Calculate ΔS⦵ for N2(g) + 3H2(g) → 2NH3(g). S⦵ in J K−1 mol−1: N2 = 191.6; H2 = 130.6; NH3 = 192.3.
- Products: 2 × 192.3 = 384.6.
- Reactants: 191.6 + (3 × 130.6) = 191.6 + 391.8 = 583.4.
- ΔS⦵ = 384.6 − 583.4 = −198.8 J K−1 mol−1.
- Check the sign: 4 moles of gas become 2 moles of gas, so a negative value is expected.
Answer: ΔS⦵ = −198.8 J K−1 mol−1