Consider the following reactions in which all the reactants and products are present in gaseous state $2\mathrm{xy} \rightleftharpoons \mathrm{x}_{2}+\mathrm{y}_{2}$ $\mathrm{K}_{1}=2.5 \times 10^{5}$ $\mathrm{xy}+\frac{1}{2}\mathrm{z}_{2} \rightleftharpoons \mathrm{xyz}$ $\mathrm{K}_{2}=5 \times 10^{-3}$ The value of $\mathrm{K}_{3}$ for the equilibrium $\frac{1}{2}\mathrm{x}_{2}+\frac{1}{2}\mathrm{y}_{2}+\frac{1}{2}\mathrm{z}_{2} \rightleftharpoons \mathrm{xyz}$ is :

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Equilibrium

Consider the following reactions in which all the reactants and products are present in gaseous state

2 xy ⇌ x_{2} + y_{2}

K_{1} = 2.5 \times 1 0^{5}

xy + \frac{1}{2} z_{2} ⇌ xyz

K_{2} = 5 \times 1 0^{- 3}

The value of

K_{3}

for the equilibrium

\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ xyz

is :

Answer: (C)

For the first equilibrium, reversing it and taking the square root gives

\frac{1}{2} x_{2} + \frac{1}{2} y_{2} ⇌ x y

with equilibrium constant

(\frac{1}{K _{1}})^{1/2} = (\frac{1}{2.5 \times 1 0 ^{5}})^{1/2} = 2 \times 1 0^{- 3}

. Adding this to the second equilibrium

x y + \frac{1}{2} z_{2} ⇌ x y z

, whose constant is

K_{2} = 5 \times 1 0^{- 3}

, reproduces the target reaction

\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ x y z

. When equilibria are added their constants multiply, so

K_{3} = (2 \times 1 0^{- 3}) (5 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 5}

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option