Calculate ΔG° for reaction with ΔH°=-92kJ and ΔS°=-198 J/K at 298K — is it spontaneous

Question

For a reaction, $\Delta H° = -92$ kJ/mol and $\Delta S° = -198$ J/K/mol at 298 K. Calculate $\Delta G°$ and predict whether the reaction is spontaneous at this temperature.

Solution — Step by Step

The Gibbs-Helmholtz equation relates $\Delta G$ , $\Delta H$ , and $\Delta S$ :

\Delta G° = \Delta H° - T \Delta S°

This equation tells us that spontaneity depends on the balance between enthalpy (energy released) and entropy (disorder change) at a given temperature.

$\Delta H° = -92$ kJ/mol = $-92{,}000$ J/mol

$\Delta S° = -198$ J/K/mol (already in J, not kJ)

$T = 298$ K

We must convert $\Delta H°$ to J (or $\Delta S°$ to kJ) before substituting — mixing kJ and J is the most common mistake in this type of problem.

T \Delta S° = 298 \times (-198) = -59{,}004 \text{ J/mol} = -59.004 \text{ kJ/mol}

\Delta G° = \Delta H° - T\Delta S° = -92{,}000 - (-59{,}004)

\Delta G° = -92{,}000 + 59{,}004 = -32{,}996 \text{ J/mol}

\Delta G° \approx -33 \text{ kJ/mol}

Since $\Delta G° < 0$ (negative), the reaction is spontaneous at 298 K under standard conditions.

Why This Works

The sign of $\Delta G°$ tells us the direction of spontaneous change at constant temperature and pressure:

$\Delta G° < 0$ : spontaneous (forward reaction favoured)
$\Delta G° > 0$ : non-spontaneous (reverse reaction favoured)
$\Delta G° = 0$ : system at equilibrium

In this problem, both $\Delta H°$ (negative = exothermic, favours spontaneity) and $\Delta S°$ (negative = decrease in disorder, opposes spontaneity) work against each other. The enthalpy effect wins at 298 K, making $\Delta G°$ negative.

At higher temperatures, the $T|\Delta S°|$ term grows larger, and eventually $\Delta G°$ will become positive — the reaction becomes non-spontaneous above a crossover temperature $T^* = \Delta H° / \Delta S°$ .

T^* = \frac{-92{,}000}{-198} \approx 465 \text{ K}

Above 465 K, this reaction is non-spontaneous.

Alternative Method — Finding Crossover Temperature

If asked “at what temperature does the reaction stop being spontaneous?”:

Set $\Delta G° = 0$ :

0 = \Delta H° - T \Delta S° \Rightarrow T = \frac{\Delta H°}{\Delta S°} = \frac{-92000}{-198} \approx 465 \text{ K}

The reaction is spontaneous for $T < 465$ K.

Common Mistake

Unit mismatch is the #1 error here. Students write $\Delta G° = -92 - 298 \times (-198) = -92 + 58,904$ , forgetting to convert $\Delta H°$ from kJ to J. They get $+58,812$ kJ — wildly wrong. Always check: if $\Delta H°$ is in kJ, convert $\Delta S°$ to kJ/K (divide by 1000), or convert $\Delta H°$ to J (multiply by 1000). Never mix units.

In JEE Main, this type of problem often adds “find the temperature at which equilibrium is attained” — that means find $T$ where $\Delta G° = 0$ , i.e., $T = \Delta H° / \Delta S°$ . This appeared in JEE Main 2022 Session 2.

Question

Solution — Step by Step

Why This Works

Alternative Method — Finding Crossover Temperature

Common Mistake

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