JEE Main 2021 — Chemical Equilibrium Question with Solution

In the given equilibrium, the solid substances do not contribute to the equilibrium constant expression since their activities are considered to be 1.

For the reaction:

$\text{A(s)}\rightleftharpoons \text{M(s)}+\frac{1}{2}{\text{O}}_2(\text{g})$

The equilibrium constant, $K_p$, is given by:

$K_p=P_{{\text{O}}_2}^n$

where $P_{{\text{O}}_2}$ is the partial pressure of ${\text{O}}_2$ and $n$ represents the stoichiometric coefficient of ${\text{O}}_2$ in the balanced chemical reaction, which is $\frac{1}{2}$. Therefore, the expression for $K_p$ becomes:

$K_p={\left(P_{{\text{O}}_2}\right)}^{\frac{1}{2}}$

Given that $K_p=4$, substituting into the equation gives:

$4={\left(P_{{\text{O}}_2}\right)}^{\frac{1}{2}}$

To find $P_{{\text{O}}_2}$, square both sides of the equation:

$4^2=P_{{\text{O}}_2}$

$16=P_{{\text{O}}_2}$

Thus, the partial pressure of ${\text{O}}_2$ at equilibrium is $\boxed{16}$ atm.