METOENCE 434

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LECTURE 8

• METO/ENCE 434
• AIR POLLUTION
• RUSSELL R. DICKERSON

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RECAP KINETICS

• First Order
• A ? Products
• N2O5 ? NO2 NO3 (1)
• Rate equation
• Second Order
• A B ? Products
• NO O3 ? NO2 O2 (2)
• Rate equation
• dA/dt dB/dt - kAB

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• Third Order
• A B C ? Products
• O O2 M ? O3 M (3)
• Rate equation
• dA/dt dB/dt dC/dt -kABC
• Units of the rate of a reaction are conc./time.
  Rate constants for reactions of order n must have
  units

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• Lifetime
• If t 1/k then ,
  then the time corresponding to 1/k t, the
  lifetime. This is similar to half life
  concept
• Pseudo first order rate constants
• If B is approximately constant then k kB.

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III KINETICS b) Activation Energy

• The energy hill that reactants must climb in
  order to produce products a barrier to
  thermodynamic equilibrium

ENERGY DIAGRAM
AB
Ea
C D
A B
?H
A B ? AB AB M ? AB M AB ? C D
(Activate complex) (Quenching) (Reaction)
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• c) Arrhenius Expressions
• The Arrhenius or pre-exponential factor, A
  is related to the number of collisions the
  reactants make, and has a maximum value of about
  2x10?¹º cm³ s?¹ for a second order reaction. If
  the orientation of the reactants is important,
  and it usually is, then A ltlt 10?¹º.
• is the activation energy in units of
  J/mole. For an endothermic reaction if
  is less than or equal to ?Hºrxn. If is
  positive then the reaction will go faster at
  higher temperature if is negative, then
  the reaction will go slower at high temperature.
  For example, for Reactions (1) to (3)

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• The rate constant for Reaction (3), k3, holds
  only if the reaction proceeds in an argon or
  helium atmosphere. If the third body (M) is a
  diatomic or triatomic, the rate constant goes up.
  For a N2O atmosphere k is 1.6 time faster, and
  water vapor it is fifteen times faster than in
  air.
• A recent evaluation (NASA, 1997) of the rate
  constant k3 for the Earths atmosphere is given
  in a different form
• GENERAL EXAMPLES
• (2)
• What happen if dilute car exhaust mixes with
  clean air? Let NO0 1.0 ppm and O30 50
  ppb. Assume P 1.0 atm and T 25º C. What is
  the rate of loss of ozone? Note that we have to
  match the units of the concentrations to the rate
  constant.

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• Ozone disappears very quickly, but this is only
  the initial rate of loss. At the end of one
  second the concentration of the reactants, ozone
  especially, will be less than the initial
  concentration. To calculate a realistic rate of
  loss, we can assume that dNO/dt 0. This is a
  pretty good assumption because NOgtgtO3. Let
• kNO k
• k 0.45 s?¹
• t 2 s
• In 2.0 s the ozone concentration is down to e?¹
  or to about 37 of the initial concentration. In
  10 s the concentration is down to e?5 or 0.67.

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Equilibrium and Rate Constants

• At equilibrium production equals loss.

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Equilibrium and Rate Constants

• For a reaction of arbitrary order
• aA bB R cC dD

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Photolysis Reactions

• Sunlight drives photochemical smog. How do we
  deal with this special class of reactions?
• AB hv ? AB Photoexecution
• AB ? A B Photodissociation or
  photolysis
• M AB ? AB M Quenching
• AB ? AB h? Reemission
• The rate depends on the intensity of sunlight and
  chemical characteristics of AB. The rate
  constant is represented as j(AB) to distinguish
  it from ordinary first order rate constants.

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• Where
• s absorption cross-section
• F quantum yield
• I solar intensity
• EXAMPLE
• NO2 h? ? NO O
• j(NO2) 1.0 x 10?² s?¹ (at noon)
• The lifetime of NO2 with respect to photolysis is
  about 100 s.
• Steady State
• When a molecule is in steady state its
  concentration is constant. This means that the
  production and destruction rates are matched, it
  does not necessarily mean that the system is in
  thermodynamic equilibrium, but a system in
  equilibrium is also in steady state.

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• EXAMPLE
• A B C D (1)
• A B M ? AB M (2)
• A AB ? 2A B (3)
• Rate Equations

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• At steady state dAB/dt 0.0 and R2 R3.
• k2ABM k3AAB
• Which means
• The concentration of AB depends only on the ratio
  of the rate constant for reactions 2 and 3, and
  on the concentration of B. It is independent of
  A. What is the steady state concentration of
  B?
• Any new reaction that are discovered can be added
  to the scheme in a similar manner.

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Photostationary State

• Here is a real-world example.
• NO2 hv ? NO O (1)
• NO O3 ? NO2 O2 (2)
• O O2 M ? O3 M (3)
• If we assume ozone is in steady state then
  production equals loss.
• But what is ?
• R1 R3
• j(NO2)NO2 k3OO2M

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• Substituting
• What happens when the sun comes up on a polluted
  air mass? Let NO 50 ppb and NO2 500 ppb.
  At noon j(NO2) 10?² s?¹, but near sunrise
• Later in the day j(NO2) 1.0x10?², if the ratio
  of NONO2 remains the same