Question

For the dissociation reaction, $N_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$, the equilibrium constant $K_p$ is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of $N_2{O}_4$.

• A. $0.586$
• B. $0.121$
• C. $0.150$
• D. $0.254$

Answer 1

The correct option is B $0.121$

Let the degree of dissociation of $N_2{O}_4$ be $\alpha$.
Let the final pressure be P.
$N_2{O}_4\rightleftharpoons 2N{O}_2$
initial moles $1$ 0
At equilibrium $1-\alpha$ $2\alpha$
So, total moles = $1+\alpha$

$K_p$ = $\frac{P_{\left[N{O}_2\right]}^2}{P_{\left[N_2{O}_4\right]}}$
$P_{N{O}_2}$ = $\frac{2\alpha }{1+\alpha }.P$
$P_{N_2{O}_4}$ = $\frac{1-\alpha }{1+\alpha }.P$
Put in the formual of $K_p$
$K_p=\frac{(\frac{2\alpha }{1+\alpha }{)}^2.P^2}{\frac{1-\alpha }{1+\alpha }.P}$
0.120 = $\frac{4(\alpha {)}^2}{1-(\alpha {)}^2}.P$
On solving for $\alpha$, we get $\alpha$ = 0.121