Page 197 - Essentials of physical chemistry

More Kinetics and Some Mechanisms 159

Now, are you ready for the clever step? In your mind mull over the idea that however
‘‘frequently’’ an event occurs that can be described as its ‘‘frequency.’’ Thus, the Eyring method
uses a frequency factor for the occurrence of the reaction event:

k B T
,
h
f ¼ n ¼
which leads to the combined expression for the Eyring rate constant as

k B T DS z DH z
e þ R e RT :
h
K rate ¼
Now we can apply Eyring’s equation for the rate constant to the solvolysis reaction we are
considering. There are at least two slightly different ways to solve this problem for the values of
DH and DS . We expect that DH will be close in value to the Arrhenius value of E* but we might
z
z
z
not know what to expect for DS , although it might be negative since the separate reactant species
z
are combined to form a more ordered transition state. Thus, this system can be solved as a problem
with two equations in two unknowns if we use two values of the experimental rate constant. Let us
try the simplest method first using the method of substitution.
EXAMPLE
3
4
Given K ¼ 3.19 10 =sat 258C and K ¼ 2.92 10 =sat458C, we choose data points reasonably
far apart to avoid small irregularities in closely spaced points yet avoid the data at 08C because we
want to use one of the equations to get results for 258C. We then rearrange the natural logarithm of
the equation, so we can cast it into two equations in two unknowns:

K rate h DH z
R ln ¼ DS ,
z
k B T T
so, we find for the two temperatures:

4 34
(3:19 10 )(6:6260693 10 ) DH z
(8:314)ln ¼ 311:8406407 ¼ DS
z
(1:3806505 10 23 )(298:15) 298:15
3 34
(2:92 10 )(6:6260693 10 ) DH z
(8:314)ln ¼ 293:4322159 ¼ DS ( 1)
z
(1:3806505 10 23 )(318:15) 318:15
_______________________________________________________________________________

1 1
18:40842479 ¼ DH z
318:15 298:15
We subtracted the second equation from the first equation to eliminate DS temporarily and
z
find DH .
z
So, we find that
18:40842478
¼ 87:30786592 kJ=mol ¼ 20:86708077 kcal=mol:
z
(298:15 318:15)
DH ¼
(318:15)(298:15)