Question

$A+B\rightleftharpoons C+D$. If initially the concentraion of $A$ and $B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$, then what will be the equilibrium constant of the reaction?

• A. $\frac{4}{9}$
• B. $\frac{9}{4}$
• C. $\frac{1}{9}$
• D. $4$

Answer 1

Answer: (D)

$4$

$A+B\rightleftharpoons C+D$. If initially the concentraion of $A$ and $B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$, then the equilibrium constant of the reaction will be $4$.

$A+B\rightleftharpoons C+D$
Initial concentration (M) $t=0$ $a$ $a$ $0$ $0$
Equilibrium concentration (M) $t_{eq.}$ $a-x$ $a-x$ $x$ $x$
At equilibrium, concentration of $D$ will be twice of that of $A$.
$x=2(a-x)$
$x=2a-2x$
$3x=2a$
$x=\frac{2}{3}a$
The equilibrium constant
$K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

$K=\frac{x\times x}{(a-x)(a-x)}$

$K=\frac{\frac{2}{3}a\times \frac{2}{3}a}{(a-\frac{2}{3}a)(a-\frac{2}{3}a)}$

$K=\frac{\frac{4}{9}{a}^2}{\left(\frac{1}{9}{a}^2\right)}$

$K=4$