Consider the following reactions in which all the reactants and products are present in gaseous state 2 xy x2 + y2 K1 = 2.5 x 10^5 xy + 1/2 z2 xyz K2 = 5 x 10^-3 The value of K3 for the equilibrium 1/2 x2 + 1/2 y2 + 1/2 z2 xyz is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q55 – JEE-MAIN 2026

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Home Chemistry EquilibriumQuestion #431

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 10^{5}

x y + \frac{1}{2} z_{2} ⇌ x y z K_{2} = 5 \times 10^{- 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:

Clue 1 of 3: Hess's Law for Equilibrium Constants

The overall equilibrium constant for a reaction that is the sum of two or more reactions is the product of the equilibrium constants for the individual reactions.

🧠 Full Solution Path

Step 1: Manipulate the first given reaction✦ Active

The target reaction has $\frac{1}{2} x_{2} + \frac{1}{2} y_{2}$ as reactants, while the first given reaction is $2 x y ⇌ x_{2} + y_{2}$ with $K_{1} = 2.5 \times 10^{5}$ . To obtain the desired reactants, we need to reverse the first reaction and divide it by 2.

x_{2} + y_{2} ⇌ 2 x y K_{1}^{'} = \frac{1}{K_{1}} = \frac{1}{2.5 \times 10^{5}} = 4 \times 10^{- 6}

\frac{1}{2} x_{2} + \frac{1}{2} y_{2} ⇌ x y K_{1}^{″} = \sqrt{K_{1}^{'}} = \sqrt{4 \times 10^{- 6}} = 2 \times 10^{- 3}

Step 2: Combine with the second given reaction○ Expand

The second given reaction is $x y + \frac{1}{2} z_{2} ⇌ x y z$ with $K_{2} = 5 \times 10^{- 3}$ . We can add the modified first reaction ( $K_{1}^{″}$ ) and the second reaction ( $K_{2}$ ) to get the target reaction.

(\frac{1}{2} x_{2} + \frac{1}{2} y_{2} ⇌ x y) + (x y + \frac{1}{2} z_{2} ⇌ x y z)

\frac{1}{2} x_{2} + \frac{1}{2} y_{2} + \frac{1}{2} z_{2} ⇌ x y z

Step 3: Calculate the overall equilibrium constant

K_{3}

○ Expand

When reactions are added, their equilibrium constants are multiplied.

K_{3} = K_{1}^{″} \times K_{2}

K_{3} = (2 \times 10^{- 3}) \times (5 \times 10^{- 3})

K_{3} = 10 \times 10^{- 6}

K_{3} = 1 \times 10^{- 5}

This value matches option 3.

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Chemistry Question 55 – JEE-MAIN 2026

Clue 1 of 3: Hess's Law for Equilibrium Constants

🧠 Full Solution Path