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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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answer is B.

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Detailed Solution

$\underset{t = e q u}{\underset{t = 0}{}} \underset{1 - x}{\underset{1}{A}} + \underset{1.5 - 2 x}{\underset{1.5}{2 B}} \Leftrightarrow \underset{2 x}{\underset{0}{2 C}} + \underset{x}{\underset{0}{D}}$

At equilibrium $[A] = [B]$

$a_{0} - x = 1.5 a_{0} - 2 x \Rightarrow x = 0.5 a_{0}$

$[C] = 2 \times 0.5 a_{0} = a_{0}, [D] = [A] = [B] = 0.5 a_{\circ}$

$K_{C} = \frac{[C]^{2} [D]}{[A] [B]^{2}} = \frac{{(a_{0})}^{2} (0.5 a_{0})}{(0.5 a_{0}) {(0.5 a_{0})}^{2}} = 4$

Therefore, the correct option is (B).

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