Question

The value of $K_p$ for the equilibrium reaction $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$ is $2$.
The percentage dissociation of $N_2{O}_4(g)$ 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.
$=\left(1-\alpha \right)+2\alpha$
$=\left(1+\alpha \right)$
$p_{N_2{O}_4}=\frac{\left(1-\alpha \right)}{\left(1+\alpha \right)}\times P$
$p_{N_2}=\frac{2\alpha }{\left(1+\alpha \right)}\times P$
$K_p=\frac{{\left(p_{N_2{O}_2}\right)}^2}{p_{N_2{O}_4}}=\frac{\left(\frac{2\alpha }{\left(1+\alpha \right)}\times P\right)}{\left(\frac{1-\alpha }{1+\alpha }\right)\times P}=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$
Given, $K_P=2,P=0.5atm$
$\therefore K_P=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$
$=\frac{4{\alpha }^2\times 0.5}{1-{\alpha }^2}$
$\alpha =0.707\approx 0.71$
$\therefore$ Percentage dissociation
$=0.71\times 100=71$