JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

$\mathrm{X}_{2}(\mathrm{~g})+\mathrm{Y}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g})$

$\mathrm{X}_{2}(\mathrm{~g})$ and $\mathrm{Y}_{2}(\mathrm{~g})$ are added to a 1 L flask and it is found that the system attains the above equilibrium at $\mathrm{T}(\mathrm{K})$ with the number of moles of $\mathrm{X}_{2}(\mathrm{~g}), \mathrm{Y}_{2}(\mathrm{~g})$ and $\mathrm{Z}(\mathrm{g})$ being 3,3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of $\mathrm{Z}(\mathrm{g})$ is added to the flask and the temperature is maintained at $\mathrm{T}(\mathrm{K})$ . Then the number of moles of $\mathrm{Z}(\mathrm{g})$ in the flask when the new equilibrium is established is $\_\_\_\_$ . (Nearest integer)

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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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.

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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.

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

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