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