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In a chemical reaction, , the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is :
a.) 16
b.) 4
c.) 1
d.) $\dfrac{1}{4}$


Answer
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Hint : The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
It can be calculated by the formula-
K = $\dfrac{{{{[C]}^2}[D]}}{{[A]{{[B]}^2}}}$

Complete step by step answer :
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. Now, let us write what is given to us and what we need to find.
Given :
Initial concentration of B = 1.5 Initial concentration of A
Equilibrium concentrations of A = Equilibrium concentrations of B
To find :
Equilibrium constant (K)
We know that the equilibrium constant can be given by the formula-
K = $\dfrac{{{{[C]}^2}[D]}}{{[A]{{[B]}^2}}}$
We have the equation -
Let the initial concentration of A = ‘a’.
So, the initial concentration of B = 1.5a
At equilibrium, the concentration of A = a - x
At equilibrium, the concentration of B = 1.5a - x
At equilibrium, the concentration of C = 2x
At equilibrium, the concentration of D = x
Further, we have a - x = 1.5a - 2x
On solving, x = 0.5a
So, At equilibrium, the concentration of A = a - 0.5a = 0.5a
At equilibrium, the concentration of B = 1.5a - 0.5a = 1a
At equilibrium, the concentration of C = 2 (0.5a) = a
At equilibrium, the concentration of D = 0.5a
Thus, the equilibrium constant K = $\dfrac{{{{[C]}^2}[D]}}{{[A]{{[B]}^2}}}$
K = \[\dfrac{{{a^2} \times 0.5a}}{{0.5a \times {{[0.5a]}^2}}}\]
K = 4

So, the correct answer is option b.).

Note: The equilibrium constant describes the relationship between products and reactants of a reaction at equilibrium with respect to specific units. We can measure the equilibrium constant at specific concentration if the concentrations of the compounds are given and also in pressure if the value of their pressure is given.