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Chem 1310: Introduction to physical chemistry Part 2b: rates

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Average and instantaneous rates. The average rate over a larger interval. differs significantly from the instantaneous rates. Following the progress. of the reaction ... – PowerPoint PPT presentation

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Title: Chem 1310: Introduction to physical chemistry Part 2b: rates


1
Chem 1310 Introduction to physical chemistry
Part 2b rates
2
Kinetics
  • How fast does a reaction go? (and why?)
  • Solution of crystal violet poured into solution
    of NaOH.

3
What is crystal violet?
  • Large organic molecule, deep purple.
  • Also known as "Gram's stain", used to distinguish
    types of bacteria ("Gram-negative" and
    "Gram-positive").
  • Disinfectant and toxic! Do not get on skin!

4
What is crystal violet?

CV
5
How does it react with OH- ?
-


CV
OH-
6
How does it react with OH- ?
-


CV
OH-
CVOH
deep purple
colorless
7
Following the progressof the reaction
  • Visually color changespurple pink
    colorless
  • Quantitative colorimetrymeasure transmitted
    light

8
Colorimetry
  • Law of Lambert-Beer
  • more convenient
  • use Absorbance A

I0
It
9
Following the progressof the reaction
t CV
s mol/L

0.0 5.00E-05

10.0 3.68E-05

20.0 2.71E-05

30.0 1.99E-05

40.0 1.46E-05

50.0 1.08E-05

60.0 7.93E-06

80.0 4.29E-06

100.0 2.32E-06
10
What is a rate ?
  • The rate of a chemical reaction is the speed at
    which it transforms reagents into products.
  • It is a measure of how fast things change with
    time.
  • Compare with the speed of a car measures how
    fast its position changes with time.
  • In chemistry we look at concentrations and
    changes in them.

A heterogeneous reaction
11
Calculating rates
note the "-" sign!
Dt large average rate over interval Dt Dt very
small instantaneous rate
average rate overt 10 .. 20 s
12
Average and instantaneous rates
  • Decrease of concentration follows a smooth curve.
  • At each point, the rate is the (negative of the)
    slope of the curve.

13
Average and instantaneous rates
  • We cannot measure rates directly.
  • We can only measure concentrations.
  • Rates can be estimated
  • by drawing a smooth curve and estimating the
    slopes (tangents)
  • by using average rates over real (but small) time
    intervals

14
Average and instantaneous rates
  • The instantaneous rate is givenby the slope
    (tangent)

15
Average and instantaneous rates
  • The average rate over a larger intervaldiffers
    significantly from the instantaneous rates.

16
Following the progressof the reaction
t CV av rate
s mol/L mol L-1s-1

0.0 5.00E-05
1.32E-06
10.0 3.68E-05
9.70E-07
20.0 2.71E-05
7.20E-07
30.0 1.99E-05
5.30E-07
40.0 1.46E-05
3.80E-07
50.0 1.08E-05
2.87E-07
60.0 7.93E-06
1.82E-07
80.0 4.29E-06
9.85E-08
100.0 2.32E-06
17
Rates depend on concentrations
  • From the table for Crystal Violetwe can plot
    rate vs concentration

Looks pretty linear. The rate law is rate k
CV with k 2.6810-2 s-1 (the rate
constant).
Another rate law determination
18
It is not always this easy
  • Here we could get a set of rates from a single
    experiment.
  • Often this cannot be done you have to stop the
    reaction to analyze the results.
  • Start reaction, stop it after a short interval,
    analyze.
  • Do this for a number of initial concentrationsÞ
    obtain a number of initial rates.
  • Need lots of experiments to get accurate curves!

19
Using initial rates
Initial conc Conc after 1 s Initial rate
5.00E-05 4.89E-05 1.14E-06
4.00E-05 3.91E-05 9.10E-07
3.00E-05 2.93E-05 6.82E-07
2.50E-05 2.44E-05 5.68E-07
2.00E-05 1.95E-05 4.55E-07
1.50E-05 1.47E-05 3.41E-07
1.00E-05 9.77E-06 2.27E-07
20
More complicated rate laws
  • 2 NO O2 2 NO2

expt NO O2 initial rate
1 0.02 0.01 0.028
2 0.02 0.02 0.057
3 0.02 0.04 0.114
4 0.04 0.02 0.227
5 0.01 0.02 0.014
21
More complicated rate laws
  • 2 NO O2 2 NO2

expt NO O2 initial rate
1 0.02 0.01 0.028
2 0.02 0.02 0.057
3 0.02 0.04 0.114
4 0.04 0.02 0.227
5 0.01 0.02 0.014
linear in O2
rate µ O2
22
More complicated rate laws
  • 2 NO O2 2 NO2

expt NO O2 initial rate
1 0.02 0.01 0.028
2 0.02 0.02 0.057
3 0.02 0.04 0.114
4 0.04 0.02 0.227
5 0.01 0.02 0.014
definitely not linear in NO !
quadratic ???
23
More complicated rate laws
  • 2 NO O2 2 NO2

expt NO O2 initial rate
1 0.02 0.01 0.028
2 0.02 0.02 0.057
3 0.02 0.04 0.114
4 0.04 0.02 0.227
5 0.01 0.02 0.014
quadratic (second-order) in NO
rate µ NO2
24
Putting it all together
  • So we have
  • rate µ O2 (NO constant)
  • rate µ NO2 (O2 constant)
  • Combining these gives
  • rate k O2NO2
  • with
  • k 7069 L2mol-2s-1
  • Reaction is
  • first-order in O2
  • second-order in NO
  • third-order overall
  • 2 NO O2 2 NO2

expt NO O2 initial rate est k
1 0.02 0.01 0.028 7000
2 0.02 0.02 0.057 7125
3 0.02 0.04 0.114 7125
4 0.04 0.02 0.227 7094
5 0.01 0.02 0.014 7000
est k rate/(O2NO2)
25
What do we mean by "the rate" ?
  • 2 NO O2 2 NO2
  • For every single molecule of O2, two molecules of
    NO are consumed and two molecules of NO2 are
    produced.
  • "The rate" of the reaction is defined
    ascontains every component divided by its
    coefficient in the balanced reaction equation.

26
Stoichiometry and rate laws
  • Rate laws can not be deduced from the balanced
    reaction equation!
  • They must be measured experimentally.
  • The results are not always intuitive.
  • NO2 CO NO CO2
  • rate k NO22

Consequences of a rate law
27
Stoichiometry and rate laws
  • You can even have non-integer orders

trans-2-butene
cis-2-butene
rate(cistrans) k cis-2-buteneI2½ first-ord
er in cis-2-butene half-order in I2
A more complicated rate law
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