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NVAMediumJEE 2026Equilibrium Basics

JEE Chemistry 2026 Question with Solution

X2(g)+Y2(g)⇌2Z(g)X_2(g) + Y_2(g) \rightleftharpoons 2Z(g). Equilibrium moles of X2,Y2,ZX_2, Y_2, Z are 33, 33, 9 mol9 \, \text{mol} (in 1 L1 \, \text{L}). 10 mol10 \, \text{mol} of Z(g)Z(g) is added. New equilibrium moles of Z(g)Z(g) is _____ . (Nearest integer)

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: X2(g)+Y2(g)⇌2Z(g)X_2(g) + Y_2(g) \rightleftharpoons 2Z(g). Initial equilibrium moles in 1 L1 \, \text{L} are X2=3X_2 = 3, Y2=3Y_2 = 3, and Z=9Z = 9. Then 10 mol10 \, \text{mol} of ZZ is added.

Find: New equilibrium moles of ZZ.

First, calculate the equilibrium constant:

Kc=[Z]2[X2][Y2]=923×3=9K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = 9

After adding 10 mol10 \, \text{mol} of ZZ, the concentrations become:

[X2]=3,[Y2]=3,[Z]=19[X_2] = 3, \quad [Y_2] = 3, \quad [Z] = 19

Now,

Qc=1923×3=3619≈40.11Q_c = \frac{19^2}{3 \times 3} = \frac{361}{9} \approx 40.11

Since Qc>KcQ_c > K_c, the reaction proceeds towards the reactants to re-establish equilibrium.

Let the backward shift be xx:

X2+Y2←2ZInitial:3319Change:+x+x−2xEquilibrium:3+x3+x19−2x\begin{aligned} X_2 + Y_2 &\xleftarrow{} 2Z \\ \text{Initial:} &\quad 3 \qquad 3 \qquad 19 \\ \text{Change:} &\quad +x \qquad +x \qquad -2x \\ \text{Equilibrium:} &\quad 3+x \qquad 3+x \qquad 19-2x \end{aligned}

Apply the equilibrium constant expression again:

Kc=(19−2x)2(3+x)2=9K_c = \frac{(19-2x)^2}{(3+x)^2} = 9

Taking square root,

19−2x3+x=3\frac{19-2x}{3+x} = 3

So,

19−2x=9+3x19 - 2x = 9 + 3x 5x=105x = 10 x=2x = 2

Therefore, new equilibrium moles of ZZ are:

19−2x=19−4=1519 - 2x = 19 - 4 = 15

So the required answer is 1515.

Common mistakes

  • Using the added amount of ZZ directly as the final answer. This is wrong because adding product disturbs equilibrium and the system shifts left. Recalculate the new equilibrium using QcQ_c and an ICE-type setup.

  • Assuming KcK_c changes after adding ZZ. This is wrong because at constant temperature the equilibrium constant remains the same. Only the reaction quotient QcQ_c changes immediately after disturbance.

  • Writing the change in ZZ as −x-x instead of −2x-2x. This is wrong because the stoichiometry is 2Z2Z, so for a backward shift of xx, moles of ZZ decrease by 2x2x. Use stoichiometric coefficients carefully.

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