Question

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

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

Answer 1

Answer: (D)

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

$2H{I}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+I_{2\left(g\right)}Initial:CEquilibrium:C(1-2\alpha )C\alpha C\alpha$

Where $\alpha =$ degree of dissociation.

Let $P_t$ be total pressure,

$P_{H_2}=x_{H_2}\times P_2=\frac{C\alpha \times P_t}{C-2c\alpha +C\alpha +C\alpha }=\alpha \times P_t$

$P_{I_2}=\frac{C\alpha }{C}\times P_t=\alpha P_t$

$P_{H_I}=\frac{C-2C\alpha }{C}\times P_t=(1-2\alpha )P_t$

$K_P=\frac{\left(P_{H_2}\right)\left(P_{I_2}\right)}{{\left(P_{HI}\right)}^2}=\frac{{\alpha }^2\times P_t^2}{(1-2\alpha {)}^2\times P_t^2}$

$\therefore \frac{\alpha }{1-2\alpha }=\sqrt{K_P}$

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