LECTURE 4
METO/CHEM 637
ATMOSPHERIC CHEMISTRY
RUSSELL R. DICKERSON
 
Wayne Chapt. 3 (kinetics)
Wark & Warner Chapt. 8 & 9
Seinfeld and Pandis Ch. 4 kinetics; Ch. 9 Thermodynamics. 
Pitts and Pitts Ch. 5 (kinetics)

   OUTLINE   
   II THERMODYNAMICS
     B. Free energy
  III KINETICS
    A. Rates, rate constants and order of rxns
    B. Lifetime & halflife

 IIB. 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.

EXAMPLE 1.

Everyday experience tells us that water evaporates at room temperature, but this is uphill in terms of the total energy, E.

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.

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
 
At room temperature:
 
H298 =  + 43.14 kcal/mole
                           
    H1500 = H298  +  Integral from 298 to 1500 Cp dT
 
 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 T 2 - 0.547x10-9 T 3
  
Cp /R  = 0.7525 - 2.7199x10-3 T + 26.5448x10-7 T 2 - 1.094x10-9 T 3
   
 size 22 { int } from 298 to 1500 Cp /R dT  = 0.7525(1500 - 298) - 1/2(2.7199x10-3)
(1500 2 - 298 2) + 1/3(26.5448x10-7)(1500 3 - 298 3) - 1/4(1.094x10-9)(15004 - 2984)
 
Cp | pile {1500 above 298}  = -454.176 R (cal/mole K)
 
Cp  = -903.81 cal/mole
 
H1500  = 43.41 - 0.904 = + 42.506  kcal/mole

But we know that combustion produces large quantities of NO!!! Therefore, these reactions must be driven by some force other than internal energy or heat! Note: There is little change in H with T.
 
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

DFN: dS = DQ/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 = DQ - PdV

dS  = DQ/T

dS  = dE/T + PdV/T
 
For one mole of an ideal gas,
 
P/T  = R/V
 
At constant volume,
 
dE  = Cv dT
 
Thus
 
dS  = {Cv dT}/T + {RdV}/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.
 
GIBBS FREE ENERGY
 
The Second Law states that for a reversible reaction:
 
dS = DQ/T
 
For an irreversible reaction,
 
dS > DQ/T
 
At constant temperature and pressure:
 
dE  = DQ - PdV - VdP

dE  <= DQ - PdV - VdP

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
 
 We define G as (E + PV + TS) or (H + TS)
 
dG = dH - TdS

G = H - TS
 
G is the Gibbs free energy, the criterion of feasibility. G tends toward the lowest values, and if G for a reaction is positive, the reaction cannot proceed!

IIC. 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,

 2		P2
     dG  = 	    nRT/P  dP =  G = nRTln(P2/P1)
 1		P1
 
Consider the following reaction,
 
aA + bB  cC + dD
 
Where small case letters represent coefficients.
 
G  = nRT ln(P2/P1)
 
 G  = nRT ln [{(PA)a (PB)b}/{ (PC)c (PD)d }] = -nRT ln [{ (PC)c (PD)d }
/ { (PA)a (PB)b } ]
 
Remember what an equilibrium constant is:
 
Keq  =  { (PC)c (PD)d }/{ (PA)a (PB)b }
 
thus
 
G = -nRT ln (Keq)
 
or, for one mole
 
Keq  = exp left ({- G}/RT right)

NOTE:  Keq = 1 if and only if  G = 0.00
 
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. 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.
 
aA(PA = 1)  aA(PA)
G A  = aRTln (PA/1)
 
Similarly for B.
For the products:
cC(PC)  cC(PC = 1)
G C   = cRTln (1/{ PC})
 
Similarly for D.
 
Go   = Grxn + GA  + GB  + GC  + GD		I

= Grxn  + aRTln(PA /1) + bRTln(PB /1) + cRTln(1/PC) + dRTln(1/PD)
 
Combining the log terms,
 
Go   = Grxn  + RT ln {(PA)a (PB)b}/{(PC)c (PD)d}
 
For the equilibrium partial pressures where Grxn = 0,
  
Go   = RT ln [{(PA)a (PB)b}/{(PC)c (PD)d}]  = -RT ln [{(PC)c (PD)d }/{(PA)a (PB)b }]
 
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.

III. KINETICS
A. RATES, RATE CONSTANTS AND REACTION ORDER
 
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.
 
1.  FIRST ORDER
 
A  PRODUCTS
 
  EXAMPLES

222Rn  ->  218Po + 

N2O5 = NO2 + NO3
 
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.

 Also
 
d[NO2] /dt  = k [N2O5]
 
and
 
d[N2O5]/dt  = -k [N2O5]
 
Separating
 
1/{ [ N2O5 ] } d[ N2O5 ]  = k dt
 
Integrating

 		    1/[N2O5] d[N2O5]  = 	     k dt

ln ([ N2O5 ]t }/{ [ N2O5 ]o)   = -k t
 
If we define the starting time as zero:
 
[N2O5]t / [N2O5]o = exp(-kt)
 
Rate constants are:
 
kRn  = 0.182 days-1
 
kN2O5  = 0.26 s-1

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  = 1.8x10-14 cm3 s-1
  
kHCl-OH  = 8.0x10-13  cm3 s-1
 
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:

 M = Mo {P}/{Po} {To }/{T} = 2.69x10-19  (P/1) (273/T) ~ Mo{P}
 
Where Mo is the molecular number density at STP.

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
 
USEFUL IDEA:
 
For the following reaction:

A + B = C + D
 
d[C]/dt  = kf [A][B] - kr [C][D]
 
 At steady state d[C]/dt = 0, by definition. Thus:
 
{ kf }/{ kr } = { [C][D] }/{ [A][B] } = Keq 

III B. HALFLIFE AND LIFETIME

Definition: Halflife, t1/2 the time such that:

[A]t1/2 / [A]0 = 1/2
 
Definition of lifetime or residence time,  , comes from kinetics, where k is the forst order rate constant with units of time-1.  We know that:
 
[A]t/[A]0 = exp(-k t)

The lifetime, , is when t = 1/k so 

  1/k
  
We can link halflife and lifetime:

t1/2 = ln(2)/k  0.69/k
 
 For radon 222 (222Rn) the lifetime is 5.5 days, but the halflife is only 3.8 days.

 = (k[ A bar ])-1
 
For example:
 
NO + O3  NO2 + O2

  k = 1.8 x 10-14 cm3 s-1
 
 ASSUME: [O3] >> [NO] and d[O3]/dt ~ 0.0
 
LET: [{O3} bar ] = 50 ppb (a reasonable value for air near the surface).
  
 NO  = 1/{ k[ {O3} bar ] } = 1/{1.8 x 10-14  x 50x10-9 x 2.5x1019}  = 44 s
 
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.  For third order reactions we must assume that two components are constant.
 
 = { (k[ A bar ][ B bar ]) }-1
 
For example:

O + O2 + M   O3 + M
k = 4.8x10-33 cm6s-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
 
 = left [ {4.8x 10-33 x 0.21 x 1.0 x (2.5x1019) 2 } right ]-1
 
             = 1.6x10 sup-6 s VERY SHORT!
 
EXAMPLE 2
Same reaction at stratospheric temperature and pressure.
 
 P sub 30km  = P0 exp(-30/7) =  0.014 atm
 
   = left [ 4.8x10-33 x 0.21 x 1.0 x (0.014 x 2.5x1019) 2 right ]-1 = 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.
 
Definition: HALFLIFE

 {[A]t}/{[A]0} = 1/2 
 
From kinetics we know that:
 
{ [A] sub {t 1/2} }/{ [A]0 } = exp(-kt)
 
 t1/2 = { ln(1/2) }/k approx 0.69/k
 
For radon 222 (222Rn) the lifetime is 5.5 days, but the halflife is only 3.8 days.

VIEWGRAPH

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
 
int } from 0 to t [A] sup {-2} d[A] = size 22 { int } from 0 to t -(2k[B])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.