Question

The enthalpy change for the conversion of $\frac{1}{2} Cl_2(g)$ to $Cl^{-}(aq)$ is (-)_____________ $KJ mol^{-1}$ (Nearest integer)
Given : $\Delta_{\text {dis }} H _{ Cl _{2( g )}}^{\ominus}=240\, kJ \,mol ^{-1}$, $\Delta_{ eg } H _{ Cl }^{\ominus}=-350\, kJ \,mol ^{-1}$, $\Delta_{\text {hyd }} H ^\theta Cl _{( g )}^{-}=-380 \,kJ \,mol ^{-1}$

Answer 1

$\frac{1}{2}​Cl_{2(g)}​→Cl_{(g)}​→Cl_{(g)}^{−}​→Cl_{(aq.)}^−​$
$ΔH\degree=\frac{1}{2}​\times240+(−350)+(−380)$
$=−610$

The correct answer is -610