JEE Main 2021 — Chemical Equilibrium Question with Solution

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

$\boxed{2}$