For the reaction, A + B⇌C + D, the initial concentration of ‘A’ and ‘B’ are equal, but the equilibrium concentration of ‘C’ is twice that of equilibrium concentration of ‘A’. Theequilibrium constant is _______

To solve this problem, we need to understand the concept of equilibrium constant (Kc) for a chemical reaction. The equilibrium constant is given by the ratio of the product of concentrations of the products to the product of concentrations of the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

For the given reaction, A + B ⇌ C + D, the equilibrium constant expression is:

Kc = [C][D] / [A][B]

Given that the initial concentrations of A and B are equal, let's denote this common initial concentration as [A]0 = [B]0 = x.

Also, it's given that at equilibrium, the concentration of C is twice that of A. So, [C]eq = 2[A]eq.

Since the reaction consumes equal amounts of A and B to produce C and D, the decrease in concentration of A (or B) is equal to the increase in concentration of C (or D). Therefore, we can write the equilibrium concentrations in terms of the initial concentration x:

[A]eq = x - y [B]eq = x - y [C]eq = y [D]eq = y

where y is the amount of A and B that reacted to form C and D.

Substituting [C]eq = 2[A]eq into the above equations, we get y = 2(x - y), which simplifies to 3y = 2x, or y = 2x/3.

Substituting these equilibrium concentrations into the equilibrium constant expression, we get:

Kc = [C][D] / [A][B] = (2x/3) * (2x/3) / ((x - 2x/3) * (x - 2x/3)) = 4/9.

So, the equilibrium constant for the given reaction under the given conditions is 4/9.