JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
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 \(\dfrac{1}{3}A_2(g) + \dfrac{1}{3}B_2(g) \rightleftharpoons \dfrac{1}{3}C(g)\) at the same temperature?
- A \((2.7 \times 10^{-5})^3\)
- B \(6 \times 10^{-2}\)
- C \(\sqrt{2.7 \times 10^{-5}}\)
- D \(3 \times 10^{-2}\)
Answer & Solution
Correct Answer
(D) \(3 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
For the reaction \(A_2(g) + B_2(g) \rightleftharpoons C(g)\), the equilibrium constant is \(K_1 = 2.7 \times 10^{-5}\). When a reaction is multiplied by a factor \(n\), the new equilibrium constant becomes \(K^n\). The given reaction is multiplied by \(\dfrac{1}{3}\) to obtain…
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