Question

Concentrations of two reactants $A$ and $B$ are $0.8mol{L}^{-1}$ each. On mixing the two, the reaction sets in at a slow rate to form $C$ and $D$.
$A+B\rightleftharpoons C+D$
At equilibrium, concentration of $C$ was found to be $0.60mol{L}^{-1}$. Calculate the equilibrium constant.

Answer 1

The initial concentrations of A and B are 0.8 M each. The equilibrium
concentration of C is 0.60 M.
Thus, 0.60 M of A will react with 0.60 M of B to form 0.60 M of C and 0.60
M of D.
$0.80-0.60=0.20$ M of A and $0.80-0.60=0.20$ M of B will remain at equilibrium.
The equilibrium constant
$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$
$K_c=\frac{0.60\times 0.60}{0.20\times 0.20}$
$K_c=9$