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AP Chemistry Practice Quiz: Absolute Entropy and Entropy Change

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Which of the following expressions correctly defines the standard entropy change for a chemical reaction (ΔS°reaction) based on the standard molar entropies (S°) of the species involved?

All Questions (7)

Which of the following expressions correctly defines the standard entropy change for a chemical reaction (ΔS°reaction) based on the standard molar entropies (S°) of the species involved?

A) ΔS°reaction = ΣS°products − ΣS°reactants

B) ΔS°reaction = ΣS°reactants − ΣS°products

C) ΔS°reaction = ΣS°products + ΣS°reactants

D) ΔS°reaction = (ΣS°products) / (ΣS°reactants)

Correct Answer: A

The standard entropy change for a reaction is calculated by summing the standard molar entropies of the products and subtracting the sum of the standard molar entropies of the reactants, with each value multiplied by its stoichiometric coefficient.

Consider the decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g). Given the standard molar entropies below, calculate the standard entropy change (ΔS°) for the reaction. S°[CaCO₃(s)] = 92.9 J/mol·K S°[CaO(s)] = 39.8 J/mol·K S°[CO₂(g)] = 213.7 J/mol·K

A) -160.6 J/mol·K

B) +160.6 J/mol·K

C) +346.4 J/mol·K

D) -81.0 J/mol·K

Correct Answer: B

Using the formula ΔS°reaction = ΣS°products − ΣS°reactants: ΔS° = (S°[CaO(s)] + S°[CO₂(g)]) - (S°[CaCO₃(s)]) ΔS° = (39.8 J/mol·K + 213.7 J/mol·K) - (92.9 J/mol·K) ΔS° = (253.5 J/mol·K) - (92.9 J/mol·K) ΔS° = +160.6 J/mol·K

Calculate the standard entropy change (ΔS°) for the synthesis of ammonia from its elements, shown in the balanced equation below, using the provided standard molar entropy values. N₂(g) + 3H₂(g) → 2NH₃(g) S°[N₂(g)] = 191.6 J/mol·K S°[H₂(g)] = 130.7 J/mol·K S°[NH₃(g)] = 192.8 J/mol·K

A) +129.5 J/mol·K

B) -129.5 J/mol·K

C) +198.1 J/mol·K

D) -198.1 J/mol·K

Correct Answer: D

Using the formula ΔS°reaction = ΣS°products − ΣS°reactants, and accounting for stoichiometric coefficients: ΔS° = (2 × S°[NH₃(g)]) - (1 × S°[N₂(g)] + 3 × S°[H₂(g)]) ΔS° = (2 × 192.8) - (191.6 + 3 × 130.7) ΔS° = 385.6 - (191.6 + 392.1) ΔS° = 385.6 - 583.7 ΔS° = -198.1 J/mol·K

For the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), the standard entropy change (ΔS°) is -243.0 J/mol·K. Using the standard molar entropies provided below, what is the absolute entropy (S°) of CH₄(g)? S°[O₂(g)] = 205.1 J/mol·K S°[CO₂(g)] = 213.7 J/mol·K S°[H₂O(l)] = 69.9 J/mol·K

A) 186.3 J/mol·K

B) 243.0 J/mol·K

C) 353.5 J/mol·K

D) 596.5 J/mol·K

Correct Answer: A

Start with the formula: ΔS° = ΣS°products − ΣS°reactants. -243.0 = (1 × S°[CO₂] + 2 × S°[H₂O]) - (1 × S°[CH₄] + 2 × S°[O₂]) -243.0 = (213.7 + 2 × 69.9) - (S°[CH₄] + 2 × 205.1) -243.0 = (213.7 + 139.8) - (S°[CH₄] + 410.2) -243.0 = 353.5 - S°[CH₄] - 410.2 -243.0 = -56.7 - S°[CH₄] S°[CH₄] = -56.7 + 243.0 S°[CH₄] = 186.3 J/mol·K

Calculate the standard entropy change (ΔS°) for the formation of liquid water from hydrogen and oxygen gas. 2H₂(g) + O₂(g) → 2H₂O(l) S°[H₂(g)] = 130.7 J/mol·K S°[O₂(g)] = 205.1 J/mol·K S°[H₂O(l)] = 69.9 J/mol·K

A) +326.7 J/mol·K

B) -326.7 J/mol·K

C) -163.3 J/mol·K

D) +163.3 J/mol·K

Correct Answer: B

Apply the formula ΔS°reaction = ΣS°products − ΣS°reactants. ΔS° = (2 × S°[H₂O(l)]) - (2 × S°[H₂(g)] + 1 × S°[O₂(g)]) ΔS° = (2 × 69.9) - (2 × 130.7 + 205.1) ΔS° = 139.8 - (261.4 + 205.1) ΔS° = 139.8 - 466.5 ΔS° = -326.7 J/mol·K

The rusting of iron is represented by the equation 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). The standard entropy change (ΔS°) for this reaction is -549.5 J/mol·K. Given that S° for Fe(s) is 27.3 J/mol·K and S° for Fe₂O₃(s) is 87.4 J/mol·K, calculate the standard molar entropy (S°) of O₂(g).

A) 137.5 J/mol·K

B) 68.3 J/mol·K

C) 205.0 J/mol·K

D) 245.3 J/mol·K

Correct Answer: C

Set up the equation: ΔS° = ΣS°products − ΣS°reactants. -549.5 = (2 × S°[Fe₂O₃]) - (4 × S°[Fe] + 3 × S°[O₂]) -549.5 = (2 × 87.4) - (4 × 27.3 + 3 × S°[O₂]) -549.5 = 174.8 - (109.2 + 3 × S°[O₂]) -549.5 = 174.8 - 109.2 - 3 × S°[O₂] -549.5 = 65.6 - 3 × S°[O₂] -549.5 - 65.6 = -3 × S°[O₂] -615.1 = -3 × S°[O₂] S°[O₂] = -615.1 / -3 = 205.0 J/mol·K

For a particular process, the sum of the standard molar entropies of the products, taking into account their stoichiometric coefficients, is 450 J/mol·K. The sum of the standard molar entropies for the reactants is 620 J/mol·K. What is the standard entropy change (ΔS°) for this process?

A) +1070 J/mol·K

B) -1070 J/mol·K

C) +170 J/mol·K

D) -170 J/mol·K

Correct Answer: D

The question provides the summed values for products and reactants directly. Using the formula ΔS°reaction = ΣS°products − ΣS°reactants: ΔS° = (450 J/mol·K) - (620 J/mol·K) ΔS° = -170 J/mol·K