The equilibrium constant for the reaction $\mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{M}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$ In $\mathrm{K}_{\mathrm{p}}=4$. At equilibrium, the partial pressure of $\mathrm{O}_{2}$ is $\_\_\_\_$ atm. (Round off to the Nearest Integer).

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Equilibrium

The equilibrium constant for the reaction

A (s) ⇌ M (s) + \frac{1}{2} O_{2} (g)

K_{p} = 4

. At equilibrium, the partial pressure of

O_{2}

____

atm.
(Round off to the Nearest Integer).

Answer: 16

Solution

For the reaction A(s) ⇌ M(s) + ½O₂(g)

The equilibrium constant Kₚ involves only gaseous species:
Kₚ = (P_O₂)^(1/2)

Given Kₚ = 4:
4 = (P_O₂)^(1/2)

Squaring both sides:
16 = P_O₂

Therefore, the partial pressure of O₂ at equilibrium is 16 atm.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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