Page 204 - Essentials of physical chemistry

166 Essentials of Physical Chemistry

(Add the equations.)

d[Br . ] 2
¼ k 1 [Br 2 ] k 2 [Br . ][H 2 ] þ k 3 [H . ] [Br 2 ] þ k 4 [H . ] [HBr] k 5 [Br . ]
0 ffi
dt
d[H . ]
¼ k 2 [Br . ][H 2 ] k 3 [H . ] [Br 2 ] k 4 [H . ] [HBr]
dt
0 ffi
_______________________________________________________________________________

0 ffi k 1 [Br 2 ] k 5 [Br . ] 2

This example is very good for teaching because it illustrates the method but has an easy solution
because of a fortuitous substitution. So, substitute the steady-state equation for [H.] into the
equation for the [Br.] and note the cancelation due to k 2 [Br .][H 2 ] ¼ k 3 [H.][Br 2 ] þ k 4 [H.][HBr].
2
Thus we find 0 ffi k 1 [Br 2 ] k 5 [Br.] , which can be solved for
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

k 1
[Br 2 ]:
[Br.] ss ffi
k 5
Now substitute the expression for [Br.] ss into the [H.] equation to find

1
2
k 1 1
k 2 [H 2 ] [Br 2 ] 2 ¼ (k 3 [Br 2 ] þ k 4 [HBr])[H . ] ss ,
k 5
and we can solve for [H.] ss as follows:

1
2
k 1 1
k 2 [H 2 ] [Br 2 ] 2
k 5
:
(k 3 [Br 2 ] þ k 4 [HBr])
[H . ] ss ¼
Now we are ready to write the overall rate equation as the appearance of HBr, and we substitute the
steady-state concentrations we have found.
d[HBr]
¼ k 2 [Br.] ss [H 2 ] þ k 3 [H . ] ss [Br 2 ] k 4 [H . ] ss [HBr]:
dt

(The next step is amazing!)

2 1 3
2
k 1 1
!
1 k 2 [H 2 ] [Br 2 ] 2 (k 3 [Br 2 ] k 4 [HBr])
d[HBr] k 1 2 1 6 k 5 7
k 2 [H 2 ] [Br 2 ] 2 þ 6 7:
6
7
¼
dt k 5 4 (k 3 [Br 2 ] þ k 4 [HBr]) 5
Put this expression over a common denominator and the numerator terms in k 4 cancel:
1
2
k 1 3
2k 2 k 3 [Br 2 ] 2 [H 2 ]
d[HBr] k 5
:
dt ¼ (k 3 [Br 2 ] þ k 4 [HBr])