State whether the entropy change is positive or negative for 2H_2(g) + O_2(g) → 2H_2O(l), and why. [2 marks]

A reaction has H = +20\ kJ mol^-1 and S = +50\ J K^-1mol^-1. Find the temperature above which it is feasible. [2 marks]

OCR OCR 2022-style practice: A reaction has H = -92\ kJ mol^-1 and S = -199\ J K^-1mol^-1. Calculate G at 298\ K and state whether the reaction is feasible.

G = H - T S (1). Converting S to kJ: S = -0.199\ kJ K^-1mol^-1. G = -92 - (298 × -0.199) = -92 - (-59.3) = -92 + 59.3 = -32.7\ kJ mol^-1 (1). G is negative, so the reaction is feasible at 298\ K (1). Markers reward the Gibbs equation with consistent units, the value of G, and the feasibility judgement from its sign.

EnglandChemistrySyllabus dot point

Why do some endothermic reactions still happen, and how do we predict feasibility?

Entropy as a measure of disorder, calculating entropy change of reaction, and the Gibbs free energy equation to decide feasibility and find the temperature at which a reaction becomes feasible.

An OCR H432 module 5 answer on entropy and free energy: entropy as disorder, calculating entropy change of reaction, and using the Gibbs free energy equation to decide feasibility and find the temperature of feasibility.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this topic is asking
Entropy
Gibbs free energy
Feasibility versus rate
Examples in context
Try this

What this topic is asking

OCR specification point 5.2.2 wants you to explain entropy as a measure of disorder, calculate the entropy change of a reaction, and use the Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$ to decide whether a reaction is feasible and to find the temperature at which it becomes feasible. This explains why some endothermic reactions still happen.

Entropy

Gibbs free energy

Finding the temperature at which a reaction becomes feasible

For $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ , $\Delta H = +178\ \text{kJ mol}^{-1}$ and $\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}$ . Find the minimum temperature for feasibility.

step 1: Set the feasibility boundary

The reaction just becomes feasible when $\Delta G = 0$ , so $0 = \Delta H - T\Delta S$ , giving $T = \Delta H/\Delta S$ .

step 2: Make the units consistent

Convert $\Delta H$ to joules: $178\ \text{kJ mol}^{-1} = 178000\ \text{J mol}^{-1}$ . (Or convert $\Delta S$ to kilojoules; either works as long as both match.)

step 3: Calculate the temperature

T = \frac{178000}{161} = 1106\ \text{K}\ (\text{about }1110\ \text{K}).

step 4: Interpret

Above about $1106\ \text{K}$ , $\Delta G$ becomes negative and the decomposition is feasible, which is why a lime kiln must reach a high temperature.

Feasibility versus rate

Examples in context

Example 1. Why ice melts above zero. Melting is endothermic ( $\Delta H > 0$ ) but increases disorder ( $\Delta S > 0$ ); above $273\ \text{K}$ the $T\Delta S$ term outweighs $\Delta H$ , so $\Delta G$ is negative and melting is feasible, a direct use of the Gibbs equation.

Example 2. Extracting metals by heating. Reducing a metal oxide with carbon becomes feasible only above a temperature where the entropy gain from producing gas makes $\Delta G$ negative, which is why smelting needs high temperatures.

Try this

Q1. State whether the entropy change is positive or negative for $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$ , and why. [2 marks]

Cue. Negative, because three moles of gas become two moles of liquid, decreasing disorder.

Q2. A reaction has $\Delta H = +20\ \text{kJ mol}^{-1}$ and $\Delta S = +50\ \text{J K}^{-1}\text{mol}^{-1}$ . Find the temperature above which it is feasible. [2 marks]

Cue. $T = \Delta H/\Delta S = 20000/50 = 400\ \text{K}$ .

Exam-style practice questions

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

OCR 20204 marksFor the decomposition

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

\Delta H = +178\ \text{kJ mol}^{-1}

and

\Delta S = +161\ \text{J K}^{-1}\text{mol}^{-1}

. (a) Calculate the temperature above which the reaction is feasible. (b) Explain why

\Delta S

is positive.

Show worked answer →

(a) At the feasibility boundary $\Delta G = 0$ , so $T = \dfrac{\Delta H}{\Delta S}$ (1). Converting $\Delta H$ to joules: $T = \dfrac{178000}{161} = 1106\ \text{K}$ (about $1110\ \text{K}$ ) (1)(1).

(b) $\Delta S$ is positive because a gas ( $\text{CO}_2$ ) is produced from a solid, greatly increasing the disorder (more ways to arrange the energy and particles) (1).

Markers reward setting $\Delta G = 0$ , the unit conversion and value, and the gas-production reason for the entropy increase.

OCR 20223 marksA reaction has

\Delta H = -92\ \text{kJ mol}^{-1}

and

\Delta S = -199\ \text{J K}^{-1}\text{mol}^{-1}

. Calculate

\Delta G

298\ \text{K}

and state whether the reaction is feasible.

Show worked answer →

$\Delta G = \Delta H - T\Delta S$ (1). Converting $\Delta S$ to $\text{kJ}$ : $\Delta S = -0.199\ \text{kJ K}^{-1}\text{mol}^{-1}$ .

$\Delta G = -92 - (298 \times -0.199) = -92 - (-59.3) = -92 + 59.3 = -32.7\ \text{kJ mol}^{-1}$ (1).

$\Delta G$ is negative, so the reaction is feasible at $298\ \text{K}$ (1).

Markers reward the Gibbs equation with consistent units, the value of $\Delta G$ , and the feasibility judgement from its sign.

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)