lecture 4 photochemistry
What is Photochemistry?
• Photochemistry: a chemical interaction involving radiation
• Why mention photochemistry?
Photochemistry plays an important role in atmospheric processes.
• The wavelengths in important atmospheric photochemical reactions are shortwave – radiation from the sun
lecture 4 photochemistry
Nuclear Energy = neutrons + protons minus binding energy
Atomic Weight/(Neutrons+Protons) does not equal 1 because of the binding energy. More binding energy, lower ratio AW/(N+P), lower total energy.
If two light nuclei join to form a heavier nucleus with higher binding energy (lower total energy), the extra energy is released as radiation – fusion.
What is the Sun’s Source of Energy?
lecture 4 photochemistry
Fusion is a thermonuclear reaction. It releases energy but needs a high enough temperature to bring the two nuclei together.
2H+2H 4He+E
2x2.0144.0026+0.0254(as energy)
Requirement, high temperature for activation ~25,000,000oC
This is the temperature in the sun’s interior, but not the temperature at the sun’s surface (the temperature at which the sun emits).
Can calculate Tsurface based on a steady state assumption (heat flow from interior balances heat loss from surface):
dT/dt=C1MsTinterior-C2(S.A.)sTsurface=0
4/3R3C’1Tinterior=4R2C’2Tsurface
Tsurface=C”RTinterior=~6000oK
Solar surface temperature is determined from the ratio of surface to volume.
Fusion
lecture 4 photochemistry
Energy per photon:E=h=hc/ h = 6.626 x 10-34 J s c = 3 x 108 m s-1
Energy of Radiation
lecture 4 photochemistry
Energy Per One Mole of Photons (Einstein)
Energy per one mole of photons at:
100nm = 290 kcal/mole
400nm = 72.5 kcal/mole
700nm = 41.5 kcal/mole
1000nm = 29 kcal/mole
Comparison to chemical bond strength:
N-N = 225 kcal/mole Very strong
O-O = 120 kcal/mole Strong
C-Cl =75 kcal/mole Intermediate
O-O2 = 35 kcal/mole Weak
HO…..H=5 kcal/mole Very weak
mol
kcal12390.0
mol
kJ 1
mol
kJ
nm
101.19625
hc 106.02
hcN hN moleper
523
00
E
lecture 4 photochemistry
What Happens When Radiation Hits a Molecule?
We learned in radiative transfer that two possible outcomes are:
1. scattering (no chemical interaction)
2. absorption:
Following absorption, there are a number of possibilities.
*ABh AB
lecture 4 photochemistry
Pathways Following Absorption
lecture 4 photochemistry
i. Dissociation/photolysis: breaking a chemical bond in the molecule
Energy of radiation must be greater than bond energy.
=100-1000nm is sufficient to break any chemical bond.
ii. Ionization: removing an electron from the molecule
In general ionization energy is greater than chemical bond strength:
He = 552 kcal/mole =52.6 nm}
N2 = 398 kcal/mole = 79.6 nm}
Na = 120 kcal/mole = 250 nm}
Pathways Following Absorption cont.
*)B(A* AB
eAB* AB
lecture 4 photochemistry
Table of Ionization Energies
lecture 4 photochemistry
Pathways Following Absorption cont.iii. Luminescence: re-emission of photon
hAB* AB
In atoms, re-emited photon is of same energy as excitation: em=ex. In molecules, it can be less: em>ex.
Flourescence: visible wavelengths
Phosphorescence: non-visible wavelengths
lecture 4 photochemistry
Pathways Following Absorption cont.
iv. Intramolecular energy transfer: conversion of the absorbed energy to several forms of lower energy (vibration, rotation and eventually to heat – typical for large molecules).
§
v. Intermolecular energy transfer
vi. Quenching
vii. Reaction: conversion to more active state and undergo selective chemical reactions
AB* AB
*CDABCD)( * AB
K.E.)( MABM* AB
speciesdifferent chemicallyEF* AB
lecture 4 photochemistry
What Determines the Pathway?
1. wavelength – whether or not its possible
2. population of excited states – whether or not its probable
3. conservation of orbital angular momentum and spin – whether or not its probable
lecture 4 photochemistry
rate of formation of AB* =
J for a photochemical reaction is the equivalent of a rate constant for a chemical reaction
J can be treated as a first order rate constant (units of time-1) but it depends on light intensity and spectral distribution.
Rate of a Photochemical Reaction
*ABh AB
]AB[ [AB*]
ABJdt
d
lecture 4 photochemistry
How is J Calculated?
For a given wavelength:
J{}=PF{} {} Y{}
PF{}: PhotoFlux
{} – Absorption cross section (population of excited states)
Y{} – Quantum yield (conservation of orbital angular momentum and spin)
Y{} =
The quantum yield is sometimes also symbolized
For a range of wavelengths (solar range):
J = PF{} {} Y{} d
absorbed photons ofnumber
process undergoing molecules ofnumber
1}{Y0
lecture 4 photochemistry
NO2+h NO + O(3P) J{NO2}
The rate of O(3P) formation
d[O(3P)]/dt = J{NO2} [NO2]
The rate of O formation will change diurnally even at constant NO2.
Photolysis Rate Example
lecture 4 photochemistry
J=Jmax cos() cos(
Jmax SRI/R2 SRI: solar radiation intensity
txx – Geographical Latitude – Seasonal motion of the earth cos(2{JD} /365)
JD: Julian Day
How Does J Vary With Latitude and Season?
lecture 4 photochemistry
The Solar Spectrum