Question:
At T(K), the equilibrium constant of \(A_2(g) + B_2(g) \rightleftharpoons C(g)\) is \(2.7 \times 10^{-5}\). What is the equilibrium constant for \(\frac{1}{3} A_2(g) + \frac{1}{3} B_2(g) \rightleftharpoons \frac{1}{3} C(g)\) at the same temperature?
At T(K), the equilibrium constant of \(A_2(g) + B_2(g) \rightleftharpoons C(g)\) is \(2.7 \times 10^{-5}\). What is the equilibrium constant for \(\frac{1}{3} A_2(g) + \frac{1}{3} B_2(g) \rightleftharpoons \frac{1}{3} C(g)\) at the same temperature?
Updated On: Apr 10, 2026
- \((2.7 \times 10^{-5})^3\)
- \(6 \times 10^{-2}\)
- \(\sqrt{2.7 \times 10^{-5}}\)
- \(3 \times 10^{-2}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
When the coefficients of a balanced chemical equation are multiplied by a factor '$n$', the new equilibrium constant ($K'$) becomes the original equilibrium constant ($K$) raised to the power of that factor ($K' = K^n$).
Step 2: Key Formula or Approach:
1. Original reaction: $K = 2.7 \times 10^{-5}$ 2. New reaction: Coefficients are multiplied by $n = 1/3$. 3. $K_{new} = (K_{old})^{1/3}$
Step 3: Detailed Explanation:
1. Apply the power rule: \[ K' = (2.7 \times 10^{-5})^{1/3} \] 2. Rewrite $2.7 \times 10^{-5}$ as $27 \times 10^{-6}$ to make the cube root easier: \[ K' = (27 \times 10^{-6})^{1/3} \] 3. Calculate the cube root: \[ K' = \sqrt[3]{27} \times \sqrt[3]{10^{-6}} = 3 \times 10^{-2} \]
Step 4: Final Answer:
The new equilibrium constant is \(3 \times 10^{-2}\).
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