[777888]

1.Cu(2+) + I- -> Cu+ + I  (fast)
2I -> I2 (fast)
Cu+ + S2O8(2-) -> CuSO4+ + SO4(2-) (slow)
Cu+ + CuSO4+ -> 2Cu(2+) + SO4(2-)  (fast)

a)identify the catalyst and the reaction intermediates in this rxn.
b)write the overall rxn.
c) what's the effect of increasing [S2O8(2-)]?

My thoughts:
a)Catalyst: Cu(2+)
Rxn intermediates: Cu+, I
b)Is it OK to multiply the whole first equation by 2?
c)rate increases

Can someone see if I am right please?

2.For the elementary step 2{A2+2B->2AB}, what's the rate law equation?

My thoughts:
r=k[A2]^2
r=k[A2]^2^4

I am not sure which one is right...would the 2 in 2{..} make a difference?

Thank you!

[Demotivator]

1) a.  CuSO4+ is also intermediate
   b. yes, to obtain a balanced overall rxn.
   c. correct

2) I've never seen {...} in a step and wonder if this is a trick question. If the external 2 is just extra algebra and not part of the actual mechanistic step, then r=k[A2]^2

[777888]

1) a.  CuSO4+ is also intermediate
   b. yes, to obtain a balanced overall rxn.
   c. correct

2) I've never seen {...} in a step and wonder if this is a trick question. If the external 2 is just extra algebra and not part of the actual mechanistic step, then r=k[A2]^2

There is a series of elementary step equations...I think the 2{ } is using to balance the overall equation...

Like Q1,
2{Cu(2+) + I- -> Cu+ + I  (fast)}
2I -> I2 (fast)
Cu+ + S2O8(2-) -> CuSO4+ + SO4(2-) (slow)
Cu+ + CuSO4+ -> 2Cu(2+) + SO4(2-)  (fast)

and when it ask for what's the rate law equation for the first elementary step? do I have to care about the 2 { }?

Thanks!

[Demotivator]

No, the coefficients used for overall balancing are not involved.

[Kong]

Realistically the rate is determined by the addtion of all the steps up to and including the slowest step.  Deleting any intermediates.   For example,
A + 2B = C + D (fast)
A + C = D (slow)
E + D = Z  (fast)
In this case the 1st and second elementary steps will control the overall rate.  
Rate = k[A]^2^2    C is not included because it is an intermediate.  Every after the slowest step is irrelavent as well.
AS