kinetics macroscopic view: how fast? rates of reaction typically as concentration per time from...
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Kinetics
Macroscopic view: How fast? Rates of reaction
typically as concentration per time
From properties of materials to Reaction Chemistry
Microscopic view: What path? Mechanism
sequence of chemical steps, to control it
2 levels of study, for 2 reasons:
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:
o KI faster
o KCl no effect
o Fe(3+) faster
o Cu(2+) faster
Also:
higher [H2O2] faster
higher [KI] faster
How to express ‘faster’ and ‘slower’?
Rate = M / time (for solutions)
so units of reaction Rate in M/sec or
M sec-1 or mol L-1 sec-1conventions:
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 = M / 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 = M / 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 = + C]/t = + D]/t = - A]/t = - B]/t
In units mol / L . sec
[H2O2]o
Time, sec
conventions:
Initial Rates depend on initial concentrations
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
Plotmeans
Rate not the
same atbeginning
and at end
For small time plot of data is nearly linear, so pO2 /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
pO2 / time = Rate
(107.59 - 107.49 ) kPA / 0.10 min
Rate = 1.0 kPA / min = slope
For small time plot of data is nearly linear, so pO2 /time approaches Instantaneous Rate
Data on how the rate of H2O2 decomposition is affected by varying the initial [H2O2]
2X 2X4.1 X
4.1 X
Initial [H2O2] is related to rates.
What does a plot of [H2O2]o vs Rate look like?
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
2.9 x10-4 / .35 = 8.3 x10-4 sec-1
2.15 x10-4 / .25 = 8.6 x10-4 sec-1
1.4 x10-4 / 0.17 = 8.2 x10-4 sec-1
7.5 x10-5 M/sec /0.085 M = 8.8 x10-4 sec-1
Rate is proportional to [H2O2]o:
Data on how the rate of H2O2 decomposition is affected by varying the Initial [I-] values.
2X4.1 X
4.1 X
2X
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]o where k’= k k*
And we say the overall reaction is Second Order, 2o, First order, 1o, in H2O2 andFirst 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.
Obtaining Rate Constants from Kinetic Data
Examples of Plots of Different Reaction Orders
Integrated Rate Laws