Question
A student finds that k = 0.0027 L/mol · s, and m and n are both 1. Calculate the expected rate if 6.0 mL of 0.10 M [S2O82− ] is mixed with 4.0 mL of 0.20 M [I − ]. Assume the final volume of the solution is 10.0 mL.
I used the given molarities and multiplied them by their volumes to find the total moles then I divided by the final volume to get the molarity to multiply with k. Please help!
I used the given molarities and multiplied them by their volumes to find the total moles then I divided by the final volume to get the molarity to multiply with k. Please help!
Answers
DrBob222
And what's wrong with multiplying them by k? Looks to me as if you're on the right track.
Olereb48
Rate Law with both reactants 1st order:
Rate = (0.0027L/mol-s)[S208^2-][I^-]
[S2O8^2-] = (0.006L)(0.10M)/(0.010L)
= 0.06M
[I^-] = (0.004L)(0.20M)/(0.010L)
= 0.08M
Rate = (0.0027L/mol-s)(0.06mol/L)(0.08mol/L) = 1.3x10^-5 mol/L-s
Rate = (0.0027L/mol-s)[S208^2-][I^-]
[S2O8^2-] = (0.006L)(0.10M)/(0.010L)
= 0.06M
[I^-] = (0.004L)(0.20M)/(0.010L)
= 0.08M
Rate = (0.0027L/mol-s)(0.06mol/L)(0.08mol/L) = 1.3x10^-5 mol/L-s