Physical Chemistry Third Edition

13.1 Catalysis 579

If there is considerable experimental error in the data, it might be difficult to

locate the asymptote in Figure 13.7 accurately. To avoid this problem, Eq. (13.1-46)

is solved for the reciprocal of the initial rate. The result is called theLineweaver–Burk

equation:^15

1

ri



Km

k 2 [E]total[R]

+

1

k 2 [E]total

(13.1-47)

where the initial rate is denoted byri. Data on initial rates should give a straight

line when 1/riis plotted as a function of 1/[R]. The slope of the line is equal to

Km/k 2 [E]total, the intercept on the vertical axis is equal to 1/k 2 [E]total, and the intercept

on the horizontal axis is equal to− 1 /Km. Note the resemblance of this plot to the linear

plot of the Langmuir isotherm in Figure 13.3.

Exercise 13.11

Show that the intercept of the Lineweaver–Burk plot on the horizontal axis is equal to− 1 /Km.

EXAMPLE13.5

The following data were gathered for the myosin-catalyzed hydrolysis of ATP at 25◦C and

pH 7.0:

[ATP]/μmol L−^1 initial rate/μmol L−^1 s−^1

7.5 0.067

12.5 0.095

20.0 0.119

43.5 0.149

62.5 0.185

155.0 0.191

320.0 0.195

Determine the value of the Michaelis–Menten constant.

Solution

The Lineweaver–Burk plot of the data is shown in Figure 13.8. The line drawn in the figure is

the line determined by an unweighted linear least-squares procedure. The correlation coeffi-

cient for the least-squares fit is equal to 0.9975. The slope is equal to 76.58 s, and the intercept

on the vertical axis is equal to 4.547 Lμmol−^1 s. The intercept on the horizontal axis is equal

to− 0 .0595 Lμmol−^1 , so that the Michaelis–Menten constant is equal to 16.8μmol L−^1.

15

10

5

0 0.05 0.10

(1/

r) × i

mol L

–1

s

–1

1

[R]/mol L–1 s–1

Figure 13.8 The Lineweaver–Burk

Plot.

Exercise 13.12

An alternative linear plot is the Eadie plot,afor which Eq. (13.1-45) is put into the form

ri

[R]



ri

Km

+

k 2 [E]total

Km

(13.1-48)

(^15) H. Lineweaver and D. Burk,J. Am. Chem. Soc., 56 , 658 (1934).