When 1 mole of is dissolved in large volume of water at 298 K and 1 bar, 184 kJ/mol heat is released — Thermodynamics and Thermochemistry Chemistry Question
Question
When 1 mole of $\text{Na(s)}$ is dissolved in large volume of water at 298 K and 1 bar, 184 kJ/mol heat is released. When 1 mole of $\text{Na}_2\text{O(s)}$ is dissolved in large volume of water at 298 K and 1 bar, 238 kJ/mol heat is released. If the enthalpy of formation of water is –286 kJ/mol, then the enthalpy of formation of sodium oxide is
💡 Solution & Explanation
Reaction 1: $2\text{Na(s)} + 2\text{H}_2\text{O(l)} \to 2\text{NaOH(aq)} + \text{H}_2\text{(g)}$ ($\Delta H = 2 \times -184 = -368\text{ kJ}$). Reaction 2: $\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \to 2\text{NaOH(aq)}$ ($\Delta H = -238\text{ kJ}$). Subtracting Rxn 2 from Rxn 1 yields $2\text{Na(s)} + \text{H}_2\text{O(l)} \to \text{Na}_2\text{O(s)} + \text{H}_2\text{(g)}$ ($\Delta H = -130\text{ kJ}$). This relates to formation enthalpies: $-130 = \Delta H_f(\text{Na}_2\text{O}) - \Delta H_f(\text{H}_2\text{O}) \implies -130 = \Delta H_f(\text{Na}_2\text{O}) - (-286) \implies \Delta H_f(\text{Na}_2\text{O}) = -416\text{ kJ/mol}$.