[ILoveISO]

From the slope of the line and the stoichiometry of the reactions involved, calculate for each run the rate of reaction of (NH4)2S2O8 in moles/sec. Using these values and the avg volume of the reaction mixture during the run 50.00ml, determine for each run the rate of reaction of (NH4)2S2O8 in mole/litre/sec

For calculating the reaction rate in moles/sec do I need any fancy equations to do so? Like rate = k[A] that formula?

I'm not sure what the question is saying when it says from the slope of the line and stoichiometry of the reactions involved, for example what if the slope was 1.15x10^-5 what do I do from there? Here is a set of data from my first run
 
Time                Moles consumed

50s                  0.002
163s                 0.004
250s                 0.006
390s                 0.008
487s                 0.01

[ILoveISO]

I have the slope of S2O3 but I need to find the rxn rate of S2O8... how do I link them together thru stoich???

[Borek]

Write reaction equation.

[ILoveISO]

I have two.. which one to use?

S2O8 + 2I --> I2 + 2SO4

2S2O8 + I2 --> 2I + S4O6

[Borek]

The one which combines S₂O₃^2- with S₂O₈^2-.

Yes, you have not listed such reaction. No, I don't know why.

[ILoveISO]

It says the slope of this line corresponds to the avg number of moles of s2o3 that are being consumed per second and is proportional to the rate of rxn of s2o8 in moles per second at that time. I don't get what this means =/ can someone explain?

[ILoveISO]

I am assuming that the two compounds are related in a stoichiometric way (either x moles of s2o3 is reacting with y moles of s2o8, or x moles of s2o3 is required to produce y moles of s2o8 in a reaction). For simplicity, I will explain it in terms of s2o3 producing s2o8, so it can be said that y moles of s2o8 are produced for every x moles of s2o3 spent. Because the number of moles of s2o8 produced can be expressed as some number times the number of moles of s2o3, we can treat it as a function y = kx where y is the moles of s2o8, k is the mole ratio and x is the moles of s2o3. In this case, the slope of the line defines the rate of change of x, so by differentiating the function you should get y' = k(slope of line) where y' is the rate of change in the number of moles of s2o8 and k is still the mole ratio.

Is this correct?

[ILoveISO]

Anyone...?