Question

$N_2{O}_4$ at an initial pressure of $2atm$ and $300K$ dissociates to an extent of $20$ % at the same temperature by the time equilibrium is established. $K_p$ for this reaction $2N{O}_2\rightleftharpoons N_2{O}_4$ is:

• A. $0.4$
• B. $.088$
• C. $2.5$
• D. $1.6$

Answer 1

Answer: (C)

$2.5$

Given, $P_{initial}=2atm$, degree of dissociation $=0.2$

Let $K_p^{\prime }$ be the equilibrium constant for this reaction.

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$t=0$ $2$ $-$

$t=t$ $2(1-0.2)$ $4\times 0.2$

$K_p^{\prime }=\frac{(0.8{)}^2}{2\times 0.8}=0.4atm$

Now, we have to calculate $K_p$ or

$2N{O}_2\rightleftharpoons N_2{O}_4$

$K_p=\frac{1}{K_p^{\prime }}=\frac{1}{0.4}=2.5at{m}^{-1}$