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 $K_c$ for $A+B\rightleftharpoons C+D$ will be

• A. $1$
• B. $\frac{1}{9}$
• C. $\frac{4}{9}$
• D. $\frac{8}{9}$

Answer 1

The correct option is A $1$

$K_c=\frac{1.5\times 1.5}{1.5\times 1.5}=1$
On changing the concentrations equilibrium constant does not change.