A reaction has H = +60.\ kJ and S = +150\ J K^-1. Calculate the temperature above which it is favorable. [2 points]

State whether a reaction with H > 0 and S < 0 can ever be favorable. [1 point]

United StatesChemistrySyllabus dot point

How does the Gibbs free energy combine enthalpy and entropy to determine whether a process is thermodynamically favorable?

Topic 9.3 Gibbs Free Energy and Thermodynamic Favorability: use the equation delta G equals delta H minus T delta S to determine thermodynamic favourability and the temperature dependence of spontaneity.

A focused answer to AP Chemistry Topic 9.3, covering the Gibbs free energy equation, how the signs of enthalpy and entropy determine favourability, the temperature dependence of spontaneity, and the four sign cases, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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

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Quick answer

The Gibbs free energy change is delta G equals delta H minus T delta S. A process is thermodynamically favorable (spontaneous) when delta G is negative, at equilibrium when delta G is zero, and unfavorable when delta G is positive. The sign of delta G depends on the signs of enthalpy and entropy and on the temperature. There are four cases: if delta H is negative and delta S is positive, delta G is negative at all temperatures (always favorable); if delta H is positive and delta S is negative, delta G is positive at all temperatures (never favorable); the other two cases are temperature-dependent, because the T delta S term grows with temperature. When enthalpy and entropy oppose each other, you can find the crossover temperature where delta G equals zero by setting T equals delta H over delta S. Watch the units: enthalpy is usually in kilojoules and entropy in joules.

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What this topic is asking
The Gibbs free energy equation
The four sign cases
The crossover temperature
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What this topic is asking

The College Board (Topic 9.3) wants you to use $\Delta G = \Delta H - T\Delta S$ to determine thermodynamic favourability and the temperature dependence of spontaneity. The Gibbs free energy combines enthalpy and entropy into a single criterion for whether a process is favorable.

The Gibbs free energy equation

Free energy weighs the enthalpy change against the temperature-scaled entropy change. A favorable enthalpy (exothermic, $\Delta H < 0$ ) and a favorable entropy ( $\Delta S > 0$ ) both push $\Delta G$ negative; the balance, and the temperature, decide the outcome when they oppose each other.

The four sign cases

The two mixed cases are temperature-dependent because the $T\Delta S$ term scales with temperature. When enthalpy and entropy agree (both favorable or both unfavorable), temperature cannot change the outcome; when they disagree, temperature decides which term dominates.

The crossover temperature

When enthalpy and entropy oppose each other, there is a crossover temperature at which $\Delta G = 0$ and favourability flips. Setting $\Delta G = 0$ in the equation gives

T = \frac{\Delta H}{\Delta S}

Above or below this temperature the reaction becomes favorable, depending on which case applies. Always convert enthalpy and entropy to the same energy unit first (entropy is usually in joules, enthalpy in kilojoules), or the crossover temperature will be off by a factor of a thousand.

Favourability and temperature dependence

A reaction has $\Delta H^\circ = -90.0\ \text{kJ mol}^{-1}$ and $\Delta S^\circ = -200.\ \text{J mol}^{-1}\text{K}^{-1}$ . Find $\Delta G^\circ$ at $298$ K and the temperature above which it becomes unfavorable.

step 1 Convert units and apply the equation

$\Delta S^\circ = -0.200\ \text{kJ mol}^{-1}\text{K}^{-1}$ ; $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -90.0 - (298)(-0.200)$ .

step 2 Evaluate at 298 K

$\Delta G^\circ = -90.0 + 59.6 = -30.4\ \text{kJ mol}^{-1}$ , so favorable at $298$ K.

step 3 Find the crossover temperature

$\Delta G^\circ = 0$ when $T = \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{-90.0}{-0.200} = 450$ K.

step 4 Interpret

Both signs are negative, so the reaction is favorable only at low temperature; above $450$ K, $\Delta G^\circ > 0$ and it becomes unfavorable.

Try this

Q1. A reaction has $\Delta H = +60.\ \text{kJ}$ and $\Delta S = +150\ \text{J K}^{-1}$ . Calculate the temperature above which it is favorable. [2 points]

Cue. $T = \dfrac{60.}{0.150} = 400$ K (favorable above 400 K, since both are positive).

Q2. State whether a reaction with $\Delta H > 0$ and $\Delta S < 0$ can ever be favorable. [1 point]

Cue. No; both terms make $\Delta G$ positive at all temperatures.

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). A reaction has

\Delta H^\circ = +40.0\ \text{kJ mol}^{-1}

and

\Delta S^\circ = +120.\ \text{J mol}^{-1}\text{K}^{-1}

. (a) Write the Gibbs free energy equation. (b) Calculate

\Delta G^\circ

298

K. (c) Determine whether the reaction is thermodynamically favorable at

298

K. (d) Calculate the temperature above which the reaction becomes favorable.

Show worked answer →

A 4-point quantitative FRQ on Gibbs free energy.

(a) Equation (1 point): $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ .
(b) $\Delta G^\circ$ at 298 K (1 point): convert $\Delta S^\circ = 0.120\ \text{kJ mol}^{-1}\text{K}^{-1}$ ; $\Delta G^\circ = 40.0 - (298)(0.120) = 40.0 - 35.8 = +4.2\ \text{kJ mol}^{-1}$ .
(c) Favourability (1 point): $\Delta G^\circ > 0$ at $298$ K, so the reaction is not thermodynamically favorable at this temperature.
(d) Crossover temperature (1 point): favorable when $\Delta G^\circ < 0$ , that is $\Delta H^\circ < T\Delta S^\circ$ ; $T > \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{40.0}{0.120} = 333$ K.

Markers reward the equation, the $\Delta G^\circ$ value with the unit conversion, the unfavorable conclusion, and the crossover temperature.

AP 2021 (style)1 marksSection I (multiple choice). A reaction with

\Delta H < 0

and

\Delta S > 0

is (A) favorable at all temperatures (B) unfavorable at all temperatures (C) favorable only at high temperature (D) favorable only at low temperature. Justify your choice.

Show worked answer →

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

With $\Delta H < 0$ and $\Delta S > 0$ , both terms make $\Delta G = \Delta H - T\Delta S$ negative at every temperature, so the reaction is favorable at all temperatures. The trap is thinking temperature can ever make it unfavorable; it cannot when both signs help.

Related dot points

Sources & how we know this

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