© R. R. Dickerson 201111 LECTURE 4 AOSC 637 Atmospheric Chemistry 1Finlayson-Pitts and Pitts Ch. 5...

© R. R. Dickerson 2011 11

LECTURE 4AOSC 637

Atmospheric Chemistry

1Finlayson-Pitts and Pitts Ch. 5 (Kinetics & Lab Techniques)Wayne Chapt. 3 (Kinetics)Wark & Warner Chapt. 8 & 9Seinfeld and Pandis Ch. 4 Kinetics; Ch. 9 Thermodynamics.

Outline Thermodynamics Free energy Kinetics Rates, rate constants and order of rxns

Lifetime & half-life


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.


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).


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


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


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

)298 - )(1500101/4(1.094x - )298 - )(15008x101/3(26.544

)298 - )(1500x101/2(2.7199 - 298) - 00.7525(150 dT Cp/R

449-337-

2231500

298

-


What is H1500 ?

re. temperatuoft independennearly

and positivestrongly is H change; noAlmost

)(kcal/mole 42.506 0.904 - 43.41 0.904 - H H

(cal/mole) 903.81-

cal/moleK) (K R 454.176- |C

2981500

1500298p


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.


Entropy and the Second Law of Thermodynamics

DFN: dS = đQ/TFor 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/TdS = dE/T + PdV/T


For one mole of an ideal gas, P/T = R/V At constant volume,

dE = Cv dT Thus

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.

V

dVR

T

dTC dS v


Gibbs Free EnergydS = đQ/T

For an irreversible reaction:

dS ≥ đQ/T

At constant T & P:

dE = đQ – PdV

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

T

PdVVdPdE dS


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.


Gibbs Free Energy, ∆G, andEquilibrium constants Keq

Consider the isothermal expansion of an ideal gas.

dG = VdP From the ideal gas law,

dG = (nRT/P)dP Integrating both sides,


Now let’s apply this to a reaction.

aA + bB ↔ cC + dD

where the small letters represent coefficients.

1

2

P2

P1

2

1

P

PnRTln G

dP P

nRT dG


• 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)


For the products:

cC(PC) cC(PC = 1)

G C = cRTln(1/PC) = −cRTln(PC)

Go = Grxn + GA + GB + GC + GD I

= Grxn + aRTln(PA/1) + bRTln(PB/1) + cRTln(1/PC) + dRTln(1/PD)


Combining the log terms,

For the equilibrium partial pressures where

Grxn = 0,

dD

cC

Ba

A0

P P

P Pln RTΔGΔG

b

rxn

b

b

Ba

A

dD

cC

dD

cC

Ba

A0

P P

P Pln RT

P P

P Pln RTΔG


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.


KineticsThermodynamics 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]

kt-

0

t e ][A

[A]


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


Examples: Radioisotope decay, thermal decomposition and photolysis.

A → Products222Rn → 218Po + N2O5 = NO2 + NO3

NO2 + h → 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.


Also d[NO2] /dt = k [N2O5]

And d[N2O5]/dt = -k [N2O5]

Separating Integrating

If we define the starting time as zero:

Rate constants at 298 K are: kRn = 0.182 days-1 kN2O5 = 0.26 s-1

kdtd

]ON[

]O[N

1 52

52

kdtd ]ON[]O[N

1 52

52

tkt

052

52

]O[N

]O[Nln

ktt e052

52

]O[N

]O[N


“j-Values”: Definition

NO2 + h NO + O ( < 424 nm)

]NO[]NO[

2NO2

2j

dt

d

Actinic flux (photons cm-2 s-1 nm-1)

absorption cross section (cm2

molec-1)photolysis quantum yield (molec. photon-1)

dTTIj NO ),(),()( ONONO20 2NO2

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 + h NO + O

tj

02NO NO

NO2 ][


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:

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

02 ]NO[

]NO[~

2NO t

j

Trailer #2UMD Actinometer

UMD Actinometer

inside

on top

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


3. Third OrderA + 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: 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

T

T

P

Pxx

T

T

P

PxMM 0

0

190

00 1069.2


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:

EQr

f KBA

DC

k

k

][][

][][


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


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’.

]B[

1

e ][A

[A]

]Bk[ k'

]tBk[-

0

t

kA


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).

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.

sxxxOk

NO 42105.21050109.1

1

][

119914

3


For third order reactions we must assume that two components are constant. For example:

O + O2 + M O3 + 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.

= 1.6x10-6 s, short indeed!

1]][[ BAk

121933 )105.2(0.121.0108.4 xxxxxO


Example 2

Same reaction at stratospheric temperature and pressure. P30km ~ P0 exp(-30/7) = 0.014 atm

= 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 destructionof tropospheric and stratospheric ozone.

Problem for the student: Compare the rate of loss of O atoms to reaction with O2 vs. NO2 in the troposphere where [NO2] ~ 10-8

and in the stratosphere where [NO2] ~ 10-6 .

121933 )105.2014.0(0.121.0108.4 xxxxxxO


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

-[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.

t

0

t

0

2- (2kB)dt- dAA


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.


Lecture 4 Summary

• Changes in enthalpy andentropy,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 201111 LECTURE 4 AOSC 637 Atmospheric Chemistry 1Finlayson-Pitts and Pitts Ch. 5...

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Transcript of © R. R. Dickerson 201111 LECTURE 4 AOSC 637 Atmospheric Chemistry 1Finlayson-Pitts and Pitts Ch. 5...