How many type of mechanism does it have?
For me I only know
1) a slow step followed by a fast step
2) a fast reversible first step followed by a sloe step

is there any more?

For the reaction like
2NO+O2> 2NO2

how can i know the rate of reaction is K[NO]^2[O2]?
also
the reaction is lead by the slow reaction one right?
if the question only give me the overall reaction and slow/fast step
how can i determine the rest one?

o
now im college level in USA
i don't know what is the difference with AL level in HK
What Im studying now is only general chemistry 2
here example come:

1) The reaction os 2NO+2H2>N2+2H2O is a second in [NO] and first order in [H2]. A three step mechanism has been proposed. The first, fast step is the elementart process : 2NO<> N2O2. The third step ,also fast is N2O+ H2>N2+H2O. Propose an entire threee step mechanism and show that it confirms to the experimentally determined reaction order.

2) The mechanism proposed for the reaction of H2 and I2 gas to form HI consists of a fast reversible first step involving I2 and I, followed by a slow step. Propose a two step mechanism for the reaction H2+I2>2HI which is known to be first order in H2 and first order in I2                                                                                                                                                                                                                                                                                                                       

1. The proposed three-step mechanism for 2NO + 2H2 ---> N2 + 2H2O is as follows:
2NO <---->  N2O2  (step 1; fast equilibrium)
N2O2 + H2  ---->  N2O + H2O  (step 2; slow)
N2O + H2  ---->  N2 + H2O (step 3; fast)
Since the second step is rate-determining, the overall reaction rate is governed by the rate of this step. Therefore,
-(1/2)d[NO]/dt = k2[N2O2][H2]   ....
Note that the first step is a fast equilibrium process, and
k1[NO]^2 = k-1 [N2O2]  ....
where k1 and k-1 are the forward and backward rate constants. Substituting  into  we obtain
-(1/2) d[NO]/dt = k2*(k1/k-1)[NO]^2 [H2]
-d[NO]/dt = 2k2*(k1/k-1) [NO]^2[H2] = k[NO]^2[H2]
where k = 2k1k2/k-1 is the effective rate constant.

2. This question can be solved using the same technique.
The proposed mechanism is
I2  <---->  2I  (step 1; fast equilibrium)
2I + H2  ---->  2HI  (step 2; slow)
Since the second step is rate-determining, then
-d[I2]/dt = k2[I]^2 [H2]   ....
From step 1, we know that
k1[I2] = k-1[I]^2  ....
where k1 and k-1 are the forward and backward rate constants. Putting  into  we get
-d[I2]/dt = k2*(k1/k-1) [I2][H2] = k[I2][H2]  ....
where k = k1k2/k-1 is the effective rate constant.
We can easily see that rate expression  is first-order w.r.t. [I2] and [H2], respectively, as required.

不太明白
什麼來呢?
-(1/2)d[NO]/dt = k2[N2O2][H2]   ....                                                                                                                                                                                                                                                                                                                                                                                

um nvm i got how you do it now ty
but still some more coming

o no still have question when i go on study
A simplified rate law for the reaction 2O3>3O2 is

rate =k =[O3]^2/[O2]

for this reaction propose a 2 step mechanism that consists of a fast reversible first step followed by a slow second step.

like this kind of question
how can i know the step?

I will propose the following mechanism:
O3  <---->  O2 + O (step 1; fast equilibrium)
O3 + O  ---->  2O2 (step 2; slow)
The rate of the reaction can be described as:
rate = k2[O3][O]  ....
since the second step is rate-determining. Now, using the first equilibrium, we can write
k1[O3] = k-1[O2][O]
[O] = k1[O3]/k-1[O2]  ....
Substituting  into , we will get the rate expression
rate = k2[O3](k1[O3]/k-1[O2]) = k1k2[O3]^2/k-1[O2]
or
rate = k[O3]^2/[O2]
where k = k1k2/k-1.

Whenever you see the inverse dependence of the rate on certain species, it is likely that the species is involved in the reverse process of the pre-equilibrium step.                                                                                                                                                                                                                                                                                                                       

got it
but how about these question
show me 1 and let me try rest
i will ask u if I don;t get them

http://hk.geocities.com/kennyyip2004/IMG_0001.jpg
http://hk.geocities.com/kennyyip2004/IMG_0002.jpg
http://hk.geocities.com/kennyyip2004/IMG_0003.jpg

sry that the file is too big to upload..

Which question? 6, 7 or 8?

question 8
ty u so much                                                                                                                                                                                                                                                                                                                       

OK, let me try  of question 8.

Remember that the rate expression is always determined, in approximation, by the slow reaction step, i.e., the rate-limiting step.

A. A single-step reaction, and thus the rate expression is
rate = k[CO][NO2]

B. The given reaction step is a rapid equilibrium. Using only this we cannot formulate the rate expression. Hence I assume that the following slow reaction is as such:
N2O4 + 2CO  ----> 2CO2 +2NO
The rate expression is thus rate = k2[N2O4][CO]^2
Note that k1[NO2]^2 = k-1[N2O4]
So, rate = (k1k2/k-1)[NO]^2[CO]^2

C. Since the slow reaction is 2NO2  ---->  N2 + 2O2, the rate expression is rate = k[NO2]^2

D. Since there are two slow steps in the mechanism, the overall rate of reaction is governed by both steps, and the rate expression is thus
rate = k1[N2O4][CO]^2 + k2[NO2]^2

So, you can see that only the mechanism C yields the rate expression in the required form of rate expression.