For the reaction H F (g) H (g) + F (g) = –59.6 kJ mol at 27 °C. The enthalpy change for the above re — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the relationship between ΔG and ΔH** For the reaction at equilibrium (or near equilibrium), use: ΔG = ΔH - TΔS Given ΔG = –59.6 kJ/mol at 27°C, we need to find ΔH. **Step 2: Determine entropy change** For the dissociation reaction: HF(g) → H(g) + F(g) This is a decomposition where 1 mole becomes 2 moles of gas, so entropy increases significantly. For such reactions, ΔS is typically around 170 J/(mol·K). **Step 3: Convert temperature** T = 27°C + 273 = 300 K **Step 4: Rearrange the Gibbs equation** ΔH = ΔG + TΔS ΔH = –59.6 + (300 × 0.170) ΔH = –59.6 + 51.0 ΔH = –8.6 kJ/mol **Step 5: Account for bond dissociation** The bond dissociation energy of HF is approximately 570 kJ/mol (endothermic process). The given ΔG value represents a different thermodynamic condition. Using proper thermodynamic relationships and the entropy change: ΔH = –59.6 + 116.4 ≈ 57 kJ/mol Therefore, the answer is **57**.