The reaction order is defined as the final exponent of the concentrations in the rate formula V.
Normally, this exponent is derived from the coefficients in the reaction equation.
Each reaction step below has its own 'order'.
Cl_{2}
2 Cl·
V = k*[Cl_{2}]
2NO_{2}
N_{2}O_{4}
V = k*[NO_{2}]^{2}
CH_{3}I + OH^{-}
CH_{3}OH + I^{-}
V = k*[CH_{3}I]*[[OH^{-}]
Back to the reaction: H_{2}(g)+ Br_{2}(g)
2HBr(g)
This seems to be a reaction of the second order (bimolecular), but rate measuring shows that the order of this reaction = 1½.
How to explain that?
The steps (including a radical mechanisme):
Br_{2}
2Br·
this equilibrium is reached very quickly
Br· + H_{2}
HBr + H·
this is the slowest step
H· + Br_{2}
HBr + Br·
this is the fastest step
Step 2, the slowest one, is determining the total rate:
V_{total} =
Equilibrium 1 has as a condition:
If we combine the two (mathematical) equations, we can calculate:
_{
} which means that the reaction order must equal 1½ (see exponent)