Question

For the reaction, $N_2{O}_4{}_{\left(g\right)}\rightleftharpoons 2N{O}_2{}_{\left(g\right)}$; the relation between the degree of dissociation $\alpha$ of $N_2{O}_4{}_{\left(g\right)}$ with the equilibrium constant $K_p$ and total pressure $P$ is:

• A. $\alpha =\frac{\frac{K_p}{P}}{4+\frac{K_P}{P}}$
• B. $\alpha =\frac{K_p}{4+K_p}$
• C. $\alpha ={\left(\frac{\frac{K_p}{P}}{4+\frac{K_p}{P}}\right)}^{1/2}$
• D. $\alpha ={\left(\frac{K_p}{4+K_p}\right)}^{1/2}$

Answer 1

Answer: (C)

$\alpha ={\left(\frac{\frac{K_p}{P}}{4+\frac{K_p}{P}}\right)}^{1/2}$

The reaction is

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

Let $\alpha$ be the degree of dissociation.

Let $a$ be the initial number of moles of $N_2{O}_4$

Initial number of moles of $N{O}_2=0$.

Let $a-x$ be the initial number of moles of $N_2{O}_4$

Initial number of moles of $N{O}_2=2x$.
The equilibrium pressure of $N{O}_2=\frac{2xP}{a+x}$

The equilibrium pressure of $N_2{O}_4=\frac{(a-x)P}{a+x}$
The equilibrium constant $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$
$K_p=\frac{(\frac{2xP}{a+x}{)}^2}{\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}}$
$K_p=\frac{4x^{2}P}{(a-x)(a+x)}$

$K_p=\frac{4x^{2}P}{a^2-x^2}$
$K_p=\frac{4P}{(\frac{a^2}{x^2}-1)}$

But, $\frac{x}{a}=\alpha =$ degree of dissociation
$K_p=\frac{4P}{(\frac{1}{{\alpha }^2}-1)}$
$\frac{1}{{\alpha }^2}-1=\frac{4P}{K_p}$
$\frac{1}{{\alpha }^2}=1+\frac{4P}{K_p}$
$\frac{1}{{\alpha }^2}=\frac{K_P+4P}{K_p}$
${\alpha }^2=\frac{K_p}{K_P+4P}$
$\alpha =\sqrt{\frac{K_p}{K_P+4P}}$
$\alpha =\sqrt{\frac{\frac{K_p}{P}}{\frac{K_P}{P}+4}}$

Hence, the correct option is $\text{C}$