Q5 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry
Consider the following reactions in which all the reactants and products are present in gaseous state
$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5\times 10^5$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5\times 10^{-3}$
The value of $K_3$ for the equilibrium $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$ is:
$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5\times 10^5$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5\times 10^{-3}$
The value of $K_3$ for the equilibrium $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$ is:
- A$2.5\times 10^{-3}$
- B$2.5\times 10^{3}$
- Correct answerC$1.0\times 10^{-5}$
- D$5\times 10^{-3}$
Solution
The given reactions are:
$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$
We need to find the equilibrium constant $K_3$ for the reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$
Reversing the first reaction and multiplying it by $\dfrac{1}{2}$, we get:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy$
The equilibrium constant for this modified reaction is:
$K' = \left(\dfrac{1}{K_1}\right)^{\dfrac{1}{2}} = \dfrac{1}{\sqrt{K_1}}$
Adding this modified reaction to the second reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy \quad (K')$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad (K_2)$
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$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$
The equilibrium constant $K_3$ for the overall reaction is the product of the equilibrium constants of the added reactions:
$K_3 = K' \times K_2 = \dfrac{K_2}{\sqrt{K_1}}$
Substituting the given values:
$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$
$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}}$
$K_3 = \dfrac{5 \times 10^{-3}}{5 \times 10^2}$
$K_3 = 1.0 \times 10^{-5}$
Answer: $1.0\times 10^{-5}$
$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$
We need to find the equilibrium constant $K_3$ for the reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$
Reversing the first reaction and multiplying it by $\dfrac{1}{2}$, we get:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy$
The equilibrium constant for this modified reaction is:
$K' = \left(\dfrac{1}{K_1}\right)^{\dfrac{1}{2}} = \dfrac{1}{\sqrt{K_1}}$
Adding this modified reaction to the second reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy \quad (K')$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad (K_2)$
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$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$
The equilibrium constant $K_3$ for the overall reaction is the product of the equilibrium constants of the added reactions:
$K_3 = K' \times K_2 = \dfrac{K_2}{\sqrt{K_1}}$
Substituting the given values:
$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$
$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}}$
$K_3 = \dfrac{5 \times 10^{-3}}{5 \times 10^2}$
$K_3 = 1.0 \times 10^{-5}$
Answer: $1.0\times 10^{-5}$
Hence, the correct answer is option: 'C'.