Question

In a gaseous reaction $A+2B\rightleftharpoons 2C+D$ the initial concentration of $B$ was $1.5$ times that of $A$. At equilibrium the concentrations of $A$ and $D$ were equal. Calculate the equilibrium constant $K_c$.

Answer 1

The given reaction is :-

$A+2B\rightleftharpoons 2C+D$

Initial moles : $C$ $1.Sc$ $0$ $0$

At eqm : $(c-x)$ $(I.Sc$ $2x$ $x$ (let)

$-2x)$

Now, given that, At eqm, $\left[A\right]=\left[D\right]$

$\Rightarrow c-x=x$

$\Rightarrow x=c/2$

Now, At eqm,

$\left[A\right]=C-C/2=C/2$

$\left[B\right]=\frac{3}{2}C-2\times \frac{C}{2}=C/2$

$\left[C\right]=C;\left[D\right]=C/2$

Now, $K_C=\frac{{\left[C\right]}^2\left[D\right]}{\left[A\right]{\left[B\right]}^2}$

$=\frac{C^2\times \frac{C}{2}}{\frac{C}{2}\times {\left(\frac{C}{2}\right)}^2}$

$\Rightarrow K_C=4$