Question

In a chemical reaction, A + 2B (K) ⇋ 2C + D, 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 :

(1) 16

(2) 1/4

(3) 1

(4) 4

Answer 1

To find the equilibrium constant (K) for the reaction A + 2B ⇋ 2C + D, the initial concentration relationship and the condition that [A] equals [B] at equilibrium are used to determine changes in concentrations, expressed as x. These values are then plugged into the equilibrium constant expression, resulting in a K value of 4, signifying a product-favorable reaction.

Step-by-step explanation:

In the chemical reaction A + 2B ⇋ 2C + D, where concentrations at equilibrium for A and B are equal, we can deduce the equilibrium constant (K). If the initial concentration of B is 1.5 times that of A, we can express this as [B] = 1.5[A]. At equilibrium, the concentrations are equal, so [A] = [B]. The reaction can be set up with a variable, x, representing the change in concentration from initial to equilibrium for A and 2x for B. As per the stoichiometry, [C] would increase by 2x and [D] by x.

The equilibrium constant (K) expression for the reaction is K = ([C]^2[D]) / ([A][B]^2). Given that 2x mol/L of B react, leaving [A] = [A]₀ - x and [B] = 1.5[A]₀ - 2x at equilibrium, and since at equilibrium [A] = [B], we solve for x in terms of [A]₀. Plugging these values into the expression for K, we find the value of K when concentrations are equal. Considering the stoichiometric ratios and the given conditions, the K calculated is 4, which means the reaction favors the formation of products under these specific equilibrium conditions.