Question

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

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

	$2HI$	$H_2$	$I_2$
Initial	1	0	0
change	$-\alpha$	$\frac{\alpha }{2}$	$\frac{\alpha }{2}$
equilibrium	$1-\alpha$	$\frac{\alpha }{2}$	$\frac{\alpha }{2}$

The expression for the equilibrium constant becomes :

$K_p=\frac{\left[\right(\frac{\alpha }{2})\times P_T{]}^2}{(1-\alpha {)}^2\times P_T^2}$

or $\sqrt{K_P}=\frac{\frac{\alpha }{2}\times P_T}{(1-\alpha )\times P_T}$
or $2\sqrt{K_P}=\frac{\alpha }{1-\alpha }$
Thus, $\alpha =\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$.