Chemistry · EQUILIBRIUM

For the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \) at \( 400\text{ K} \), the equilibrium constant \( K_p \) is \( 85.87 \). Taking \( R = 0.083\text{ bar L mol}^{-1}\text{K}^{-1} \), what is the value of \( K_c \) for this equilibrium?

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Correct answer · Option A

\( 2.59 \)

\( K_p = K_c(RT)^{\Delta n} \implies K_c = \frac{K_p}{RT} = \frac{85.87}{0.083 \times 400} = \frac{85.87}{33.2} \approx 2.59 \).