Value of K_P for the equilibrium reaction N_2O_4 (g) leftharpoons 2 NO_2 (g) at 288 K is 47.9. The K_C for this reaction at same temperature is \underline\hspace2cm . (Nearest integer) (R = 0.083 L bar K - 1 mol - 1)

Chemical Equilibrium PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q19 with Solution

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Correct answer: 2

Given equilibrium

$N_2O_4(g) \rightleftharpoons 2NO_2(g)$

Relation between $K_P$ and $K_C$

For gaseous equilibria,

$K_P = K_C (RT)^{\Delta n}$

where

$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

Here,

$\Delta n = 2 - 1 = 1$

So,

$K_P = K_C (RT)$

Hence,

$K_C = \frac{K_P}{RT}$

Substitute values

Given:

$K_P = 47.9, \quad R = 0.083\, \text{L bar K}^{-1}\text{mol}^{-1}, \quad T = 288\, \text{K}$

First calculate $RT$ :

$RT = 0.083 \times 288 = 23.904$

Now,

$K_C = \frac{47.9}{23.904} \approx 2.00$

Nearest integer

$K_C \approx 2$

Therefore, the required integer is:

$\boxed{2}$

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