Question

In the dissociation of $HI$, $20\%$ of $HI$ is dissociated at equilibrium. Calculate $K_P$ for:
$HI\left(g\right)\rightleftharpoons {\frac{1}{2}H}_2\left(g\right)+{\frac{1}{2}I}_2\left(g\right)$.

• A. $K_P=K_C=0.625$
• B. $K_P=K_C=2.5$
• C. $K_P=K_C=0.025$
• D. $K_P=K_C=0.125$

Answer 1

Answer: (D)

$K_P=K_C=0.125$

$2HI\rightleftharpoons {\frac{1}{2}H}_2+{\frac{1}{2}I}_2$
$1$ $0$ $0$ Initial moles
$(1-\alpha )$ $\alpha /2$ $\alpha /2$ Moles at equilibrium

where $\alpha /2$ is degree of dissociation and volume of container is $VL$.

Since, $\Delta n_g=0$ so $K_P=K_C$

$K_P=K_C=\frac{{\left(\frac{\alpha }{2V}\right)}^{\frac{1}{2}}{\left(\frac{\alpha }{2V}\right)}^{\frac{1}{2}}}{\frac{(1-\alpha )}{V}}$
$⟹K_P=K_C=\frac{\alpha }{2(1-\alpha )}$

$\alpha =0.2$

$K_P=K_C=\frac{0.2}{2(1-0.2)}$ $⟹K_P=K_C=0.125$