Question

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

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

Answer 1

The correct option is C $\frac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$

For the reaction:

$\underset{\underset{\left(atequilibrium\right)1-\alpha }{\left(initial\right)1}}{2HI\left(g\right)}\rightleftharpoons \underset{\underset{\frac{\alpha }{2}}{0}}{H_2\left(g\right)}+\underset{\underset{\frac{\alpha }{2}}{0}}{I_2\left(g\right)}$
$\because$ Partial pressure $=$ Mole fraction $\times {\rho }_t$ and $K_p=\frac{{\rho }_{H_2}\times {\rho }_{N_2}}{{\rho }_{\left(HI\right)}^2}\Rightarrow {\rho }_{H_2}=\left(\frac{\alpha }{2}\right){\rho }_T$
${\rho }_{N_2}=\left(\frac{\alpha }{2}\right){\rho }_T\Rightarrow {\rho }_{HI}=(1-\alpha ){\rho }_T$

Hence, $K_P=\frac{{[\left(\frac{\alpha }{2}\right)\cdot {\rho }_T]}^2}{(1-\alpha {)}^2\cdot {\rho }_T^2}\Rightarrow \frac{\alpha }{1-\alpha }=2\sqrt{K_p}$

On doing cross multiplication

$\alpha =2\sqrt{K_p}(1-\alpha )$
or $2\sqrt{K_p}-2\sqrt{K_p}\cdot \alpha =\alpha$

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

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

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