Question

The Born-Haber cycle for $KCl$ is evaluated with the following data :
${\Delta }_f{H}^{\ominus }$ for $KCl=-436.7kJmo{l}^{-1}$
${\Delta }_{\text{sub }}{H}^{\ominus }$ for $K=89.2kJmo{l}^{-1}$
${\Delta }_{\text{ionization }}{H}^{\ominus }$ for $K=419.0kJmo{l}^{-1}$;
${\Delta }_{\text{electron gain }}{H}^{\ominus }$ for $C{l}_{(g)}$
$=-348.6kJmo{l}^{-1};{\Delta }_{\text{bond}}{H}^{\ominus }$ for $C{l}_2=243.0kJmo{l}^{-1}$
The magnitude of lattice enthalpy of $KCl$ in $kJmo{l}^{-1}$ is _______(Nearest integer)

Answer 1

Answer: 718

${\Delta }_f{H}_{KCl}^{\ominus }={\Delta }_{\text{sub }}{H}_{(K)}^{\ominus }+{\Delta }_{\text{ionization }}{H}_{(K)}^{\ominus }+\frac{1}{2}{\Delta }_{\text{bond }}{H}_{\left(C{l}_2\right)}^{\ominus }+{\Delta }_{\text{electron gain }}{H}_{(Cl)}^{\ominus }+{\Delta }_{\text{lattice }}{H}_{(KCl)}^{\ominus }$
$\Rightarrow -436.7=89.2+419.0+\frac{1}{2}(243.0)+\{-348.6\}+{\Delta }_{\text{lattice }}{H}_{(KCl)}^{\ominus }$
$\Rightarrow {\Delta }_{\text{lattice }}{H}_{(KCl)}^{\ominus }=-717.8kJmo{l}^{-1}$
The magnitude of lattice enthalpy of $KCl$ in $kJmo{l}^{-1}$ is $718$ (Nearest integer).