Question

Consider the following reaction :
$HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)$

At equilibrium, 40% of $HI$ is dissociated.
The value of $K_x$ (equilibrium constant in terms of mole fraction) is :

• A. 1.25
• B. 0.75
• C. 0.67
• D. 0.25

Answer 1

The correct option is C 0.67

We know,
$K_p=K_x\times (P_{total,eq.}{)}^{\Delta n_g}=Kc\times (RT{)}^{\Delta n_g}$

Calculating $\Delta n_g$ for the given reaction:

$HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)$

$\Delta n_g=0.5+0.5-1=0\therefore K_p=K_x=K_c$
Let inital moles of $HI$ be $a$, and $\alpha$ be the no. of moles dissociated at equilibrium. Total volume is V

$HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)Initial:a00Equilibrium:a(1-\alpha )\frac{a.\alpha }{2}\frac{a.\alpha }{2}$
We know,
$Concentration=\frac{Moles}{Volume}$
$K_c=\frac{\left[H_2{]}^{0.5}\right[I_2{]}^{0.5}}{\left[HI\right]}=\frac{{\left[\frac{a.\alpha }{2V}\right]}^{0.5}\times {\left[\frac{a.\alpha }{2V}\right]}^{0.5}}{\left[\frac{a(1-\alpha )}{2V}\right]}{K}_c=\frac{{\left[\frac{a.\alpha }{2V}\right]}^1}{\left[\frac{a(1-\alpha )}{2V}\right]}=\frac{\alpha }{1-\alpha }$
Given, $\alpha =0.4$
$K_c=K_x=\frac{0.4}{1-0.4}=\frac{0.4}{0.6}=\frac{2}{3}\therefore K_x=\frac{2}{3}=0.67$


Theory:

Relation between $K_{x}and{K}_p$ :

$aA\left(g\right)+bB\left(g\right)\rightleftharpoons cC\left(g\right)+dD\left(g\right)$

Equilibrium constant in terms of pressure $K_p$:

$K_p=\frac{\left[P_c{]}_{eq}^c\right[P_d{]}_{eq}^d}{\left[P_a{]}_{eq}^a\right[P_b{]}_{eq}^b}$

Equilibrium constant in terms of mole fraction $K_x$:

$K_x=\frac{\left[x_c{]}_{eq}^c\right[x_d{]}_{eq}^d}{\left[x_a{]}_{eq}^a\right[x_b{]}_{eq}^b}$

$x_c=\frac{P_c}{P_{total,eq}}$

$K_p=\frac{\left[x_c{]}_{eq}^c\right[x_d{]}_{eq}^d(P_{total}{)}_{eq}^{c+d}}{\left[x_a{]}_{eq}^a\right[x_b{]}_{eq}^b(P_{total}{)}_{eq}^{a+b}}$

So $K_p=K_x(P_{Total,eq}{)}^{\Delta n_g}$