JEE Main 2026 — Chemical Equilibrium Question with Solution

The equilibrium constant for the reaction $X_2(g)+Y_2(g)\rightleftharpoons 2Z(g)$ is calculated from the initial equilibrium state:
$K_c=\frac{[Z{]}^2}{[X_2][Y_2]}=\frac{9^2}{3\times 3}=9$ mol/L.
When 10 mol of Z(g) is added, the new concentrations are: $[X_2]=3$, $[Y_2]=3$, $[Z]=19$ mol.
The reaction quotient is $Q=\frac{19^2}{3\times 3}=40.11$.
Since $Q>K_c$, the equilibrium shifts left (toward reactants).
Let $x$ mol of Z decompose. At the new equilibrium: $[X_2]=3+\frac{x}{2}$, $[Y_2]=3+\frac{x}{2}$, $[Z]=19-x$.
Applying the equilibrium expression: $\frac{(19-x{)}^2}{(3+\frac{x}{2}{)}^2}=9$.
Taking the square root: $\frac{19-x}{3+\frac{x}{2}}=3$.
Solving: $19-x=9+\frac{3x}{2}$, which gives $10=\frac{5x}{2}$, so $x=4$.
The moles of Z at new equilibrium = $19-4=15$ mol.