   Chapter 17, Problem 107CP

Chapter
Section
Textbook Problem

# a. Using the free energy profile for a simple one-step reaction, show that at equilibrium K = kf/kr, where kf and kr, are the rate constants for the forward and reverse reactions. Hint: Use the relationship ∆G° = −RT ln(K) and represent kf and kr, using the Arrhenius equation. ( K = A e − E a / R T ) b. Why is the following statement false? “A catalyst can increase the rate of a forward reaction but not the rote of the reverse reaction.”

(a)

Interpretation Introduction

Interpretation: It is to be shown that, K=kfkr by using the free energy profile for a simple one-step solution. The reason of falsity in the statement “A catalyst can increase the rate of a forward reaction, but not the rate of the reverse reaction” is to be stated.

Concept introduction:

To show: At equilibrium, K=kfkr by using the free energy profile for a simple one-step solution.

Explanation

Explanation

The expression for Arrhenius equation is,

k=AeEaRT

Where,

• k is the rate constant.
• A is the Arrhenius parameter.
• Ea is the activation energy.
• R is the universal gas constant.
• T is the absolute temperature.

The rate constants are expressed using the Arrhenius equation,

kf=AeEaRTkr=Ae(EaΔG°)RT

Where,

• kf is the rate constant of forward reaction.
• kr is the rate constant of forward reaction.

The overall rate constant of a reaction is,

k=kfkr (1)

Substitute the values of kf and kr in the above equation

(b)

Interpretation Introduction

Interpretation: It is to be shown that, K=kfkr by using the free energy profile for a simple one-step solution. The reason of falsity in the statement “A catalyst can increase the rate of a forward reaction, but not the rate of the reverse reaction” is to be stated.

Concept introduction:

To show: At equilibrium, K=kfkr by using the free energy profile for a simple one-step solution.

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