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.
(b) Draw a qualitative graph that shows how the rates of the forward and reverse reactions vary with time.

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.
(b) Draw a qualitative graph that shows how the rates of the forward and reverse reactions vary with time.

(b) Draw a qualitative graph that shows how the rates of the forward and reverse reactions vary with time.

Accepted Answer

Hello. In this problem we are told the inter conversion of molecules X and Y is a first order reversible reaction. We have X produces Y in the four direction and Y produces X in the reverse direction. Where K. F. Is the rate constant for the ford reaction. The value of 4.5 times 10 to minus four per second and K. R. Is the rate constant for the reverse reaction with the value of 1.5 times 10 to minus four per second. The rate of the reaction does not go to completion but reaches a state of equilibrium when K4 and K reverse are virtually similar with comparable reacting and product concentrations. We wonder which of the following plots depict how forward and reverse reaction rates change over time. So recall that the rate for the forward reaction is going to be equal to the reaction rate constant for the forward reaction times the concentration of X to the one power. The first order of reaction. And the rate for the reverse reaction is equal to the reaction rate constant for the reverse reaction times the concentration of why to the first order. So if we initially have just X present, then the rate of the forward reaction is going to be fast because we are going to have higher concentrations of X. As the amount of X gets reduced and converted to Y. In the forward reaction. The rate of the forward reaction will go down. So this will go down over time. And the rate of the reverse reaction initially starts off at zero because we have initially no products as we start to produce y. Then the rate of the reverse reaction will increase due to an increase in the concentration of rye. So the reverse reaction will increase over time. So this is true. The rate of the forward reaction will down and that of the reverse will increase. They will reach equilibrium when the rate of the forward reaction is equal to the rate of the reverse. So if we look at our plots that were provided, we have then a shows the forward reaction right? We have rate on the y axis as a function of time on the X. So initially we have a high rate for the forward reaction. Initially the reverse reaction is zero. The rate of the forward reaction decreases as time proceeds and that of the reverse increases until we reach a point in time where the rate of the ford is equal to the rate of the reverse and at this point we've reached equilibrium. Whereas B then shows that the reverse reaction initially is high and decreases over time and that of the forward reaction is zero and increases over time. And see says that they both start off at equilibrium. Right. We have the same rate for the ford and reverse reaction and then the way the forward reaction increases and that of the reverse decreases over time. And he shows us something similar initially are at equilibrium. The rate of the forward reaction is equal to the rate of the reverse and then the rate of the reverse reaction increases and that of the four decreases, So C. And D. Are eliminated because we don't begin at equilibrium. We initially start off with reactant, snow products and then reach a state of equilibrium. If we look at A and B again, we initially start with our reactant and so we have initially a high rate for the ford reaction, which decreases over time until we have a rate of the forward reaction equal to the rate of the reverse. And the reverse reaction starts out at zero and then increases over time until the rate of the forward reaction is equal to the rate of the reverse. So this is our correct answer. Be shows just the opposite of that. It shows that the reverse reaction decreases over time and the forward reaction increases over time until we reach a state of chemical equilibrium. So we also eliminate that. So, thanks for watching Help. This helped

Verified Solution

Key Concepts

Reversible Reactions

Rate Constants

Equilibrium