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!
2 answers
And what's wrong with multiplying them by k? Looks to me as if you're on the right track.
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
2 answers