The Born-Haber cycle for

Question:

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 $\mathrm{K}=89.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\text {inination }} \mathrm{H}^{\ominus}$ for $\mathrm{K}=419.0 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_{\text {clatra }}$ gin $\mathrm{H}^{\ominus}$ for $\mathrm{Cl}_{(\mathrm{g})}$

$=-348.6 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_{\text {band }} \mathrm{H}^{\ominus}$ 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)

Solution:

$\Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{KCl}}^{\ominus}=\Delta_{\text {sub }} \mathrm{H}_{(\mathrm{K})}^{\ominus}+\Delta_{\text {ionization }} \mathrm{H}_{(\mathrm{K})}^{\ominus}+\frac{1}{2} \Delta_{\text {bond }} \mathrm{H}_{\left(\mathrm{Cl}_{2}\right)}^{\Theta}$

$+\Delta_{\text {electron gain }} \mathrm{H}_{(\mathrm{Cl})}^{\ominus}+\Delta_{\text {lattice }} \mathrm{H}_{(\mathrm{KCl})}^{\ominus}$

$\Rightarrow-436.7=89.2+419.0+\frac{1}{2}(243.0)+\{-348.6\}$

$+\Delta_{\text {lattice }} \mathrm{H}_{(\mathrm{KCl})}^{\ominus}$

$\Rightarrow \Delta_{\text {lattice }} \mathrm{H}_{(\mathrm{KCl})}^{\ominus}=-717.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The magnitude of lattice enthalpy of $\mathrm{KCl}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is 718 (Nearest integer).

Leave a comment

None
Free Study Material