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JEE MAIN 2026 · Equilibrium

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 55

Updated
Single correct+4 / -1Answer matches NTA’s final key

Consider the following reactions in which all the reactants and products are present in gaseous state

2 xy⇌x2+y2K1=2.5×1052\,xy \rightleftharpoons x_2 + y_2 \qquad K_1 = 2.5 \times 10^5

xy+12z2⇌xyzK2=5×10−3xy + \tfrac12 z_2 \rightleftharpoons xyz \qquad K_2 = 5 \times 10^{-3}

The value of K3K_3 for the equilibrium 12x2+12y2+12z2⇌xyz\tfrac12 x_2 + \tfrac12 y_2 + \tfrac12 z_2 \rightleftharpoons xyz is:

Choose one

Solution

Solution

Build the target from the two given reactions.

  1. Reverse the first and halve it: 12x2+12y2⇌xy\frac12 x_2 + \frac12 y_2 \rightleftharpoons xy, with K=1K1=1500=2×10−3K = \dfrac{1}{\sqrt{K_1}} = \dfrac{1}{500} = 2 \times 10^{-3}.
  2. Add the second: xy+12z2⇌xyzxy + \frac12 z_2 \rightleftharpoons xyz, K2=5×10−3K_2 = 5 \times 10^{-3}.

Adding reactions multiplies their constants:

K3=2×10−3×5×10−3=1.0×10−5.K_3 = 2 \times 10^{-3} \times 5 \times 10^{-3} = 1.0 \times 10^{-5}.

Answer: 1.0×10−51.0 \times 10^{-5}.

Answer checked against the official answer key. The solution is our own working.

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