Question

The molar concentration of $A$ and $B$ are $0.80mol{litre}^{-1}$ each. On mixing them, the reaction starts to proceed as: $A+B\rightleftharpoons C+D$, and attains equilibrium. At equilibrium, molar concentration of $C$ is $0.60mol{litre}^{-1}$.

Calculate $K_c$ of the reaction:

Answer 1

Given $\left[C\right]$ at equilibrium $=0.6$; $\therefore \left[D\right]=0.6$.

Thus, the concentration of $A$ and $B$ which reacts are $0.6mol{litre}^{-1}$ at equilibrium.
$A+B\rightleftharpoons C+D$

Initial concentration $0.80.800$

Equilibrium concentration $(0.8-0.6)(0.8-0.6)0.60.6$
$=0.2=0.20.60.6$

The equilibrium concentration expression is as given below:

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

Substitute values in the above expression.

$K_c=\frac{0.6\times 0.6}{0.2\times 0.2}=9$

Hence, the value of the equilibrium constant is 9.