Reaction Rates

[D-chris91]

Can someone please explain how orders of reaction w.r.t different reactants can be predicted by the speed of a reaction?
Imagine this question
In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt
and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide
ions in acidic solution.
H2O2 + 2I– + 2H+ 2H2O + I2
This reaction is considered to go by the following steps.
step 1 H2O2 + I– IO– + H2O
step 2 IO– + H+ HOI
step 3 HOI + H+ + I– I2 + H2O
The general form of the rate equation is as follows.
rate = k[H2O2]a[I–]b[H+]c
(b) Suggest values for the orders a, b and c in the rate equation for each of the following
cases.
CASE                                  a                        b                      c
step 1 is the slowest overall
step 2 is the slowest overall
step 3 is the slowest overall
[3]
Thanks. This is the 2nd question from Oct/Nov 2008 past Chemistry P4 paper.

[Gaz]

I tried working it out by the way I have learned this and had some problems. When I checked from the marking scheme, I noticed a 'pattern'. The answer is:
Case 1    1   1   0  
Case 2    1   1   1                            
Case 3    1   2   2
 
In the first case, its is easy to determine the values as only H2O2 and I- are involved and are 1 moles.
In the second case, H+ is obvious, and the I- and O from H2O2 are present together in the form IO-. Hence all 3 have values of 1.
In the third case, the same rule applies as in the second case. 2H are present as H- and in HOI. One O and 2I. Hence a ratio of 1 2 2.
I hope this helps.

[abuelzouz]

edexcel or cie?

[Light]

welll actually it is easy 2 see.for case 1,step 1 has H2O2 AND I- but no H+ .so 110.
if step 2 is slow,step 1 will also be slow.from both equations,H202, I- appear once in step 1 and H+ appear once in step 2.so 111.
if step 3 is slow, step 1 and 2 wil be slow. by seeing all equations,H+ and I- appear twice but H2O2 appear only once.so 122.