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In a chemical reaction, $A + 2 B \overset{K}{⇌} 2 C + D,$ the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is:

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$\frac{1}{4}$

answer is A.

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Detailed Solution

$(a) A + 2 B ⇌ 2 C + D t = 0 2 3 0 0 t_{eq} 2 - x 3 - 2 x 2 x x Given 3 - 2 x = 2 - x \Rightarrow x = 1 ∴ [C] = 2, [D] = 1, [A] = 1, [B] = 1 ∴ K_{c} = \{\frac{2^{2} \times 1}{1^{2} \times 1}\} = 4$

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