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Chemistry · THERMODYNAMICS

हिन्दी

In the Born-Haber cycle for the formation of crystalline \( \text{NaCl}(s) \) shown in the given figure, which thermochemical step corresponds to the energy value of \( +495.6\text{ kJ mol}^{-1} \) (approximately \( 496\text{ kJ mol}^{-1} \))?

  1. Sublimation enthalpy of sodium metal, \( \Delta_{\text{sub}} H^\ominus \)
  2. Bond dissociation enthalpy of chlorine, \( \frac{1}{2}\Delta_{\text{bond}} H^\ominus \)
  3. Ionization enthalpy of gaseous sodium atoms, \( \Delta_i H^\ominus \)
  4. Lattice enthalpy of sodium chloride, \( \Delta_{\text{lattice}} H^\ominus \)

Correct answer · Option 3

Ionization enthalpy of gaseous sodium atoms, \( \Delta_i H^\ominus \)

Explanation

As shown in Fig. 5.9, the transition from \( \text{Na}(g) + \frac{1}{2}\text{Cl}_2(g) \) to \( \text{Na}^+(g) + \frac{1}{2}\text{Cl}_2(g) \) involves \( \Delta_i H^\ominus = +495.6\text{ kJ mol}^{-1} \), which is the ionization enthalpy of sodium.
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