Question

For the reaction: $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
The degree of dissociation (A) of $HI$ (g) 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 $\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$

$\begin{array}{ccccc}2HI\left(g\right)& \rightleftharpoons & H_2\left(g\right)& +& I_2\left(g\right)\\ 1-\alpha & & \frac{\alpha }{2}& & \frac{\alpha }{2}\end{array}$

$K_p=\frac{{\left(\frac{\alpha }{2}{P}_T\right)}^2}{(1-\alpha {)}^2{P}_T^2}\Rightarrow \frac{\alpha }{1-\alpha }=2\sqrt{K_p}$
Inverting both sides
$\frac{1-\alpha }{\alpha }=\frac{1}{2\sqrt{K_p}}\frac{1}{\alpha }-1=\frac{1}{2\sqrt{K_p}}$

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