Question

The value of $K_p$ for the equilibrium reaction $N_2O_4(g)\rightleftharpoons 2NO_2(g)$ is $2$.
The percentage dissociation of $N_2O_4(g)$ at a pressure of $0.5$ atm is

• A. $25$
• B. $88$
• C. $50$
• D. $71$

Answer 1

Answer: (D)

($\alpha$ = degree of dissociation)
Total number of moles at equil.
$=\left(1-\alpha \right)+2\alpha$
$=\left(1+\alpha \right)$
$p_{N_2O_4}=\frac{\left(1-\alpha \right)}{\left(1+\alpha \right)}\times P$
$p_{N_2}=\frac{2\alpha}{\left(1+\alpha\right)}\times P$
$K_{p}=\frac{\left(p_{N_2O_2}\right)^{2}}{p_{N_2O_4}}=\frac{\left(\frac{2\alpha}{\left(1+\alpha\right)}\times P\right)}{\left(\frac{1-\alpha}{1+\alpha}\right)\times P}=\frac{4\alpha^{2}P}{1-\alpha^{2}}$
Given, $K_{P}=2, P=0.5\,atm$
$\therefore K_{P}=\frac{4\alpha^{2}P}{1-\alpha^{2}}$
$=\frac{4\alpha^{2}\times0.5}{1-\alpha^{2}}$
$\alpha=0.707\approx0.71$
$\therefore $ Percentage dissociation
$=0.71\times100=71$