Question

The reaction, $A+2B\rightleftharpoons 2C+D$ was studied using an initial concentration of $B$ which was 1.5 times that of $A$. But the equilibrium concentration of $A$ and $C$ were found to be equal. Then the $K_c$ for the equilibrium is :

• A. 0. 41
• B. 0. 83
• C. 0. 66
• D. 0. 32

Answer 1

The correct option is D 0. 32

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

At $t=0$ $\alpha$ $\frac{3}{2}\alpha$ $0$ $0$

At eq. $\alpha -x$ $\frac{3}{2}\alpha -x$ $2x$ $x$

Given, $\left[A\right]=\left[C\right]$

$a-x=2x$

$\therefore x=\frac{a}{3}$

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

$=\frac{{\left(\frac{2a}{3}\right)}^2\times \frac{a}{3}}{(a-\frac{a}{3}){\left(\frac{3a}{2}-\frac{2a}{3}\right)}^2}=0.32$