ERT 108 – Physical Chemistry Semester II- 2010/2011
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Subtopics
 Experimental Chemical and Kinetics Reactions
 First Order Reactions
 Second Order Reactions
 Reaction Rates and Reaction Mechanisms
 Light Spectroscopy and Adsorption Chemistry
(Experimental methods for fast reactions).
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Experimental Chemical and Kinetics
Reactions
 Rates of chemical Reactions:
 the rate of speed with which a reactant disappears
or a product appears.
 the rate at which the concentration of one of the
reactants decreases or of one of the products
increases with time.
 typically, mol L-1 s-1.
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The rates of reactions
 Depends on composition and temperature of reaction
mixture.
(a) Definition of rate
- as the slope of the tangent drawn to the curve
showing the variation of conc with time
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Rate of Reaction: A variable quantity
 Rate of reaction is expressed
as either:
 reac tan t 
Re action rate 
t
[ Negative value ]
or
 product 
Re action rate 
t
[ Positive value ]
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Consider a reaction:
A + 2B
3C + D
 Rate of consumption – (one of the reactants, A or
B) at a given time is d[R]/dt, R is A/B.
 Rate of formation – (products, C/D denotes as P) at
a given time is d[P]/dt.
 These rates are positive values. The +/- sign- indicate
that conc is increasing/decreasing.
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 From the stoichiometry:
 Rate of reaction is related to rates of change of
concentration of products and reactants
 Undesirability of having different rates to describe the
same reaction- using the extent of reaction, ξ.
 vJ is the stoichiometric coefficient for species J.
 Rate of reaction, r
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 For homogenous reaction the V can be taken inside
the differential, [J]=n /V
J
 For heterogeneous reaction, use the surface area
(constant), A as substitution to V, σJ=n /A
J
 Common units for r = mol dm-3 s-1 or related units for
homogenous reaction
 For heterogeneous reaction= mol m-2 s-1
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 Example 1
 The rate of change of molar conc of CH3 radicals in the
reaction 2CH3(g)
CH3CH3(g) was reported as
d[CH3]/dt= -1.2 mol dm-3 s-1 under particular
conditions. What is
(a) the rate of reaction
(b) the rate of formation of CH3CH3?
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Rate laws and rate constants
 Rate of reaction is often proportional to conc of
reactants raised to a power.
r = k[A] [B]
Each conc raised to first power.
 k is the rate constant for the reaction
 k- independent of conc but depends on temp.
 Experimentally determined equation of this kindrate law of the reaction
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 Rate law- an equation that expresses the rate of


-
reaction as a function of conc of all species present in
overall chemical equation
r = f([A], [B],….)
Rate law is determined experimentally- cannot be
inferred from the stoichiometry of balanced chemical
equation.
Application of rate law:
To predict the rate of reaction from the composition of
mixture
Guide to the mechanism of the reaction- for any
proposed mechanism- must be consistent with the
observed rate law.
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Reaction order
 Many reactions are found to have rate laws of the form
r = k [A]a[B]b…….
 The power to which the conc of a species is raised in a
rate law- the order of the reaction with respect to that
species.
 A reaction with rate law r = k [A] [B]
is first-order in A and first-order in B.
 The overall order of a reaction, second-order overallthe sum of the individual orders.
 Some reactions obey zero-order rate law- rate that is
independent of conc of the reactant. Thus,
r=k
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Integrated rate laws
 First-order rate law
is
or [A] = [A]0 e-kt
Where [A]0 is the initial conc of A at t = 0
 If ln ([A]/[A]0) is plotted against t- first-order reaction
will give a straight line of slope= -k.
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Half lives and time constants
𝜏
 Useful indication of the rate of a first-order chemical
reaction- half-life, t1/2 of a substance.
 Half-life: time taken for the conc of reactant to fall to half
its initial value.
 Time for [A] to decrease from [A]0 to ½[A]0 in first-order
reaction:
kt1/2 = -ln
= -ln ½ = ln 2
∴ t1/2 =
 Another indication of the rate of a first-order reactiontime constant, τ
 time required for the conc of reactant to fall to 1/e of its
initial value.
kτ = -ln
= -ln 1/e = 1 ∴ time constant, τ = 1/k
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Second-order reactions
 Second-order rate law:

is
or
 Where [A]0 is the initial conc of A (at t = 0)
 To plot a straight line for second order reaction- plot
1/[A] against t. the slope= k.
 The half life for second order reaction is
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Half-life of nth-order reaction
 In general for nth-order reaction (with n > 1) of the
form A  products, the half-life is related to the rate
constant and the initial conc of A by
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Zero-order, First-order, Second-order Reactions
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Zero-order, First-order, Second-order Reactions
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Zero-order,
First-order,
Secondorder
Reactions
ERT 108 – Physical
Chemistry Semester II2010/2011
Zero order
First order
Second order
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Example 2
(a) When [N2O5] =0.44M, the rate of decomposition
of N2O5 is 2.6 x 10-4 mol L-1 s-1.
 what is the value of k for this first-order reaction?
(b) N2O5 initially at a concentration of 1.0 mol/L in
CCl4, is allowed to decompose at 450C. At what
time will [N2O5] be reduced to 0.50M?
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Example 3
Time, min
[A], M
log [A]
1/[A]
0
1.00
0.00
1.00
5
0.63
-0.20
1.59
10
0.46
-0.34
2.17
15
0.36
-0.44
2.78
25
0.25
-0.60
4.00
 The data of the above table were obtained for the
decomposition reaction: A → 2B + C.
(a) Establish the order of the reaction.
(b) What is the rate constant, k?
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Answer (Example 3)
(a) Plot graph based on the data given in the Table.
Not Straight line – Not
Zero order
Not Straight line – Not
First order
(b) The slope of the 3rd graph:
Straight line – 2nd order

4.00  1.00 L / mol
k
 0.12 L mol 1 min 1
22
25 min
ERT 108 – Physical Chemistry Semester II- 2010/2011
Determination of the rate law
 Experimental data gives species conc at various times
during the reaction.
 a few methods to determine the rate law from
experimental conc vs. time data.
 Consider the following:
r = k [A]a[B]b…….
it is usually the best to find the order of a, b, … first and
then the rate constant, k.
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Determination of rate law- half-life
method
 Applies when the rate law has the form r = k[A]n. then
this equation
and
apply.
 If n = 1, then t1/2 is independent of [A]0. If n ≠1, then
gives:
 A plot of log10 t1/2 vs. log10 [A] – gives straight line of
slope = n-1.
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 To use the half-life method for determination of rate
-
-
law:
Plot [A] vs. t
Pick any [A] value, eg. [A]’ and finds the point where
[A] has fallen to ½ [A]’. The time interval between this
2 points is t1/2 for the initial conc [A]’.
Pick another point [A]” and determines the t1/2 for
this A conc.
Repeat this process several times
Plot log10 t1/2 vs. the log of the corresponding initial A
conc and measures the slope.
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Example 4
 Data for the dimerization 2A A2 of certain nitrile
oxide (compound A) is ethanol solution at 40oC
follow:
[A]/(mmol
/dm-3)
t/min
68.0
50.2
40.3
33.1
28.4
22.3
18.7
14.5
0
40
80
120
160
240
300
420
Find the reaction order using the half-life method
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Method of Initial Rates
 This simple method of establishing the exponents in a
rate equation involves measuring the initial rate of
reaction, ro for different sets of initial
concentration.
 Suppose we measure r0 for the 2 different initial A conc
[A]0,1 and [A]0,2 while keeping [B]0, [C]o,… fixed.
 With only [A]0 changed and with the rate law assumed
to have form r = k[A]a[B]b…[L]l, the ratio of initial rates
for run 1 and 2 is
ro,2/ro,1 = ([A]o,2/[A]0,1)n
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Example 5
 The data of three reactions involving S2O82- and I- were
given in the below table.
(i) Use the data to establish the order of reaction:
S 2O82 (aq)  3I  (aq)  2SO42 (aq)  I 3 aq 
with respect to S2O82-, the order with respect to I- & the
overall order.
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Example 5
(ii) Determine the value of k for the above reaction.
(iii) What is the initial rate of disappearance of in a
S2O82- reaction in which the initial concentrations
are [S2O82- ] =0.050M & [I-]=0.025M?
(iv) What is the rate of formation of SO42- in
Experiment 1?
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The temperature dependence of
reaction rates
 Chemical reaction begins with a collision between
molecules of A and molecules of B.
 Chemical reactions tend to go faster at higher
temperature.
 slow down some reactions by lowering the
temperature.
 Increasing the temperature increases the fraction
of the molecules that have energies in excess of the
activation energy.
 this factor is so important that for many chemical
reactions it can lead to a doubling or tripling of the
reaction rate for a temperature increase of only 100C.
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Reaction rates: Effect of temperature
 In 1889, Arrhenius noted that the k data for many
reactions fit the equation:
k  Ae
 Ea RT
where A & Ea are constants characteristics of the reaction
& R = the gas constant.
 Ea – the Arrhenius activation energy (kJ/mol or kcal/mol)
 A – the pre-exponential factor (Arrhenius factor).
 the unit of A is the same as those of k.
 Taking log of the above equation:
Ea
Ea
 log 10 A
ln k  
 ln A log 10 k  
2.303RT
RT
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Reaction rates: Effect of temperature
 If the Arrhenius equation is obeyed:
 a plot of ln k versus 1/T is a straight line with slope =
(-Ea/R) and A is the intercept of the line at 1/T = 0.
 This enables Ea and A to be found.
Ea  T2  T1 
k2
• Another useful equation:


log 
k1
2.303R  T2T1 
(eliminate the constant A).
 T2 and T1 - two kelvin temperatures.
 k2 and k1 - the rate constants at these temperatures.
 Ea – the activation energy (J/mol)
 R – the gas constant (8.314 Jmol-1 K-1).
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 The activation energy, Ea: the minimum kinetic
energy that reactants must have in order to form
products.
 The pre-exponential factor, A: a measure of the rate
at which collisions occur irrespective of their energy.
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Reaction Mechanisms
 The mechanism of reaction is the sequence of elementary





steps involved in a reaction
Most reactions occur in a sequence of steps called
elementary reactions.
A mechanism is a hypothesis about the elementary steps
through which chemical change occurs.
A typical elementary reaction is
H + Br2 HBr + Br
Chemical equation for elementary reaction: equation only
represents the specific process occurring to individual
molecules
Molecularity of an elementary reaction is the no of
molecules coming together to react in an elementary
reaction.
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Reaction Mechanisms
 Elementary processes in which a single molecule
dissociates (unimolecular) or two molecules collide
(bimolecular) much more probable than a process
requiring the simultaneous collision of three bodies
(termolecular).
 All elementary processes are reversible and may reach a
steady-state condition. In the steady state the rates of
the forward & reverse processes become equal. The
concentration of some intermediate becomes constant
with time.
 One elementary process may occur much more slower
than all the others. In this case, it determines the rate at
which the overall reaction proceeds & is called the ratedetermining step.
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Rate laws and equilibrium
constants for elementary reactions
 An overall reaction occurs as a series of elementary




steps
These steps constituting the mechanism of reaction
This section consider the rate law for elementary
reaction.
Consider a bimolecular elementary reaction
A + B  products, the rate of reaction,
will be
proportional to ZAB, the rate of A-B collisions per unit
of time.
∴ r for an elementary bimolecular ideal-gas reaction
will be r = k[A][B]
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 For unimolecular ideal-gas reaction B products,




fixed probability that any particular B molecule will
decompose/isomerize to products per unit time.
The rate of reaction, r = k[B]
Similar considerations apply to reactions in
ideally/ideally dilute solution.
In summary, in an ideal system, the rate law for the
elementary reaction aA + bB  products is
r = k[A]a[B]b, where a + b is 1,2 or 3.
For an elementary reaction, the orders in the rate
law equal the coefficients of the reactants.
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 Relation between the equilibrium constant for a
reversible elementary reaction and rate constants for
forward and reverse reactions.
 Consider the reversible elementary reaction
kf
aA + bB ⇌ cC + dD
kb
 rate laws for forward (f) and back (b) elementary
reactions are rf = kf [A]a[B]b and rb = kb [C]c[D]d.
 At equilibrium, these opposing rates are equal:
rf,eq = rb,eq or
kf([A]eq)a([B]eq)b = kb([C]eq)c([D]eq)d and

= Kc = kf/kb
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(b) The rate-determining-step approximation
 In rate-determining step approximation- reaction
mechanism assumed to consist 1 or more reversible
reactions that stay close to equilibrium during most
of the reaction.
 followed by relatively slow rate-determining step
then in turn followed by 1 or more rapid reactions.
 As an example:
k1
k3
k2
A⇌B⇌C⇌D
k-1
k-2 k-3
where step 2 (B⇌C)- assumed to be rate-determining
step.
 For this assumption to be valid- k-1 >> k2
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 The slow rate of B  C compared with B  A –
ensures that most B molecules go back to A rather
than to C- ensuring that step 1 (A ⇌ B) remain close
to equilibrium.
 k3 >> k2 and k3 >> k-2, to ensure that step 2 acts as
‘bottleneck’ and product D is rapidly formed from C.
 The overall rate is controlled by the rate-limiting
step B  C.
 Since we are examining rate of the forward reaction
A  D, we further assume that k2[B] >> k-2[C].
 During early stage- the conc of C will be lower than Bthis condition will hold. Thus, we neglect reverse
reaction for step 2.
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 The relative magnitude of k1 compared with k2 is





irrelevant to the validity of the rate-determining-step
approximation.
∴ the rate constant k2 of the rate-determining step
might be larger than k1.
However, the rate r2 = k2[B] of the rate-determing step
must be smaller than r1=k1[A] of the first step.
This follows from k2<<k-1 and k1/k-1 [B]/[A] (the
conditions for step 1 –near equilibrium).
For reverse overall reaction, the rate-determining step
is the reverse of that for forward reaction.
For example: the rate-determining step is C  B. So,
k-2 << k3 (ensures that step D⇌C is in equilibrium) and
k-1>> k2 (ensures that B  A is rapid).
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The rate determining-step approximation
 The rate law for the Br- - catalyzed aqueous reaction
H+ +HNO2 + C6H5NH2  C6H5N2+ + 2H2O
is observed to be
r = k[H+][HNO2][Br-]
…..(1)
A proposed mechanism is
k1
+
H + HNO2⇌ H2NO2+
rapid equilib
k-1
2
Br- k
H2NO2+ +
ONBr + H2O
slow ……(2)
k3
ONBr + C6H5NH2  C6H5N2+ +H2O + Br- fast
Deduced the rate law for this mechanism and relate the
observed rate constant, k in (1)to the rate constants in
assumed mechanism (2).
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 The second step in (2) is rate limiting. Since step 3 is much
faster than 2, we can take d[C6H5N2+]/dt as = rate of
formation of ONBr in step 2. therefore, the reaction rate is
r = k2[H2NO2+][Br-]
…..(3)
(since step 2 is an elementary reaction, its rate law is
determined by its stoichiometry. The species H2NO2+ in (3)
is a reaction intermediate and we want to express r in terms
of reactants and products. Since step 1 is in near equilib,
gives:
Kc,1 = k1/k-1 =[H2NO2+]/[H+][HNO2]
And
[H2NO2+] = (k1/k-1) [H+][HNO2]
Substitute in (3) gives
r = (k1k2/k-1)[H+][HNO2][Br-]
k = k1k2/k-1 = Kc,1k2
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The steady-state approximation
 Multistep reaction mechanisms usually involve 1 or
more species that do not appear in overall equation.
a
b
A k
Ik
P
 After an initial induction period,[I] will start at 0, rise
to max, [I]max, and then fall back to 0.
 During the major part of the reaction, the rates of
change of conc of all reaction intermediates are
negligibly small, therefore d[I]/dt = 0 for each reaction
intermediate.
 The steady-state approximation assumes that rate of
formation of reaction intermediate = rate of
destruction.
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Steady-state approximation
 Apply steady-state approximation to the mechanism
for
Br
H+ + HNO2 + C6H5NH2  C6H5N2+ + 2H2O
Given in the preceding exercise to find the predicted law.
-
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 To apply for rate-determining step approximation:
(a) Take the reaction rate, r = rate of determining step
(divided by the stoichiometric no srds of rate-determining
step, if srds ≠ 1.
(b) Eliminates the conc of any reaction intermediates that
occur in the rate expression obtained in (a) by using
equilibrium-constant expressions.
 To apply steady-state approximation:
(a) Take the reaction rate, r = rate of formation of product
(b) Eliminate the conc of any reaction intermediates that
occur in (a) by using d[I]/dt = 0 to find the conc of each I
(c) If step (b) introduces conc of other I, apply d[I]/dt = 0 to
eliminate their conc.
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The Hydrogen-Iodine Reaction
H2 (g) + I2 (g) → 2HI (g)
 Rate of formation of HI = k [H2][I2]
 The hydrogen-iodine reaction is proposed to be a twostep mechanism [Sullivan J. (1967). J.Chem.Phys.46:73].
 1st step: iodine molecules are believed to dissociate
into iodine atoms.
 2nd step: simultaneous collision of two iodine atoms
and a hydrogen molecule.
(this termolecular step is expected to occur
much more slowly – the rate-determining step).
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The Hydrogen-Iodine Reaction
1st step:
I 2 g 
k1


k2
[Fast]
2I ( g )
k3
2nd step: 2I g   H 2 ( g ) 
2HI ( g )
Net:
[Slow]
I 2 ( g )  H 2 ( g )  2HI g 
 If the reversible step reaches a steady state condition:
 rate of disappearance of I2 = rate of formation of I2
k1[ I 2 ]  k2 [ I ]
2
I 
2
k1
 I 2 
k2
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The Hydrogen-Iodine Reaction
 For the rate-determining step:
Rate of formation of HI
= k3 [I]2[H2]
k1
 k3 H 2 I 2 
k2
= K[H2][I2] (K=k1k3/k2)
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Example 6
 The thermal decomposition of ozone to oxygen:
2O3 (g) → 3O2 (g)
 The observed rate law:
2


O
Rate of disappearance of O3 = k 3
O2 
 Show that the following mechanism is consistent with
this experiment rate law.
k1
1st:

O3
O2  O

k2
2nd:
k3
O  O 3 
2O 2
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Experimental methods for fast reactions
 Many reactions are too fast to follow by the classical
methods.
 Several ways to study fast reactions :
1. Rapid flow methods:
(i) Continuous flow
(ii) Stopped flow
2. Relaxation methods:
(i) Temperature jump (T-jump)
method
(ii) Pressure jump method
(iii)Electric field jump method
3. Flash photolysis
4. Shock tube
5. Nuclear-magnetic-resonance (NMR) spectroscopy
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Experimental methods for fast reactions
 Continuous flow system
 Liquid phase:
 Reactant A & B are rapidly drive into the mixing
chamber M by pushing in the plungers of the syringes.
 Mixing occurs in 0.5 – 1ms.
 The reaction mixture then flows through the narrow
observation tube, where one measures the light
absorption at a wavelength (at which one species
absorbs to determine the concentration of that
species).
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Experimental methods for fast reactions
Figure: A continuous flow system with rapid mixing of reactants.
 For gas phase reaction, the syringes are replaced by bulbs of
gases A & B.
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Experimental methods for fast reactions
 Stopped flow method:
 the reactants mixed at M & rapidly flow through the
observation tube into the receiving syringe, driving its
plunger against a barrier & thereby stopping the flow.
 this plunger hits a switch which stops the motor driven
plungers & triggers the oscilloscope sweep.
 One observes the light absorption at P as a function of
time.
• The continuous flow & stopped flow methods are
applicable to reactions with half-lives in the range of
0.001 to 10s.
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Experimental methods for fast reactions
Figure:
A sloppedflow system
Figure:
A flashphotolysis
experiment.
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Experimental methods for fast reactions
 Relaxation methods:
 Take a system in reaction equilibrium & suddenly
change one of the variables that determine the
equilibrium position.
Relaxation methods
1. Temperature jump (T-jump)
method
2. Pressure jump method
3. Electric field jump method
Descriptions
A sudden change in T shifts the equilibrium.
A sudden change in P shifts the equilibrium.
A sudden applied electric field shifts the
equilibrium (a change in total dipole moment).
 A limitation on relaxation methods – the reaction
must be reversible, with detectable amounts of all
species present in equilibrium.
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Experimental methods for fast reactions
 Rapid flow & relaxation method have been used to
measure the rates of proton transfer (acid-base)
reactions, complex-ion-formation reactions, ion-pairformation reactions & enzyme-substrate-complex
formation system.
 Relaxation methods apply rather small perturbations
to a system & do not generate new chemical species.
 The flash-photolysis and shock tube methods apply
a large perturbation to a system, thereby generating
one or more reactive species whose reactions are
then followed.
 NMR spectroscopy is used to measure the rates of
certain rapid isomerization & exchange reactions.
ERT 108 – Physical Chemistry Semester II- 2010/2011
57
Answer (Example 6a)
 Tabulate the data as follows.
Temp, 0C
Temp, K
1/Temp, 1/K
k, s-1
log10 k
25
298
0.0034
0.001
-3
 Construct the Arrhenius plot of log10k versus 1/T for the reaction.
 Intercept (log10A)=13.5
A = 3x1013s-1
 Slope=-5500K,
Ea
 5500 
2.303R
Ea=25kcal/mol
=105 kJ/mol
ERT 108 – Physical
58 Chemistry Semester II2010/2011
Figure: Arrhenius plot of log10 k versus 1/T for this reaction.
Note: the long extrapolation needed to find A.
Answer (Example 6b)
 Based on the given info:
 k2 = 2k1 ,
T1 = room temperature (298K),T2=298+10 = 308K,
 The Arrhenius equation:
 Substitute:
log
2 k1
k1
log
k2
k1
Ea  T2  T1 



2.303R  T2T1 
Ea  (308)  298 



2.303R  308(298) 
Ea = 53 kJ/mol
59
ERT 108 – Physical Chemistry Semester II- 2010/2011
Answer (Example 7)
 Assume the 1st step reaches the steady state condition:
Rate of formation of O = Rate of disappearance of O
k1 [O3] = k2 [O2] [O]
k1 O3 
O  
k 2 O2 
 Assume the 2nd step is the rate-determining step:
Rate of disappearance of O3 = k3 [O][O3]

k1k3 O3 O3 
O3 

k
O2 
k 2 O2 
2
(where k = k1k3/k2)
60
ERT 108 – Physical Chemistry Semester II- 2010/2011
Apparatus of
the
determining
the rate of
decomposition
of N2O5
61
ERT 108 – Physical Chemistry Semester II- 2010/2011