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.
(c) What are the relative concentrations of B and A when the rates of the forward and reverse reactions become equal?

Question

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.
(c) What are the relative concentrations of B and A when the rates of the forward and reverse reactions become equal?

(c) What are the relative concentrations of B and A when the rates of the forward and reverse reactions become equal?

Accepted Answer

Hello. In this problem we are told the following reaction is a reversible reaction of first order. So why produces E. In the forward direction and Z produces Y in the reverse direction is forward and reverse reactions have corresponding rate constants of 1. minus four per second and one times 10 to minus four per second. When K. four and K reverse have comparable values. The reaction does not proceed to completion but instead reaches equilibrium. Where the concentration of reactant and products are equal. When the rates of the forward reaction and reverse reaction are equal. We wanna what are the relative concentrations of Y. N. C. So we begin then by re writing the rate of the forward reaction and put subscript S. F. To stanford forward. It's unequal to our reaction rate constant for the forward reaction and this is dependent on the concentration Y. And it is first order. So it's to the one power. The rate of the reverse reaction to subscript R is equal to K. Subscript. Our reaction rate constant for the reverse reaction. And this is dependent on the concentration of Z. It's the first order. So it also has a power of one. So when we reach equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. So that means that K4 times the concentration of Y. Is equal to K. Reverse times the concentration of Z. We get our reaction rate constants on one side and our concentrations to the other side. So we divide both sides by Kr and we divide both sides by why? So now we have K forward over K. Reverse is equal to Z over Y. We are told. The reaction rate constant for the forward reaction is 1.5 times 10 to the minus four per second, And that of the reverse is one times 10 to the - per second. So this works out to 1.5 over one. So our concentration of Z to Y Is 1.5 - one. So that means then the relative concentration of Z and why Our 1. and one respectively. Hope this helps. Thanks for watching.

Verified Solution

Key Concepts

Equilibrium in Chemical Reactions

Rate Constants (k_f and k_r)

Concentration Ratios at Equilibrium