ΔG = ΔH - TΔS. Spontaneous when ΔG < 0. Temperature dependence.

Gibbs Free Energy — When Is a Reaction Spontaneous?

Question

A reaction has $\Delta H = -120 \text{ kJ/mol}$ and $\Delta S = -200 \text{ J/mol·K}$ . At what temperature does the reaction shift from spontaneous to non-spontaneous?

This question type appears regularly — JEE Main 2024 Shift 2 had a nearly identical setup asking students to identify the crossover temperature.

Solution — Step by Step

The fundamental relation is:

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

Spontaneity requires $\Delta G < 0$ . Our goal: find the temperature where $\Delta G$ flips sign.

$\Delta H = -120 \text{ kJ/mol} = -120{,}000 \text{ J/mol}$

$\Delta S = -200 \text{ J/mol·K}$

Always convert both to the same unit before substituting. Mixing kJ and J in one equation is the single most common error on this topic.

At the boundary between spontaneous and non-spontaneous:

0 = \Delta H - T\Delta S

T = \frac{\Delta H}{\Delta S} = \frac{-120{,}000}{-200} = 600 \text{ K}

Below 600 K: the $\Delta H$ term dominates (it’s more negative than $T\Delta S$ ), so $\Delta G < 0$ → spontaneous.

Above 600 K: $T\Delta S$ becomes more negative than $\Delta H$ , so $\Delta G > 0$ → non-spontaneous.

The reaction is spontaneous below 600 K.

Why This Works

The sign of $\Delta G$ is a tug-of-war between enthalpy and entropy, weighted by temperature. At low temperatures, $T$ is small so $T\Delta S$ barely matters — enthalpy wins. At high temperatures, the entropy term $T\Delta S$ dominates.

In our problem, both $\Delta H$ and $\Delta S$ are negative. The negative $\Delta H$ favours spontaneity; the negative $\Delta S$ works against it. As $T$ increases, the unfavourable entropy penalty ( $-T\Delta S$ becomes more positive) eventually outweighs the enthalpy benefit.

This is why many exothermic reactions with decreasing entropy (like gas-phase combination: $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$ ) become non-spontaneous at high temperatures — exactly the kind of real-world chemistry JEE loves to ask about.

$\Delta H$	$\Delta S$	Spontaneous?
$-$	$+$	Always ( $\Delta G$ always negative)
$+$	$-$	Never ( $\Delta G$ always positive)
$-$	$-$	Only at low T
$+$	$+$	Only at high T

This table is worth memorising cold — it directly appears as MCQ options.

Alternative Method

Instead of finding the crossover temperature algebraically, you can check $\Delta G$ at a specific temperature to confirm the trend.

At $T = 300 \text{ K}$ :

\Delta G = -120{,}000 - (300)(-200) = -120{,}000 + 60{,}000 = -60{,}000 \text{ J/mol}

$\Delta G < 0$ → spontaneous at 300 K. ✓

At $T = 900 \text{ K}$ :

\Delta G = -120{,}000 - (900)(-200) = -120{,}000 + 180{,}000 = +60{,}000 \text{ J/mol}

$\Delta G > 0$ → non-spontaneous at 900 K. ✓

This verification approach is useful in MCQs when you need to quickly eliminate options rather than solve the full equation.

Common Mistake

The unit mismatch trap. Students substitute $\Delta H = -120$ and $\Delta S = -200$ directly, getting $T = 0.6 \text{ K}$ — a physically absurd answer. The fix: always check whether $\Delta H$ is in kJ or J before dividing by $\Delta S$ (which is almost always in J/mol·K). Convert $\Delta H$ to joules first, every time.

A second mistake: interpreting “non-spontaneous” as “the reaction cannot happen.” It means the reaction won’t proceed spontaneously under standard conditions — it can still be driven by external energy (electrolysis, etc.).

When both $\Delta H$ and $\Delta S$ have the same sign, there’s always a crossover temperature. When they have opposite signs, the reaction is either always or never spontaneous — no calculation needed, just identify which case you’re in.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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