Physical Chemistry , 1st ed.

Solution

First, we need to determine the values that we will be plotting: they aren’t the

values right from the table above! We need the following pairs of numbers:

ln k 1/T(K^1 )

29.829 0.00336

29.641 0.00319

29.476 0.00303

See Figure 20.16 for a plot of ln kversus 1/T. Although the three points do

not all lie exactly on a straight line, it is a pretty good approximation of one

(again, indicative of the natural variations in experimental measurements).

The slope, which would equal EA/R, is about 1070, so by multiplying the

ideal gas law constant Rthrough, we get an estimate ofEAas 1070 8.314

J/mol 8900 J/mol. (The temperature unit in the denominator ofRis at-

tached to the 1/Tterm.) The y-intercept of the plot is about 26.3, which is

equal to ln A. Therefore,Ais about 3.78 10 ^12 cm^3 /(moleculesecond).

Notice that Ahas the same units as the given rate constants.

Example 20.9

Use the information derived from Example 20.8 to estimate the rate con-

stant at 370 K. Compare it with the experimentally determined value of

2.10 10 ^13 cm^3 /(moleculesecond).

Solution

With an activation energy of 8900 J/mol, we can use the value ofAand the

given temperature and equation 20.50 directly:

kAeEA/RT

k3.8 10 ^12 

molecu

c

l

m

e

3

second

exp





Note how the units cancel in the exponential, as they should. Solving, we

find that

k2.09 10 ^13 

molecu

c

l

m

e

3

second



This is very close to the experimentally determined value, showing that the

Arrhenius equation is a good model for this reaction.

If we have two sets of conditions, two versions of equation 20.51, labeled

with 1 and 2 subscripts, can be subtracted to get the following expression:

ln 

k

k

1

2



E

R

A

T

1

1



T

1

2

 (20.52)

which eliminates the need to know the pre-exponential factor A.

For some reactions, using the van’t Hoff equation as the starting point to

develop the Arrhenius equation is a little simplistic. Rather than assuming that

8900 

m

J

ol





8.314 mo

J

lK

(370 K)

704 CHAPTER 20 Kinetics

 26

 30

0

1/T (K^1 )

lnk

0.001 0.002 0.003

 27

 28

 29

Figure 20.16 See Example 20.8. According to
the plot of ln kversus 1/T, the slope is equal to
EA/Rand the y-intercept is equal to the natural
logarithm of the pre-exponential constant.
According to this plot,EAis 8900 J/mol and Ais
about 3.78 10 ^12 cm^3 /(moleculesecond). See
text for details.