Chemical Thermodynamics: Exam-Pattern Drill (4)

Question

For the reaction $2H_2(g) + O_2(g) \to 2H_2O(l)$, $\Delta H = -572 \, \text{kJ/mol}$ and $\Delta S = -327 \, \text{J/(K·mol)}$ at $298 \, \text{K}$. Compute $\Delta G$ and predict whether the reaction is spontaneous.

Solution — Step by Step

Why This Works

$\Delta G = \Delta H - T\Delta S$ packages two driving forces (enthalpy and entropy) into one number. Negative $\Delta G$ guarantees the reaction can proceed forward without external energy input.

Here $\Delta H$ is strongly negative (exothermic), and $-T\Delta S$ is positive (entropy of system decreases as gases combine into a liquid). Enthalpy wins at room temperature.

Alternative Method — Test Both Driving Forces

Without computing: $\Delta H < 0$ favours forward; $\Delta S < 0$ disfavours forward. Mixed signs — depends on temperature.

Cross-over temperature: set $\Delta G = 0$, get $T = \Delta H/\Delta S = 572{,}000/327 = 1750 \, \text{K}$.

Below $1750 \, \text{K}$, the reaction is spontaneous. At $298 \, \text{K}$, well below crossover, so spontaneous.

Common Mistake

Students often forget to convert $\Delta H$ from kJ to J before subtracting $T\Delta S$ (which is in J). Mixed units give wildly wrong answers.

The four sign cases: $\Delta H < 0, \Delta S > 0$ → always spontaneous. $\Delta H > 0, \Delta S < 0$ → never spontaneous. Mixed signs → temperature-dependent.