In reaction $$A + 2B \rightleftharpoons 2C + D$$, initial concentration of B was 1.5 times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is :

Consider the reaction: $$A + 2B \rightleftharpoons 2C + D$$

Let the initial concentration of A be $$a$$ mol/L. According to the problem, the initial concentration of B is 1.5 times that of A, so [B]_initial = $$1.5a$$ mol/L. Initially, assume no products are present, so [C]_initial = 0 and [D]_initial = 0.

At equilibrium, let the amount of A reacted be $$x$$ mol/L. Using the stoichiometry of the reaction:

• For every mole of A reacted, 2 moles of B react, so B reacted = $$2x$$ mol/L.
• For every mole of A reacted, 2 moles of C are produced, so C produced = $$2x$$ mol/L.
• For every mole of A reacted, 1 mole of D is produced, so D produced = $$x$$ mol/L.

Equilibrium concentrations:

• [A] = initial A - reacted A = $$a - x$$
• [B] = initial B - reacted B = $$1.5a - 2x$$
• [C] = initial C + produced C = $$0 + 2x = 2x$$
• [D] = initial D + produced D = $$0 + x = x$$

The problem states that at equilibrium, [A] = [B]. So: $$a - x = 1.5a - 2x$$

Solve for $$x$$: $$a - x = 1.5a - 2x$$ Subtract $$a$$ from both sides: $$-x = 0.5a - 2x$$ Add $$2x$$ to both sides: $$-x + 2x = 0.5a$$ $$x = 0.5a$$ So, $$x = \frac{1}{2}a$$

Substitute $$x = \frac{1}{2}a$$ into the equilibrium concentrations:

• [A] = $$a - \frac{1}{2}a = \frac{1}{2}a$$
• [B] = $$1.5a - 2 \times \frac{1}{2}a = \frac{3}{2}a - a = \frac{1}{2}a$$
• [C] = $$2 \times \frac{1}{2}a = a$$
• [D] = $$\frac{1}{2}a$$

The equilibrium constant $$K$$ for the reaction is: $$K = \frac{[C]^2 [D]}{[A] [B]^2}$$

Substitute the equilibrium concentrations: $$K = \frac{(a)^2 \times \left(\frac{1}{2}a\right)}{\left(\frac{1}{2}a\right) \times \left(\frac{1}{2}a\right)^2}$$

Simplify step by step: Numerator: $$(a)^2 \times \frac{1}{2}a = a^2 \times \frac{a}{2} = \frac{a^3}{2}$$ Denominator: $$\left(\frac{1}{2}a\right) \times \left(\frac{1}{2}a\right)^2 = \frac{1}{2}a \times \frac{1}{4}a^2 = \frac{1}{2} \times \frac{1}{4} \times a \times a^2 = \frac{1}{8}a^3$$

So, $$K = \frac{\frac{a^3}{2}}{\frac{1}{8}a^3} = \frac{a^3}{2} \times \frac{8}{a^3} = \frac{8}{2} = 4$$

Hence, the equilibrium constant is 4.

So, the answer is Option B.