Question

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

Answer 1

Formula to find the equilibrium constant of the reaction is given as below.
$K=\dfrac{[C][D]}{[A][B]}$
We can find the new concentration of A and B at equilibrium in order to find the equilibrium constant.

Complete step by step answer:
We are given that at equilibrium, concentration of C is 0.60 $mol{{L}^{-}}$.
- As we see the reaction, we can say that if one mole of C gets formed, one mole of D will also get formed. Thus, as concentration of C is 0.60$mol{{L}^{-}}$ at equilibrium, concentration of D will also be 0.60 $mol{{L}^{-}}$.
- Now, we will also need to find the concentration of A and B at equilibrium in order to find the equilibrium constant.
- We can say from the reaction that one mole each of A and B will give one mole of C and D on complete reaction. We know the initial concentration of A and B. We also know the concentration of C and D at equilibrium.

- So, we can say that at equilibrium, concentration of A will be
[A]= Initial concentration – concentration of C or D = 0.80 – 0.60 = 0.20 $mol{{L}^{-}}$
And concentration of B will be
[B]= Initial concentration – Concentration of C or D = 0.80 – 0.60 = 0.20 $mol{{L}^{-}}$

Now, we know that equilibrium constant for any reaction is
$K=\dfrac{[\text{Product}]}{[\text{Reactant }\\!\\!]\\!\\!\text{ }}$
So, we can write that $K=\dfrac{[C][D]}{[A][B]}$ ……..(1)
So, put the available values into equation (1)
$K=\dfrac{[0.60][0.60]}{[0.20][0.20]}$
$K=\dfrac{0.36}{0.04}$
Thus, equilibrium constant, K = 9

Note: Do not forget to subtract the amount of A or B decreased because of production of C or D in this reaction. As the equilibrium constant involves the concentration of two products and two reactants, here it will be unitless.