JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
The Born-Haber cycle for \(\mathrm{KCl}\) is evaluated with the following data : \(\Delta_{f} \mathrm{H}^{\ominus}\) for \(\mathrm{KCl}=-436.7 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}\) \(\Delta_{\text {sub }} \mathrm{H}^{\ominus}\) for \(\mathrm{K}=89.2 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}\) \(\Delta_{\text {ionization }} \,\mathrm{H}^{-}\) for \(\mathrm{K}=419.0\, \mathrm{~kJ}\, \mathrm{~mol}^{-1}\) \(\Delta_{\text {electron gain }} \mathrm{H}^{\ominus}\) for \(\mathrm{Cl}_{(\text {e) }}=-348.6 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}\) \(\Delta_{\mathrm{bond}} \mathrm{H}^{-}\) for \(\mathrm{Cl}_{2}=243.0 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}\) The magnitude of lattice enthalpy of \(\mathrm{KCl}\) in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is ..... . (Nearest integer)
- A \(718\)
- B \(951\)
- C \(632\)
- D \(521\)
Answer & Solution
Correct Answer
(A) \(718\)
Step-by-step Solution
Detailed explanation
\(\Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{KCl}}^{\Theta}=\Delta_{\text {sub }} \mathrm{H}_{(\mathrm{K})}^{\Theta}+\Delta_{\text {ionization }} \mathrm{H}_{(\mathrm{K})}^{\Theta}+\frac{1}{2} \Delta_{\mathrm{bond}} \mathrm{H}_{\left(\mathrm{Cl}_{2}\right)}^{\Theta}\)…
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