Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s x x x x Rate Law Determination For the reaction: IO I - + 6H +  3I 2 + 3H 2 O The following data were collected:

Determine the Rate Law Recall: Rate = k[IO 3 - ] m [I - ] n [H + ] p Compare trials, changing the concentration of one substance at a time. Compare the change in concentration to the change in rate.

Trial 1 & 2 When the concentration is doubled (x2) the rate increases by a factor of 2 1 Therefore, in rate law: [IO 3 - ] 1 Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s x x x x 10 -3

Trial 1 & 3 When the concentration is increased by a factor of 3 (x3), the rate increases by a factor of 3 1 Therefore, in rate law: [I - ] 1 Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s x x x x 10 -3

Trial 3 & 4 When the concentration is doubled (x2) the rate increases by a factor of 4 = 2 2 Therefore, in rate law: [H + ] 2 Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s x x x x 10 -3

Rate law expression: Rate = k[IO 3 - ] 1 [I - ] 1 [H + ] 2 What about k? Select one of the trials and calculate k. Since only the concentrations changed, k will be a constant. Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s x x x x 10 -3