UNIT 19 PHOTOCHEMISTRY

Structure
19.1 Introduction
Obj,jectives

19.2 Laws of Photochemistry

19.3 IERperimental Determination of Quantuin Efficiency
19.4 Photochemical Dissociation
19.5 Some Photochemical Reactions
19.6 Photophysical Processes
19.7 Photosensitisation
19.8 AppIications of Photochemistry
19.9 Chemiluminescence
19.10 Summary
19.11 Terminal Questions
19.12 Answers

19.1 INTRODUCTION
In Unit 9 of this course, we stated that free energy must decrease for a physical or a
chemical process to be feasible. This is applicable to thermal reactions (also known as dark
reactions). The conversion of carbon dioxide and water into starch is a reaction in which
free energy increases. Under laboratory conditions, this reaction is not possible. But plants
carry out starch synthesis using sunlight. Ozonisation of oxygen and decomposition of
an~moi~ia are also nonspontaneous reactions under laboratory conditions; but these could
take pIace by using light of proper energy. Such possibiIities of new reactions and new
synthetic methods encouraged the scientists to probe into light-initiated reactions

In photochemistry, we study the absorption and emission of light by matter. It consists of

the study of various pbotophysical processes and photochemical reactions. Two importailt
photophysical processes are fluorescence and phosphorescence. During fluorescence, light
emission takes place in the presence of exciting radiation; but the light emission stops, once
the exciting radiation is removcd. In contrast to this, during phosphorescence, light emission
takes place even after the removal of the exclting radiation.
In photochelnical reactions, the substances acquire the necessary activation energy through
light absorption. Again this is in contrast to the tberinal reactions in which the reactants
acquire their activation energy through collisions between molecules.
111 this unit, we shall discuss the laws of photochemistry. We shall tben describe some
phofochernical reactiolls and photophysical processes. We shall also explain some
photosensitised reactions. Finally we shall discuss the applications of photochemical
studies.

Objectives
After studying this unit, you should be able to

explain the mathematical form of Beer-Lambert law,

state Grotthuss-Draper law and Stark-Eiilstein law,

calculate the quantuin yield of a reaction using the given exyeriinental data,

explain the reasoll for the higher energy requirement for the photochen~icaldissociatior~
of a molecule as compared to its thermal dissociatiou,
Dynamics and Macromolecules
derive rate expressio~~s
for simple photochemical reactions,

explain some photophysical properties such as fluorescence and phosphorescence,

list some photosensitised reactions,

state the applications of photochemical studies, and

define chemiluminescence..
-4
-

19.2 LAWS OF PHOTOCHEMISTRY

Grotthuss-Draper law and Stark-Einstein law are the two laws concerning the interaction
between matter and light Before discussing these two laws, let us state the mathematical form of
Beer-Lambert law discussed in Unit 8 of Atoms and Molecules course.
Beer-Lambert law is useful in calculating the concentration of a solution on the basis of its
light absorption. This law relates the intensity of the transmitted monochromatic light to the
concentration of the solution and the thickness of the cell in which the solution is kept.
Mathematical form of Beer-Lambert law is given below:
log Io/I = A = ~ c l . . . (19.1)
where l o = Intensity of the incident radiation
I = Intensity of the transmitted radiation
A = log Io/I = Absorbance or optical density of the solution
c = Concentration of the solutio~texpressed in mol m-3 units
I = thickness of the cell
E = Molar extinctio~tcoefficient; expressed in m2 mol-'
The molar extinctio~~ coefficient of a substance can be determined using a colorirneter or a
syectrophotoineter as follows. The absorbances of a solution are measured at different
using a cell of knownthickness (I ). The plot o f A against c gives a
~ I I O W I Icol~ce~ltrations
straight line (Fig. 19.1) and its slope is equal to E 1.

Hence, E =
9
1
Since 1 is known, E can be calculated. Using this E value, the concentration of a solution can
be determined by measuring its absorbance.

Calorimeters and spectrophotometers are commercially available for the measurement of

absorbances. In colorimeters, colour filters are used to get the incident radiation in a
particular wavelength range. Spectrophotometers have arrangements for obtaining nearly
monochromatic incident radiation. Both colorimeters and spectrophotometers have devices
for converting the light transmitted inte suitable signal through detector devices (see Sec.
19.3). The signal generated from transmitted light is directly read as absorbance values.
Study the following example carefully. The calculation shown below is used in chemical
actinometers (Example 3 in Sec. 19.3) for estimating the intensity of the light absorbed
during a photochemical reaction. This example illustrates the use of Eq. 19.1 in calculating
the concentration of a solution.

Example 1
A solution of the red complex formed by ~ e ion~ with + 1,lO-phenanthroline is taken in a
cell of thickness 1.00 cm.If E for the complex is 1.11x lo3 m2 mol-'and the absorbance of
the solution is 0.391, calculate the concentration of the co~nplexin mol m" units.

Solution

A 0.391
Rearranging Eq. 19.1, c = -= mol m"
€1 1.11Xio3xi.ooxio-2

= 0.0352 mol m"

We shall now state the two laws of photochemistry.

Grotthuss-Draper Law
According to this law, only the light that is absorbed by a ~noleculecan produce a
photochemical change in it. This means that it is not sufficient to pass light through a
substance to bring about a chemical reaction; but the light must be absorbed by it.
Stark-Einstein law of photochemical equivalence provides a quantum mechanical form to
Grotthuss-Draper law.

Stark-Einstein Law of Photochemical Equivalence

Stark-Einstein law,of photochemical equivalence can be stated as follows:

Each molecule taking part in a photochemical reaction absorbs one quantum of radiation
which causes the reaction.
Stark-Einstein law is applicabkonly
This law is applicable to the primary act of excitation of a molecule by light absorption. if the intensity of light is not very
high.
This law helps in calculating the quantum efficiency ( ) which is a measure of the
efficiency of the use of light in a photochemical reactioa.

The quantum efficiencies for the formation of a product and for the disappearance of the
reactant are defined below:

The quantum efficiency

for the formation of a
product ( $P 1 I=[
Number of molecules of
product formed in 1-seco~ld
Number of quanta absorbed
in 1 second I ...

N Pare
In this equation, dNp/dt denotes the rate of formati011of the product. The units for -
dt
molecules per second. Ia refers to the number of photons absorbed per second. Similarly we
can define the quantum efficiency for the disappearance of a reactant.

I1 I
Number of molecules of reactant
The quantum efficiency for
the disappearance of the
consumed in 1second . . . (19.3)
Number of quanta absorbed
Itactant ( h ) = in 1 second
D p a n i c s a d Macromolecules where - WR/dt is the rate of consumptio~~ of the reactant. The minus sign in - N ~ / d is
t
due to the decrease in concentration of the reactant with time.

As per Stark - Einstein law, each reacting nlolecule absorbs one quantum of light. Hence,
the energy absorbed by one
mole of a substailce undergoing
photochemical reaction t = NAh v =
N ~ h c
-
A
This unit of energy is also called einstein and, as apparent from Eq. 19.4, the value of
einstein depends on v or A.
... (19.4)

c - 3 x 1 0 ~ms-'
Lr
A = wavelei~gthof the light in m unit

v = frequency in s-' u~iit:

To calculate the quantum efficiency, use the following steps:

i) Calculation of Energy Absorbed per Quantum
Energy Per quantum hc
of radiation absorbed
Unit for energy
per quantum ii) Calculation of I.
Usually intensity of light absorbed is given in J s-' units. From this, Ia could be
Unit for I. - joule s t c o n e
joule
calculated using the follown~gequatioil :

- photon second-' Intensity in tenlis of

number of photons Intensity in J s-'units
absorbed in 1 seroltd ( I, ) Energy per quantum

iii) Calculation $f - ~ N or
- R WP
-
dt dt
First we have to obtain the number of moles of reactant consumed or of product formed in 1
second by dividing the respec+i\rr amount by the time in second unit

In order to calculate the rate of Ebrination or rate of disappearance in molecule per second
units, we have to use the following equations:
Number of ~nolesof reactant
decomposed411 1 second

Also,

iv)
Number of moles of product
formed in 1 second
Calculation of Quantum Efficiency
I NA

Using Eq. 19.2 or 19.3, @p or @R can be calculated.

Let us illustrate the calculation of quantum efficiencv for a reaction.

In photochemical decompositio~~ of acetone uslng 313 nm light, 7.57 x lo4 mol of carbon
monoxide is formed in 20 minutes. If the light absorbed corresponds to 2.41 x J s-',
calculate the quantum efficiency for the fonnatidn of carbon monoxide.

= 6.35 x 10-l9 joule photon-'

ii) I,= Intensity in joule secoi~d-I
6.35 x 10-l9 joule photon-'

- 2.14 x joule second-'

6.35 x 10-l9 joule photon-1

i . e . , I , = 3.80 x 10'' photons secondT'

iii) Amount of carbon monoxide

fonned in 1 second
_ 7.57 x lo4 mol
20 x 60 s

= 6.31 x mol s-'

-dNco
- - 6.31 x x 6.022 x loz3 molecule second-'
dt

= 3.80 x 10'' molecule second-'

Quantum efficiency for the
jv) Using Eq. 19.2' ko = formation of carbon monoxide

In the next skction, we shall discuss the experiinental method of deternlining the quantunl
efficiency of a reaction. Before studying this, work out the following SAQ.

SAQ 1

A radiation of 250 iun incident oil HI results in the decomposition 1.85 x mol within a
particular time intezval. Light energy absorbed during this time interval is 4.18 k J.
Calculate the quantum efficiency for the decomposition of HI.

19.3 EXPERIMENTAL DETERMINATION OF

QUANTUM EFFICIENCY
For the determination of quantum efficiency of a photocheinical reaction, we inust measure
the following:

the rate of decohposition of the reactants or the rate of formation of the products ; this
can be done using any of the methods discussed in Unit 18.

the number of quanta absorbed in the given time.

In short, we need a method for measuring the amount of light absorbed. If we can devise a
method :or estimating the intensity of the incident light ( I. ) and the intensity of the light
of the light absorbed ( I, ) can be
transmitted ( I ) by the reaction mixture, the i~ltei~sity
calculated using the relationship ;

For measuring I0 and I, same cell is used. I0 corresponds to the measurement of intensity
with empty cell (known as blank) and I, to the measurement using the cell containing the
reaction mixture.
The experimental set-up used for measuring the intensity of light is shown in Fig. 19.2.
As a source (A) of visible light, a high intensity tungsten lamp or quartz halogen lamp or
xenon arc lamp can be used. For obtaining ultraviolet light, hydrogen or deuterium lamps
are good sources. The light is first passed through a lens, B (to get a parallel beam of light),
and then through a monochromator (C). The mo~~ochromator yields a narrow band of light
in the desired wavelength. The monochromatic light the11 passes through the reaction cell
(D).The light transinitted by the reaction cell reaches the detector (E).
Dynamics and Macromolecules

C D E

Fig. 19.2 : Measurement of intensity of Light during a photochemicalreaction

While using ultraviolet light as a The detector is used for measuring the intensity of light. The quantitative measurement of
source.of radiation, quartzcuvettes
(cells) and lenses must be used.
light intensity based on energy conversion or on chemical reaction is called actinometry.
Detectors such as thermopiles and the photoelectric cells function on energy conversion
Sincepyrex giass absorbs light
below 300 nm, it can be used for principle. We discuss each of them in brief below :
radiation above 300 nrn only.
The visible light has the wavelength Thennopile
region, 800-380 urn. The ultraviolet
light has wavelength below 380 nrn Thermopile (Fig. 19.3) is a device useful for converting light into heat, and then, heat into
and above 10 nm. electricity. I t has a set of junctions of different metals having a blackened surface. This
black coating is to ensure absorption of all radiation falling on it. The energy s o absorbed
increases the temperature of the metals. Thermopile converts this temperature increase into
a potential difference. Light intensity is related to electricity generated by it. The detector
Thermopile consists of device must be calibrated against a standard light source. Thermopile could be used as a
thermocou~lesconnectedinseries detector for light of any wavelength.
and i t generates electricity on

Fig. 19.3 :Thermopile :I. light radiation, 11. blackened surface exposcd to radiation; 111. unexposed surface
at the back; IV galvanometer.
Photoelectric Cells
Photoelectric cell converts light directly into electricity. The current generated is directly
proportional to light intensity. Photoelectric cell is an evacuated bulb with a photosensitive
cathode. Light radiation results in the emission of electrons from the cathode which get
collected at the anode. Thus the flow of current is caused. The photoelectric cell is sensitive
to the wavelength of the light used and must be calibrated using a thermopile.

Chemical Actinometers
Light intensity could be estimated by a quantitative estimation of a substance formed ox
 decomposed by light. The photochemical decoinposition of K3Fe( C2O4)3 is a widely used Photochemistry
chemical method for estimating intensity of light between the waveleiigth range 250 nm and
577 nm. Light within this wavelength range causes K3Fe( C204 )3 dissolved in sulphuric
acid to react as follows:

The concentration of Fe2+is estimated colorimetrically through complex formation with

1,lO-phenanthroline. (See Example 1). This complex has a deep red colour. The
concentration of the complex is equal to the concentration ofFe2+present before
complexation. The following example is worked out to illustrate this method.

Exan~ple3
in a potassium ferrioxalate actinometer, the quantum efficiency for Fe" ( )mn
at 480 mn wavelength is 0.95. After irradiating the potassium ferrioxalate solution for 20

I
minutes, it is completely transferred to a 200 cm3volumetric flask, mixed with required
quantity of 1,lO-phenanthroline for complex formation and made up to mark with a buffer
solution. For colorimetric estimation, a sample of this complex is taken in a cell of 1.00 cm
thickness. The complex has an absorbance value of 0.391 and its E is 1.11 x lo3m2 mol-l.
Calculate

I i) the concentration of the complex,

I ii) the number of Fe2+ions formed in the actinometer due to irradiation,

iii) the rate of formation of Fe2+ions ( cW~s/dt)

iv) Ia (number of photons absorbed per second).

Assume that one Fe2+ion forms one complex molecule with 1,lO-phenanthroline.

I Solution

i) Note that the co~~centration

of the complex is saine as that calculated in Example 1
I using Eq. 19.1; c = concentration of complex = 0.0352 mol mS.
1 Also, concentration of Fe2+ion = co~~centration of the complex

1 1
I
Number of moles of Fe2+ion volume of the
I
I

I
present in 200 cm3 of the
solution
- CollcelltntiolI
(of Fe" ion }x :;ti011 in m3
I

= 0.0352 x 200 x 10" mol

because, 200 cm3

= 7.04 x 10" mol
= 200 x 10" m3

Hence the number of moles

of ~ e ion+ fonned
actinometer

The number of Fe2+ions

~

formed in the actinometer

in the

=
I = 7.04 x 10" mol

7.04 x 10" x 6.022 x ials

= 4.24 x 1018 ions

iii) -WFS
=
Number of Fe2+ions formed 4.24 x 1018
E
ions second-'
dt Time in seconds 1200

= 3.53 x 1015 ions second-'

dNn/dt
iv) Using Eq. 19.2, & = -
Ia
Dynamics and Macromolecules Given that $IFS = 0.95

NF
S 1
Hence I, = -x -
dt 0.95

= 3.53 lo" photons second-'

0.95

= 3.72 x 1015 photons second-'

The Quantum Eff~cienciesof Some Reactions

If a photochemical reaction follows Stark-Einstein law, then $I ust be unity. Studies on a
large number of photoche~nicalreactions indicate that while some of them follow
Stark-Einstein law, many others do not. The quantum efficiencies of some photochemical
reactions along with their effective wavelengths are given in Table 19.1.
Table 19.1: Quantum ETTIciencies and Effective Wavelengths of Some Photochemical Reactions

Reaction Effective wavelength I nm 4'

The photochemical reactions for which quantum efficiency is a small integer are said to
follow Stark-Einstein law. Examples are the dissociation of H2S or ozonisatioi~of oxygel~
etc. Reactions like the formation of hydrogell chloride or hydrogen bromide do not follow
Stark-Einstein law. The quantum efficiency for the former reaction ( H2 + Cl2 -r 2 HCl ) is
very high while it is much low for the latter ( HZ+ Br2 -r 2 HBr ). In order to explain these
discrepancies, Bodellstein pointed out that photochemical reactions involve two distinct
processes:

i) In the primary process, absorption of one quantum of light results in the excitation of
one molecule or an atom (A ) to form the excited species, A*,

Atom or Excited
molecule species

ii) In the secondary process, the epcited species undergoes a chemical change
/

A* -r Products
According to Bodenstein, the deviations from Stark-Einstein law are due to the secondary
processes.

In the case of reactiolls having low quantum yields, the number of molecules decomposed
by the absorption of one quantum of radiation is less than one. The probable reasol~sfor the
low quantum yields are given bk~dw:

The excited species formed at the first step may be deactivated by collisions before the
product could be formed. This process is called quenching. Sometimes the presence of
paramagnetic substances like oxygen or nitric oxide could result in quenching leading
to low quantum yield for a cheinical reaction.

One or more of the reactions in the secondary processes may be endothermic. High
energy requirement could decrease the reaction rate. In Sec. 19.5, we will see that the
quantum yield for the formation of hydrogen bromide is low due to the endothermic
nature of the reaction between Br and Hz (see Example 5 in Sec. 19.5).
 Photochemistry
The excited species nlay recombine to give the reactant ~nolecule.In the hydrogen
bronlide forination, another reasoll for the low quailtun1 yield is the recc>lnbinationof
broilline atoms which are formed in the primary step.
In thc case of reactions with high quantum yields, the excited species fc3rined ill the first step
could initiate a series of chain reactions. This causes a large number of inolecules to react
by absorptioil of one quantum of light, as in the case of hydrogen chloride fonnation (see
Sec. 19.5).

The following SAQ could be answered correctly, if you have uilderstood the materials in
this section.

SAQ 2
The ql~antiiinyield for the photocheinical forination of hydrogen chloride is high in the
absence of oxygen but low when oxygen is present. Exylain.

19.4 PHOTOCHEMICAI . DISSOCIATION

In Unit 3 of Atoins and Molecules course, it has been stated that the bond enthalpy gives an
estiinate of the average energy required to break a particular bond. These bond enthalpies
are derived Goin therinochemical calculations (see Sec. 7.8 of Unit 7 of this course). It is
ge~llcrallyseen that the energy required for the dissociation of a particular bond by liglit
absorption is nluch greater than its bond cnthalpy value. In Table 19.2, we illustrate this for
some diatoinic n~olecules.
Table 19.2: Comparisou of Pholucbcrnical Dissociatiou Energies aud Boud Entlaalpics

Dissociating h/nn~' (Photochemicaldissociation" Boud cntlralpy"'

n~olecule energy ) 1 kJ mol-' k~ ~IOI-' Bond enthalpy of a diatomic
molecule indicates the energy
requirenlent for its thermal
dissociation.

* h value corresponds to maximum wavelength of light required for decomposition by direct

irradiation.
* * Photochemical dissociation energy for one mole is calculated by substituting for h in Eq. 19.4.
* * * Bond enthalpies except for HI are taken from Table 3.9 of Unit 3 of Atoms and Molecules
course.
The reasoils for the higher photocheinical dissociation energies as coinpared to bond
enthalpies are given below:

For dissociating a inolecule through light absorption, there inust be an upper electronic
state with appropriate energy levels. There is no such restriction for thermal
decomposition.

During photochemical dissociation, the product spe.cies could be in an excited state

and / o r in ground state. For example, decomposition of bromine by absorption of light
of wavelength 511 xu11can be shown as below:

The atom with asterisk sign indicates excited state. Thus, photochemical decoinposition of
bronliile needs 235 kJ mol-'. But thernlal decomposition of bro~nineneeds 193 kJ mol-'
only, since both the broilline atoms are formed in ground state.
Dp.mics . ~ l d
M.eromolrmles The photddhemical dissociation of molecules is also known as photolysis. The photo1ysis
can be understood using potential energy diagrams of the type discussed in Unit 4 of Atoms
and Molecules course. In Fig. 19.4, the ground state and the excited state are represented
using potential energy diagrams I and 11, respectively. The quantised vibrational sublevels
in each state are shown by horizontal lines such as AB, EF etc.

F
ig.19.4: Electronic excitation; E @deutisl energy) vs r (internuclear distance).

When a molecule is excited from zero vibrational level AB of ground electronic state to any
of the vibrational levels below GH in the upper electronic state, the resultant electronic
spectrum shows an absorption band with vibration -rotation fine structure. The fine
structure is due to numerous transitions (such as 111, N etc.) possible from the zero
vibrational level (AB) in the lower electronic state to any of the quantised vibrational levels
in the upper electronic state. This can further be understood from the fact that each
vibrational level has its own rotational sublevels.

When a molecule absorbs sufficient energy such that it is transferred from the ground state
to or above GH in the upper electronic state, then the molecule undergoes photochemical
dissociation. The spectrum shows a continuum ( lack of discrete lines ), once the molecule
dissociates. The difference in energy between the levels AB and GH ( AEp ) is the
photochemical dissociation energy. The thermal dissociation energy ( AET ) is equal to
the bond enthalpy in the case of diatomic molecules and it is the energy difference between
the lowest and uppermost vibrational levels (AB and CD) in the ground state. Note that
AEp > AET.

We have discussed photolysis in detail so far because it is the initial reaction in many
photochemical reactions. The excited atom or radical formed due to photolysis of a
molecule often starts a chain reaction. We shall use this principle in the next section in the
study of some photochemical reactions. Before studying that, answer the following SAQ
which will help you in understanding Example 4.
 SAQ 3 Photochemistry

111the photoche~nicaldissociatioi~of HI, the first step is given below:

H+hv -+ H+I
The excess energy ( AE, - AE, ) that
Assume that H atom formed is in excited state while I atom is in ground state. Calculate the a photochemical decomposition
demands as compared to thermal
excess energy that the excited hydrogen atom carries as ( use Table 19.2) compared to a
decomposition is given to one of the
ground state hydrogen atom. atoms formed. This atom is said to
be in the excited state.

19.5 SOME PHOTOCHEMICAL REACTIONS

In this section, we discuss the mechanism of some photochemical reactions and then explain
flash photolysis. For the first two reactions given below, we derive rate equations also. You
go through thes'e derivations carefully. These two examples could give you an idea as to
how the photochemical rate expressions are written.

The first step in both the examples is pt~iolysis.The rate of the photolysis step is expressed
as I, which is the rate of absorption of light (number of quanta absorbed per second). The
initial photolysis is followed by thennal (or dark) reactions for which kinetic expressions
are similar to those discussed in Unit 18.

Example 4

Photochemical Decomposition of Hydrogen Iodide

Let us derive expressions useful in calculati~~g
the rate of decoinposition of HI and the
quantum efficiency for this reaction. HI undergoes photochemical deco~npositionbelow 327
nm. The mechanism is given below:
Rate of
HI+hv -, H+I = I,
photolysis

While deriving rate expressions, we

Such steps are written based on energy considerations. If you have answered SAQ in the try to eliminate terms containing
last section correctly, you could follow why in the second step H atom attacks HI whereas I active species using steady state
does not. principle.

HI is consumed in two ways as per Eqs. 19.7 and 19.8. The rate of disappearance of HI can
C be written as follows :

As per steady state approximation discussed in Unit 18, the concentration of the active
species H is constant. In other words, its co~~centration
does not vary with time.

Note that H is formed as per Eq. 19.7 and is used as per Eq. 19.8.

Using Eqs. 19.10 and 19.12, we get,

D ~ . m i e f ~ d M . v ~ ~Experimentally
I ~ ~ it has been observed that the quantum efficiency for HI decomposition is 2
(Have you worked out SAQ I?).It is worth noting that the rate of decomposition of HI
depends on the intensity of the absorbed light as per Eq. 19.13.

Example 5

Photochemical Reaction between Hz and Brz

Hydrogen and bromine combine at wavelengths below 511 nm to give HBr as per the
mechanism given below. Let us calculate the rate of formation of HBr and alsb the quantum
efficiencv for this reaction.
Br,+hv+Br+~r Rate of photolysis = /a
- .,
...(19.14)

k
H + HBr --&H~ + BT

k,
Br+Br-----,~r,
The rate of formation of HBr can be written as,
d [ HBr 1 .. (19.19)
= k2[Br][H2]+ks[H][Br2]-h[H ( [ H B r l
dt
Note that Brand H are active species and we can apply steady state approximation for these
two.

The coefficient 2 in 21, and 2k5 [ Br l2 are due to the formation or disappearance of two
bromine atoms in the respective steps.

Adding Eqs. 19.20 and 19.21 and rearranging we get

2 k 5 [ ~ r 1 2= 21,

or [Br] (1a/k5lH
Rearranging Eq. 19.21 and using Eq; 19.23 we get,
k2 [ B r ] [Hz I
k2 [ H2 ] ( I a / b 1%
LIHBr]+k3[Ih2] h[HBr]+b[Br2]
Using Eqs. 19.19,19.23 and 19.24,

Dividing the numerator and the denominator of R.H.S by k3 [ Br2 ] ,

i.e., d[HBrl fi ...(19.26)

dt

This shows that the rate of hydrogen bromide formation is proportional to the square root of
 absorbed intensity. This has been proved experimentally. Photochemistry

In the above two cases, we derived rate expressions. Next, we shall study the reaction *
between Hz and C12 in a qualitative way, and then, study the principles of flash photolysis.

Hydrogen-chlorineReaction
The quantum efficiency for the photochemical combination of hydrogen and chlorine is
very high. When exposed io light of wavelength 478 nm, the primary process in the
photochemical reaction is the decomposition of chlorine,
C12 + hv -+ 2 CI ...(19.273
This is f~llowedby the following secondary processes:
C1+H2 -+ HCl+H ... (19.28)
H + C12 -+ HCI + CI ... (19.29)
The reaction between CI and Hz (Eq. 19.28) is exothermic and much fast. This results i11
the propagation o i chain reaction with high quantum efficiency. The chain-terminating step
is the recombination of chlorine atoms on the walls of the vessel to form chlorine molecule.
CI + C1 -+ Cl2 ... (19.30)
Next we shall explain briefly the main features of flash photolysis. Flash photolysis is useful
in detecting the short-lived intermediates in a reaction sequence. The reaction mechanisn~s
are proposed based 011 flash photolysis results.

Flash Photolysis
Flash photolysis was developed by Norrish and Porter in 1949. In ordinary photolysis, the
steady state coi~centrationsof the intermediates are so small that these cannot be detected by
, absorption spectrophotometers. In flash photolysis, a high-intensity flash of microsecond
duration is used for photolysing the substance and the products are identified using
absorption spectrophotometers. The flash duration must match the decay rate of the
intermediates. Flash lamps work for a time of around 15 ys. This restricts their use to the
study of intermediates of life time around 10C ys. In recent years, laser flash sources have
been developed. The flash duration is around s.

As far as this section is concerned, make sure that you understand the derivations for
calculating the rate of decomposition of hydrogen iodide and the rate of formation of
hydrogen bromide. This could help you in arriving at the rate expressions for simple
photochemical reactions for which reaction sequence (like Eq. 19.14 to Eq. 19.18) is
known. You can build up confidence by answering the SAQ given below. The aim of this
SAQ is to make you derive expression for the rate of formation of carbon monoxide in the
photolysis of acetaldehyde. You are guided through a series of steps with helpful hints. This
guidance has added to the length of the problem. Don't mind it!

SAQ 4
Look at the reaction sequence for the photolysis of acetaldehyde:
Rate of
CH3CHO + kv -+ CH3 + CHO
photolysis
= ra

lais the absorbed light intensity and it represents the rate of photochemical excitation (as
per Eq. (1)).
Using the above mechanism, derive expressions for the following:
I Dynamics and Macromolecules
i)
d [ CH3CO 1
dt
--

ii
d [ CH3 .I
dt

(Hints: CH3C0 and CH3 are active species. CH3C0 is formed as per Eq. (2) and
consumed as per Eq. (3). CH3 is formed as per Eqs. (1)add (3), and used up as per Eqs.
(2) and (4). The answers for (i) and (ii) arc to be arrived at using Eqs. 19.21 and 19.22
as models. Note that as per Eq. (4), two CH3 radicals are consumed for every molecule
of ethane formed).
d[CHJCOl+d[CH3]
iii) Derive the combined ertpres~ionfor
For a 2-electron system, the fust . dt dt

II three singlet stat& are represented

below :

-
(Hint: Add up the expressions you have got as answers for (i)and (ii) ).

iv) Find the relationship between [ CH3 ] and I..

(Hint: Rearrange the answer for (iii)).

v) State the relationship between and [ CH3CO 1.

dt

~ Note thatEY > ES1> E w whereE

(Hint: Use Eq. (3)).
Find the relationship between the rate of formqtion of CO and the absorbed light
-.
stands for,be enere" of a . Darticular vi)
level. intensity.
Similarly, the first three triplet states
are represented below for a (Hints: First combine answers for (v) and (i). Then use the answer for (iv)).
2-electron system: .'
.............................................................................................................................................

19.6 PHOTOPHYSICAL PROCESSES

T, T2 T, The light absdrbed by a molecule is not always used up in producing a chemical reaction.
The absorbed energy can be lost thrgugh various physical processes also. In this section, we
E T >~ET? >ETI examine such physical processes.

The absorption of ultraviolet or visible light results in the increase of electronic energy from
the ground state to the excited state. Usually electronic excitation is also accompanied by an
increase in the rotational and vibrational energy levels. In our discussion, for convenience,
Normally excitation of a ground we depict only transitions between electronic energy levels. In order to understand the
leads One
nature of electronic transitions, it is essential to know the concept of spin multiplicity. A
the excited singlet states. But in
some specified cases, direct molecule with electrons paired and with anti-parallel spins is said to be in the singlet ground
excitation from Soto a triplet level is state ( So). An excited molecule with two of its electrons unpaired a n 4 with
~ossible.Werestrictourdiscussion anti-parallel spins is said to be in the excited singlet stateasuchas S1, S2, S3, ... etc. An
to S, -.S,, S, -.S, etc. transitions
only. excited molecule with two of its electrons unpaired, but with parallel spins is said to be
in the excited triplet state such as TI, Tz,T3, .... etc.
1
state is given by Ihe Normally a molecule in the Sostate on absorbing a quantum of light gets two of its paired
expression 2S t 1, where S (note the
,.
italicised tvDe, is the sum of sDin electroils unpaired and gets transferred to S1 or S2 or S3 etc. levels, but not to ....
T1 or T2 or T3 .... etc levels. Tn,t is, due to excitation, spin multiplicity is not generally
r
1
valuesot electrons. This symbols
should not be confused with S altered. This condition is called the selection rule for electronic transition. In other words,
(Roman type) for singlet state.
absorptioii of energy by the molecule in the ground state leads to allowed transition such as
So 4 S1, So S2, So + S3 and so on. Such excitations and the subsequent energy loss
+

while reaching the ground state are shown by Jablonski diagram (Fig. 19.5). The solid
Imagine a molecule in the singlet
state. It has two electrons with arrows pointing upwards refer to absorption of energy. The solid arrows pointing
anti-parallelsp-ins(such as1 k). downwards refer to energy emission as'light, known as radiative transitions. The wavy
Tken, sum of the electron spins horizontal arrows stand for transition between excited singlet and triplet states without
- -s + -1- - 1 0
2 2
- energy'loss, while wavy vertical arrows stand for transitions between singlet-singlet or
triplet-triplet levels with energy loss as heat (the later is not shown in Fig. 19.5). These
Hence multiplicity wavy arrows stand for nonradiative transitioils (transitions without light emission ).
-2S+1- (2x0)+1.- 1
In Fig. 19.5, the excitation from the singlet ground state to the excited singlet levels
 Photochemistry

Thus the singlet state has two

electrons with anti-parallel spins and
its (2S + 1 ) value is equal to 1. In
the presence of a magnetic field, a
singlet state does not split further.

A molecule in the triplet state has

T3 two electrons with parallel spins
(such as 11).The sum of eledron
spins

IC - s - 2l + 12. - 1

Elq El
Hence the multiplicity
=(2S+1)=3
So,a mdlecule in the triplet state has
* T2 two electrons with parallel spins and
its (2s + 1 ) value is equal to 3. In
the presence of a magwtic field, a
IC triplet state splits into three enp.rgy
levels.

B ,, -7
Isc
T1

l A IC P

F
1
v

Fig. 19.5: Jnblonsh diagram

S1, Sz and S3 are shown by vertical arrows marked A. The excited species at S2 and S3
have very short lifetimes and these species, quickly lose heir energy as heat to the medium
in about lo-'' second and reach S1 level. Such a singlet-singlet transition is called an
internal conversion (IC). The molecule at S1 state has a life time of lo4 - lo-'' second. The
system at S1 may undergo any of the following transitions:

A = Absorption of quanta lending to excitntion to S1, Sz and SJ levels. N d c the mtipudd spies a t various
S levels.

-
IC = Internal conversion from Sp to Sz, from Sz to S1 and from S1 to So.
F = Fluwrscence; n h r o s i t i w from S1 Sowith light emission
ISC'= Intersystem crossing ; S1 -.T1 transition. Note the parallel spins nt v d o & T levels.
The name "fluorescence" is derived
from the name of the mineral,
"fluprite", which emits visible light
P = Phosphorescence; n transition h m TI -. Sowith light emission pn exposure to ultraviolet radiation.
F o r clarity, transitions to and IroT2 and Tp are not shown.

i) Fluorescence
The excited molecule wuld undergo the transition, S1 -.So ,with the emission of light.
This ph'enomenon is called fluorescence 0.Since S1 -.'So transition is allowed by
selection rule, it is very fast. In other words, substances fluoresce in the presence of the
exciting radiation. Once the exciting radiation is stopped, fluorescence stops.
ii) Internal Conversion
o!-Q
The excess energy may be lost as heat while S1 -+ Sotransition takes place which is again a Bewphenone
case of internal conversion and a radiationless transition.
" iii) Intersystem crossing'
The excited molecule could cross over to the first triplet state through SI -. T I transition.
Such a transition involves spin inversion. For such intersystem crossing (ISC) to be
efficient, the energy gap between S1 and Tl levels must be low.
Dynamics and Macron~olecules Ketones have very low energy gap between S1 and TI levels and have high efficiency for
intersystem crossing. Thus benzopheilone has 100% efficiency for intersystem crossing.
Cornpared to ketones, aromatic hydrocarbons are less efficient in intersystem crossing and
olefins are still less efficient. Let us now study two of the processes by which the molecule
in the triplet state could reach the ground state.
1) Phosphorescence
An interesting physical process by which an excited species at Ti level may undergo
transition to So level is by emitting light ; TI Sotransition with emission of light is called
phosphorescence (P). This is a process with a spin change and is a forbidden transition.
"Phosphorescene" is derived from Hence, in contrast to fluorescence, light emission during phosphorescence is slow and it
"phosphorus" which glows in dark
although it is not strictly lasts even after the removal of exciting radiation.
phosphorescent (see Sec. 19.9). 2) Energy Transfer
Another process by which a molecule in the triplet state ( called a donor molecule ) may
lose its excess energy is by energy transfer to an acceptor molecule.This is an instance of
sensitisation which we shall discuss in Sec. 19.7.
Fluorescence and phosphorescence
are exhibited by those substances
In this section, we have dealt with some of the important physical processes only. Let us
which do not undergo chemical now see some of the applications of the study of the physical processes. Study of fluorescent
reaction at the wavelength of behaviour of substances has led to the development of fluorescence spectroscopy. Using
radiation. spectrofluorometers,it is possible to identify several fluorescing substances present in the
same solution, provided they have sufficiently different fluorescent spectra. This method is
3
both sensitive and selective. Even substances with a low concentration such as g/cm
could be detected. For example, this method is quite useful in the analysis of drugs,
pesticides and atmospheric pollutants which are present in trace amounts. Studies based on
fluorescence and phosphorescence provide in~portantdata on the properties of excited states
such as lifetime, energy and electronic configuration.
On the commercial side, fluorescent lamp is one of the applications of the phenomenon of
fluorescence. A fluorescent lamp consists of a glass tube with

a small amount of mercury,

two electrodes, and

a coating of phosphor.
A phosphor is a solid substance which emits fluorescent light when excited by ultravioiet
radiation. The electrodes initiate an electric arc which helps in vaporizing and exciting
mercury atoms. The excited mercury atoms emit ultraviolet radiation. The phosphor, being
excited by ultraviolet radiation, emits fluorescence.
Some other commercial applications of fluorescence are given below:

Optical brighteners are added to detergents to give extra-brightness to the clothes.

Optical brighteners fluoresce in sunlight.

Fluorescent paints are manufactured using suitable additives.

TV screens of different colours are produced using phosphors.

Although molecules in the excited singlet and triplet states could show interesting chemical
behaviour, their studies are included in Unit 14 of the course on Organic ~ e a c t i o n
Mechanisms. In the next section,we shall discuss photosensitisation ;this discussion could
make you understand how a substance gains photochemical activity in the presence of
another photochemically excited substance. Based on the materials of this section, answer
the following SAQ.

SAQ 5
State two differences between fluorescence and phosphorescence. '
 Photochemistry
19.7 BHOTBSENSITISA'XION
Photosensitisation is the process of exciting a n~oleculeby energy transfer from an excited
molecule. In this process, a donor molccule ( D ) absorbs a quantum light and forms an
excited rnolecule 0'). The excited donor molecule then transfers its excitation energy to an
acceptor molecule (Aj in the ground state in order to excite it. This can be explained using
the following reaction sequence:
Light absorption r D + IN 4 D*

The dorlor lnolecule is called the sensitiser. In Sec. 19.6, we mentioned about such energy
transfer by thc excited molecule. The excited acceptor molecule A* could take part either in
chemical reactions or in physical processes which we shall discuss now.

I'hotosensitised Chemical Reactions

IS the excited rnolecule A* has obtained sufficient energy, it will get dissociated and start a
chelnical reaction. The advantage in photosensitised dissociation of a molecule is that it is
enough to transfer energy equivalent to its bond enthalpy to dissociate it. It is so since the
~~hotosensitised rnolecule gets dissociated in the ground state. This is in contrast to direct
photocheinical decornposjtion for which much higher energy is required due to the necessity
of exciting a molecule to upper electrc..rc state (see Sec. 19.4). Thus the energy required for
photosensitised dissociation of a molecule 1s !ess :han that for photochemical dissociation.
Let us consider an example. When irradiated with 253.7 nin light, hydrogen and oxygen
react in prrsence of inercury vapour but not in its absence. Mercury vapour acts as a
sensitiser. The reaction sequence is given below:
Hg + hv 4 Hg*

The cham reaction continues further. The energy transferred to hydrogen molecule by
excited mercury atom is equal to 472 kJ mol-' (as per Eq. 19.4 where h'= 253.7 nm). This
energy is sufficient for thermal dissociation of hydrogen since its bond enthalpy is
430 kJ mol-'(see Table 19.2). Note that excited mercury atom cannot directly dissociate
oxygen molecule since its bond enthalpy is higher ( 497 kJ mol-' ). Further in the absence
of mercury vapour, light of 253.7 nm (or 472 kJ mol-' energy) cannot photolyse hydrogen
or oxygen directly, since the energies needed for their photochemical dissociation are much
higher ( 1420 kJ mol-' and 682 kJ mol-', respectively ) . Hence mercury vepour is
essential as a sel~sitiserfor H2-02 photochemical reaction at h = 253.7 nm.
A well-known photosensitised reaction is photosynthesis. Chlorophyll ("chl") and other
plant pigments act as photosensitisers in the synthesis of starch from carbon dioxide and Chlorophyll is the name given for a
group of compounds with minor
water. A simplified reaction sequence is as follows: variation in structure. Chlorophyll-a
absorbs effectively in the redregion
of sunlight; the red light is in
abundance in sunlight.

The reaction mechanism is very complex. Energy calculations show that apart kom
chlorophyll, there must be other coloured light-absorbing materials (pigments) which also
provide energy required for the synthesis of starch.
Photosensitisation is frequently used by chemists for preparing compounds which cannot be
formed by thermal or direct irradiation methods.

Sensitised Fluorescence
Let us now discuss a physical prncess which proceeds through sensitisation. Thallium
vapour does not give rise to fluorescence when directly irradiated with light of wavelength
253.7 nm. But if mercury vapour is also included in the reaction vessel, thallium exhibits
fluorescence. Mercury atoms get excited first and transfer energy to thallium atoms to excite
them. The excited thallium atoms emit fluorescence as they g o down to the ground level.
- Using the ideas learnt in this section, answer the following SAQ.
 SAQ 6
Oxalic acid is not decomposed directly by light of wavelength, 335 nm.Irradiation in
presence of uranyl ( UOi ) ion results in the decomposition of oxalic acid to carbon
monoxide and carbon dioxide.
i) How do you explain this reaction?
ii) Suggest a reaction sequence.
(Note: The above procedure was used in the earlier chemical actinometric method of
.determining the light intensity. Oxalic acid solution of known concentration was irradiated
in presence of uranyl ion and the unreacted oxalic acid was estimated using potassium
permanganate solution. From the amount of oxalic a d d reacted, the intensity of light was
calculated using I$ for a particular wavelength.)

19.8 APPLICATIONS OF PHOTOCHEMISTRY

We have earlier indicated the applications of the study of fluorescence, phosphorescence,
flash photolysis and photosensitisation. Apart from the above, a few are worth mentioning.
Synthetic organic chemists have increasingly started using photochemical methods for
synthesis due to greater efficiency and selectivity as compared to dark reactions.
Photochemistry offers a method of conducting reactions which are not possible
Three common types of cells used thermodynamically.
for converting light into electricity In the analysis of pollutants, photochemistry plays an important role. For example,
are given below :
photochemical studies have indicated how ozone layer is affected by chlorofluorocarbons
i) Photoelectric cells or photo (freon) used as refrigerants, solvents and spray- propellants. The simplest freon is CFzC12.
cells convert light into Freon is chemically inert and remains as such for years. But when it reaches stratosphere
electricity using ' (10 km to 50 km above the earth's surfaw), freon decomposes and gives out free chlorine
photosensitive cathode. (see atoms. These ehlorine atoms can react with ozone decomposing it. This can cause depletion
Sec .19.3). of ozone layer. It is a matter of serious concern since ozone layer protects our planet from
low wavelength. portion of sun's rays (290 nm-320 nrn wavelength ). Irradiation with such
ii) Photovoltaic'cellshave two high energy radiation could produce skin cancer. As a result of photochemical studies,
dissimilar silicon (or alternatives are tried for freon.
germanium) crystals in close The present energy crisis has compelled the scieatists to look for alternatives. Solar energy,
contact. Irradiation of light if properly utilised through suitable photochemical reactions, could offer a solution for this
causes flow of electrons from energy problem. In this connection we now discuss the functioning of a photogalvanic cell
one crystal to other. Solar which is helpful for solar energy conversion through chemical reactions.
cells used in calculators are
photovoltaic cells.
Photognlvanic Cell
A reversible cell which converts light into electricity through an intermediate chemical
iii) Photogalvaniccellsconvert
reaction is called photogalvanic cell. It undergoes cyclic charging and recharging processes
light into electricity through
as explained below:
chemical reactions.
Charging in presence of light
Two substances undergo oxidation -reduction in presence of light

In this OA1 and RA1 refer to oxidised and reduced forms of substance 1,while
OA2 and RA2 to those of substance 2. In this process, light is used in conducting a
chemical reaction. This photochemical reaction is thermodynamically non-spontaneous
( A G > 0 ). By the above photochemical process, the cell is charged.

Discharging in the absence of light

In the absence of light, the reaction reverses sponhneously ( A G < 0 ):
+
d ~ 2RA1 OA+
~ RA2
- -
tris (52' bipyridine)
-+

ruthenium (11) ion The chemical energy gained during charging is converted into electrical energy and the cell
is discharged.
An example of such a system is given below:
74
 Pbotochromism is the light induced
reversible cobur change. The wlour
change is brought about by
reversible reactions. Photochromic
sunglasses darken in sunlight and
protect the eyes from thd excess
~ e "+ Ru ( bipy )3
[ ]2r "
dark [lIu(bipy)3]l*+~eh light. The sunglassescontain Ag'
and CL' ions and react reversibly as
follows :
In this system, tris- (2,2'-bipyridine) ruthenium @I) - trii- (2,2'-bipyridine) ruthenium (111)
In presence dsunlight :
and iron (11) - iron(II1) are the two redox pairs. The platinum electrodes serve as electrical
contacts (Fig. 19.6). Ag+ (s) + c"+ (s) -.. Ag (s) + c'2+
(s)

In the next section, we shall discuss chemiluminescence, which in a way, is the reverse of Silver formed is responsible for the
photochemical reaction. darkening of the glasses.
In tbe absence d sunlight r
SAQ 7
Ag ( s ) + Cu: (s) -.. ~ g (s)+ + Cu+(s)
Suggest a basic difference between galvanic cells (discussed in Unit 17) and photogalvanic cells.
\.........................................................................................................................
............\............
...................................................................................................................................................

19.9 OHEMILUMINESCENCE
Chemiluminescence is the e ission of light as a result of chemical reaction at room
'P
temperature. It must be clear, y understood that chemiluminescence is not due to any
photophysical process like fluorescence or phosphorescence. Some examples of
chemiluminescence are given below :

Glow of phosphorus is due to slow oxidation ;it is not due to phosphommnce as name suggests.
' Oxidation of Grignard compounds by air or oxygen results in greenish-blue
luminescence.
Will-o-the-wisp(mistaken as light produced by evil spirits) is the glow caused by the
oxidation of decaying wood in marshy places.
Emission of light by fuefly is due to oxidation of some proteins in its body (this is also
called bioluminescence).

19.10 SUMMARY -
In this unit, we have discussed the physical and chemical processes accompanying ligh
absorption.
The laws of photochemistry have been stated and explained.

Quantum yield has been.defmed and the method of determining it has been explained.

The variation in energy requirements for photochemical and thermal dissociation has
been discussed.

The rate expresr;ionshave been derived for some photochemical reactions.

Dynamics and Macromolecules
e The photophysical processes have been explained. The applications of fluorescence and
phosphorescence have been stated.
Photosensitisation has been explained with examples.

e Applications of photochemical studies have been listed.

Chemiluminescence has been defined.

19.11 TERMINAL QUESTIONS

1) State two differences between a chemical reaction and a photochemical reaction.

2) If 10% of the energy of a 100 W incandescent bulb generates visible light having
average wavelength 600 nm, how many quanta of light are emitted in 1 0 minutes?

(Hint :A tungsten filament bulb is an example of incandescent bulb. Incandescence is

light emission due to intense heat.)

b) The extinction coefficient ( E ) value of a substance is 4.66 m2 mol-'. Calculate the

I
coilcentration of its solution (in molarity) which has - = 0.2. The thickness of the
10
cell is 1.00 cm.

4) At 478 nm, hydrogen and chlorine combine to give hydrogen chloride with a quantum
efficiency of 1.00 x lo6. If the absorbed intensity is 2.50 x J ss-', calculate the
amount of hydrogen chloride formed in 1 0 minutes.
(Hint: The unit for amount is mole.)

5j State some of the applications of fluorescence and phosphorescence studies.

6) In the photoche~nicalreaction between hydrogen and bromine at 511 nm wavelength,

the first step is the photolysis of bromine but not of hydrogen. Why?
(Hint: Use Table 19.2.)
7) Fill in the blanks:

i) Light emission through S1 + So transition is called ..............

ii) Light emission through T1 + So transition is called ...............
iii) The energy required for photolysis of bromine is ......than its bond enthalpy.
8) Suggest a condition that should be fulfilled in order to photolyse a molecule using a
sensitiser.

19.12 ANSWERS
Self-assessment Questions
6.626 x
1) Energy absorbed per photon = x 3 x 10' joule pboton-l
250 x
= 7.95 x 10-l9 joule photon-'

Assume that the reaction is conducted fort seconds.

Intensity in joule second-'
Hence, I, =
7.95 x 10-l9 joule photon-'

- 4.18 x lo3 joule/r second

-
7.95 x 10-l9 joule photon-1

-- 5'26 lo2'photon second-'

t
 Photochemistry
1.85 x 10" mol
Amount of HI decomposed in 1second =
t second

-am
= 1"5 x*6.022x id3 molecule semnd-'
dt t

- l4 1022 molecule sicond-'

t
4 d d t =
1.114x10~~/t
Using Eq. 1 9 . 3 , b =
Ia 5.26 x 102'/t

2) Oxygen molecules could quench the excited species. Hence in presence of oxygen,
+Ha is low.

3) For HI, the difference between photochemical dissociation energy and bond enthalpy
is 68 kJ mol-'and the excited hydrogen atom carries this excess enetgy.

ii) d[:F1 = 0 = Ia - k2[CH3][CH3CHO] + k3[CHCO] - = [ c H ~ ] ~

iii) d [ C H f O l
dt
+ d[CH31 = O = I a _ & [ C H 3 l 2
dt

iv) [CH3 ] = (&) 45

5) -
The light emission during fluorescence is due to S1 Sotransition. Fluorescence,
being due to an allowed transition, is quite fast and stops as soon as exciting source is
removed.
The light emission during phosphorescence is due to T1 -,Sotransition. Since this
transition is forbidden, it is slow and it persists even if the exciting source is removed.

i) UO?ion acts as a photosensitiser and facilitates the decomposition of oxalic acid.

6)
ii) U& + ttv + ( UOi )*

( UO? )' + H2C204 UO? + H a + CO + (207.

7) In a galvanic cell, the potential difference is due to conversion of chemical energy into
electrical enetgy .
In a photogalvanic cell, light energy is converted into chemical energy, which is then,
converted into electrical enetgy.

Terminal Questions
1) A photochemical reaction could take place even if AG is positive. For a chemical
reaction to take place, AG must be negative at a given temperature and pressure.
In a p h o t o c h ~ i c areaction,
l the reacting molecules get the excitation energy through
light,absorption. Ip a chemical reaction, the reacting molecules get the activation ener-
gy through collidons.
2) A 100 W bulb emits 100 J s-' of which 10 J s-' comes out as light.
 Dynamics and Maeromoleculn 6.626 x 10"~x 3 x log
Energy per quantum = joule photon-'
600 x
= 3.31 x 10-l9 joule photon-'
-
Number of quanta
emitted per second

=
I -
Energy emitted per second

10 joule second-'
Energy per photon

3.31 x 10-l9 joule photon-'

= 3.02 x 1019 photons second-'

Number of quanta emitted = 3.02 1019 600 photons

in 10 minutes
= 1.81 x lon photons

I0
I/Io = 0.2; so, - = 5
I
A - log Io/I = 0.699 = E c1

=
=
15 mol m-3
15 x mol dm" - 0.015 M
6.626 x 10"' x 3 x log jouk photoil
1M - 1 dmJ and
mol
Energy per quantum =
478 x
1 rn? - (10dm )-)
- 10JdmJ = 4.16 x 10-l9 joule photon-'

Number of quanta
per sewnd
I =

-
Intensity in joule sewnd-'
Energy per photon
2.50 x 10" joule sewnd-'
.4.16 x 10-l9 joule photon-'
= 6.01 x 1015 photons second-'
CWHa
dt W H ~
h a = -; hence, -
1, dt
- h a .I,

dNHa - 6.01 x 1015x 1.00 x lo6 molecules second-'

i.e.,' -
dt
= 6.01 x lo2' molecules second-'
Number of of HC1
= 6-01 1012x60() molecules
formed in 10 minutes
= 3.61 x molecules

Amount of HCl formed 3.61 x

in 10 minutes } = 6.022 x loz
mole

= 5.99 mole.
See Sec. 19.6.
Light of wavelength 511 n m is just sufficient to photolyse bromine molecule. To
photolyse hydrogen molecule directly, light of higher energy is required.
i) fluorescence ii) phosphorescence
iii) higher
The energy emitted by the excited sensitiser during its return to ground state should be
atleast equal to the energy required for the thermal decomposition of the molecule.