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 =  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 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
 ___________

 E a 

 ___________ _ _ _ _ _ _ _ _ _ 
 REACTANTS

 D H 

 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ________
 PRODUCTS

 SECOND ORDER REACTION COORDINATE       

Molecularity: The number of molecules in the activated complex. Thermodynamics tells only that Ea > DH.
 
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 - D H f = E ar 

 ____________ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 
 O3 + NO 
 D H f =
 _ ____ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ________
 NO2 + O2 

 k f = A f exp(- E a /RT)

k r = A r exp left ( {-(E af + D H r ) } over { RT } right )

 2
We need the enthalpy:

DHrxn = DH f sup o prod. - DHf sup o react.

DH f = {8.1 + 0.0 - 34.0 - 21.6} = -47.5 kcal/mole

DH r = - DH 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(-DG/RT)

 DG = 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 + DHr  = 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!