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. The $K_c$ for the equilibrium is:

• A. $4$
• B. $8$
• C. $6$
• D. $0.32$

Answer 1

The correct option is D $0.32$

$A+2B\rightleftharpoons 2C+D$
Initial $a$ $\frac{3}{2}a$ $0$ $0$
Final $(a-x)\left(\frac{3}{2}a-2x\right)$ $2x$ $x$
Given, $a-x=2x$
$\therefore x=a/3$
Now, $K_c=\frac{{\left[C\right]}^2\left[D\right]}{\left[A\right]{\left[B\right]}^2}$
$=\frac{{\left(\frac{2a}{3}\right)}^2\times \frac{a}{3}}{(a-\frac{a}{3}){\left(\frac{3a}{2}-\frac{2a}{3}\right)}^2}=0.32$