These worked examples bring together entropy changes and Gibbs free energy calculations.
Example 1: Calculating ΔS
N₂(g) + 3H₂(g) → 2NH₃(g)
Given S° values (J K⁻¹ mol⁻¹):
- S°[N₂] = 191.6
- S°[H₂] = 130.7
- S°[NH₃] = 192.5
\( \Delta S = \sum S(\text{products}) - \sum S(\text{reactants}) \)
\( = [2(192.5)] - [191.6 + 3(130.7)] \)
\( = 385.0 - 583.7 \)
\( = -198.7 \text{ J K}^{-1}\text{mol}^{-1} \)
Negative ΔS makes sense: 4 moles of gas → 2 moles of gas (decrease in disorder).
Example 2: Calculating ΔG
Is the Haber process spontaneous at 298 K?
Given: ΔH = −92.2 kJ mol⁻¹, ΔS = −198.7 J K⁻¹ mol⁻¹
Convert ΔS to kJ: −198.7 ÷ 1000 = −0.1987 kJ K⁻¹ mol⁻¹
\( \Delta G = \Delta H - T\Delta S \)
\( = -92.2 - (298)(-0.1987) \)
\( = -92.2 + 59.2 \)
\( = -33.0 \text{ kJ mol}^{-1} \)
ΔG < 0, so the reaction is spontaneous at 298 K.
Example 3: Finding the Crossover Temperature
At what temperature does the Haber process become non-spontaneous?
At the crossover, ΔG = 0:
\( 0 = \Delta H - T\Delta S \)
\( T = \frac{\Delta H}{\Delta S} = \frac{-92.2}{-0.1987} \)
\( T = 464 \text{ K} \approx 191°\text{C} \)
Above 464 K, the TΔS term dominates and ΔG becomes positive — the reaction is non-spontaneous. This is why the Haber process operates at a compromise temperature of ~450 °C (to increase rate despite lower yield).
Exam Strategy
The most common error is unit mismatch: ΔH is usually in kJ mol⁻¹ while ΔS is in J K⁻¹ mol⁻¹. Always convert ΔS to kJ (÷1000) before substituting into ΔG = ΔH − TΔS.