Question

In reaction $A + 2B {\rightleftharpoons} 2C + D$, initial concentration of $B$ was $1.5$ times of $[A]$, but at equilibrium the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is :

• A. 8
• B. 4
• C. 12
• D. 6

Answer 1

Answer: (B)

4

Answer 2

$\begin{matrix}A\quad+&2B&{\rightleftharpoons}&2C\quad+&D\\\ a&1.5a&&0&0\\\ \left(a-x\right)&\left(1.5a-2x\right)&&2x&x\end{matrix}$
Hence $K_{c} = \frac{\left(2x\right)^{2}\times x}{\left(a-x\right)\left(1.5a-2x\right)^{2} }$
Given, at equilibrium
$\therefore\quad\left(a-x\right)\left(1.5a-2x\right)$
$\therefore\quad a = 2x$
On solving $K_{c} = 4$