Question

500 ml vessel contains 1.5 M each of $A,B,C$ and $D$ at equilibrium. If 0.5 M each of $C$ and $D$ are taken out, the value of $K_C$ for $A+B\rightleftharpoons C+D$ will be:

• A. 1
• B. 1/9
• C. 4/9
• D. 8/9

Answer 1

Answer: (C)

4/9

$A+B\rightleftharpoons C+D$

$1.5$ $1.5$ $1.5$ $1.5$

$\frac{1.5}{0.5}$ $\frac{1.5}{0.5}$ $\frac{1}{0.5}$ $\frac{1}{0.5}$

$K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{\left[\frac{1}{0.5}\right]\left[\frac{1}{0.5}\right]}{\left[\frac{1.5}{0.5}\right]\left[\frac{1.5}{0.5}\right]}=\frac{2\times 2}{3\times 3}=\frac{4}{9}$