Question

In an equilibrium, $A+B\rightleftharpoons C+D$, $A$ and $B$ are mixed in a vessel at temperature $T$. The initial concentration of $A$ was twice the initial concentration of $B$. After the equilibrium has reached, a concentration of $C$ was thrice the equilibrium concentration of $B$. Calculate $K_c$.

(Nearest integer)

Answer 1

Let the inital concentration of $B=a$ mol litre${}^{-1}$ and thus, $\left[A\right]=2a$ (given)

$A+B\rightleftharpoons C+D$
Initial concentration $2aa00$
Equilibrium concentration $(2a-x)(a-x)xx$

The expression for the equilibrium constant is as given blow.
$K=\frac{x^2}{(2a-x)(a-x)}$ .......(i)

$\because$ At equilibrium $\left[C\right]=3\left[B\right]$

or $x=3(a-x)$

$\therefore$ $x=\frac{3a}{4}$ .......(ii)
By equations (i) and (ii) $K=\frac{{\left(3a/4\right)}^2}{[2a-\left(3a/4\right)][a-3a/4]}=1.8$

The nearest integer is 2. So, the correct answer is 2