LECTURE 4
AOSC 637
Atmospheric Chemistry
1Finlayson-Pitts and Pitts Ch. 5 (Kinetics & Lab Techniques)
Wayne Chapt. 3 (Kinetics)
Wark & Warner Chapt. 8 & 9
Seinfeld and Pandis Ch. 4 Kinetics; Ch. 9 Thermodynamics.
Outline
Thermodynamics
Free energy
Kinetics
Rates, rate constants and order of rxns
Lifetime & half-life
© R. R. Dickerson 2011
1
FREE ENERGY
We have a problem, neither energy (E) nor
enthalpy (H) is the "criterion of feasibility".
Chemical systems generally tend toward the
minimum in E and H, but not always.
Everyday experience tells us that water
evaporates at room temperature, but this is
uphill in terms of the total energy, E.
© R. R. Dickerson 2011
2
Example 1
H2O(l)  H2O(g)
P = 10 torr, T = 25°C
E = +9.9 kcal/mole
The enthalpy, H, is also positive, about 10
kcal/mole, and PdV is too small to have an
impact (R =1.99 cal/mole K; V2/V1 ~ 1000).
© R. R. Dickerson 2011
3
Example 2.
The formation of nitric oxide from nitrogen and
oxygen occurs at combustion temperature.
We know that H >>0 at room temperature, but what
about at combustion temperature? We can calculate
H as a function of temperature with heat capacities,
Cp, found in tables. Remember that
R = 1.99 cal/moleK and dH = Cp dT
N2 + O2  2NO
© R. R. Dickerson 2011
4
N2 + O2  2NO
At room temperature:
H298 = + 43.14 kcal/mole
The reaction is not favored, but combustion
and lightning heat the air, and dH = CpdT
© R. R. Dickerson 2011
5
What is ∆H at 1500K?
H1500 = H298 + Integral from 298 to 1500 Cp dT
We can approximate Cp with a Taylor expansion.
 Cp = 2Cp (NO) - Cp (N2) - Cp(O2)
Cp (O2)/R = 3.0673 + 1.6371x10-3 T - 5.118x10-7 T 2
Cp (N2)/R = 3.2454 + 0.7108x10-3 T - 0.406-7 T 2
Cp (NO)/R = 3.5326 - 0.186x10-3 T - 12.81x10-7 T2 - 0.547x10-9 T3
Cp /R = 0.7525 - 2.7199x10-3 T + 26.5448x10-7 T2 - 1.094x10-9 T3
1500
-3
2
2

Cp/R
dT

0.7525(150
0
298)
1/2(2.7199
x10
)(1500
298
)

298
1/3(26.544 8x10 -7 )(1500 3 - 2983 ) - 1/4(1.094x 10 -9 )(1500 4 - 2984 )
© R. R. Dickerson 2011
6
What is H1500 ?
C p | 1500
298  - 454.176 R (K  cal/moleK)
 - 903.81 (cal/mole)
H1500  H 298 - 0.904  43.41 - 0.904   42.506 (kcal/mole )
Almost no change; H is strongly positive and
nearly independen t of temperatu re.
© R. R. Dickerson 2011
7
But we know that
high temperature
combustion and
lightning produce large
quantities of NO!
Therefore, these
reactions must be
driven by some force
other than internal
energy or heat.
Note: The
independence of H
with T will be useful.
© R. R. Dickerson 2011
8
Entropy and the Second Law of
Thermodynamics
DFN: dS = đQ/T
For a reversible reaction the change in entropy,
S, is a function of state of system only. To
get a feel for what entropy is, let us derive an
expression for the entropy of an ideal gas.
General Relations:
dE = đQ - PdV
dS = đQ/T
dS = dE/T + PdV/T
© R. R. Dickerson 2011
9
For one mole of an ideal gas,
P/T = R/V
At constant volume,
dE = Cv dT
Thus
C v dT R dV
dS 

T
V
Integrating
S = Cv ln(T) + R ln(V) + So
Where So is the residual entropy. This equation lets you calculate entropy
for an ideal gas at a known T and V.
© R. R. Dickerson 2011
10
Gibbs Free Energy
dS = đQ/T
For an irreversible reaction:
dS ≥ đQ/T
At constant T & P:
dE = đQ – PdV
dS 
dE  VdP  PdV
T
But VdP = 0
dE – TdS + pdV ≤ 0
Because dP and dT are zero we can add VdP and SdT to the equation.
dE – TdS – SdT + PdV – VdP ≤ 0
d(E + PV – TS) ≤ 0
© R. R. Dickerson 2011
11
We define G as (E + PV - TS) or (H - TS)
dG ≡ dH – TdS
∆G ≡ ∆H – T∆S
∆G for a reaction is the Gibbs free energy, and it is the criterion of
feasibility. G tends toward the lowest value.
If ∆G is greater than zero then the reaction will not proceed
spontaneously; the reactants are favored over the products.
© R. R. Dickerson 2011
12
Gibbs Free Energy, ∆G, and
Equilibrium constants Keq
Consider the isothermal expansion of an ideal gas.
dG = VdP
From the ideal gas law,
dG = (nRT/P)dP
Integrating both sides,
© R. R. Dickerson 2011
13
2
P2
nRT
1 dG  P1 P dP
 P2 
G  nRTln  
 P1 
Now let’s apply this to a reaction.
aA + bB ↔ cC + dD
where the small letters represent coefficients.
© R. R. Dickerson 2011
14
• If a + b is not the same as c + d, we can get into trouble trying
to take the log of an expression with units. (Note error in
Hobbs’ book.) For this type of reaction, the Gibbs free energy
is the sum of the G for the chemical reaction and the G for
the change in pressure. Assuming that the reactants start at 1.0
atm and go to an equilibrium pressure and assuming that the
products finish at 1.0 atm. The units will always cancel.
aA(PA = 1)  aA(PA)
GA = aRTln (PA/1)
© R. R. Dickerson 2011
15
For the products:
G C
cC(PC)  cC(PC = 1)
= cRTln(1/PC) = −cRTln(PC)
Go = Grxn + GA + GB + GC + GD
= Grxn + aRTln(PA/1) + bRTln(PB/1) +
cRTln(1/PC) + dRTln(1/PD)
© R. R. Dickerson 2011
I
16
Combining the log terms,
a
b

PA  PB 
0

ΔG  ΔG rxn  RT ln  c
d 
 PC  PD 
For the equilibrium partial pressures where
Grxn = 0,
c
d
a
b




P

P
P

P
0
C
D
A
B
  RT ln  a

ΔG  RT ln  c
d 
b
P P 
P

P
 C D 
 A B 
© R. R. Dickerson 2011
17
If you remember that each of the partial pressures was a ratio
with the initial or final pressure taken as 1.0 and that the ln(1) = 0
are left out you can see that Keq is always dimensionless.
© R. R. Dickerson 2011
18
Kinetics
Thermodynamics tells us if a reaction can proceed and gives
equilibrium concentration. Kinetics tells us how fast reactions
proceed. If thermodynamics alone controlled the atmosphere
it would be dissolved in the oceans as nitrates - we would be
warm puddles of carbonated water.
Reaction Rates and Order of Reactions
1. FIRST ORDER
A  PRODUCTS
dA/dt = -k[A]
[A] t
 e -kt 
[A ]0
© R. R. Dickerson 2011
19
The red line
describes first order
loss with a rate
constant of 1 min-1
The blue line
describes the rate of
formation of the
product.
minutes
© R. R. Dickerson 2011
20
Examples: Radioisotope decay, thermal
decomposition and photolysis.
A → Products
222Rn → 218Po + 
N2O5 = NO2 + NO3
NO2 + hn → NO + O
Radon is important source of indoor air pollution, and N2O5 is nitric acid
anhydride, important in air pollution nighttime chemistry. The rate equations
take the form:
d[prod.]/dt = -k[A] = -d[react.]/dt
For example:
d[Po]/dt = kRn [Rn] = -d[Rn]/dt
Where k is the first order rate constant and k has units of time-1 such as s-1,
min-1, yr-1. We usually express concentration, [Rn], in molecules cm-3 and k
in s-1.
© R. R. Dickerson 2011
21
Also d[NO2] /dt = k [N2O5]
And d[N2O5]/dt = -k [N2O5]
Separating
Integrating
 1 

d [ N 2O5 ]  kdt
 [N 2O5 ] 
1
 [N 2O5 ]d[ N 2O5 ]    kdt
 [N 2O5 ]t 
  kt
ln 
 [N 2O5 ]0 
If we define the starting time as zero:
[N 2 O5 ]t
 e  kt
[N 2 O5 ]0
Rate constants at 298 K are: kRn = 0.182 days-1
kN2O5 = 0.26 s-1
© R. R. Dickerson 2011
22
“j-Values”: Definition
NO2 + hn  NO + O ( < 424 nm)
d [ NO2 ]

 jNO2 [ NO2 ]
dt

jNO2   I ( ) NO 2 ( , T )NO2NOO ( , T )d
0
Actinic flux (photons cm-2 s-1 nm-1)
absorption cross section (cm2 molec-1)
photolysis
quantum yield (molec. photon-1) 23
© R. R. Dickerson
2011
International Photolysis Frequency
Measurement and Modeling
Intercomparison (IPMMI)
NCAR Marshall Field Site, 39°N 105°W, elevation: 1.8 km; June 15–19, 1998
Objectives: j [NO2  NO + O], j [O3  O2 + O(1D)], spectral actinic flux.
Measurements by 21 researchers from around the world. Photolysis Frequency of
NO2: Measurement and Modeling During the International Photolysis Frequency Measurement and
Modeling Intercomparison (IPMMI), R. E. Shetter, W. Swartz, et al., J. Geophys. Res., 108(D16),
10.1029/2002JD002932, 2003.
UMD jNO2
Actinometer
Schematic
NO2 + hn  NO + O
j NO2 
NO
[NO 2 ]0 t
Problem for the student:
Show that for 1.00 ppm NO2, 1.00 atm pressure, exposure times of 1.00 s,
and j(NO2) values of 10-2 s-1 the errors to:
jNO2
[ NO]
~
t[ NO2 ]0
from complicating reactions are less than 1%.
1. O + O2 + M → O3 + M
k1 = 6.0 x10-34 cm6 s-1
2. O3 + NO → NO2 + O2
k2 = 1.9×10–14 cm3 s-1
3. O + NO2 → NO +O2
k3 = 1.04×10–11 cm3 s-1
© R. R. Dickerson 2011
26
Trailer #2
UMD Actinometer
UMD Actinometer
on top
inside
quartz photolysis tube
UMD jNO2 Actinometer Data
2. Second Order
A + B  PRODUCTS
Examples
NO + O3  NO2 + O2
HCl + OH  H2O + Cl
Examples of the rate equations are as follows:
d[NO]/dt = -k[NO][O3]
d[Cl]/dt = k[OH][HCl]
Units of k are conc-1 time-1.
1/(molecules/cm3) (s-1) = cm3 s-1
Rate constants have the following values:
kNO-O3 = 1.9x10-14 cm3 s-1
kHCl-OH = 8.0x10-13 cm3 s-1
© R. R. Dickerson 2011
30
3. Third Order
A + B + C  PRODUCTS
d[A]/dt = -k[A][B][C]
Examples
2NO + O2 = 2NO2
O + O2 + M = O3 + M†
M is any third body (usually N2) needed to dissipate excess energy. From the ideal
gas law and Avg's number:
P T0
P T0
M  M0x
 2.69 x1019 x
P0 T
P0 T
Where Mo is the molecular number density at STP in molecules cm-3.
Third order rate constants have units of conc-2 time-1. These are usually (cm-3)-2 s-1.
kNO-O2 = 2.0 x 10-38 cm6 s-1
kO-O2 = 4.8 x 10-33 cm6 s-1
© R. R. Dickerson 2011
31
Useful idea:
For the following reversible reaction:
A+B↔C+D
d[C]/dt = kf [A][B] - kr [C][D]
At steady state d[C]/dt = 0, by definition.
Thus:
[C ]  [ D]

 K EQ
kr [ A]  [ B]
kf
© R. R. Dickerson 2011
32
Half-life and Lifetime
Definition: Half-life, t1/2, is the time t such that:
[A]t / [A]0 = 1/2
Definition of e-folding lifetime or residence time,  , comes from kinetics,
where k is the first order rate constant with units of time-1. We know that:
[A]t/[A]0 = exp(-kt)
The lifetime, , is when t = 1/k so
  1/k
We can link half-life and lifetime:
t1/2 = ln(2)/k  0.69/k
For radon 222 (222Rn) the lifetime is 5.5 days, but the half-life is only 3.8 days.
 = (k[Ā])-1
© R. R. Dickerson 2011
33
For second and third order reactions we can sometime approximate first order
conditions – or use pseudo first order kinetics.
A + B → Prod.
If [B] >> [A]
Then k[B] is approximately constant. We call this pseudo first order rate
constant k’.
k'  k[B]
[A] t
 e -k[B]t 
[A ]0
1
A 
k[B]
© R. R. Dickerson 2011
34
For second and third order reactions we can sometime approximate first order
conditions – or use pseudo first order kinetics.
For example:
NO + O3  NO2 + O2
k = 1.9 x 10-14 cm3 s-1
Assume: [O3] >> [NO] and d[O3]/dt ~ 0.0
Let: mean [O3] = 50 ppb (a reasonable value for air near the surface).
 NO
1
1


 42s
14
9
19
k[O3 ] 1.9 x10  50 x10  2.5x10
CONCLUSION: any NO injected into such an atmosphere (by a car for
example) will quickly turn into NO2 , if there are no other reactions that play
a role. We will call k[O3] the pseudo first order rate constant.
© R. R. Dickerson 2011
35
For third order reactions we must assume that two components are constant.
  k[ A][ B]
1
For example:
O + O2 + M  O 3 + M
k = 4.8x10-33 cm6 s-1
ASSUME:
d[O2]/dt = d[M]/dt = 0.0
We know that [O2] = 0.21 and that [M] ~ [O2] + [N2] ~ 1.00.
At RTP P02 = 0.21 atm and PM ~ 1.0 atm. Therefore the lifetime of O atoms is
1/k[O2][M]M02 where M0 is the conversion to molecules per cm3.
 O  4.8 x10
33

19 2 1
x0.21x1.0 x(2.5 x10 )
= 1.6x10-6 s, short indeed!
© R. R. Dickerson 2011
36
Example 2
Same reaction at stratospheric temperature and pressure.
P30km ~ P0 exp(-30/7) = 0.014 atm
 O  4.8 x10
33

19 2 1
x0.21x1.0 x(0.014 x 2.5 x10 )
= 2.1x10-3 s
This is still short, but it is a thousand times longer than in the troposphere!
The pressure dependence has a major impact on the formation and destruction
of tropospheric and stratospheric ozone.
Problem for the student: Compare the rate of loss of O atoms to reaction with O 2 vs. NO2 in the
troposphere where [NO2] ~ 10-8 and in the stratosphere where [NO2] ~ 10-6 .
© R. R. Dickerson 2011
37
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:
d[A]/dt = -(2k[B])[A]2
[A]-2 d[A] = -(2k[B])dt
t
t
0
0
-2
A
 dA   - (2kB)dt
-[A]t-1 + [A]0-1 = -(2k[B])t
[A]t-1 = 2k[B]t + [A]0-1
Now we can calculate the concentration at any time t in terms of the
initial concentration and the rate constant k.
© R. R. Dickerson 2011
38
The method works for the self reaction of nitric oxide:
NO + NO + O2 → 2NO2
and shows that this reaction can ruin NO in N2 calibration
standards if any air gets into the cylinder.
d[O2]/dt = -k[NO]2[O2]
d[NO]/dt = -2k[NO]2[O2] = 2d[NO2]/dt
k = 3.3x10-39 exp (530/T) cm6 molecules-2 s-1
In the atmosphere it is only important for highly concentrated
exhaust gases.
© R. R. Dickerson 2011
39
Lecture 4 Summary
• Changes in enthalpy and entropy, H and S, are nearly independent of
temperature.
• Gibbs free energy provides the criterion of feasibility.
• The residence time (lifetime), t, is the inverse of the first order rate
constant, k.
• If second or third order reactions can be approximated as first order then
lifetimes can be estimated.
• For reversible reactions, kf/kr = Keq
© R. R. Dickerson 2011
40