OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीThe lattice enthalpy of \( \text{NaCl}(s) \) is \( +788\text{ kJ mol}^{-1} \) and its enthalpy of hydration (\( \Delta_{\text{hyd}} H^\ominus \)) is \( -784\text{ kJ mol}^{-1} \). What is the standard enthalpy of solution (\( \Delta_{\text{sol}} H^\ominus \)) when \( 2\text{ moles} \) of \( \text{NaCl}(s) \) are completely dissolved in excess water?
Correct answer · Option 1
\( +8\text{ kJ} \)
Explanation
Per mole: \( \Delta_{\text{sol}} H^\ominus = \Delta_{\text{lattice}} H^\ominus + \Delta_{\text{hyd}} H^\ominus = 788 - 784 = +4\text{ kJ mol}^{-1} \). For 2 moles: \( 2 \times 4 = +8\text{ kJ} \).
