240 likes | 317 Views
Kinetics. From properties of materials to Reaction Chemistry. 2 levels of study, for 2 reasons:. Macroscopic view: How fast? Rates of reaction typically as D concentration per time. Microscopic view: What path? Mechanism sequence of chemical steps, to control it.
E N D
Kinetics From properties of materials to Reaction Chemistry 2 levels of study, for 2 reasons: Macroscopic view: How fast? Rates of reaction typically as Dconcentration per time Microscopic view: What path? Mechanism sequence of chemical steps, to control it Let’s try it…
H2O2 decomposition reaction 2 H2O2 (aq) --> O2(g) + 2 H2O
H2O2 decomposition reaction 2 H2O2 (aq) --> O2(g) + 2 H2O • reaction progress affected by: • KI faster • KCl no effect • Fe(3+) faster • Cu(2+) faster • Also: • higher [H2O2] faster • higher [KI] faster
How to express ‘faster’ and ‘slower’? • Rate = D M / D time (for solutions) • so units of reaction Rate in M/sec or • M sec-1 or mol L-1 sec-1 • conventions: • Rate is positive – so disappearance of peroxide has • negative rate 2 H2O2 (aq) --> O2(g) + 2 H2O
How to express ‘faster’ and ‘slower’? conventions: Rate = D M / D time describes Average Rate [H2O2]o Time, sec
How to express ‘faster’ and ‘slower’? conventions: Instantaneous Rate: measured over infinitely small times, a differential function: Rate = d M / d time more precise [H2O2]o Time, sec
How to express ‘faster’ and ‘slower’? conventions: Rate depends on stoichiometry 2 H2O2 (aq) --> O2(g) + 2 H2O More generally, for: A + B C + D Rate = + D[C]/Dt = + D[D]/Dt = - D[A]/Dt = - D[B]/Dt In units mol / L . sec
conventions: Initial Rates depend on initial concentrations [H2O2]o Time, sec
Experiment to obtain kinetic data to measure H2O2 decomposition
Let’s use these conventions and look at some real data for the peroxide decomposition
Data: O2 pressure as H2O2 decomposes over 10 min Note: Non-linear Plot means Rate not the same at beginning and at end
For small time plot of data is nearly linear, so D pO2 /D time approaches Instantaneous Rate In first 0.10 sec, the pressure goes from 102.74 to 102.91 kPA D pO2 / D time = Rate (102.91 - 102.74) kPA / 0.10 min Rate = 1.7 kPA / min = slope
For small time plot of data is nearly linear, so D pO2 /D time approaches Instantaneous Rate In first 0.10 sec, the Rate = 1.7 kPA / min After 4 min, a 0.10 sec interval shows the pressure goes from 107.49 to 107.59 kPA DpO2 / Dtime = Rate (107.59 - 107.49 ) kPA / 0.10 min Rate = 1.0 kPA / min = slope
Data on how the rate of H2O2 decomposition is affected by varying the initial [H2O2] 2X 2X 4.1 X 4.1 X Initial [H2O2] is related to rates. What does a plot of [H2O2]o vs Rate look like?
7.5 x10-5 M/sec/0.085 M = 8.8 x10-4 sec-1 1.4 x10-4 / 0.17 = 8.2 x10-4 sec-1 2.15 x10-4 / .25 = 8.6 x10-4 sec-1 2.9 x10-4 / .35 = 8.3 x10-4 sec-1 Rate is proportional to [H2O2]o: Then: Rate = k[H2O2]o or Rate / [H2O2]o = k units: M / s M-1 = sec-1 Average k = 8.5 x10-4 sec-1 from a line fitting of data: Rate constant, k = 8.3 x10-4 sec-1
2X 2X 4.1 X 4.1 X Data on how the rate of H2O2 decomposition is affected by varying the Initial [I-] values.
So Rate depends on [H2O2]o : Raterxn = k [H2O2]o AND Rate depends on [KI] o : Raterxn = k* [KI]o Overall, Rate depends on two parameters: Raterxn = k’ [H2O2]o [KI]owhere k’= k k* And we saythe overall reaction is Second Order, 2o, First order, 1o, in H2O2 and First order, 1o, in KI
This expression where both dependences are written: Raterxn = k’ [H2O2]o [KI]o is the Rate law. The Rate Law is the reason Kinetics studies are done: It shows us the slowest step in reaction sequence: the Rate Determining Step, r.d.s.