AP Chemistry Entropy

Predicting the sign of ΔS, calculating ΔS° from standard molar entropies, and ranking substances, in four worked examples.

Entropy questions come in three flavors: decide whether a change increases or decreases entropy, compute ΔS° for a reaction from a table, and use the sign of ΔS to explain why something happens. The first two are quick once you know the rules, and the third connects to Gibbs free energy. The Unit 9 review introduces the idea; this page adds the sign-prediction rules, the unit handling, and the comparisons the exam likes to ask for.

What Entropy Measures and the Rules for Predicting ΔS

Entropy (S) describes how dispersed the matter and energy of a system are. Particles that are free to move through a larger volume, or to occupy more arrangements, have more entropy. Use these rules, in this order:

1. Count moles of gas. More moles of gas in the products than the reactants means ΔS > 0; fewer means ΔS < 0. This rule decides most reaction questions.
2. If the gas moles match, check phases. Solid → liquid → gas increases entropy, and the reverse decreases it.
3. Dissolving usually increases entropy as ions or molecules spread through the solution, and precipitation decreases it.
4. Higher temperature and larger, more complex molecules both mean higher entropy.

Worked Example 1: Predicting the Sign of ΔS for Six Changes

Predict the sign of ΔS for each change.

1. H₂O(s) → H₂O(l)
2. H₂O(g) → H₂O(l)
3. NaCl(s) → Na⁺(aq) + Cl⁻(aq)
4. 2SO₂(g) + O₂(g) → 2SO₃(g)
5. CaCO₃(s) → CaO(s) + CO₂(g)
6. N₂(g) + 3H₂(g) → 2NH₃(g)

1. Positive. A solid becomes a liquid, which is more dispersed.
2. Negative. Condensation takes a gas to a liquid.
3. Positive. The crystal separates into ions that move freely in solution.
4. Negative. Three moles of gas become two.
5. Positive. No gas on the left, one mole of gas on the right.
6. Negative. Four moles of gas become two.

In every case the answer comes from counting gas moles or changing phase, with no calculation needed. On a free-response question, say it in those terms: “the number of gas molecules decreases, so entropy decreases.”

Worked Example 2: Calculating ΔS° for the Formation of Ammonia

Use S°(N₂) = 191.6, S°(H₂) = 130.7, and S°(NH₃) = 192.5, all in J/(mol·K), to find ΔS° for N₂(g) + 3H₂(g) → 2NH₃(g).

Products: 2(192.5) = 385.0.
Reactants: 191.6 + 3(130.7) = 583.7.
ΔS° = 385.0 − 583.7 = −198.7 J/(mol·K).

The negative sign matches Example 1: four moles of gas became two. Multiply each S° by its coefficient before adding, which is the same bookkeeping used for enthalpy in Hess's law problems. The values here are rounded from standard tables; on the exam, use the values the question provides.

Worked Example 3: Using ΔS° and ΔH° to Find When N₂O₄ Decomposition Is Favored

For N₂O₄(g) → 2NO₂(g), S°(N₂O₄) = 304.4 and S°(NO₂) = 240.1 J/(mol·K), and ΔH° = +57.2 kJ/mol. Find ΔS°, then the temperature above which the reaction is thermodynamically favored.

ΔS° = 2(240.1) − 304.4 = +175.8 J/(mol·K), positive because one mole of gas becomes two.
Convert units: 175.8 J/(mol·K) = 0.1758 kJ/(mol·K).
Crossover: ΔG° = 0 when T = ΔH°/ΔS° = 57.2 ÷ 0.1758 ≈ 325 K.

Since ΔH° > 0 and ΔS° > 0, the reaction is favored only at high temperature, above about 325 K. At 298 K, ΔG° = 57.2 − (298)(0.1758) ≈ +4.8 kJ/mol, so it is not favored. The most common error here is mixing joules and kilojoules, covered in AP Chemistry Gibbs free energy.

Worked Example 4: Ranking Substances by Standard Molar Entropy

For each pair, choose the one with the higher standard molar entropy at 298 K, and say why: (a) H₂O(l) or H₂O(g); (b) CH₄(g) or C₂H₆(g); (c) Fe(s) or H₂(g); (d) 1 mol of N₂ at 500 K or at 300 K.

(a) H₂O(g) (188.8 vs 70.0): gases are far more dispersed than liquids.
(b) C₂H₆(g) (about 230 vs 186): a larger molecule has more ways to store energy and more motions.
(c) H₂(g) (130.7 vs 27.3): a gas beats a solid.
(d) N₂ at 500 K: the molecules have a broader spread of kinetic energies, so entropy rises with temperature.

The chart above shows the same pattern in real values: solids and liquids sit lowest, small gases in the middle, and the larger N₂O₄ molecule highest. The temperature effect is the Maxwell-Boltzmann broadening discussed in the kinetic molecular theory page.

Five Entropy Practice Questions With Answers

1. Predict the sign of ΔS for 2H₂(g) + O₂(g) → 2H₂O(l).
   Show answer

   Negative. Three moles of gas become zero moles of gas, and the product is a liquid.
2. Ammonium nitrate dissolves in water even though the process is endothermic. What does this suggest about ΔS?
   Show answer

   ΔS is positive. The ions become dispersed in solution, and the entropy increase outweighs the unfavorable ΔH, so ΔG is negative at that temperature.
3. Use S° = 92.9 for CaCO₃(s), 39.8 for CaO(s), and 213.8 for CO₂(g), all in J/(mol·K), to find ΔS° for CaCO₃(s) → CaO(s) + CO₂(g).
   Show answer

   ΔS° = (39.8 + 213.8) − 92.9 = +160.7 J/(mol·K).
4. For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔS° is only about −5 J/(mol·K). Why is it so small?
   Show answer

   There are three moles of gas on each side, so the gas-mole rule predicts almost no change. The small remainder comes from differences in molecular complexity.
5. A student writes ΔS° = −198.7 kJ/(mol·K) for ammonia formation. What is wrong?
   Show answer

   The unit. Entropy values are in J/(mol·K), so the answer is −198.7 J/(mol·K), or −0.1987 kJ/(mol·K) if converted for use with ΔH in kJ.

Common Entropy Mistakes

• Mixing J and kJ. ΔS° is in J/(mol·K) and ΔH° in kJ/mol; convert one before using ΔG = ΔH − TΔS.
• Treating the S° of an element as zero. That is true for the enthalpy of formation, not for entropy; every S° is positive.
• Forgetting coefficients when summing S° for products and reactants.
• Using the wrong phase. S°(H₂O(l)) and S°(H₂O(g)) differ by more than a factor of two.
• Assuming a negative ΔS means the reaction cannot happen. It can, if ΔH is negative enough at a low enough temperature.
• Judging by the number of molecules instead of gas molecules. Solids and liquids barely count in the sign rule.

Thermodynamics is a steady source of points on both sections of the exam. To see how a stronger Unit 9 would shift your overall estimate, the AP Chem Score Calculator turns section scores into a 1–5 range.

Related Resources

• AP Chem Score Calculator
• Unit 9 Review: Thermodynamics and Electrochemistry
• AP Chemistry Gibbs Free Energy
• AP Chemistry Endothermic vs Exothermic
• AP Chemistry Hess's Law
• AP Chemistry Kinetic Molecular Theory
• AP Chemistry Heating Curve
• AP Chemistry Study Guide

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