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Chemistry · Chemical Equilibrium

JEE Main 2026 — 8 April, Evening Shift — Question 47

Consider the following reactions in which all the reactants and products are present in gasesous state 2xy⇌x2+y2 K1=2.5×1052 \mathrm{xy} \rightleftharpoons \mathrm{x}_{2}+\mathrm{y}_{2} \mathrm{~K}_{1}=2.5 \times 10^{5} xy+12z2⇌xyzK2=5×10−3\mathrm{xy}+\frac{1}{2} \mathrm{z}_{2} \rightleftharpoons \mathrm{xyz} \quad \mathrm{K}_{2}=5 \times 10^{-3} The value of K3\mathrm{K}_{3} for the equilibrium 12x2+12y2+12z2⇌xyz\frac{1}{2} \mathrm{x}_{2}+\frac{1}{2} \mathrm{y}_{2}+\frac{1}{2} \mathrm{z}_{2} \rightleftharpoons \mathrm{xyz} is :

  1. Option A:

    2.5×10−32.5 \times 10^{-3}

  2. Option B:

    2.5×1032.5 \times 10^{3}

  3. Option C:

    1.0×10−51.0 \times 10^{-5}

    Correct
  4. Option D:

    5×10−35 \times 10^{-3}

Answer: C

Step-by-step solution

x2+y2⇌2xy,K′=1 K1=12.5×105=125×104\mathrm{x}_{2}+\mathrm{y}_{2} \rightleftharpoons 2 \mathrm{xy}, \mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}_{1}}=\frac{1}{2.5 \times 10^{5}}=\frac{1}{25 \times 10^{4}}

\frac{1}{2} x_{2}+\frac{1}{2} y_{2} \rightleftharpoons x y, K^{\prime \prime}=\left(\frac{1}{25 \times 10^{4}}\right)^{1 / 2}=\frac{1}{5 \times 10^{2}} \end{gathered}$$ $$\begin{gathered} \mathrm{xy}+\frac{1}{2} \mathrm{z}_{2} \rightleftharpoons \mathrm{xyz}, \mathrm{~K}_{2}=5 \times 10^{-3} \end{gathered}$$ On adding equation \& $\frac{1}{2} \mathrm{x}_{2}+\frac{1}{2} \mathrm{y}_{2}+\frac{1}{2} \mathrm{z}_{2} \rightleftharpoons \mathrm{xyz}$, $\mathrm{K}_{3}=\frac{1}{5 \times 10^{2}} \times 5 \times 10^{-3}=1 \times 10^{-5}$

Answer key and solution verified before publishing.

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

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