Question

Standard heat of formation of $KI$ is $-78.31{\text{kcal mol}}^{-1}$. Calculate the lattice energy from following information:
$I_1\left(K\right)=4.3\text{eV}{E}_1\left(I\right)=73.4{\text{kcal mol}}^{-1}\text{Bond dissociation energy of}{I}_2=36.1{\text{kcal mol}}^{-1}\text{Sublimation energy of}K=21.51{\text{kcal mol}}^{-1}$

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

Answer 1

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

Let,
$I_1\left(K\right)=4.3\text{eV}=x_1{E}_1\left(I\right)=73.4{\text{kcal mol}}^{-1}=x_2\text{Bond dissociation energy of}{I}_2=36.1{\text{kcal mol}}^{-1}=x_3\text{Sublimation energy of}K=21.51{\text{kcal mol}}^{-1}=x_4\text{Lattice energy}=x_5$
$\therefore x_1+\frac{1}{2}{x}_3+x_2++x_4+x_5=-78.31\Rightarrow 4.3\times 23.06+\frac{36.1}{2}-73.4+21.51+x_5=-78.31[\because 1\text{eV}=23.06{\text{kcal mol}}^{-1}]\Rightarrow x_5\approx -143{\text{kcal mol}}^{-1}$