Kinetics
Reaction Chemistry
2 levels of study, for 2 reasons:
From properties of materials to
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
The Bombardier beetle in action
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 = D M / D time (for solutions)
so units of reaction Rate in M/sec or
conventions:
sec-1
M sec-1 or mol L-1
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]
4.1 X
2X
2X
4.1 X
Initial [H2O2] is related to rates.
What does a plot of [H2O2]o vs Rate look like?
Rate is proportional to [H2O2]o:
Then:
Rate = k[H2O2]o
or
Rate / [H2O2]o = k
units: M / s M-1 = sec-1
7.5 x10-5 M/sec /0.085 M = 8.8 x10-4 se
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
Average k = 8.5 x10-4 sec-1
from a line fitting of data:
Rate constant, k = 8.3 x10-4 sec
Data on how the rate of H2O2 decomposition is affected by varying the
Initial [I-] values.
4.1 X
2X
2X
4.1 X
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 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.
Obtaining Rate Constants from Kinetic Data
Examples of Plots of Different Reaction Orders
Integrated Rate Laws