At T(K), the equilibrium constant of A2(g) + B2(g) C(g) is 2.7 x 10^-5. What is the equilibrium constant for 1/3A2(g) + 1/3B2(g) 1/3C(g) at the same temperature?

The correct answer is Option 4. 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 #420

At T(K), the equilibrium constant of

A_{2} (g) + B_{2} (g) ⇌ C (g)

2.7 \times 10^{- 5}

. What is the equilibrium constant for

\frac{1}{3} A_{2} (g) + \frac{1}{3} B_{2} (g) ⇌ \frac{1}{3} C (g)

at the same temperature?

Clue 1 of 3: Relationship between Equilibrium Constants

Understand how the equilibrium constant changes when a chemical reaction is modified (e.g., reversed, multiplied by a factor, or added to another reaction).

🧠 Full Solution Path

Step 1: Identify the relationship between the two reactions✦ Active

The initial reaction is $A_{2} (g) + B_{2} (g) ⇌ C (g)$ with an equilibrium constant $K_{1} = 2.7 \times 10^{- 5}$ . The target reaction is $\frac{1}{3} A_{2} (g) + \frac{1}{3} B_{2} (g) ⇌ \frac{1}{3} C (g)$ . This target reaction is obtained by multiplying the initial reaction by a factor of $\frac{1}{3}$ .

Step 2: Apply the rule for modifying equilibrium constants○ Expand

When a chemical reaction is multiplied by a stoichiometric factor $n$ , the new equilibrium constant $K_{n e w}$ is equal to the original equilibrium constant $K_{o l d}$ raised to the power of $n$ . In this case, $n = \frac{1}{3}$ .

K_{n e w} = K_{o l d}^{n}

💡 Teacher's Secret Hint

Remember that multiplying a reaction by a factor $n$ means raising the equilibrium constant to the power of $n$ , not multiplying it by $n$ .

Step 3: Calculate the new equilibrium constant○ Expand

Substitute the given value of $K_{o l d}$ and the factor $n$ into the formula:

K_{n e w} = (2.7 \times 10^{- 5})^{1 / 3}

To simplify the cube root, rewrite $2.7 \times 10^{- 5}$ as $27 \times 10^{- 6}$ :

K_{n e w} = (27 \times 10^{- 6})^{1 / 3} = \sqrt[3]{27} \times \sqrt[3]{10^{- 6}}

K_{n e w} = 3 \times 10^{- 6 / 3} = 3 \times 10^{- 2}

Comparing this result with the given options, option 4 matches the calculated value.

💡 Teacher's Secret Hint

When taking roots of numbers with powers of 10, adjust the exponent to be a multiple of the root index for easier calculation (e.g., $10^{- 5}$ to $10^{- 6}$ for a cube root).

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Chemistry Question 55 – JEE-MAIN 2026

Clue 1 of 3: Relationship between Equilibrium Constants

🧠 Full Solution Path