Consider the reversible, first-order interconversion of two molecules A and B: where k_f = 3.0⨉10-3 s^-1 is the rate constant for the forward reaction and kr = 1.0⨉10^-3 s^-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants k_f and k_r have comparable values.

(a) What are the rate laws for the forward and reverse reactions?

Video transcript

Hello. In this problem we are told the following reaction is a reversible reaction of the first order. So why produces Z in the four direction? And Z produces Y in the reverse direction is forward. And reverse reactions have corresponding rate constants of 1.5 times 10 to minus four per second and one times 10 minus four per second. When K four and K reverse have comparable values. The reaction does not proceed to completion but instead reaches equilibrium or the concentration of reactions and products are equal. We want to give the rate laws for both the forward and the reverse reactions. The rate for the forward reaction to the right rate subscript F is equal to the reaction rate constant for the forward reaction times the concentration of why? So the rate for the forward reaction is dependent on the concentration of why and it is the first order. So that's the one power and the rate of the reverse reaction. So subscript R. Is equal to the reaction rate constant for the reverse reaction times the concentration of Z. So the reverse reaction depends on the concentration of Z. And it is first order. So it's to the one power. So these are the great law expressions for both the ford and the reverse reaction. Thanks for watching. Hope this helps