In reaction A + 2B 2C + D, initial concentration of B was 1.5 , t… - Vidyadip

MCQ

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
$8$
✓
$4$
C
$12$
D
$6$

✓

Answer

Correct option: B.

$4$

b
$\mathop {\mathop A\limits_a }\limits_{(a - x)} + \mathop {\mathop {2B}\limits_{1.5a} }\limits_{(1.5a - 2x)} \leftrightarrow \mathop {\mathop {2C}\limits_0 }\limits_{2x} + \mathop {\mathop D\limits_0 }\limits_x $

Hence ${K_C} = \frac{{{{(2x)}^2} \times x}}{{(a - x){{(1.5a - 2x)}^2}}}$

Given, at equilibrium

$\therefore \,\,(a - x) = (1.5a - 2x)$

$\therefore \,\,a = 2x$

On solving ${K_C} = 4$

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