Page 195 - Instant notes

Formulation of rate laws 181

Steady state assumption

The steady state assumption presumes that the concentrations of all intermediate
species, I, in the reaction mechanism remain constant and small during the reaction
(except right at the beginning and right at the end), or in other words that the net change
in concentration [I] with time is zero, d[I]/dt≈0. The assumption is equivalent to equating
the total rate of removal of each intermediate with its total rate of formation.
Application of the steady state assumption effectively converts the differential
equation for each intermediate into an algebraic expression from which the steady state
concentrations can be derived for substitution into the overall rate law. The steady state
assumption is widely applied in kinetics to formulate rate laws of complex reactions.

Formulating rate laws

The first step in formulating a rate law is to write down the proposed mechanism in
terms of individual unimolecular and bimolecular elementary reaction steps. Since a
rate law is an experimentally derived property of a reaction (see Topic F1) the
combination of the individual first and second order rate laws must yield an overall rate
law that is consistent with observation. For example, the gas phase oxidation of nitrogen
monoxide, NO:
2NO(g)+O 2(g)→2NO 2(g)

is observed experimentally to be third order overall:
2
rate of formation of NO 2=k[NO] [O 2]
Although one explanation for the third order behavior might be a single termolecular
(three molecule) elementary reaction, the simultaneous collision between three molecules
is extremely unlikely so the reaction is likely to be complex. The additional observation
that rate of NO oxidation decreases with increasing temperature is further evidence of a
complex reaction because rates of elementary reactions increase with temperature (Topic
F3).
A reaction mechanism can be postulated which proceeds through the formation of an
N 2O 2 dimer, which may dissociate back into two NO molecules or undergo reactive
collision with an O 2 molecule to produce two NO 2 molecules. The decomposition of
N 2O 2 into NO is a unimolecular reaction step, the other two are bimolecular.
2
NO+NO→N 2 O 2 rate of formation of N 2 O 2 =k 1 [NO]
N 2 O 2 →NO+NO rate of decomposition of N 2 O 2 =k −1 [N 2 O 2 ]
N 2 O 2 +O 2 →NO 2 +NO 2 rate of consumption of N 2 O 2 =k 2 [N 2 O 2 ][O 2 ]

(the subscripts on the rate constants merely represent labels for the corresponding
reaction.) The overall rate at which NO 2 is formed is controlled by the third reaction in
the mechanism: