Question

For the reaction $2 HI (g) \rightleftharpoons H _{2}(g)+ I _{2}(g)$ the degree of dissociation $(\alpha)$ 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

Answer: (D)

For the reaction :



$\because$ Partial pressure $=$ Mole fraction $\times p_{T}$

and $K_{p}=\frac{p_{ H _{2}} \times p_{ N _{2}}}{p_{( HH )}^{2}} $

$\Rightarrow p_{ H _{2}}=\left(\frac{\alpha}{2}\right) p_{T}$

$p_{ N _{2}}=\left(\frac{\alpha}{2}\right) p_{T}$

$ \Rightarrow \, p_{ HI }=(1-\alpha) p_{T} $

Hence, $ K_{p}=\frac{\left[\left(\frac{\alpha}{2}\right) \cdot p_{T}\right]^{2}}{(1-\alpha)^{2} \cdot p_{T}^{2}}$

$\Rightarrow \frac{\alpha}{1-\alpha}=2 \sqrt{K_{p}}$

On doing cross multiplication

$\alpha=2 \sqrt{K_{p}}(1-\alpha)$

or $2 \sqrt{K_{p}}-2 \sqrt{K_{p}} \cdot \alpha =\alpha$

$2 \sqrt{K_{p}}=\alpha+2 \sqrt{K_{p}}$

$2 \sqrt{K_{p}}=\alpha\left(1+2 \sqrt{K_{p}}\right) $

$\therefore \, \alpha =\frac{2 \sqrt{K_{p}}}{1+2 \sqrt{K_{p}}}$