Question

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

• A. $4/9$
• B. $9/4$
• C. $1/9$
• D. 4

Answer 1

The correct option is D 4

The given reaction is :-

$A+B\rightleftharpoons C+D$

Initial conc. : $M$ $M$ $0$ $0$ (given ${\left[A\right]}_0={\left[B\right]}_0$)

At eqm : $M-m$ $M-m$ $m$ $m$

given that, $m=2\left[A\right]$ (at eqm)

$\Rightarrow m=2\times (M-m)$

$\Rightarrow m=2M-2m$

$\Rightarrow 3m=2M$

$\Rightarrow \frac{M}{m}=\frac{3}{2}$

Now, $K_{eq}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

$=\frac{m.m}{(M-m)(M-m)}$

$=\frac{m}{(\frac{M}{m}-1)(\frac{M}{m}-1)}$

$\Rightarrow K_{eq}=\frac{1}{(\frac{3}{2}-1)\times (\frac{3}{2}-1)}=\frac{1}{\frac{1}{2}\times \frac{1}{2}}=4$