[Sis290025]

The following set of data was obtained by the method of initial rates for the reaction:
S2O8(2-) (aq) + 3I- (aq) --> 2SO4(2-) + I3- (aq)

What is the initial rate when S2O82- is 0.15 M and I- is 0.15 M?

Exp--------[S2O8(2-)], M----------------------[I-], M--------------Initial Rate, M/s
1-----------0.25--------------------------------0.10------------------9.00E-3
2-----------0.10--------------------------------0.10------------------3.60E-3
3-----------0.20--------------------------------0.30------------------2.16E-2



  a. 4.10 x 10-6 M s-1    
  b. 8.10 x 10-3 M s-1    
  c. 1.22 x 10-2 M s-1    
  d. 5.40 x 10-2 M s-1    


I think that the correct choice is d, but I am not even certain that it is the correct answer!

This is what I get:

The Rate Law is Rate = k[S2O8(2-)] ?

So find k.

k = rate/[S2O8(2-)] = (9.00E-3 M/s)/(0.25 M) = 0.0365 1/s

Rate = (0.0365 !/s)(0.15 M) = 0.0054 M/s = 5.40E-3


Did I do an incorrect step?

Thanks.

[plu]

The Rate Law is Rate = k[S2O8(2-)] ?

Look at experiments 2 and 3 again.  You are doubling the concentration of thiosulfate and tripling the concentration of iodate.  The rate increases by a factor of 6.  Since you can see from experiments 1 and 2 that the rate is first order in thiosulfate, there is no way the rate is zeroth order in iodate.

[Sis290025]

B is the answer?

Rate = k[a]