Chemistry · Chemical Equilibrium

JEE Main 2026 — 23 January, Evening Shift — Question 62

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

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{X}_{2}(\mathrm{~g})+\mathrm{Y}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g})$ $\mathrm{K}_{\mathrm{C}}=\frac{(9)^{2}}{3 \times 3}=9$ Now 10 moles of $Z$ are added then reaction will move in backward direction. $\mathrm{X}_{2}(\mathrm{~g})+\mathrm{Y}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g})$ $3+\mathrm{X} 3+\mathrm{X} 19-2 \mathrm{X}$ $K_{C}=\frac{(19-2 X)^{2}}{(3+X)(3+X)}=9$ $\frac{19-2 X}{3+X}=3$ $19-2 X=9+3 X$ $10=5 \mathrm{X}$ $\mathrm{X}=2$ At equilibrium ⇒ moles of $\mathrm{Z}=19-2 \times 2$

=15 \text { moles }

Answer key and solution verified before publishing.

Practise Chemical Equilibrium

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient