determing rate law

B

The reaction rate increases with the concentration of the reactant side, which is the left hand side of the equations.
The overall reaction rate depend on both the rates step 1 and step 2

The 1st step is significantly faster than the 2nd one as given by the question.
As the first step is already very fast, varying the conc. of reactants of step 1 won't have significant change of the whole process, remember the overall reaction is determined by both step 1 and step 2.
The 2nd step is slow which brings the overall reaction slow. Therefore, increase if you can speed up the 2nd step, you can speed up the overall reaction significantly.

To speed up step 2, of course you will incease the conc. of the reactants of step 2, which is [HO2(g)] and [O -].
So the overall rate is determine by option B.

2013-06-18 01:33:00 補充：
So the overall rate is determined* by option B.

2013-06-18 01:34:26 補充：
.... is [HO2(g)] and [O]*. typo

The answer should be C

For step 1, an equilibrium is established.

Equilibrium constant Kc = [O2][HO2]/ [O3][HO] (hope you understand this
expression; it is inconvenient to type on the Internet)

For step 2 rate-determining step,

rate = k[O][HO2]

Since there is no [O] in Kc expression, we only need to substitute [HO2] in terms of Kc and other chemicals

[HO2] = Kc[O3][HO]/[O2]

So that comes C

Mind you, the 'K' in A to C is NOT Kc or rate constant k

It means ANY CONSTANT!!!

Got it? :)