OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीGiven the following thermochemical data for \( \text{NaCl} \): Sublimation enthalpy of sodium: \( +108.4\text{ kJ mol}^{-1} \) Ionization enthalpy of sodium: \( +496\text{ kJ mol}^{-1} \) Dissociation enthalpy of \( \text{Cl}_2 \): \( +242\text{ kJ mol}^{-1} \) Electron gain enthalpy of chlorine: \( -348.6\text{ kJ mol}^{-1} \) Standard enthalpy of formation of \( \text{NaCl}(s) \): \( -411.2\text{ kJ mol}^{-1} \) What is the lattice enthalpy (\( \Delta_{\text{lattice}} H^\ominus \)) of \( \text{NaCl}(s) \)?
Correct answer · Option 3
\( +788.0\text{ kJ mol}^{-1} \)
Explanation
\( \Delta_f H^\ominus = \Delta_{\text{sub}} H + \Delta_i H + \frac{1}{2}\Delta_{\text{bond}} H + \Delta_{eg} H - \Delta_{\text{lattice}} H \). Therefore, \( \Delta_{\text{lattice}} H = 108.4 + 496 + 121 - 348.6 - (-411.2) = +788.0\text{ kJ mol}^{-1} \).
