Question

At equilibrium 500mL vessel contains 15 M of each $A,B,C,D$. What would be the value of $K_C$?

• A. $1$
• B. $\frac{1}{9}$
• C. $\frac{4}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is A $1$

The reaction taking place is $A+B\rightleftharpoons C+D$

A=B=C=D=15M

$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

Since concentration is same for all, $K_c=1$

Hence, the correct option is $\text{A}$