The enthalpy of neutralization of a strong acid by a strong base is –57.32 kJ mol. The enthalpy of f — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The neutralization reaction is $\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \to \text{H}_2\text{O(l)}$ ($\Delta H = -57.32\text{ kJ/mol}$). By convention, $\Delta H_f(\text{H}^+\text{,aq}) = 0$. $\Delta H_{rxn} = \Delta H_f(\text{H}_2\text{O}) - \Delta H_f(\text{OH}^-) \implies -57.32 = -285.84 - \Delta H_f(\text{OH}^-) \implies \Delta H_f(\text{OH}^-) = -285.84 + 57.32 = -228.52\text{ kJ/mol}$.