Question

When the reaction, $2N{O}_2\left(g\right)\rightleftharpoons N_2{O}_4\left(g\right)$ reaches equilibrium at 298 K. The partial pressure of $N{O}_2$ and $N_2{O}_4$ are $0.2$ $KPa$ and $0.4$ $KPa$, respectively. What is the equilibrium constant $K_p$ of the above reaction at $298K$?

• A. $0.1$
• B. $0.5$
• C. $1.0$
• D. $2$
• E. $10$

Answer 1

The correct option is E $10$

At equilibrium $\underset{0.2kPa}{2N{O}_2}\rightleftharpoons \underset{0.4kPa}{N_2{O}_4}$
$K_p=\frac{\left[p_{N_2{O}_4}\right]}{[p_{N{O}_2}{]}^2}=\frac{0.4}{0.2\times 0.2}=10$