Question

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

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

$2HI\rightleftharpoons H_2+I_2$ Total moles

at $t=0$ $1$ - - $1$

at $Eq{b}^m$ $1-\alpha$ $\alpha /2$ $\alpha /2$ $1$

$K_P=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}[RT{]}^{\Delta ng}$

$\Rightarrow K_P=K_n{\left[\frac{P}{n}\right]}^{\Delta ng}$ as $\Delta ng=0$

$\Delta ng=nP-nR$

$K_P=K_n$

$\Rightarrow K_P=\frac{[\alpha /2{]}^2}{(1-\alpha {)}^2}$

$\Rightarrow \frac{\alpha }{1-\alpha }=2\sqrt{K}P$

$\Rightarrow \alpha =2{\sqrt{K}}_P-2{\sqrt{K}}_P\alpha$

$\Rightarrow 2\sqrt{K_P}=\alpha (1+2\sqrt{K_P)}$

$\Rightarrow \alpha =\left(\frac{2{\sqrt{K}}_P}{1+2{\sqrt{K}}_P}\right)$