Gibbs-Helmholtz equation ΔG = ΔH - TΔS — predict spontaneity at different temperatures

Question

For a reaction, $\Delta H = +50$ kJ/mol and $\Delta S = +150$ J/(mol·K). Determine the temperature above which the reaction becomes spontaneous. Also, classify the spontaneity behaviour at (a) 300 K and (b) 400 K.

(JEE Main 2022, similar pattern)

Solution — Step by Step

\Delta G = \Delta H - T\Delta S

A reaction is spontaneous when $\Delta G < 0$ . So we need:

\Delta H - T\Delta S < 0

At the crossover point, $\Delta G = 0$ :

T = \frac{\Delta H}{\Delta S} = \frac{50 \times 10^3}{150} = 333.3 \text{ K} \approx 60.3°\text{C}

Above this temperature, the $T\Delta S$ term dominates, making $\Delta G$ negative (spontaneous). Below it, $\Delta G$ is positive (non-spontaneous).

\Delta G = 50000 - 300 \times 150 = 50000 - 45000 = +5000 \text{ J/mol}

$\Delta G > 0$ — non-spontaneous at 300 K. Makes sense, since 300 K < 333.3 K.

\Delta G = 50000 - 400 \times 150 = 50000 - 60000 = -10000 \text{ J/mol}

$\Delta G < 0$ — spontaneous at 400 K. The entropy term has now overtaken the enthalpy term.

Why This Works

The Gibbs equation balances two competing drives: enthalpy ( $\Delta H$ ) and entropy ( $\Delta S$ ). When both are positive — endothermic with entropy increase — the reaction is entropy-driven but needs sufficient temperature for the $T\Delta S$ term to overcome $\Delta H$ .

Think of it like melting ice: it’s endothermic ( $\Delta H > 0$ ) and increases disorder ( $\Delta S > 0$ ). Below 0°C, it’s non-spontaneous. Above 0°C, entropy wins and ice melts on its own. The crossover temperature is exactly 273 K for ice melting at 1 atm.

The four possible sign combinations give a neat classification:

$\Delta H$	$\Delta S$	Spontaneity
$-$	$+$	Always spontaneous
$+$	$-$	Never spontaneous
$-$	$-$	Spontaneous at low $T$
$+$	$+$	Spontaneous at high $T$ (our case)

Alternative Method

You can rearrange the condition $\Delta G < 0$ directly:

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

This gives the critical temperature in one step. No need to compute $\Delta G$ at individual temperatures if the question only asks for the crossover.

Watch the units. $\Delta H$ is usually given in kJ/mol and $\Delta S$ in J/(mol·K). Convert $\Delta H$ to J first (multiply by 1000), or you’ll get the temperature wrong by a factor of 1000. This is the most common calculation error in this type of problem.

Common Mistake

Students memorise “negative $\Delta G$ means spontaneous” but forget that $\Delta G$ changes with temperature. A reaction that is non-spontaneous at room temperature can become spontaneous at a higher temperature (or vice versa). Always calculate $\Delta G$ at the specific temperature asked — don’t assume it’s the same at all temperatures.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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