JEE Main 2026 Apr 8 Shift 2, Chemistry Q5: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Apr 8, Shift 2Chemistry

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

$2 x y ⇌ x_{2} + y_{2}$ , $K_{1} = 2.5 \times 1 0^{5}$

$x y + \frac{1}{2} z_{2} ⇌ x y z$ , $K_{2} = 5 \times 1 0^{- 3}$

The value of $K_{3}$ for the equilibrium $\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ x y z$ is :

Solution

Build the target reaction by combining the given equilibria.

Step 1. Reverse reaction 1: $x_{2} + y_{2} ⇌ 2 x y$ , with $K = \frac{1}{K _{1}} = \frac{1}{2.5 \times 1 0 ^{5}} = 4 \times 1 0^{- 6}$ .

Step 2. Halve the stoichiometry: $\frac{1}{2} x_{2} + \frac{1}{2} y_{2} ⇌ x y$ , with $K^{'} = 4 \times 1 0^{- 6} = 2 \times 1 0^{- 3}$ .

Step 3. Add reaction 2 ( $x y + \frac{1}{2} z_{2} ⇌ x y z$ , $K_{2} = 5 \times 1 0^{- 3}$ ). Adding equilibria multiplies their constants:

$K_{3} = K^{'} \cdot K_{2} = (2 \times 1 0^{- 3}) (5 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 5}$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.