QJEEM21Chdd86: JEE Main 2021 Chemistry - Class 11 JEEM21Chdd86 Solution: Value of K P for the equilibrium reaction N 2 O 4(g) 2NO 2(g)…

For the equilibrium reaction

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$

the relationship between $K_P$ and $K_C$ is given by the equation:

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

where $\Delta n$ is the change in the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

For the given reaction:

$\Delta n = 2 - 1 = 1$

Given:

$K_P = 47.9$

$R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1}$

$T = 288 \, \text{K}$

Substitute these values into the equation:

$47.9 = K_C (0.083 \times 288)^{1}$

$K_C = \frac{47.9}{0.083 \times 288}$

Calculate:

$0.083 \times 288 = 23.904$

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

$K_C \approx 2.0033$

Rounding to the nearest integer, the value of $K_C$ is

$\boxed{2}$