A reaction's products have total S^ = 400 and its reactants S^ = 520\ J mol^-1K^-1. Calculate S^. [2 points]

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How do we calculate the standard entropy change of a reaction from absolute entropies?

Topic 9.2 Absolute Entropy and Entropy Change: use standard molar entropies to calculate the standard entropy change of a reaction as the sum for products minus the sum for reactants.

A focused answer to AP Chemistry Topic 9.2, covering absolute (standard molar) entropy, why it is positive for all substances, and calculating the standard entropy change of a reaction as products minus reactants, with full worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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The absolute (standard molar) entropy of a substance is its entropy per mole under standard conditions. Unlike enthalpies of formation, absolute entropies are positive for every substance above absolute zero, because there is always some dispersal of energy and matter; only a perfect crystal at absolute zero has zero entropy. To find the standard entropy change of a reaction, take the sum of the standard molar entropies of the products minus the sum for the reactants, each multiplied by its stoichiometric coefficient: delta S standard equals products minus reactants. A negative result means the products are more ordered (often fewer moles of gas); a positive result means they are more dispersed. Gases have much larger molar entropies than liquids or solids, so the change in moles of gas usually dominates the sign.

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What this topic is asking
Absolute entropy
The products-minus-reactants formula
Interpreting the sign
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What this topic is asking

The College Board (Topic 9.2) wants you to use standard molar entropies to calculate the standard entropy change of a reaction as the sum for the products minus the sum for the reactants. This is the quantitative companion to the qualitative sign prediction of Topic 9.1.

Absolute entropy

This is a key contrast with enthalpy: there is no absolute zero of enthalpy, so we tabulate enthalpies of formation (relative to elements), but entropy has a true zero (a perfect crystal at $0$ K), so we tabulate absolute entropies. Every real substance has a positive $S^\circ$ . Gases have much larger values than liquids or solids because of their greater dispersal.

The products-minus-reactants formula

The structure is identical to the enthalpy-of-formation calculation (Topic 6.8), but here you use absolute entropies and, crucially, elements are not zero (every substance contributes a positive $S^\circ$ ). Sum the products, sum the reactants, and subtract. The sign of the result classifies the reaction's entropy change.

Interpreting the sign

A negative $\Delta S^\circ$ means the products are more ordered than the reactants (often because gas moles decrease), and a positive $\Delta S^\circ$ means they are more dispersed (gas moles increase, a gas or solution forms). Because gaseous entropies are so much larger than condensed-phase ones, the change in the number of moles of gas usually controls the sign, matching the qualitative rule of Topic 9.1. The calculated value then feeds into the free-energy equation of Topic 9.3.

Standard entropy change of a reaction

For $2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$ , use $S^\circ$ values: $\text{H}_2\text{O}_2(l)\ 110$ , $\text{H}_2\text{O}(l)\ 70$ , $\text{O}_2(g)\ 205\ \text{J mol}^{-1}\text{K}^{-1}$ .

step 1 Sum the products

$2 \times 70 + 1 \times 205 = 140 + 205 = 345\ \text{J mol}^{-1}\text{K}^{-1}$ .

step 2 Sum the reactants

$2 \times 110 = 220\ \text{J mol}^{-1}\text{K}^{-1}$ .

step 3 Subtract

$\Delta S^\circ = 345 - 220 = +125\ \text{J mol}^{-1}\text{K}^{-1}$ .

step 4 Check the sign

A mole of gas ( $\text{O}_2$ ) is produced where there was none, so the entropy increases, consistent with the positive value.

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Q1. A reaction's products have total $S^\circ = 400$ and its reactants $S^\circ = 520\ \text{J mol}^{-1}\text{K}^{-1}$ . Calculate $\Delta S^\circ$ . [2 points]

Cue. $\Delta S^\circ = 400 - 520 = -120\ \text{J mol}^{-1}\text{K}^{-1}$ (negative; more ordered products).

Q2. Explain why absolute entropies are positive for all substances, unlike enthalpies of formation. [2 points]

Cue. Entropy has a true zero (a perfect crystal at absolute zero), so every substance above that has a positive absolute entropy; enthalpy has no absolute zero, so it is tabulated relative to elements.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For

\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)

, the standard molar entropies (in

\text{J mol}^{-1}\text{K}^{-1}

) are:

\text{N}_2\ 192

\text{H}_2\ 131

\text{NH}_3\ 193

. (a) Write the formula for the standard entropy change. (b) Calculate the sum for the products. (c) Calculate

\Delta S^\circ

for the reaction. (d) Justify the sign of your answer using the change in moles of gas.

Show worked answer →

A 4-point quantitative FRQ on entropy change.

(a) Formula (1 point): $\Delta S^\circ = \sum n S^\circ(\text{products}) - \sum n S^\circ(\text{reactants})$ .
(b) Products sum (1 point): $2 \times 193 = 386\ \text{J mol}^{-1}\text{K}^{-1}$ .
(c) Entropy change (1 point): reactants sum $= 192 + 3(131) = 192 + 393 = 585$ ; $\Delta S^\circ = 386 - 585 = -199\ \text{J mol}^{-1}\text{K}^{-1}$ .
(d) Justify (1 point): four moles of gas (1 + 3) become two moles of gas, so the gas moles decrease; fewer gas particles means less dispersal, consistent with the negative $\Delta S^\circ$ .

Markers reward the formula, the products sum, the entropy change as products minus reactants, and the moles-of-gas reasoning for the sign.

AP 2021 (style)1 marksSection I (multiple choice). The standard molar entropy of a pure crystalline element at

25\ ^\circ\text{C}

is (A) zero (B) negative (C) a positive value (D) equal to its enthalpy of formation. Justify your choice.

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A 1-point conceptual MCQ. The answer is (C).

Unlike enthalpies of formation, absolute (standard molar) entropies are positive for all substances at temperatures above absolute zero, because there is always some dispersal of energy and matter. Only a perfect crystal at absolute zero has zero entropy. The trap is (A): that applies to enthalpy of formation of elements, not to absolute entropy.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)