Page 184 - Essentials of physical chemistry

146 Essentials of Physical Chemistry

over 100 reactions with Arrhenius factors to model the details of all the free radicals and inter-
mediate species in the simple combustion of the methane flame [8].
CH 4 þ 2O 2 ! CO 2 þ 2H 2 O

The details of such complex mechanisms require a simple model capable of parameter fitting, so the
Arrhenius model is still used for studies of combustion of jet fuel and various rocket engine fuels.
A later chapter will explore more detailed treatments but here we can use the equation in natural

E* E* 1
logarithm form. If K rate ¼ Ae RT , we have ln (K(T)) ¼ ln A and this leads to
R T
another graph with (1=T) along the x-axis. A side issue in notation is that we will have to use
K rate for the rate constant to distinguish it from the Boltzmann constant ‘‘k’’ and of course we will
use ‘‘K’’ for Kelvin temperature so this author favors using 8K for the temperature when all three
variations of (k, K rate , 8K) are used in the same equation.

Example 5
An example from Ref. [7] will be treated here with the simpler Arrhenius equation and will be used
again later to compare this method to more sophisticated treatments. The data consists of measure-
ment of the solvolysis of an alkyl halide (1-chloro, 1-methyl cycloheptane) in a solution of 80%
ethanol for what is nominally an SN1 replacement of the chloride ion. In a later chapter, we will
examine this reaction in more detail as a model of a steric hindrance in a carbocation substitution
reaction but here we are just interested in the measured reaction rates at various temperatures.

CH 3

Cl

Temp, 8C 0 25 35 45
1 5 4 4 3
K rate ,s 1.06 10 3.19 10 9.86 10 2.92 10
1=T (8K) 3.660992 10 3 3.354016 10 3 3.245173 10 3 3.143172 10 3
11.454656 8.050319 6.921854 5.836172
ln K rate

Clearly the reaction speeds up as the temperature increases, so the question is whether the data
agrees with the equation proposed by Arrhenius. Thus, we plot the data versus (1=T)as

E* 1
þ ln A in linear y ¼ mx þ b form.
ln (K rate ) ¼ ðÞ
R T
2
The best-fit line yields a value of R which indicates a very good linear fit. The variable ‘‘x’’ on
the graph refers to the x-axis which is 1=T (8K) here and the slope has a value of 10872. Using

E* E*
¼ 10,872
m ¼ we can calculate the activation energy according to Arrhenius as
R R
and we have to recognize that the (1=T) cancels the temperature units in R.
(10, 872)R ¼ E* ¼ (10872)(8:314 J=mol) ¼ 90:3898 kJ=mol ¼ 21:604 kcal=mol:
Next, we can ask what value is predicted for the A parameter of the Arrhenius plot. Since ‘‘y’’ on the
best-fit line refers to ln (K rate ), the ‘‘b’’ term is really ln (A) ¼ 28:365, so we can find A by taking
12 1
the anti-ln to obtain A ¼ exp (28:365)=s ¼ 2:08335 10 s , which is a very large number. In the
rate equation, the actual rate could be 2.08335 10 12 s 1 (L=mol). Rough calculations with actual