Chemical Thermodynamics: Step-by-Step Worked Examples (2)

Question

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$ at 298 K, $\Delta H° = -92.4$ kJ/mol and $\Delta S° = -198.3$ J/(K·mol). Find $\Delta G°$ and predict whether the reaction is spontaneous at this temperature.

Solution — Step by Step

Final answer: $\Delta G° = -33.31\text{ kJ/mol}$, reaction is spontaneous

Why This Works

The Gibbs free energy combines enthalpy (energy) and entropy (disorder). At constant $T$ and $P$, a process is spontaneous iff $\Delta G < 0$.

Here $\Delta H < 0$ favours the reaction (energy released), but $\Delta S < 0$ opposes it (gas molecules decrease from 4 to 2). At low enough temperatures, the enthalpy term dominates and the reaction proceeds. At high temperatures, the entropy term takes over and the reaction reverses — this is why the Haber process is run at moderate temperatures.

Alternative Method

Find the temperature at which $\Delta G° = 0$ (equilibrium):

$T = \frac{\Delta H°}{\Delta S°} = \frac{-92.4}{-0.1983} = 466\text{ K}$

Below 466 K, the reaction is spontaneous. Above 466 K, it reverses. At 298 K, well below the threshold — spontaneous. Same conclusion, different angle.

Common Mistake