Editing Rate-determining step (section)

===Pre-equilibrium: if the second step were rate-determining===
The other possible case would be that the second step is slow and rate-determining, meaning that it is slower than the first step in the reverse direction: ''r''<sub>2</sub> ≪ ''r''<sub>−1</sub>. In this hypothesis, ''r''<sub>1</sub> − r<sub>−1</sub> ≈ 0, so that the first step is (almost) at [[chemical equilibrium|equilibrium]]. The overall rate is determined by the second step: ''r'' = ''r''<sub>2</sub> ≪ ''r''<sub>1</sub>, as very few molecules that react at the first step continue to the second step, which is much slower.  Such a situation in which an intermediate (here {{NOx|x=3}}) forms an equilibrium with reactants ''prior'' to the rate-determining step is described as a ''pre-equilibrium''<ref name=Atkins>[[Peter Atkins]] and Julio de Paula, ''Physical Chemistry'' (8th ed., W. H. Freeman 2006) p. 814–815. {{ISBN|0-7167-8759-8}}.</ref> For the reaction of {{NOx|x=2}} and CO, this hypothesis can be rejected, since it implies a rate equation that disagrees with experiment.

# {{NOx|x=2}} + {{NOx|x=2}} → NO + {{NOx|x=3}} ''(fast step)''
# {{NOx|x=3}} + CO → {{NOx|x=2}} + {{CO2}} ''(slow step, rate-determining)''

If the first step were at equilibrium, then its [[equilibrium constant]] expression permits calculation of the concentration of the intermediate {{NOx|x=3}} in terms of more stable (and more easily measured) reactant and product species:

:<math chem>K_1 = \frac{\ce{[NO][NO3]}}{\ce{[NO2]^2}},</math>
:<math chem>[\ce{NO3}] = K_1 \frac{\ce{[NO2]^2}}{\ce{[NO]}}.</math>

The overall reaction rate would then be
:<math chem>r = r_2 = k_2 \ce{[NO3][CO]} = k_2 K_1 \frac{\ce{[NO2]^2 [CO]}}{\ce{[NO]}},</math>
which disagrees with the experimental rate law given above, and so disproves the hypothesis that the second step is rate-determining for this reaction. However, some other reactions are believed to involve rapid pre-equilibria prior to the rate-determining step, [[#Composition of the transition state|as shown below]].