Lecture 5
AOSC/CHEM 637
Atmospheric Chemistry
R. 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
Copyright © R. R. Dickerson 2010
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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
A+B
A + B → AB†
AB† + M → AB + M†
AB† → C + D
ΔH
C+D
(Transition state or activated complex)
(Sometimes there can be quenching)
(Reaction)
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ACTIVATION ENERGY
Remember the Van't Hoff (or Gibbs-Helmholtz) equation.
dlnKeq/dT = DH/(RT2)
This suggests:
dlnk/dT = Ea /(RT2)
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 e(-Ea/RT)
This is the Arrhenius Equation in which A is the pre-exponential factor, also
Copyright
© R./RT)
R. Dickerson
2010
called the Arrhenius factor, and
exp(-E
is the Boltzmann
factor.
a
3
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 << 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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Copyright © R. R. Dickerson 2010
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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.
1/ 2
 8kT 
2
A

d
 Collision Frequency

  
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 Molecules
dN2 = 3.2x10-8 cm
MN2 = 28/6.023x1023 g
For N2 + N2
1/ 2
 8 1.38x10 -16  298 
-8 2
A

(
3
.
2
x
10
)  Collision Freq.

- 23
   2.3x10

 2x10 -10 s -1 cm3
This is a good estimate for the maximum rate constant for any reaction.
Note A is proportional to d2, -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).
Copyright © R. R. Dickerson 2010
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Example calculation of a rate constant
O3 + NO ↔ NO2 + O2
ENERGY DIAGRAM ↑
O=O-O ∙ ∙ ∙ N=O
Ea
O3 + NO
ΔH
NO2 + O2
O=O-O ∙ ∙ ∙ N=O is the activate complex
Ea is the activation energy, unknown.
ΔH is the enthalpy of the reaction, known from
thermodynamics.
Copyright © R. R. Dickerson 2010
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Rate const for the forward reaction:
kf = A exp(-Ea/RT)
Rate const for the reverse reaction :
kr  Ar e
 - E a - DH 
 RT 


We need the enthalpy:
∆Horxn = ∆H fo prod. - ∆Hf o react.
∆H f = {8.1 + 0.0 - 34.0 - 21.6} = -47.5 kcal/mole
∆Hr = - ∆Hf = + 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
f = 18.5/6.023x1023 g
r = 18.9/6.023x1023 g
Af < Forward collision rate = 3.4x10-10 s-1 cm3
Ar < Reverse collision 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!
Rate const for the reverse reaction :
 - 47500 
k r  2.6x10 -10 exp 

1.98 x 298 
 6.35 x10 - 45 cm 3 s -1
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
 47700 
34
K eq  exp 

5
.
7
x
10
 RT 
The products are heavily favored.
kf = Keq x kr
K eq x k r  5.7x1034 x 6.35x10 -45
k f  3.6 x10 -10 cm -3s -1
But we knew that much from the collision rate already.
The measured rate constant for this reaction is:
 - 1500 
3 -1
k f  3.0x10 -12 exp 
cm
s

 T 
3 -1
k f (298)Copyright
 1.9x10©-14
cm
s
R. R. Dickerson 2010
E a  2750 cal/mole
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The measured "A" is 113 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) = 113
Only one collision in 113 has the proper
orientation. Now lets try to calculate a
better value for kr. Assume same steric
factor.
Ear = Eaf + DHr = 3000 + 47500
kr = 3.0x10-12 exp(-25500/T) cm3 s-1
= 2.0x10-49 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.
[NO 2 ][O 2 ] 
-34
K eq  

5
.
7
x
10

[NO][O
]
3 

[NO 2 ][O 2 ] 
-34
K eq  

5
.
7
x
10

[0.5][0.5]


[NO]  [O 3 ]  2.1x10 -18 atm  10-6 ppt
Pretty small, so we can say that all the NO and O3 react away.
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Stedman’s Kinetic Kidney
Enthalpy ↑
+428 KJ/mole
-216KJ/mole
+286 KJ/mole
O2 → 2O
DH = +498 KJ/mole
k = A e(-Ea/RT)
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Lecture 5 Summary
• For simple bimolecular (second order) reactions, the Arrhenius expression
describes the temperature dependence of the rate constants.
• Collision theory gives some insight into the limits of rate constants.
• The enthalpy of a reaction provides insight into the activation energy.
Molecular oxygen, O2, is not generally an important oxidant in atmospheric
reactions.
• The radicals OH and HO2 are the “hit men” of atmospheric chemistry.
Copyright © R. R. Dickerson 2010
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