Calculate the average value for the rate constant from the given data.

experiment 1
[I-] = 0.075 M
[S2O82-] = 0.0375 M
[S2O32-] = 0.00125 M
reaction time = 78 seconds
rate (mol/L*s) = 0.000008013

experiment 2
[I-] = 0.075 M
[S2O82-] = 0.025 M
[S2O32-] = 0.00125 M
reaction time = 108 seconds
rate (mol/L*s) = 0.000005787

How would I start the problem?
I know that [I-] and [S2O82-] are first order reactions. I assume that I'm supposed to use the formula ln([A]/[A]initial) = -kt but I don't know [A] final.

Question

Calculate the average value for the rate constant from the given data.

experiment 1
[I-] = 0.075 M
[S2O82-] = 0.0375 M
[S2O32-] = 0.00125 M
reaction time = 78 seconds
rate (mol/L*s) = 0.000008013

experiment 2
[I-] = 0.075 M
[S2O82-] = 0.025 M
[S2O32-] = 0.00125 M
reaction time = 108 seconds
rate (mol/L*s) = 0.000005787

How would I start the problem?
I know that [I-] and [S2O82-] are first order reactions. I assume that I'm supposed to use the formula ln([A]/[A]initial) = -kt but I don't know [A] final.

aaronq · Answer

what is the reaction?

lissiewalkin · Answer

initial reaction 2I- + S2O82- -> I2 + SO42-

final reaction is I2 + 2 S2O32- -> 2I- + S4O62-

aaronq · Answer

you can find the final amount of a given reactant (use the rate and subtract the amount that was used) then use the formula you posted (integrated first order) to find k.

lissiewalkin · Answer

Oh ok I get what you're saying. But how do I find the amount used?

aaronq · Answer

reaction time = 78 seconds
rate (mol/L*s) = 0.000008013

reaction time*rate= 78 seconds*0.000008013 mol/L*s= amount consumed.

lissiewalkin · Answer

ohhh so I got 0.000625014mol/L as the amount consumed.

Do I just plug this number into the integrated first order formula to find K?

aaronq · Answer

you would subtract that from the initial amount, then plug into the formula.

lissiewalkin · Answer

i got ln (0.000625014)/(0.000008013) = - k(78)

k= -0.055855241

Does that sound correct?

aaronq · Answer

no, you're not using the right parameters. You need to subtract the amount consumed from the initial amount (of the reactant of your choice).

lissiewalkin · Answer

Oh i see what i did wrong thanks for being patient with me.

so i chose 0.075 M of [I-] as my initial amount 

 it would be 0.075M - 0.000625014M = 0.074374986M

and i would plug it in so its ln (0.074374986M/0.075M) = -k(78)?

aaronq · Answer

yes, exactly. Though it seems that you would have to multiply it by 2 (look at the stoichiometric coefficient) before you subtract it.

aaronq · Answer

0.075M - (2)*0.000625014M

lissiewalkin · Answer

ohh it all makes sense now! Thank you so so much aaronq!

aaronq · Answer

glad it makes sense. no problem !