Question

For the reaction: $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$, the degree of dissociated $\left(\alpha \right)$ of $HI\left(g\right)$ is related to equilibrium constant $K_P$ by the expression:

• A. $\frac{1+2\sqrt{K_P}}{2}$
• B. $\sqrt{\frac{1+2K_P}{2}}$
• 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 $\frac{2\sqrt{K_P}}{1+2\sqrt{K_P}}$

$2HI\leftrightharpoons H_2+I_2$

$C(1-a)$ $\frac{Ca}{2}$ $\frac{Ca}{2}$

$K_p=\frac{{\left(\frac{Ca}{2}\right)}^2}{C(a-1)}{K}_p=\frac{a^2}{4({1-a)}^2}\frac{a}{1-a}=\sqrt{4K_p}a=2\sqrt{K_p}-2\sqrt{K_p}aa[1+2\sqrt{K_p}]=2\sqrt{K_p}$

$a=\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$

Degree of dissociation is given by $\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$