JEE Main 2026 — Chemical Equilibrium Question with Solution

The given reactions are:
$2xy\rightleftharpoons x_2+y_2{K}_1=2.5\times 10^5$
$xy+\frac{1}{2}{z}_2\rightleftharpoons xyz{K}_2=5\times 10^{-3}$

We need to find the equilibrium constant $K_3$ for the reaction:
$\frac{1}{2}{x}_2+\frac{1}{2}{y}_2+\frac{1}{2}{z}_2\rightleftharpoons xyz$

Reversing the first reaction and multiplying it by $\frac{1}{2}$, we get:
$\frac{1}{2}{x}_2+\frac{1}{2}{y}_2\rightleftharpoons xy$
The equilibrium constant for this modified reaction is:
$K^{\prime }={\left(\frac{1}{K_1}\right)}^{\frac{1}{2}}=\frac{1}{\sqrt{K_1}}$

Adding this modified reaction to the second reaction:
$\frac{1}{2}{x}_2+\frac{1}{2}{y}_2\rightleftharpoons xy(K^{\prime })$
$xy+\frac{1}{2}{z}_2\rightleftharpoons xyz(K_2)$
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$\frac{1}{2}{x}_2+\frac{1}{2}{y}_2+\frac{1}{2}{z}_2\rightleftharpoons xyz$

The equilibrium constant $K_3$ for the overall reaction is the product of the equilibrium constants of the added reactions:
$K_3=K^{\prime }\times K_2=\frac{K_2}{\sqrt{K_1}}$

Substituting the given values:
$K_3=\frac{5\times 10^{-3}}{\sqrt{2.5\times 10^5}}$

$K_3=\frac{5\times 10^{-3}}{\sqrt{25\times 10^4}}$

$K_3=\frac{5\times 10^{-3}}{5\times 10^2}$

$K_3=1.0\times 10^{-5}$

Answer: $1.0\times 10^{-5}$