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Trial [IO 3 - ] 0 mmol/L [I - ] 0 mmol/L [H + ] 0 mmol/L Rate I 2 production mmol/L/s 10.10 5.0 x 10 -4 20.200.10 1.0 x 10 -3 30.100.300.101.5 x 10 -3 40.100.300.206.0 x 10 -3 Rate Law Determination For the reaction: IO 3 - + 5I - + 6H + 3I 2 + 3H 2 O The following data were collected:
![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:](http://images.slideplayer.com/17/3389115/slides/slide_1.jpg "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:")
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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.
![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.](http://images.slideplayer.com/17/3389115/slides/slide_2.jpg "Compare the change in concentration to the change in rate..")
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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 10.10 5.0 x 10 -4 20.200.10 1.0 x 10 -3 30.100.300.101.5 x 10 -3 40.100.300.206.0 x 10 -3
![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](http://images.slideplayer.com/17/3389115/slides/slide_3.jpg "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")
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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 10.10 5.0 x 10 -4 20.200.10 1.0 x 10 -3 3 0.100.300.101.5 x 10 -3 40.100.300.206.0 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](http://images.slideplayer.com/17/3389115/slides/slide_4.jpg "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")
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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 10.10 5.0 x 10 -4 20.200.10 1.0 x 10 -3 30.100.300.101.5 x 10 -3 40.100.300.206.0 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](http://images.slideplayer.com/17/3389115/slides/slide_5.jpg "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")
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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 10.10 5.0 x 10 -4 20.200.10 1.0 x 10 -3 30.100.300.101.5 x 10 -3 40.100.300.206.0 x 10 -3
![Rate law expression: Rate = k[IO 3 - ] 1 [I - ] 1 [H + ] 2 What about k.](http://images.slideplayer.com/17/3389115/slides/slide_6.jpg "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")
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