Question

For the reaction $2HI(g)\rightleftharpoons H_2(g)+I_2(g)$ 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 $2\sqrt{K_p}-2\sqrt{K_p}\cdot \alpha =\alpha$

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

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

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