The reaction 2 NO2(g)--> 2NO(g)+O2(g) is postulated to occur via the mechanism below:

NO2)g)+ NO2(g)-->NO(g)+NO3(g) ;K1,slow NO3(g)--+O2(g);K2 , fast
What is the rate law for this mechanism

1 answer

The rate law for the slowest step (which is the rate-determining step) will give us the overall rate law.

The slow step is: NO2(g) + NO2(g) --> NO(g) + NO3(g)

The rate law for this step can be written as:

rate = k1[NO2][NO2]

The fast step is: NO3(g) + O2(g) --> NO2(g) + O3(g)

The rate law for this step can be written as:

rate = k2[NO3][O2]

Now, we need to eliminate NO3 from the rate law of the overall reaction. We can do this by using the fast step to express [NO3] in terms of [NO2] and [O2]:

k2[NO3][O2] = rate of NO3 consumption = k1[NO2][NO2]

[NO3] = (k1/k2)[NO2]^2[O2]

Substituting this expression for [NO3] into the rate law for the fast step:

rate = k2[(k1/k2)[NO2]^2[O2]][O2] = k1k2[NO2]^2[O2]^2

Therefore, the rate law for the overall reaction is:

rate = k1k2[NO2]^2[O2]^2