Question

For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $K_p = 0.492 \, \text{atm}$ at 300 K. $K_c$ for the reaction at the same temperature is _____ $\times \, 10^{-2}$.
(Given: $R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1}$)

Answer 1

To convert $K_p$ to $K_c$:

$K_p = K_c \cdot (RT)^{\Delta n_g}$

For the reaction $\text{N}_2\text{O}_4(g) \leftrightharpoons 2\text{NO}_2(g)$,

$\Delta n_g = 2 - 1 = 1$

Therefore:

$K_c = \frac{K_p}{RT} = \frac{0.492}{0.082 \times 300} = 2 \times 10^{-2}$

So, the correct answer is: $2 \times 10^{-2}$

Answer 2

To convert $K_p$ to $K_c$:

$K_p = K_c \cdot (RT)^{\Delta n_g}$

For the reaction $\text{N}_2\text{O}_4(g) \leftrightharpoons 2\text{NO}_2(g)$,

$\Delta n_g = 2 - 1 = 1$

Therefore:

$K_c = \frac{K_p}{RT} = \frac{0.492}{0.082 \times 300} = 2 \times 10^{-2}$

So, the correct answer is: $2 \times 10^{-2}$