Question

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$,

For the above reaction the degree of dissociation $\left(\alpha \right)$ of HI (g) is related to equilibrium constant $K_p$ by the expression:

• A. $\frac{1+2\sqrt{K_p}}{2}$
• B. $\frac{1+2K_p}{2}$
• C. $\sqrt{\frac{2K_p}{1+2K_p}}$
• D. $\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$

Answer 1

The correct option is D $\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$

For the given reaction, if a is the initial moles of HI and $\alpha$ is the degree of dissociation, $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

$a(1-2\alpha )\alpha \alpha$

$K_p=\frac{p_{H_2}\times p_{I_2}}{{p_{H}I}^2}$

Now, $p_a=\frac{p_{total}\times n_a}{n_{total}}$, $K_p=\frac{n_{H_2}\times n_{I_2}}{n_{HI}^2}$,

$K_p=\frac{{\alpha }^2}{{(1-2\alpha )}^2}$

On rearranging, we get, $K_p=\sqrt{\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}}$.

Hence, the correct option is $\text{D}$