For an equilibrium $A + 2 B ⇌ 2 C + D; A$ and B are mixed in a reaction vessel at 300K. The initial concentration of B was 1.5 times the initial concentration of A. After the equilibrium, the equilibrium concentration of A and D are same. Calculate $K_{c}$ .

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answer is 4.

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Detailed Solution

$A + 2 B ⇌ 2 C + D$

$\begin{array}{ccccc} t = 0 & a & 1.5 a & - & - \\ At eq. & a - x & 1.5 a - 2 x & 2 x & x \end{array}$

$\begin{array}{l} a - x = x (given) \\ \Rightarrow x = \frac{a}{2} \\ K_{C} = \frac{a^{2} \times \frac{a}{2}}{\frac{a}{2} \times {(\frac{a}{2})}^{2}} = 4 \end{array}$

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