Question

Standard heat of formation of $KI$ is $-78.31kcalmo{l}^{-1}$. Calculate its lattice energy from following information:
$IE\left(K\right)=4.3eV$
$EA\left(I\right)=-73.4kcalmo{l}^{-1}$
Bond dissociation energy of $I_2$ is $36.1kcalmo{l}^{-1}$, sublimation energy of $K$ is $21.51kcalmo{l}^{-1}$.

(1$eV$ $=$ $23.06kcal$)

• A. -143 kcal mol${}^{-1}$
• B. 143 kcal mol${}^{-1}$
• C. 14.3 kcal mol${}^{-1}$
• D. -14.3 kcal mol${}^{-1}$

Answer 1

The correct option is A -143 kcal mol${}^{-1}$

$\text{Enthalpy of formation of KI from its element}=$$IE\left(K\right)+\frac{1}{2}$ $\text{[Bond dissociation energy of}$ $I_2$] $+EAof$ $I$ $+\text{Sublimation energy of K solid + lattice energy of}$ $K^{+}$ and $I^{-}$

$1eV=$ $23.06$ kcal

$\therefore \Delta H_f=4.3\times 23.06+\frac{36.1}{2}-73.4+21.51+latticeenergy=-\mathrm{78.31.}$

$\therefore Latticeenergy\approx -143kcal/mol$