Physical Chemistry Third Edition

13.2 Competing Mechanisms and the Principle of Detailed Balance 583

[E]tot 1. 67 × 10 −^5 mol L−^1 ,Km 4. 0 ×

10 −^6 mol L−^1 andk 2  4. 36 × 10 −^4 s−^1.

a.Find the initial rate for a fumarate concentration of

0 .0000250 mol L−^1.

b.Find the initial rate for a fumarate concentration of

0.00500 mol L−^1.

c.Construct a graph of the initial rate as a function of

fumarate concentration.

13.17The reaction

CO 2 +H 2 O−→H++HCO− 3

is catalyzed by bovine carbonic anhydrase. At

pH 7.1 and 0. 5 ◦C,Km 8. 23 × 10 −^3 mol L−^1 and

k 2  7. 54 × 104 s−^1.

a.Find the initial rate for an initial concentration of CO 2

equal to 5. 00 × 10 −^3 mol L−^1 and an enzyme

concentration equal to 2. 8 × 10 −^9 mol L−^1.

b.Find the maximum rate for the same enzyme

concentration.

13.18For a fixed total enzyme concentration, a certain
enzymatically catalyzed reaction has an initial
rate of 0.0015 mol L−^1 min−^1 for an initial reactant
concentration of 0.100 mol L−^1 and an initial rate of

0 .0020 mol L−^1 min−^1 with a reactant concentration of

0 .200 mol L−^1.

a.Find the value of the Michaelis–Menten constantKm

and the maximum rate.

b.Construct a graph of the rate as a function of the

reactant concentration.

c.Construct an Eadie plot for this reaction.

13.19For the enzymatically catalyzed hydrolysis of

AT P a t 2 5◦C and pH 7.0, the Michaelis–Menten

constant,Km, was found to equal 16. 8 μmol L−^1 ,

and the value of k 2 [E]totalwas found to be 0.220μmol

L−^1 s−^1.

a.Write the formula for the initial rate as a function of

[ATP] with values specified for the constants.

b.Construct a graph of the initial rate as a function [ATP]

for the given data, including scales on the axes.

c.Find the initial rate at an initial ATP concentration of

30.0μmol L−^1.

d.Write a formula for the reciprocal of the initial rate

with values for the constants.

e.Construct a Lineweaver–Burk plot for the given data.

Give values for the slope and the intercept.

13.2 Competing Mechanisms and the Principle

of Detailed Balance

For any reaction that proceeds by two competing mechanisms, theprinciple of detailed

balancegoverns the rate constants for the two mechanisms:Different mechanisms for

the same reaction must give the same value of the equilibrium constant at the same

temperature. Another statement is:At equilibrium, each mechanism must separately

be at equilibrium, with canceling forward and reverse rates.Because of this principle,

a catalyst cannot change the equilibrium constant for a reaction, and a catalyst must

catalyze the reverse reaction if it catalyzes the forward reaction.

Assume that a gaseous system containing H 2 ,I 2 , and HI has come to equilib-

rium. Consider two mechanisms for the reaction of H 2 and I 2 : (1) the mechanism

of Eq. (12.5-15) with inclusion of a reverse reaction in step 2, and (2) the elementary

mechanism:

(1a)H 2 +I 2 
2 HI (13.2-1)

Figure 13.10 shows the two pathways. Since the principle of detailed balance requires

both mechanisms separately to be at equilibrium, it is not possible at equilibrium for

the forward reaction of one mechanism to be canceled by the reverse reaction of the

other mechanism.

Two-step mechanism

One-step mechanism

H 2 + 2I

H 2 + I 2 2HI

k 1

k (^91)
k 9
k 1 a
k (^92)
k 2
1 a
Figure 13.10 Two Mechanisms for
the H 2 +I 2 Reaction.