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LECTURE 5
METO/CHEM 637
ATMOSPHERIC CHEMISTRY
RUSSELL R. DICKERSON
OUTLINE
III. KINETICS
C. Activation Energy
D. Kinetic Theory of Gases
E. Calculation of Rate Constants w/Collision Theory
Finlayson-Pitts Ch. 4
Seinfeld Ch. 4.8
III C. ACTIVATION ENERGY
Remember the Van't Hoff (or Gibbs-Helmholtz) equation.
{dln Keq}/dT = H/(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 preexponential factor, also called the Arrhenius
factor, and exp(-E a/RT) is the Boltzman factor.
{VIEWGRAPH}
For a reaction to occur the molecules must surpass some critical energy - the activation energy,
Ea. The Boltzman factor gives us the energy distribution for an ensemble of molecules at a given
temperature. The following diagram depicts the activation energy for an exothermic, two-body
reaction.
ACTIVATED
COMPLEX
___________
Ea
___________ _ _ _ _ _ _ _ _ _
REACTANTS
H
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ________
PRODUCTS
SECOND ORDER REACTION COORDINATE

Molecularity: The number of molecules in the activated complex. Thermodynamics tells only that
Ea > H.
III D. 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 temperature and independent of gas, i.e., the same for all gases.
BOLTZMAN 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.
III E. 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.
A = left ( { 8kT } over { pi  } right ) sup 1/2 pi d 2 = COLLISION FREQ.
WHERE:
k = Boltzman const. = 1.38x10 -16 erg/K
T = Abs. Temp. (K)
d = Diameter of molecules.
A has units of cm3 s-1
For N2
d = 3.2x10 sup -8 cm
 = reduced mass = {M1 x M2} / {M1 + M2}
M sub N2 = 28 over { 6.023x10 23} g
again for N2
For N2 + N2
A mark = left ( { 8*1.38x10 -16 * 298 } over { pi * 2.3x10 sup -23 } right )
sup 1/2 pi * ( 3.2x10 sup -8 ) 2
= 2 x 10 -10 s-1 cm3
This is the maximum rate constant for any reaction. (For units, remember ergs are g cm2 s sup -2 ).
Note that "A" is proportional to: d2 , -1/2, and T sup 1/2 . One would expect the Arrhenius factor to
have a T sup 1/2 , but this temperature dependence is generally swamped by the exponential
temperature dependence.
{VIEWRGAPH}
EXAMPLE
CALCULATION OF REACTION RATE CONSTANT
{It ain't easy}
O3 + NO = NO2 + O2
5
______________________
E af = ?
E af -  H f = E ar
____________ _ _ _ _ _ _ _ _ _ _ _ _ _ _
O3 + NO
Hf=
_ ____ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ________
NO2 + O2
k f = A f exp(- E a /RT)
k r = A r exp left ( {-(E af +  H r ) } over { RT } right )
2
We need the enthalpy:
Hrxn =
H f sup o prod. - Hf sup 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 preexponential 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
REDUCED MASSES
mu f = 18.5/6.023x10 23 g
mu r = 18.9/6.023x10 23 g
Af < COLLISIONAL RATE = 3.4x10 -10 s-1 cm3
Ar < 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!
kr <= 2.6x10 -10 exp left ( { -47500 } over { 2.00 x 298 } right )
<= 6.35x10 sup -45 cm3 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.
To get at kf lets look to thermodynamics.
kf/k r = Keq = exp(-G/RT)
G = 0 + 12.4 - 21.0 - 39.1 = -47.7 kcal/mole
Keq = exp left ( { 47700 } over { 2 x 298 } right ) = 5.7 x 10 sup 34
The products are heavily favored.
kf = Keq x kr
Keq x k r <= { 5.7x10 sup 34 } x { 6.35x10 sup -45 }
k f <= 3.6x10 -10 cm3 s-1
But we knew that much from the collision rate already. The measured rate constant for this
reaction is:
kf = 2.0x10 -12 exp(-1400/T) cm3 s-1
kf (298) = 1.8x10 -14 cm3 s-1
Ea = 2800 cal/mole
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 + Hr = 2800 + 47500
kr = 2.0x10-12 exp(-25150/T) cm3 s-1
= 4.4x10 sup -49 cm3 s-1 at 298 K
Thermodynamics says:
kr = kf/Keq
= 3.2x10 sup -49 cm3 s-1
The agreement is not too bad, less than a factor of two difference! The thermodynamic value
more likely correct. We cannot measure the reverse rate constant because it is too slow. For
example if we took a mixture of 50% NO2 and 50% O2 at equilibrium the ozone and nitric oxide
concentrations would be much too small to measure.
Keq = { [ NO2 ][ O2 ] } over { [NO][O3] } = 5.7x10 sup 43
Keq = { [ 0.5 ][ 0.5 ] } over { [NO][O3] } = 5.7x10 sup 43
[ O3 ] = [NO] = 2.1x10 -18 atm = 10 sup -6 ppt
Pretty small!