A reaction mixture initially contained 0.1 mol/L of reactant A and 0.2 mol/L of reactant B. When a catalyst was added, the rate of conversion of A and B to the product increased significantly. Calculate the new equilibrium concentration of reactant B if the equilibrium constant of the reaction is 0.1 and the initial equilibrium concentration of the product is 0.05 mol/L. Assume that the catalyst does not change the equilibrium constant.

Question

A reaction mixture initially contained 0.1 mol/L of reactant A and 0.2 mol/L of reactant B. When a catalyst was added, the rate of conversion of A and B to the product increased significantly. Calculate the new equilibrium concentration of reactant B if the equilibrium constant of the reaction is 0.1 and the initial equilibrium concentration of the product is 0.05 mol/L. Assume that

Answer 1

Let's denote the reactants A and B as A and B, and the product as C. The reaction can be represented as:A + B  CThe initial concentrations are as follows:[A] = 0.1 mol/L[B] = 0.2 mol/L[C] = 0.05 mol/LLet x be the change in concentration of A and B at equilibrium. At equilibrium, the concentrations will be:[A] = 0.1 - x mol/L[B] = 0.2 - x mol/L[C] = 0.05 + x mol/LThe equilibrium constant  K  is given as 0.1. The expression for the equilibrium constant is:K = [C] /  [A] * [B] Substitute the equilibrium concentrations into the expression:0.1 =  0.05 + x  /   0.1 - x  *  0.2 - x  Now, we need to solve for x. This is a quadratic equation, but we can make an approximation since K is small, which means that x will also be small. Therefore, we can assume that  0.1 - x  and  0.2 - x  are approximately equal to 0.1 and 0.2, respectively.0.1 =  0.05 + x  /  0.1 * 0.2 0.1 * 0.1 * 0.2 = 0.05 + x0.002 = 0.05 + xNow, solve for x:x = 0.002 - 0.05x = -0.048Since x is the change in concentration of A and B, the new equilibrium concentration of reactant B is:[B] = 0.2 - x[B] = 0.2 -  -0.048 [B] = 0.248 mol/LSo, the new equilibrium concentration of reactant B is 0.248 mol/L.