Question

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

The equilibrium reaction and the equilibrium number of moles of various species are shown below.
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
$1-\alpha \frac{\alpha }{2}\frac{\alpha }{2}$
The expression for the equilibrium constant is $K_p=\frac{P_{H_2}{P}_{I_2}}{P_{HI}^2}.$
Substitute values in the above expression.
$K_p=\frac{(\frac{\alpha }{2}{P}_T{)}^2}{(1-\alpha {)}^2{P}_T^2}\Rightarrow \frac{\alpha }{1-\alpha }=2\sqrt{K_p}.$

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