to cell potential. Electrochemistry covers galvanic and electrolytic cells and Faraday's law. Questions often ask which factor drives a reaction, or how a cell changes under nonstandard conditions. Students tend to plug numbers into the Gibbs equation without reasoning about whether enthalpy or entropy dominates. Many also assume a favorable reaction must be fast, which mixes up thermodynamics and kinetics.

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White belt · 60 questionsRegular · 60 questionsBlack belt · 60 questionsFree response

White belt covers core terms and foundational ideas, Regular is set at exam level, and Black belt has harder multi-step questions.

Topics in this unit

Free-response questions

Sample questions

These are taken from the Regular set, with the answer and explanation shown.

Topic 9.2 Absolute Entropy and Entropy Change

Standard molar entropies at 298 K
SubstanceS° (J/(mol·K))
NH₃(g)192.8
O₂(g)205.2
N₂(g)191.6
NO(g)210.8
NO₂(g)240.1
N₂O₄(g)304.4
H₂O(g)188.8
H₂O(l)69.9

1. According to the table, S° of N₂O₄(g) is greater than S° of NO₂(g). Yet ΔS° for 2 NO₂(g) → N₂O₄(g) is negative. Which statement best explains both facts?

  1. AEach N₂O₄ molecule has more ways to move, but 2 mol of gas become 1 mol
  2. BN₂O₄ has a lower molar mass than NO₂, so it moves faster
  3. CThe S° values are for 1 mol, so ΔS° should be positive after all
  4. DS° depends only on the number of atoms, so ΔS° must be zero

Answer: A. Per mole, the larger N₂O₄ molecule has more atoms and more ways to vibrate and rotate, so its S° is higher. But the reaction turns 2 mol of gas into 1 mol: 304.4 − 2(240.1) = −175.8 J/(molrxn·K).

Common trap: Thinking a larger S° for the product guarantees a positive ΔS°. You must multiply by coefficients first.

Topic 9.6 Free Energy of Dissolution

2. At 298 K, ΔS°soln is +43.4 J/(molrxn·K) for NaCl but −114.7 J/(molrxn·K) for MgCl₂. Which statement best explains why dissolving MgCl₂ decreases entropy?

  1. AMgCl₂ produces fewer ions in solution than NaCl does
  2. BThe small, doubly charged Mg²⁺ ion holds many water molecules in ordered shells
  3. CMg²⁺ ions move faster in water than Na⁺ ions do
  4. DDissolving MgCl₂ is endothermic, so it absorbs energy from the water

Answer: B. Dissolving releases ions, which raises entropy. But Mg²⁺ is small and has a 2+ charge, so it pulls water molecules into tight, ordered layers. That ordering lowers entropy more than the ion dispersal raises it.

Common trap: Saying MgCl₂ gives fewer ions. It gives 3 ions per formula unit, compared with 2 for NaCl.

Topic 9.10 Cell Potential Under Nonstandard Conditions

3. A galvanic cell is based on Cu(s) + 2 Ag⁺(aq) → Cu²⁺(aq) + 2 Ag(s). Which change would decrease the cell potential?

  1. AAdding AgNO₃(s) to the silver half-cell
  2. BAdding NaCl(s) to the silver half-cell, which precipitates AgCl(s)
  3. CAdding Na₂S(s) to the copper half-cell, which precipitates CuS(s)
  4. DReplacing the copper electrode with a larger one

Answer: B. Precipitating AgCl removes Ag⁺, a reactant. Q = [Cu²⁺]/[Ag⁺]² gets larger, so the reaction is closer to equilibrium and Ecell decreases.

Common trap: Choosing the Na₂S option. Removing Cu²⁺ removes a product, which lowers Q and raises the voltage.

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