Chemical Kinetics Theory

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by  UBD

Chemical Kinetics 455

Chapter

11
Chemical Kinetics

The branch of physical chemistry which deals with the rate at x dx x x 
which the chemical reactions occur, the mechanism by which the chemical or  2 1
reactions take place and the influence of various factors such as t dt  t2 t1 
concentration, temperature, pressure, catalyst etc., on the reaction rates is where x or dx is the concentration change, i.e., (x x ) in the
2 1
called the chemical kinetics.
time interval t or dt, i.e., (t t ).
2 1
Types of chemical reactions Concentration is generally expressed in active mass, i.e., mole L–1

 The rate measured over a long time interval is called average rate
On the basis of reaction rates, the chemical reactions have been
and the rate measured for an infinitesimally small time interval is called
classified into the following three types,
instantaneous rate and
(1) Very fast or instantaneous reactions : These reactions occur at a Instantaneous ra te  (Average rate)
t0
very fast rate generally these reactions involve ionic species and known as
 For the reaction aAbBcCdD
ionic reactions. It is almost impossible to determine the rates of these
Rate of disappearance of a reactant is negative
reactions.
d[A]
Examples  Rate of disappearance of A
dt
(i) AgNO3NaCl AgClNaNO3 (Precipitation reaction) d[B]
(PPt.)  Rate of disappearance of B
dt
(ii) HClNaOHNaClH O (Neutralization reaction) Rate of formation of a product is positive
2
(acid) (base) (Salt)
d[C]
Rate of formation of C
(2) Moderate reaction : These reactions proceed with a measurable
dt
rates at normal temperature and it is these reactions are studied in chemical
d[D]
kinetics. Mostly these reactions are molecular in nature. Rate of formation of D
dt
Examples
 In terms of stoichiometric coefficient rate may be expressed as
(i) Decomposition of H2O2: 2H2O2 2H2OO2 dx 1 d[A]1 d[B]1 d[C] 1 d[D]
dt a dt b dt c dt d dt
(ii) Decomposition of N O : 2N O 2N O O
2 5 2 5 2 4 2  The rate of reaction is always positive.
(3) Very slow reactions : These reactions are extremely slow and  The rate of chemical reaction decreases as the reaction proceeds.
take months together to show any measurable change. Unit of conc.
 Unit of rate of a reaction = =mole L–1 time –1
Examples Unit of time
In term of gaseous reaction the unit is atm time and
(i) Rusting of iron : Fe O xH O Fe O .xH O -1
2 3 2 Hydrate2d fe3rric oxid2e (Rust) Rate in atm time-1= Rate in mole L1time1RT
Factors affecting rate of a reaction
(ii) 2H O Room temperature2H O
2 2 2
The rate of a chemical reaction depends on the following things
Rate of a reaction (1) Nature of reactants
The rate (speed or velocity) of a reaction is the change in (i) Physical state of reactants : This has considerable effect over rate
concentration in per unit time. of reaction.

Conc. of product
e/lit.)
ol
m
c. (
n
o
C
Conc. of reactant
Time (Sec)
Change of concentration
456 Chemical Kinetics
Where k is called rate constant or velocity constant.
Gaseous satae Liquid stateSolid state

dx
Decreasing rate of reaction When [A][B]1mol/litre, then k
dt
(ii) Physical size of the reactants : Among the solids, rate increases
Thus, rate constant k is also called specific reaction rate.
with decrease in particle size of the solid.
 The value of rate constant depends on, nature of reactant,
(iii) Chemical nature of the reactants
temperature and catalyst. It is independent of concentration of the
(a) Reactions involving polar and ionic substances including the
reactants.
proton transfer reactions are usually very fast. On the other hand, the
n1 1n
reaction in which bonds is rearranged, or electrons transferred are slow.  Unit of rate constant litre sec1 mol sec1
   
(b) Oxidation-reduction reactions, which involve transfer of mol litre
electrons, are also slow as compared to the ionic substance. Where n order of reaction.
(c) Substitution reactions are relatively much slower. Rate law : Molecularity and Order of a reaction
(2) Effect of temperature : The rate of chemical reaction generally Molecularity is the sum of the number of molecules of reactants
increases on increasing the temperature. The rate of a reaction becomes involved in the balanced chemical equation. Molecularity of a complete
almost double or tripled for every 10oC rise in temperature. reaction has no significance and overall kinetics of the reaction depends
upon the rate determining step. Slowest step is the rate-determining step.
Temperature coefficient of a reaction is defined as the ratio of rate
This was proposed by Van't Hoff.
constants at two temperatures differing by (generally 25°C and 35°C) 10°C.
Example : NH NO N 2H O (Unimolecular)
 Temperature coefficient kat (t10oC) k35oC 4 2 2 2
kat toC k NOO3NO2 O2 (Bimolecular)
25oC
2NOO 2NO (Trimolecular)
(3) Concentration of reactants : The rate of a chemical reaction is 2 2
directly proportional to the concentration of the reactants means rate of The total number of molecules or atoms whose concentration
reaction decreases with decrease in concentration. determine the rate of reaction is known as order of reaction.
(4) Presence of catalyst : The function of a catalyst is to lower down Order of reaction = Sum of exponents of the conc. terms in rate law
the activation energy. The greater the decrease in the activation energy
For the reaction xAyBProducts
caused by the catalyst, higher will be the reacRtieoacnti orna ptaeth.
Without catalyst The rate law is Rate[A]x[B[y

E Then the overall order of reaction. nxy
a
E Reaction where x and y are the orders with respect to individual reactants.
gy a pcaathta lwysitth deter min edI fb rye athctei osnlo wis eisnt tshteep f oorfm m oefc hraenacistmion. mechanism then the order is
ner Reactants Energy of Reaction 2A3BA B
al E 2 3
nti Products ABAB(fast)
e
Pot A catalyst changes the reaction path ABB2 AB3(slow) (Rate determining step)
(5) Effect of sunlight : There are many chemical reactions whose
rate are influenced by radiations particularly by ultraviolet and visible light. AB AA B (fast)
3 2 3
Such reactions are called photochemical reactions. For example, (Here, the overall order of reaction is equal to two.)
Photosynthesis, Photography, Blue printing, Photochemical synthesis of
compounds etc.  Molecularity of a reaction is derived from the mechanism of the given
The radiant energy initiates the chemical reaction by supplying the reaction. Molecularity can not be greater than three because more than three
necessary activation energy required for the reaction. molecules may not mutually collide with each other.
 Molecularity of a reaction can't be zero, negative or fractional.
Law of mass action and Rate constant
order of a reaction may be zero, negative, positive or in fraction and greater
The rate at which a substance reacts is directly proportional to its than three. Infinite and imaginary values are not possible.
active mass and the rate at which a reaction proceeds is proportional to the
 When one of the reactants is present in the large excess, the
product of the active masses of the reacting substances. second order reaction conforms to the first order and is known as pesudo
 For a reaction, aAbB product unimolecular reaction. (Table 11.1)
dx dx
Rate  [A]a[B]b;  k[A]a[B]b
dt dt
Table : 11.1 Order and molecularity of some reaction
S. Chemical equation Molecularity Rate law Order w.r.t.
No. First Second reactant Overall
reactant
1. aAbBproduct a + b dx a b a + b
 k[A]a[B]b
dt
2. aAbBproduct a + b dx 2 zero, if B is in excess 2
 k[A]2[B]0
dt 
3. 2H O Pt,2H OO 2 (Bimolecular) dx 1* ----- 1
2 2 2 2  k[H O ]
dt 2 2
Chemical Kinetics 457
4. CH3COOCC2HH5COHO2HOCHHO H 2 (Bimolecular) ddxtk[CH3COOC2H5] 1* Zero, eixf cHes2Os is in 1
3 2 5
5. C12SHucr2o2seO11H2OH 2 (Bimolecular) ddxtk[C12H22O11] 1* Zero, eixf cHes2Os is in 1
C H O C H O
6 12 6 6 12 6
Glucose Fructose
6. (CH ) CClOH  2 (Bimolecular) dx 1* Zero, if OH– does not 1
3 3  k[(CH ) CCl] take part in slow
(CH ) COHCl dt 3 3 step
3 3
7. CH3ClOH  CH3OHCl 2 (Bimolecular) ddxtk[CH3Cl][OH] 1 1 2
8. C H N Cl C H ClN 1 (Unimolecular) dx 1 ---- 1
6 5 2 6 5 2  k[C H N Cl]
dt 6 5 2
9. CH CHOCH CO 1 (Unimolecular) dx 1.5 ---- 1.5
3 4  k[CH CHO]3/2
dt 3
10. H O 2I2H  2H OI 5 dx 1 1 2
2 2 2 2 dtk[H2O2][I] (H+is medium)
11. 2O3 3O2 2 (Bimolecular) dxk[O ]2[O ] 1 -1 with respect to O2 1
dt 3 2
*Pseudo-unimolecular reactions.
Table : 11.2 Rate constant and other parameters of different order reactions
Order Rate constant Unit of rate Effect on rate by changing conc. (Half-life period) T=
50
constant to m times
0 x conc. time–1 No change a
k  (mol L s)
0 t –1 –1 2k
0
1 2.303  a  time–1 (s–1) m times 0.693
k  log  , CC ek1t
1 t 10ax 0 k1
2.303 (ax )
N N ek1t, k  log 1
0 1 (t t ) 10 (ax )
2 1 2
2 1 1 1 x conc–1 time–1 m2 times 1
k2  t (ax)a  ta(ax) (for the case when (Lm oml oLl–– 11)s s–1 –1 k2a
each reactant has equal concentration)
2.303 b(ax)
k2  t(ab)log10a(bx)(for the case when both
reactants have different concentration)
3 1  1 1  conc–2 time–1 m3 times 3
k3  2t (ax)2 a2 (mLo ml Lo–l1) –2 s s –1 2k3a2
2 –2 –1
n 1  1 1  conc(1–n) time–1 mn times 2n11
kn (n1)t (ax)n1 (a)n1; n2 (Lm(n–1)o ml Lo–l1)(1–(n1–) n) ss–1 –1 (n1)kn(a)n1

Methods for determination of order of a reaction 2.303 a
k  log (For first order reactions)
t 10 (ax)
(1) Integration method (Hit and Trial method)
(i) The method can be used with various sets of a,x and t with 11 1 
k   (For second order reactions)
 
integrated rate equations. t a ax
(ii) The value of k is determined and checked for all sets of a,x
1  1 1 
and t. k     (For third order reactions)
2t (ax)2 a2
(iii) If the value of k is constant, the used equation gives the order
(2) Half-life method : This method is employed only when the rate
of reaction.
law involved only one concentration term.
(iv) If all the reactants are at the same molar concentration, the
kinetic equations are : t a1n; t ka1n; logt logk(1n)loga
1/2 1/2 1/2
458 Chemical Kinetics
A plotted graph of logt vs log a gives a straight line with slope Subtracting equation (ii) from (i),
1/2
(1n), determining the slope we can find the order n. If half-life at dC  dC 
log  1log  2
different concentration is given then, 10 dt  10 dt 
n …..(iii)
1 1 (t ) a n1 log10C1log10C2
(t )  ; (t )  ; 1/2 1  2 
1/2 1 a1n1 1/2 2 a2n1 (t1/2)2 a1  dC1 and dC2 are determined from concentration Vs time
dt dt
log (t ) log (t ) (n1)[log a log a ]
10 1/2 1 10 1/2 2 10 2 10 1 graphs and the value of 'n' can be determined.
log (t ) log (t ) (5) Ostwald's isolation method (Initial rate method)
n1 10 1/2 1 10 1/2 2
(log a log a ) This method can be used irrespective of the number of reactants
10 2 10 1
This relation can be used to determine order of reaction ‘n’ involved e.g., consider the reaction, n1An2Bn3CProducts.
Plots of half-lives Vs concentrations (t a1–n) This method consists in finding the initial rate of the reaction taking
1/2 known concentrations of the different reactants (A, B, C).
Z ero order 1st order 2nd order 3rd order
Suppose it is observed as follows,
t 1/2 t 1/2 t 1/2 t 1/2 of A is (di)o uKbeelepdi,n gth teh er actoen ocfe nrteraacttioionns obfe cBo maneds Cfo ucro ntsimtaenst. , Tifh icso nmceeanntsr atthioant,
Rate [A]2 i.e., order with respect to A is 2
C onc. Conc. 1/a 1/a2 (ii) Keeping the concentrations of A and C constant, if concentration
(3) Graphical method : A graphical method based on the respective
of B is doubled, the rate of reaction is also doubled. This means that, Rate
rate laws, can also be used.
 [B] i.e., order with respect to B is 1
(i) If the plot of log(ax) Vs t is a straight line, the reaction
(iii) Keeping the concentrations of A and B constant, if
follows first order.
concentration of C is doubled, the rate of reaction remains unaffected. This
1
(ii) If the plot of Vs t is a straight line, the reaction means that rate is independent of the concentration of C i.e., order with
(ax) respect to C is zero. Hence the overall rate law expression will be, Rate =
follows second order. k[A] [B] [C]
2 0
1  Overall order of reaction = 2 + 1 + 0 = 3.
(iii) If the plot of Vs t is a straight line, the reaction
(ax)2 Theories of reaction rate
follows third order.
(1) Collision theory
1
(iv) In general, for a reaction of nth order, a graph of (i) The basic requirement for a reaction to occur is that the reacting
(ax)n1 species must collide with one another. This is the basis of collision theory
Vs t must be a straight line. for reactions.
Plots from integrated rate equations (ii) The number of collisions that takes place per second per unit
volume of the reaction mixture is known as collision frequency (Z). The
Conc. [A]  Z ero order l og. [A]  1st order [1A ] 2nd order [A1 ]2 3rd order vcaasluee o fo f(b iiicnio)a lrlEiysv ieocronyl l ifcsroieolqlniussie.o nnc yd oise sv enroyt hbirgihn go af tchhee moricdaelr cohfa n1g0e. 2T5hteo c1o0ll2is8io nins
t t t t that actually produce the product are effective collisions. The effective
collisions, which bring chemical change, are few in comparison to the total
Plots of rate Vs concentrations [Rate = k(conc.) ]
n number of collisions. The collisions that do not form a product are

Zero order 1st order 2nd order 3rd order ineffective elastic collisions, i.e., molecules just collide and disperse in
    different directions with different velocities.
Rate Rate Rate Rate Fcarpacatbiolen ooff bmr ionlgeicnugl e s

effective collisions
tThheu ns thf op (roC4 wt)ow neVcor. a odnfi't ff tehHreea nfcft o dninicfifeCtenioartnelr cnac.t to iiaonlnc e mnWetrthahetoriode n :s' (C n o'CTn hci.se) 2 a trhnaedt e o Cordf er ree aqocuf(t Caitoothinnoec n.v)r,3a e cr aiacentsi obanes. Fraction of molecules
written in the form, 1 2 Energy E
Distribution of energies at a definite
dC dC temperature
1 kCn and 2 kCn (iv) For a collision to be effective, the following two barriers are to be
dt 1 dt 2 cleared,
Taking logarithms, (a) Energy barrier : “The minimum amount of energy which the
colliding molecules must possess as to make the chemical reaction to occur,
dC 
log  1log knlog C …..(i) is known as threshold energy”.
10 dt  10 10 1
 In the graph 'E' corresponds to minimum or threshold energy for
dC  effective collision.
and log  2log knlog C …..(ii)
10 dt  10 10 2
Chemical Kinetics 459
 There is an energy barrier for each reaction. The reacting species must E E  E
a (Thresholdenergy) (Reactants)
be provided with sufficient energy to cross the energy barrier.
Activation energy = Threshold energy – Average kinetic energy of
(b) Orientation barrier : The colliding molecules should also have
proper orientation so that the old bonds may break and new bonds are the reacting molecules.
formed. For example, NO (g)NO (g)N O (g). During this (a) Zero activation energy = Fraction of effective collision (f) will be
2 2 2 4
reaction, the products are formed only when the colliding molecules have very large = Very fast reaction (Instantaneous reaction).
proper orientation at the time of collisions. These are called effective
(b) Low activation energies = Fraction of effective collision (f) will
collisions.
be large = Fast reactions.

Collisions not properly oriented (c) High activation energies = Fraction of effective collision (f) will

O O be small = Slow reaction.
O O
N O N aMpoplreocauclhes N O N MSeopleacrautlee s N N The activation energy (Ea) depends upon the nature of chemical
O O O O O O bporonddus cutsn.d ergoing rupture and is independent of enthalpies of reactants and
NO2 + NO2 Collision NO2 NO2 According to the concept of activation energy, the reactants do not
Pro perly oriented collisions form products No product change directly into the products. The reactant first absorb energy equal to
activation energy and form activated complex. At this state, the molecules
O O O O O O
must have energy at least equal to the threshold energy. This means that the
N N Molecules N N Bond N N reaction involves some energy barrier which must be overcome before
O O approach O O Formation O O products are formed. The enTehrrgeys hboladr rier is kAncotiwvante da s activation energy barrier.
NO2 + NO2 Collision N2O4 Product Et energy (Et) complex
(v) Thus, the main points of collision theory are as follows,
Fig. 11.1 Energy barrier
(a) For a reaction to occur, there must be collisions between the Ea (activation energy)

reacting species.

E Energy of the
effective( bin) fOornmlyi nag tcheer tparino dufrcatcst. ion of the total number of collisions is r Rea(cEta) nts E reaction
E r
p
(c) For effective collisions, the molecules should possess sufficient Products (E)
p
energy as well as orientation.
Progress of reaction
(vi) The fraction of effective collisions, under ordinary conditions (2) Transition state theory
may vary from nearly zero to about one for ordinary reactions. Thus, the (i) According to transition state theory the activated complex is
rate of reaction is proportional to : supposed to be in equilibrium with the reactant molecules.
(ii) Once the transition state is formed it can either return to the
(a) The number of collisions per unit volume per second (Collision
initial reactants or proceeds to form the products.
frequency, Z) between the reacting species
(iii) Assuming that once formed the transition state proceeds to products
(b) The fraction of effective collisions (Properly oriented and we can say that rate is proportional to concentration of transition state.
dx Mathematically, Rate Transition state
possessing sufficient energy), f i.e., Rate  fZ
dt Rate= Constant × Transition state
Where f is fraction of effective collision and Z is the collision (iv) The activation energy for the forward reaction, (Ef)and the
a
frequency.
activation energy for the reverse reaction (Er) are related to the enthalpy
a
(vii) The physical meaning of the activation energy is that it is the
minimum relative kinetic energy which the reactant molecules must possess (H)of the reaction by the equation H Ef Er.
a a
for changing into the products molecules during their collision. This means
(a) For endothermic reactions, H 0, so that Er Ef
that the fraction of successful collision is equal to eEa/RT called a a
Boltzmann factor. (b) For exothermic reaction, H0, so that Er Ef.
a a
(viii) It may be noted that besides the requirement of sufficient Arrhenius equation
energy, the molecules must be properly oriented in space also for a collision
Arrhenius proposed a quantitative relationship between rate
to be successful. Thus, if ZAB is the collision frequency, P is the orientation constant and temperature as,
factor (Steric factor) then, kPZ .eEa/RT. If we compare this k AeEa/RT …..(i)
AB
equation with Arrhenius equation. k AeEa/RT The equation is called Arrhenius equation.
In which constant A is known as frequency factor. This factor is related
We know that pre-exponential form 'A' in Arrhenius equation is, to number of binary molecular collision per second per litre.
APZ . E is the activation energy.
AB a
T is the absolute temperature and
Concept of activation energy
R is the gas constant
The excess energy (Over and above the average energy of the Both A and E are collectively known as Arrhenius parameters.
a
reactants) which must be supplied to the reactants to undergo chemical
Taking logarithm equation (i) may be written as,
reactions is called activation energy (E ),
a
460 Chemical Kinetics
logk logA Ea …..(ii) is consumed in the following two reactions,2NO2k1N2O4;
2.303RT
NO SO k2NOSO
The value of activation energy (E ) increases, the value of k 2 2 3
a The rate of disappearance of NO will be sum of the rates of the
decreases and therefore, the reaction rate decreases. 2
When log k plotted against 1/T, we get a straight line. The d[NO ]
two reactions i.e.,  2 2k [NO ]2 k [NO ][SO ]
E dt 1 2 2 2 2
intercept of this line is equal to log A and slope equal to a .
2.303R Photochemical reaction
Therefore E 2.303Rslope.
a Absorption of radiant energy by reactant molecules brings in
photophysical as well as photochemical changes. According to Einstein's law
of photochemical equivalence, the basic principle of photo processes, each
k reactant molecule is capable of absorbing only one photon of radiant
og energy. The absorption of photon by a reactant molecule may lead to any of
l
the photo process. Reactant molecule
Slope  Ea Absorpt io n of photon (As
1/T 2.303R per E instein law)
Rate constants for the reaction at two different temperatures T
1 Excitation of Knock out the electron from
and T , electronic level the reactant species
2

k E 1 1 
logk12  2.30a3RT1 T2 …..(iii) Excited mol ecule Photoelectric effect
where k and k are rate constant at temperatures T and T
1 2 1 2
respectively (T2 T1). Photophysical process Photochemical process
(i) Oxidation
Mechanism of the reaction (i) Fluorescence (ii) Reduction
(ii) Ph osp horescence (iii) Dissociation
(1) Reaction involving first order consecutive reactions (iv) Double decomposition
(i) In such reactions, the reactions form a stable intermediate (v) Isomeric transformation

compound before they are finally converted into the products. (vi) Photosensitization
(ii) For example, reactants (R) are first converted to intermediate (I)
The chemical reactions, which are initiated as a result of absorption
which is then converted to product (P) as
of light, are known as photochemical reactions. In such cases, the absorbed
Rk1Ik2P energy is sufficient to activate the reactant molecules to cross the energy
Therefore, the reaction takes place in two steps, both of which are barrier existing between the reactants and products or in other words,
first order i.e., energy associated with each photon supplies activation energy to reactant
molecule required for the change.
Step I : Rk1I ; Step II : Ik2P
(1) Characteristics of photochemical reactions
This means that I is produced by step I and consumed by step II. In these
(i) Each molecule taking part in a photo process absorbs only one
reactions, each stage will have its own rate and rate constant the reactant
concentration will always decrease and product concentration will always hc
photon of radiant energy thereby increasing its energy level by hv or
increase as shown in fig. 
P
(ii) Photochemical reactions do not occur in dark.

on (iii) Each photochemical reaction requires a definite amount of
ntrati I energy which is characteristic of a particular wavelength of photon. For
nce example, reactions needing more energy are carried out in presence of UV
Co R light (lower , more E/Photon). A reaction-taking place in UV light may
Time
not occur on exposure to yellow light (lower and lesser E/Photon)

Concentration profile of reactants (R), intermediate (I)
(iv) The rate of photochemical reactions depend upon the intensity
and products (P) as a function of time
of radiation’s absorbed.
(2) Reaction involving slow step : When a reaction occurs by a (v) The G values for light initiated reactions may or may not be
sequence of steps and one of the step is slow, then the rate determining negative.
step is the slow step. For example in the reaction
(vi) The temperature does not have marked effect on the rate of
Rk1I; Ik2P, if k k then I is converted into light initiated reactions.
1 2
products as soon as it is formed, we can say that (2) Mechanism of some photochemical reactions
d[R] d[P] (i) Photochemical combination of H and Cl : A mixture of H and
 k [R] 2 2 2
dt dt 1 Cl on exposure to light give rise to the formation of HCl, showing a
2
(3) Parallel reactions : In such type of reactions the reactants are
chain reaction and thereby producing 106to108molecules of HCl per
more reactive, which may have different orders of the reactions taking place
photon absorbed.
simultaneously. For example, in a system containing NO and SO ,NO
2 2 2
H Cl sunlight2HCl
2 2
Chemical Kinetics 461
The mechanism leading to very high yield of HCl as a result of
chemical change can be as follows. Chlorine molecules absorb radiant  Different reactions have different rates because their activation
energy to form an excited molecule which decomposes to chlorine free energies are different. Lesser the activation energy faster is the reaction.
radicals (Cl) to give chain initiation step.
1
 The reaction, NO O NO , exhibits a small negative
Light absorption step : Cl hvCl * ........(i) 2 2 2
2 2
temperature coefficient and the rate of reaction decreases with increase of
Chain initiation step : Cl2*ClCl ........(ii) temperature.
The chlorine free radical then combines with H2 molecule to form  Fuels in contact with oxygen do not burn by themselves. This is because
HCl and H free radical. The H free radical so formed again combines they need activation energy (provided by the flame) to initiate the reaction.
Thus, fuels are thermodynamically unstable (G is –ve) but kinetically
with another Cl molecule to give HCl and Cl free radical back resulting
2 stable.
into chain propagation step.
 Quantum efficiency of the photochemical reaction,
Chain propagation step : ClH2HClH ........(iii) H Cl hv2HCl is very high while that of
2 2
HCl2HClCl H Br hv2HBr, is very low. This is because after the chain
2 2
The combination of two Cl free radicals leads to chain initiating step X hv2X ClH HClH is exothermic
terminating step. 2 2
Chain terminating step : ClClCl ........(iv) whereas BrH2 HBrH is endothermic.
2
(ii) Photochemical combination of H and Br : The combination of  Drapper’s effect : H Cl 2HCl, Such reactions are
2 2
2 2
H and Br to form HBr in presence of light is also an example of chain accompanied by the increase in the volume. This is called Drapper’s
2 2
effect. The reason is that the reaction is exothermic and heat released
reaction like photochemical combination of H and Cl . Here two Br
2 2 2 raises the temperature and gas expands resulting in the increase in
molecules absorb photon, however, inspite of chain reaction only one
volume.
molecule of HBr is formed for each 100 photon absorbed by 100 molecules
 Actinometer : A device which is used to measure the intensity of
of Br .
2 radiation is konwn as actimometer. e.g., Uranyl oxalate actinometer.
H Br light2HBr A
2 2  Amount of the substance left after ‘n’ half lives  0 .
Mechanism 2n
Light absorption step : Br hvBr* ........(i)  Free energy change (G) for thermochemical reactions is always
2 2
Chain initiation step : Br*BrBr ........(ii) negative but remember, G for photochemical reaction may not always
2 be negative. It may be +ve also because a part of the light energy
Chain propagation step : Br*H HBrH ........(iii) absorbed is converted into the free energy of the products.
2
 Negative catalysts or inhibitors are those substances which decrease
H*Br HBrBr ........(iv)
2 the rate of a reaction.
Chain termination step : BrBr Br2 ........(v)  Example of fourth order reaction, 4KClO ⇌3KClO KCl
3 4
The lower values of HBr formation per photon of light absorbed has
 Grothus-Draper law : When light falls on a substance, a part of
been attributed to the fact that step (III) is highly endothermic and thus
before step (III) can take place most of the bromine free radicals recombine light is absorbed, a part is reflected and a part is transmitted. only that
part of light which is absorbed causes a particular reaction to occur.
as per step (V) to give Br molecule and thus providing less feasibility for
2  Stark’s Einstein law of photochemical equivalence According to this law,
step (IV) i.e. steps regenerating free radicals. Also the decomposition of HBr
every atom or molecule taking part in photochemical reaction absorbs only one
increases with increase in temperature.
quantum of radiaton.
(3) Quantum yield (or quantum efficiency) : The quantum efficiency
or yield  of a photochemical reaction may be expressed as,  Kinetics of fast reactions can be studied by (i) Relaxation method
(ii) Flash photolysis technique etc.
No. of molecules reacted or product formed
  Enzyme catalysed reactions are faster than metal catalysed
No.of photon absorbed
reactions, the former has lower activation energy.
(4) Application of photochemistry : Photochemistry has significant

role in our daily life. Some of the photochemical reactions commonly known
as cited below,
(i) Photosynthesis in plants
(ii) Photography

(iii) The formation and destruction of ozone layer

(iv) Photoetching in electronic industry
(v) Many polymerization reactions. Rate of a reaction
(vi) Modern printing technology
1. The rate of a chemical reaction
(vii) Free radical combinations to obtain many compounds.
[MP PMT 1973; CPMT 1982]

(a) Increases as the reaction proceeds

(b) Decreases as the reaction proceeds

(c) May increase or decrease during the reaction
(d) Remains constant as the reaction proceeds
462 Chemical Kinetics
2. The rate of a reaction that not involve gases is not dependent on [CPMT 1988; AFMC 1995]
(a) Pressure (b) Temperature
(c) Concentration (d) Catalyst
3. The rate at which a substance reacts depends on its
[MP PMT 1987; BHU 1999; KCET 2005]
(a) Atomic weight (b) Equivalent weight
(c) Molecular weight (d) Active mass
4. The rate law for the reaction
RClNaOH(aq)ROHNaCl is given by Rate
K [RCl]. The rate of the reaction will be [IIT 1988]
1
(a) Doubled on doubling the concentration of sodium hydroxide
(b) Halved on reducing the concentration of alkyl halide to one half
(c) Decreased on increasing the temperature of the reaction
(d) Unaffected by increasing the temperature of the reaction
5. If doubling the concentration of a reactant `A' increases the rate 4
times and tripling the concentration of `A' increases the rate 9
times, the rate is proportional to [AIIMS 1991]
(a) Concentration of `A'
(b) Square of concentration of `A'
(c) Under root of the concentration of `A'
(d) Cube of concentration of `A'
6. The rate of chemical reaction at constant temperature is
proportional to
(a) The amount of products formed
(b) The product of masses of the reactants
(c) The product of the molar concentration of the reactants
(d) The mean free path of the reaction
7. The concentration of a reactant decreases from 0.2 M to 0.1 M in 10
minutes. The rate of the reaction is
(a) 0.01 M (b) 102
(c) 0.01 mol dm3min1 (d) 1 mol dm3min1
8. When a reaction is progressing
(a) The rate of the reaction goes on increasing
(b) The concentration of the products goes on decreasing
(c) The concentration of the reactants goes on decreasing
(d) The reaction rate always remains constant