Page 206 - Essentials of physical chemistry

168 Essentials of Physical Chemistry

and the middle terms cancel leaving

1
2
k 1 1
[CH 3 CHO] 2
[ . CH 3 ] ss ¼
k 4
for the steady-state concentration of the methyl radical. As usual we have to choose something
which we can measure to follow the ‘‘extent of the reaction’’ and so, we choose the appearance of
methane, CH 4 . Then the rate is

1
d[CH 4 ] k 1 2 3
k 2 [CH 3 CHO] 2 ,
dt ¼ k 2 [ . CH 3 ] [CH 3 CHO] ¼ k 4

with perhaps unexpected dependence on the acetaldehyde concentration to the (3=2) power. The
original paper by Rice and Herzfeld gives approximate E* values for the various reactions in the
overall scheme (in kcal of course, since the paper was written in 1934). We can estimate the overall
activation energy from the final rate expression by combining the Arrhenius forms of the several rate
constants:

1 1 (76:0þ0) 1
2 2 10:0þ 2 Š kcal 2
k 1 A 1 ½ A 1 48, 000 cal 48, 000 cal
k 2 ¼ A 2 e RT ¼ A 2 e ð RT Þ ¼ A e RT Þ :
0 ð
k 4 A 4 A 4
Thus, even though we are not given the values of the A constants, we can estimate the E* energy.
There are many other such free radical mechanisms in thermal cracking of hydrocarbons and these
have been in use in designing petroleum refining operations for many years.

STEADY-STATE EXAMPLE NO.3:THE LINDEMANN MECHANISM
An important use of the steady-state concept is the application to nominal unimolecular reactions
that have high activation energies. This was first formulated by Frederick A. Lindemann (1886–
1957), an English physicist. Some examples are isomerizations, such as CH 3 NC ! CH 3 CN or
decompositions like CH 3 CH 2 Cl ! CH 2 ¼ CH 2 þ HCl. For simplicity, let us consider a unimole-
cular isomerization, such as what seems to be A ! B in the gas phase. Having specified that this is a
gas-phase reaction, we need to remember what we learned about the kinetic theory of gases and also
Dalton’s law of partial pressures. From Dalton’s law, consider that there might be another gas
present, M, but at the end of this discussion we will be free to let M ¼ A as well. Then, we have
r ffiffiffiffiffiffiffiffiffi
8RT
, but that is only the average velocity. The
pM
activation by collisions. Remember that v ¼
Boltzmann distribution shows there are some molecules with much higher velocities. Collisions
with other higher energy molecules can transfer energy to the A molecules and raise their energy
above the activation energy to react to form A*.
k 1 k 2
A þ M Ð A* þ M and then A* ƒƒƒƒ! B but we can consider A* as a transient steady-state
k 1
intermediate, so
d[A*]
¼ k 1 [A] [M] k 1 [A*] [M] k 2 [A*] ffi 0:
dt