Lecture 5 AOSCCHEM 637 Atmospheric Chemistry R' Dickerson

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Lecture 5AOSC/CHEM 637Atmospheric ChemistryR.
Dickerson

• OUTLINE
• KINETICS
• Activation Energy
• Kinetic Theory of Gases
• Calc. Rate Constants w/Collision Theory
• Finlayson-Pitts (2000) Ch. 5
• Seinfeld and Pandis (2006) Ch. 3

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If two of the reactants in a third order reaction
are the same, we can derive a useful expression
for the rate of loss of the reactant. A A B
? PROD For a great excess of B dA/dt
-(2kB)A 2 A -2 dA -(2kB)dt
-At-1 A0-1 -(2kB)t At-1 2kBt
A0-1 Now we can calculate the concentration at
any time t in terms of the initial concentration
and the rate constant k.
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Kinetics continued Activation Energy

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

ENERGY DIAGRAM
AB
Ea
C D
A B
?Hof
A B ? AB AB M ? AB M AB ? C D
(Activate complex) (Quenching) (Reaction)
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ACTIVATION ENERGY Remember the Van't Hoff (or
Gibbs-Helmholtz) equation. dln Keq/dT
DH/(RT2) This suggests dln k/dT Ea /RT
2 Which is the Arrhenius expression where Ea is
the activation energy. If we integrate both
sides ln(k) (-Ea / R) 1/T ln(A) Where ln(A)
is the constant of integration. Rearranging k
A exp(-Ea/RT) This is the Arrhenius Equation
in which A is the pre-exponential factor, also
called the Arrhenius factor, and exp(-E a/RT) is
the Boltzmann factor.
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KINETIC THEORY OF GASES When molecules in the
gas phase collide they sometimes rearrange their
chemical bonds to form new molecules. The rate of
formation of the new molecules is determined by
the fraction of molecules with sufficient energy
to overcome the activation energy
barrier. POSTULATES OF CHEMICAL KINETICS 1.
Pressure is the result of molecular
collisions. 2. Collisions are elastic, i.e. no
change in kinetic energy. 3. Volume of the
molecules ltlt volume occupied by gas. 4. Kinetic
energy proportional to T and independent of gas,
i.e., the same for all gases.
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Boltzmann Distribution N1/N0 e-(E1 - E0
)/kT (Also called Maxwell distribution for ideal
gases) WHERE N1 number of particles
(molecules) with energy E1 N0 number of
particles (molecules) with energy E0 M
molecular weight dN/N0 M/kT exp -Mc-2
/2kT c dc WHERE c2 V2 U2 W2 SEE
Lavenda, "Brownian Motion," Sci. Amer., 252(2),
70-85, 1985.
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• CALCULATING RATE CONSTANTS FROM COLLISION THEORY
• From thermodynamics and Arrhenius
• k A exp(-Ea/RT)
• A is a function of diameter, temperature, and
  mass its maximum possible value is the frequency
  of collisions.
• WHERE
• k Boltzmann const. 1.38x10-16 erg/K
• T Abs. Temp. (K)
• d Diameter of molecules.
• reduced mass M1 x M2 / M1 M2
• A has units of (molecules cm-3)-1 s-1 or cm3 s-1

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Example Collisions Between Nitrogen

• dN2 3.2x10-8 cm
• MN2 28/6.023x1023 g
• For N2 N2
• This is a good estimate for the maximum rate
  constant for any reaction. Note A is proportional
  to d2, m-1/2, T1/2. One would expect the
  Arrhenius factor to have a T1/2 factor, but this
  is usually swamped out by the exponential
  temperature dependence of activation energy.
  (Remember ergs are g cm2 s-2).

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Example calculation of a rate constant

• O3 NO ? NO2 O2

ENERGY DIAGRAM ?
OO-O NO
Ea
NO2 O2
O3 NO
?Hof
OO-O NO is the Activate complex Ea is the
activation energy, unknown ?Hof is the enthalpy
of the reaction, known from Thermo
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Rate const for the forward reaction kf A
exp(-Ea/RT) We need the enthalpy ?Hrxn
??H fo prod. - ??Hf o react. ?H f 8.1 0.0 -
34.0 - 21.6 -47.5 kcal/mole ?H r - ?H f
47.5 kcal/mole We need the pre-exponential
factors Af and Ar DIAMETERS d(NO) 0.40 nm d(
O3 ) 0.46 nm d( NO2 ) 0.46 nm d( O2 ) 0.296
nm
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REDUCED MASSES mf 18.5/6.023x1023 g mr
18.9/6.023x1023 g Af lt COLLISIONAL RATE
3.4x10-10 s-1 cm3 Ar lt COLLISIONAL RATE
2.6x10-10 s-1 cm3 Now we need an estimate of
activation energy, Ea Ear ? 47.5 kcal/mole
Eaf We know nothing! This is very
slow! STUDENTS Calculate the lifetime of NO2
with respect to conversion to NO at the typical
oxygen content of the atmosphere.
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To get at kf lets look to thermodynamics. kf/k r
Keq exp(-DG/RT) DGo 0 12.4 - 21.0 -
39.1 -47.7 kcal/mole The products are
heavily favored. kf Keq x kr But we knew
that much from the collision rate already. The
measured rate constant for this reaction is
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The measured "A" is 170 times smaller than the
maximum "A". Why? Not every collision with
sufficient energy results in a reaction. The
molecules must have the proper orientation.
STERIC FACTOR A(collisional)/A(actual)
170 Only one collision in 170 has the proper
orientation. Now lets try to calculate a better
value for kr. Assume same steric factor. Ear
Eaf DHr 2800 47500 kr 2.0x10-12
exp(-25150/T) cm3 s-1 4.4x10-49 cm3 s-1 at 298
K Thermodynamics says kr kf/Keq 3.2x10-49
cm3 s-1
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The agreement is not too bad, less than a factor
of two difference! The thermodynamic value is
more likely correct. We cannot measure the
reverse rate constant because it is too slow.
For example if we took a 1 atm mixture of 50 NO2
and 50 O2 at equilibrium (square brackets
represent partial pressure) the ozone and nitric
oxide concentrations would be much too small to
measure.
Pretty small, so we can say that all the NO and
O3 react away.
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