I have a reaction that has three elementary steps: 3I - +S 2 O 8 2- ---> I 3 - +

Question

I have a reaction that has three elementary steps:

3I- +S2O82- ---> I3- + 2 SO42- (this is the RLS)
I3 + 2 S2O32- ---> 3 I- + S4O62-
starch + I3- ---> starch - I3- complex. (color change to indicate rxn is over)

In multiple trials I change the concentrations to see its effect on the rate of the rxn. But the reagents I'm using in this rxn are: (1) starch, Na2S2O3 , KI, KNO3 , and (2) (NH4)2S2O8 , (NH4)2SO4

(1) and (2) are mixed together in separate vials before we mix both to start the rxn. Different trials have different volumes of the above reagents so there's a difference in concentrations between trials and we can therefore see the effect this has on rate. We time each rxn with a stopwatch.

I need to calculate the initial [I-] and [S2O82-] by using the equation M1V1 = M2V2. However I'm completely confused how to do this. V1 is initial volume and V2 is total volume. I'm also given the molarity of KI and (NH4)2S2O8. Do I need to factor in that my reagents are not actually S2O82- or I- but actually (NH4)2S2O8 and KI?

For example, trial 1's reagents are:

Starch - .2% - .10 mL
Na2S2O3 - .20 mL of .012 M
KI - .80 mL of .20 M
KNO3 - 0 mL of .20 M

(NH4)2S2O8 - .40 mL of .20 M
(NH4)2SO4 - .40 mL of .20 M

What is the intial [I-] and [S2O82-] for this trial?

Explanation / Answer

We know,

Molarity = moles / Litres of solution

Work out how many moles of each reagent you add to the initial solution.

Then use the total volume of the initial solution to determine the molarity.

eg trial 1

total volume = 0.10 ml + 0.12 ml + 0.80 ml + 0 mL + 0.40 mL + 0.40 mL

                     = 1.82 mL = 0.00182 L

moles of [S2O8^2-] = moles in the 0.40 ml (or 0.0004 L) of solution used             

                          = molarity x Litres

                          = 0.2 M x 0.0004 L

                          = 8 x 10^-5 moles S2O8^2-

Initial concentration of [S2O8^2-] = moles added / total volume initial solution

                                   = 8 x 10^-5 mol / 0.00182 L

                                   = 0.044 M [S2O8^2-]

Similarly,

moles of [I-] = moles in the 0.80 ml (or 0.0008 L) of solution used             

                          = molarity x Litres

                          = 0.2 M x 0.0008 L

                          = 1.6 x 10^-4 moles [I-]

Initial concentration of [I-] = moles added / total volume initial solution

                                   = 1.6 x 10^-4 mol / 0.00182 L

                                   = 0.088 M [I-]

Thus the intial concentrations of [S2O82-] = 0.044 M and [I-] = 0.088 M

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