Photochemical kinetics

*

* *fluorescence

quenching

AB

AB k AB M

A

A

[ ]11 q

f

k

A

M

[ ]abs

d XI

dt

• Intensity

Transmittance

Absorbance

Transmittance

Absorbance

Beer’s Law

dP(x)/dx = -κ P(x)

P(x) = exp(-κ x)

I NIA

Nx I

A xN

x IVx I

I dII

x dx

0( ) xI x I e

Beer’s law

0

xtI eI

23

1NB: cm cm

cx

m

hnDx

A

0

10x cxtI eI

NB: L mole

cmmole cm

xL

c

optical densityc

absorptivityavogadroN

2.303 avogadroN

Molar extinctionCoefficient~250 L.mol-1cm-1

Cross section~10-18 cm2

ε = 313 M-1 cm-1

b = 2 cmT = 3.16% = 0.0316

A = -logT = 1.50c = 1.5/(313 M-1) (2 cm-1) = 0.0024 M

NB 1: Beer fails when

photochemistry happens

NB 2: The photophysics

Is hidden in σ

(So we haven’t done much yet)

Absorption of a mixture

Isosbestic point

Photochemical kinetics

*

*

*

AB h AB

AB AB h

AB M AB M

** *[ ]

[ ] [ ][ ]abs q

d ABI AB k AA B M

dt

*[ ]0

d AB

dt *[ ]

[ ]abs

q

AA

A

IAB

k M

[ ]abs

fluorescenceq

II

k

A

A M

*[ ]

[ ]abs

q

IB

AA

k M

[ ]11

[ ]f q

fabs q f

I k M

I k

A

MA A

STEADY STATEHYPOTESIS

Ozone production

hn

hn

O2

O2

O2

O3

O O

Photochemical kinetics

λ ≤ 240 nm

λ ≤ 280 nm

λ ≥ 280 nm

1. Steady state hypothesis: d[O]/dt = 02. Solve for [O]3. Plug into disappearance rate of [O3]