JEE Main 2026 · 8 Apr · Shift 2 · Chemistry Q55

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ChemistrySingle correct+4 / -1Question 55 of 75

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

$2xy\rightleftharpoons x_2+y_2$ , $K_1=2.5\times10^5$

$xy+\frac12z_2\rightleftharpoons xyz$ , $K_2=5\times10^{-3}$

The value of $K_3$ for the equilibrium $\frac12x_2+\frac12y_2+\frac12z_2\rightleftharpoons xyz$ is :

2.5\times10^{-3}

2.5\times10^3

1.0\times10^{-5}

5\times10^{-3}

Answer and worked solution

Official answer: C

Combine. Half the reverse of reaction 1 gives $\tfrac12x_2+\tfrac12y_2\rightleftharpoons xy$ , with $K=\left(\tfrac1{K_1}\right)^{1/2}=2\times10^{-3}$ .

Add reaction 2.
$K_3=2\times10^{-3}\times5\times10^{-3}=1.0\times10^{-5}.$

(C)

Topics:Law of mass action and Kc, Kp

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JEE Main 2026 · 8 Apr · Shift 2, Chemistry question 55