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50 Basic physical chemistry

k 2
Br(g) + H2(g� HBr(g) + H(g) (ii) (slow)
k3
H(g) + Brz(g� HBr(g) + Br(g) (iii) (fast)
where the k's are the rate coefficients for each elementary process,
and k1r and k1r are the forward and reverse rate coefficients for (i).
d[HBr(g)]
Derive an expression for
dt in terms of [Hz(g)] and [Br2(g)] .
Solution. Since (ii) is the slowest reaction, it is the rate-determining
step. Therefore, from Reaction (ii)
d[H r (g)]
B
kz[H2(g)][Br(g)]
dt
Also, for

Br (g) � 2 Br(g)
2
[Br(g)] 2
K = ---
c [Br2(g)]

where, Kc is the equilibrium constant. Hence,

This problem illustrates how half-integer rate laws arise. Also, it
shows how the rate coefficient k for a reaction can be related to other
and Kc in this case).
parameters (k 2
Just because a series of elementary processes is consistent with a
chemical reaction and its experimentally determined rate law, these
elementary �rocesses are not necessarily responsible for the reaction.
This is beca.use more than one series of steps can satisfy a reaction
and a rate law (see Exercise . 1 3 ) . In general, information in addition
3
to the rate law is required if a reaction mechanism is to be determined.

3.3 Reaction rates and equilibria
When a chemical reaction is in equilibrium, the forward and reverse
reactions occur at the same rate. U s ing this principle we can derive a
general relationship between the equilibrium constant and the forward
and reverse rate constants for any elementary process. Let