Maxwell-Boltzmann Distribution Curves

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Maxwell-Boltzmann Distribution Curves

• Mrs. Kay
• Chem 12 A

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• In a sample gas or liquid, the molecules are in
  constant motion and collision.
• When they do this, no energy is lost from the
  system - theyre elastic collisions.
• Each molecule varies in its own energy, some have
  more, some less.

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• A Maxwell-Boltzmann distribution curve/graph
  shows how the energy was spread out over
  different molecules

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Remember

• the area under the curve the total molecules in
  the sample, and it doesnt change
• All molecules have energy
• Few molecules have high energies, but there is no
  maximum energy for a molecule.

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• If you increase the temperature of the sample,
  the distribution becomes stretched out and has a
  lower peak, but the area under the curve remains
  the same.

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• To speed up the reaction, you need to increase
  the number of the very energetic particles -
  those with energies equal to or greater than the
  activation energy. Increasing the temperature has
  exactly that effect - it changes the shape of the
  graph.

/ Energy
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Molecularity and Rate Law
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Molecularity

• The number of molecules that participate as
  reactants in an elementary step
• Unimolecular a single molecule is involved.
• Ex CH3NC (can be rearranged)
• Bimolecular Involves the collision of two
  molecules
• NO O3 ? NO2 O2
• Termolecular simultaneous collision of three
  molecules. Far less probable.

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Rate Law

• Math expression to show how rate depends on
  concentration
• Rate kreactant 1mreactant 2n
• m and n are called reaction orders. Their sum is
  called the overall reaction order.
• k is the rate constant. It is specific to a
  reaction at a certain temperature.
• The value of the exponents are determined
  experimentally, we cannot use stoichiometry!

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Sample Exercise 1

• 2NO 2H2 ? N2 2H2O is the reaction were
  studying, this is the data found during our
  experimentation

Experiment Number Conc. of NO (M) Conc. of H2 (M) Rate of N2 forming (M/s)
1 0.210 0.122 0.0339
2 0.210 0.244 0.0678
3 0.420 0.122 0.1356
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Study the data

• Exp 1 and 2, NO is unchanged, H2 is doubled
  and this causes the rate to double (0.0678/0.0339
  2), the H2s rate order is 1.
• Exp 1 and 3, H2 is unchanged, NO is doubled
  and this causes the rate to quadruple
  (0.1356/0.0339 4), NOs rate order is 2.
• Rate k NO2H2
• The overall rate order for this reaction is 3
  (12)

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Suppose

• If when we ran exp 1 and 3 , the rate didnt
  change, what rate law would we expect?
• Since it didnt change when we doubled NO, the
  rate order is 0, meaning the rate doesnt depend
  on the concentration of NO at all, so rate kH2
  and the overall reaction order is 1.

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Sample Exercise 2

• Use the kinetics data to write the rate law for
  the reaction. What overall reaction order is
  this? 2NO O2 ?2NO2

Exp NO O2 Rate forming NO2 (M/s)
1 0.015 0.015 0.048
2 0.030 0.015 0.192
3 0.015 0.030 0.096
4 0.030 0.030 0.384
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Sample Exercise 3

• Rate data for the reaction
• CH3Br OH- ? CH3OH Br-

Exp CH3Br OH- Rate of forming CH3OH
1 0.200 0.200 0.015
2 0.400 0.200 0.030
3 0.400 0.400 0.060
Use the data to find the experimental rate law.
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