Lecture 28 : Elementary Reactions and Reaction Mechanisms
b)
Oxidation of Nitric oxide
This example is given to illustrate the point that a reaction need not have a unique mechanism. Several paths could lead to the same products and rate laws. The reaction is
2 NO(g) + O2(g)
2NO2 (g)
(28.21)
Experiments reveal that the rate law is of the form
(1/ 2) d [ NO2 ] / dt = k expt [ NO ] 2 [O2]
(28.22)
This may imply that the reaction is elementary. Experiments however indicate that the reaction is not elementary. Two mechanisms have been proposed for the reaction (28.21), one involving a preequilibrium (involving NO3) and another involving a steady state approximation of an intermediate N2O2 .
Mechanism 1
NO(g) + O2 (g)
NO3 (g) rapid equilibrium
(28.23)
NO3 (g) + NO (g)
2 NO2 (g)
(28.24)
In the first step, a rapid equilibrium is established between the reactants and the NO3 radical
Keq = k 1 / k -1 = [ NO3] / [NO] [O2]
(28.25)
For the second step of the mechanism, the rate law is
(1/ 2) d [NO2] / dt = k 2 [ NO ] [ NO3]
(28.26)
The second step is the rate determining step, since the preequilibrium eq (28.23) is established rapidly. Substituting eq (28.25) in eq (28.26) we have,
(1/ 2) d [NO2] / dt = k 2 Keq[ NO ]2 [O2]
(28.27)
This is the same as eq (28.22) if kexpt is replaced by k2 Keq. The experimental rate constant is now a product of k2 and Keq and not the rate constant for an elementary reaction.
2NO2 (g) 
NO3 (g) rapid equilibrium 
2 NO
