Question

Consider the reaction: 2NOBr(g) ⇌ 2NO(g) + Br2(g)The Kc value for this reaction at a certain temperature is 0.045. A catalyst is added to the reaction mixture that increases the rate of the forward and reverse reactions to the same extent. What effect will this catalyst have on the equilibrium position of the reaction mixture? Justify your answer using calculations.

Answer 1

A catalyst speeds up the rate of both the forward and reverse reactions, but it does not affect the equilibrium constant  Kc  or the equilibrium position of the reaction mixture. This is because the catalyst lowers the activation energy for both the forward and reverse reactions equally, allowing them to reach equilibrium faster, but it does not change the relative stability of the reactants and products.To justify this using calculations, let's consider the reaction quotient  Qc  and the equilibrium constant  Kc . The reaction quotient is given by:Qc =  [NO]^2 * [Br2]  / [NOBr]^2At equilibrium, Qc = Kc. Since the catalyst does not affect the equilibrium constant, Kc remains the same at 0.045. Therefore, the equilibrium position of the reaction mixture will not change, and the ratio of the concentrations of the products  NO and Br2  to the reactants  NOBr  will remain the same.In conclusion, the catalyst will increase the rate at which the reaction reaches equilibrium, but it will not change the equilibrium position of the reaction mixture. The Kc value will remain the same at 0.045, and the ratio of the concentrations of the products to the reactants will also remain the same.