Atmospheric chemistry

Lecture 2:

Photochemistry & kinetics

Dr. David GlowackiUniversity of Bristol,UK

david.r.glowacki@bristol.ac.uk

Quick review of yesterday

• We discussed atmospheric structure• Temperature & pressure gradients, as well as Coriolis

forces are related to atmospheric transport

Today…

• We’ll gain some insight into the relationship between atmospheric structure and atmospheric chemistry

• Atmospheric chemistry depends on sunlight, temperature, and pressure; Today we’ll learn about • Photochemistry• Chemical kinetics

The Atmosphere is a low temperature chemical reactor

Troposphere

Stratosphere

Tropopause-70oC14 km

O3 layer

UV

visible

Urban Anthropogenic emissionsSurface O3

Regional and global biogenic emissions

(CH4)

Important Chemistry:

UV absorption by O3

IR absorption byGreenhouse gases(H2O, CH4, CO2)

Surface emissions resulting in O3 and aerosol formation, and acid rain

Atmospheric Chemistry starts with sunlight

€

O3 → O +O2

E = hv

• Breaking chemical bonds requires energy

• Sunlight has energy• If sufficient energy is deposited

in the bond, then it will break• O3 has a bond energy of ~105

kJ mol-1

v = c/

Energy/kJ mol-1

Red 700 170Orange 620 190Yellow 580 210Green 530 230Blue 470 250

Violet 420 280Near UV 400-200 300-600Far UV 200-50 600-2400

visi

ble

Photoexcitation gives excited molecules, A*

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A + hv → A*

A* → B +C

A* → A + hv

A* + M → A + M

A* → A+ + e-

Initial photoexcitationDissociation

FluorescenceCollisional relaxation

Ionization

• Photoexcitation may result in a number of processes:

• Photochemistry depends on temperature, pressure, and the wavelength of the absorbed light

*

Photoexcitation kinetics

• The rate of formation of A* is written:

€

€

d[A*]

dt= jA[A]

where jA is the photochemical rate constant

• Competition between subsequent processes is determined by the quantum yield, ϕ, for each process where:

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A* 1 ⏐ → ⏐ B +C

A* 2 ⏐ → ⏐ A + hv

A* + M 3 ⏐ → ⏐ A + M *

A* 4 ⏐ → ⏐ A+ + e-

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φi

i

∑ =1

Dissociation yield =Φ1

Fluorescence yield =Φ2

Collisional relaxation yield =Φ3

Ionization =Φ4

Understanding the photolysis rate

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jA = σ A (λ ,T)φA (λ ,T)∫ I(λ )dλ

= σ A (λ i,T)φA (λ i,T)i

∑ I(λ i)Δλ i

absorption cross section: number of photons

absorbed by a molecule at a particular wavelength

Spectral actinic flux: density of photons in the atmosphere

at a particular wavelength

Quantum yield: efficiency at which absorbed photons

result in the molecular process of interest

need to integrate over the entire wavelength

range

Understanding photolysis rates

Atmospheric actinic flux

• Photochemical processes depend on:• temperature (absorption cross sections & quantum yields)• Pressure (collisional relaxation)• Altitude (actinic flux)

O3 absorption cross section

Atmospheric absorption of light

• Gases absorb light• The absorption of light depends

on the concentration of the gas, N, its absorption cross section, σ, & the path length, l,through the gas

• May be described by the Beer-Lambert law

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T =I

I0

= exp(−σ l N)

€

l

€

l

€

I(l )

€

σ,N

Atmospheric absorption of light

• The Beer Lambert law: • Explains the altititude dependence of actinic flux• Is often used to measure atmospheric trace gas concentrations

DOAS (differential optical absorption spectrometry)

FTIR spectrometry

Chemical Kinetics

Kinetics depends on the potential energy surface (PES)

• What molecules do is determined by their potential energy landscapes – energy as a function of coordinates

• Stable molecules are minima on a PES

• Potential energy surfaces (PES) are multidimensional, but we usually think about their motion projected in one dimension

• T dependence of reaction rate coefficients well described by the Arrhenius equation:

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k(T) = Aexp −Ea

RT

⎛

⎝ ⎜

⎞

⎠ ⎟

First order Unimolecular kinetics

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A k(T ) ⏐ → ⏐ B

d[A]

dt= −k[A]

d[A]

[A]= −kdt

d[A]

[A]= − kdt∫∫

ln[A]0 − ln[A]t = −kt

[A]t = [A]0 exp(−kt)

Mechanisms with more than one chemical reactions: exact solutions

• Coupled chemical reactions, often result in mechanisms of the sort:

• For this system we can

write three rate equations,

one for each species:

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d[A]

dt= −k1[A]

d[B]

dt= k1[A] − k2[B]

d[C]

dt= k2[B]

A B

k1

k2

B C

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dn

dt=Mn where n =

[A]

[B]

[C]

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

and M =

-k1 0 0

k1 −k2 0

0 k2 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

In matrix form:

Chemical Mechanisms with Coupled Chemical Reactions: Coupled differential Equations

• Analytic solutions exist for this eigenvalue problem to solve for concentration vs. time

• If the initial concentration of every species but [A] is zero, the solutions are

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[A] = [A]0e−k1t

[B] = [A]0

k1

k2 − k1

(e−k1t − e−k2t )

[C] = [A]0(1+k1e

−k2t − k2e−k1t

k2 − k1

)Concentration vs time when k2/k1=10

B changes a lot;Not low or constant

Concentration vs time when k2/k1=0.5

B doesn’t change muchLow and ~constant

• Consider again the following mechanism:

• Steady state approximation: assume the

rate of change of intermediate B is zero

A B

k1

k2B C

Chemical Mechanisms with Coupled Chemical Reactions: Steady State Approximation

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d[A]

dt= −k1[A]

d[B]

dt= k1[A] − k2[B]

d[C]

dt= k2[B]

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d[B]

dt= k1[A] − k2[B] = 0

⇓

[B] =k1

k2

[A]

[B] =k1

k2

[A]0e−k1t

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[B] = [A]0

k1

k2 − k1

(e−k1t − e−k2t )

Equivalent when k2 >> k1

making [B] low & ~constant

Approximate Steady state solution

Exact solution

Chemical Lifetimes

• Often we are interested in the average lifetime of a molecule before it reacts away

• Lifetime has units of time• The interplay between chemical lifetimes and

atmospheric mixing processes determines much of atmospheric chemistry

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lifetime of intermediate =[intermediate]

sum of intermediate loss processes

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τB =[B]

k2[B]=

1

k2

A B

k1

k2B C

Collision Theory

• Molecules are constantly moving

• Molecular gases are constantly colliding with each other with a T & P dependent collision frequency

• Each collision has a particular amount of energy associated with it

• This energy may lead to chemical reaction

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KE =4kBT

π

Thresholdenergy

Bimolecular Kinetics

• Atmospheric chemistry involves both unimolecular and bimolecular processes

• Bimolecular kinetics depend on pressure, [M]• A reasonable model for a bimolecular reaction is

Visualizing bimolecular pressure dependence: O + O2 + M O3 + M

O + O2 reaction coordinate

O OO

M

M = O2 or N2

O3

Bimolecular Kinetics: The Low & High pressure Limits

• The total bimolecular process:

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d[AB]

dt= k5[AB*][M]

d[AB*]

dt= k3[A][B] − k4[AB*]− k5[AB*][M]

k3[A][B] = k5[AB*][M] + k4[AB*]

d[AB]

dt=

k3k5[A][B][M]

k4 + k5[M]

Assume AB* is in steady state

Solve for AB* and plug into the first equation

Write rate equations for AB*

We want to know the rateof AB formation

Bimolecular Kinetics: The Low & High pressure Limits

• Low Pressure Limit– [M] is very small

– k4 >> k5[M]

– k5[M] goes to zero

– Overall reaction rate depends linearly on [M]

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d[AB]

dt=

k3k5[A][B][M]

k4 + k5[M]

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d[AB]

dt=

k3k5[A][B]

k4

[M]

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d[AB]

dt= k3[A][B]

• High Pressure Limit– [M] is very large– k4 << k5[M]– k4 goes to zero– Overall reaction rate is

independent of [M]– Instantaneous

stabilization

T & P dependent kinetic effects

• Laboratory measurements of rate coefficients give rise to T & P dependences which are well described by the kinetic master equation

Quick Summary

• Atmospheric chemistry dominated by photolysis• Molecular motion on a potential energy surface (PES)

determines reactivity• In the atmosphere, simple reactions combine to form

kinetic networks (i.e., coupled sets of important reactions)

• The steady state approximation is a useful simplification for short lifetimes

• Chemical reactions depend on both pressure & temperature, and are determined through a combination of experimental & theoretical approaches