Page 95 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch07 Elementary Physical Chemistry

80 Elementary Physical Chemistry

Here, M is a third body which removes energy. The concentration of M
has been absorbed in the reaction rate constant, k d .(Thereare other
recombination reactions, such as 2H • → H 2 and H • +Br • → HBr, but
these are unimportant and will be ignored.)
The final reaction rate must not contain free radicals Br • or H .Using
•
the steady state approximation, we consider

d[Br •]/dt =2k a [Br 2 ] − k b [Br •][H 2 ]+ k [H ][Br 2 ]+ k c [H ][HBr]
•
•
b
2
− 2k d[Br •] = 0 (7.66)
d[H • ]/dt = k b [Br •][H 2 ] − k [H • ][Br 2 ] − k c [H • ][HBr] = 0 (7.67)

b
Thus, there are two equations with two unknowns. Solving them gives
[Br •]= (k a /k d ) 1/2 [Br 2 ] 1/2 (7.68)

[H • ]= {k b(k a /k d) 1/2 [H 2 ][Br 2 ] 1/2 }/{k [Br 2 ]k c[HBr]} (7.69)
b
The final expression for the rate law, which is in terms of the production
of HBr, combines Eqs. (7.62)–(7.64) to give

d[HBr]/dt = k b[Br •][H 2]+ k [H ][Br 2 ] − k c[H ][HBr] (7.70)
•
•
b
Substitution of the values for [H ]and [Br •]gives
•
1/2 3/2

d[HBr]/dt =2k b(k a /k d) [H 2][Br 2 ] /{[Br 2 ]+ (k c /k )[HBr]} (7.71)
b
which, when replacing k by 2k b (k a /k d )and k by k c /k ,gives the

b
experimental rate law.
Example 7.9. The production of HI in the reaction
H 2 +I 2 → 2HI
obeys the rate law d[HI ]/dt = k[H 2 ][I 2 ].
The following mechanism has been suggested

1) I 2 2I rapid equilibrium
2) 2I + H 2 → 2HI slow

The forward and reverse reaction constants of Eq. (1) are k 1 and k ;the
1
(forward) reaction constant of Eq. (2) is k 2. Determine the rate law d[HI]/dt
andexpress theresult interms of the reaction constants k 1 ,k ,k 2 .

1