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Chemistry · CHEMICAL BONDING AND MOLECULAR STRUCTURE
हिन्दीGiven the ionization enthalpy of \( \text{Na(g)} \) is \( +495.8\text{ kJ mol}^{-1} \), the electron gain enthalpy of \( \text{Cl(g)} \) is \( -348.7\text{ kJ mol}^{-1} \), and the lattice enthalpy of \( \text{NaCl(s)} \) is \( +788\text{ kJ mol}^{-1} \). What is the net enthalpy change for the formation of \( 1\text{ mole} \) of solid \( \text{NaCl} \) from \( \text{Na(g)} \) and \( \text{Cl(g)} \)?
Correct answer · Option 3
\( -640.9\text{ kJ mol}^{-1} \)
Explanation
\( \Delta H_{\text{net}} = \Delta_i H + \Delta_{\text{eg}} H - \Delta_{\text{lattice}} H = 495.8 - 348.7 - 788 = 147.1 - 788 = -640.9\text{ kJ mol}^{-1} \).
