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1
SOME IMPORTANT GRAPH REPRESENTATION
Use in chemical kinetics :
(1) y = mx + C
(2) y = –mx + C
(3) y = mx
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2
(4) y = –mx
Slope = –m = tan
Intercept = zero
(5) y = ex
(6) y = e–x
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3
(7) y = 1x
CHEMICAL KINETICS “Chemical kinetics involves the study of the rates and mechanism of chemical reactions.”
The rates of reactions :
(a) The definition of rate : Consider a reaction of the form
A + 2B 3C + D …(1)
in which the molar concentration of participants are [A], [B], [C] & [D].
The rate of consumption or decomposition of the one of the reactants at a given time is
[ ]d Rdt
, where R is A or B. The rate of formation of one of the products is [ ]d Pdt
, where P is C or D.
The rate of reaction can be expressed with respect to any species in equation (1).
Rate = [ ] 1 [ ] 1 [ ] [ ]2 3
d A d B d C d Ddt dt dt dt
Thus, the rate of reaction can be defined with respect to both reactants and products.
For example :
4NO2(g) + O2(g) 2N2O2(g)
find the expression for rate of reaction.
Sol. 4NO2(g) + O2(g) 2N2O2(g)
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rate = 2 52 2 [ ]1 [ ] [ ] 14 2
d N Od NO d Odt dt dt
(b) Rate laws and rate constant :
The rate of a reaction will generally depends on temperature pressure and concentration of
species involving in the reaction.
The rate of reaction is proportional to the molar concentration of reacting species.
i.e. A + B + C + D + ……. Product
then, rate of reaction = k[A]a [B]b [C]c [D]d ………..
where [A] is the concentration of reactant A, [B] is the concentration of reactant B and so on.
The constant a is known as the reaction order with respect to species A, b the reaction order with
respect to species B and so on.
The over all reaction order is equal to the sum of the individual reaction orders (a + b + c + d +
……..). Finally the constant k is rate constant for the reaction.
The rate constant dependent on concentration but also on temperature & pressure.
This relationship is known as a rate law.
(c) Order of the reaction :
A + B + C + ……… Product
The rate law = v = k[A]a [B]b [C]c ……..
The order of reaction = a + b + c + ……
For example :
if rate law = v = k[A]1/2 [B]
Then, it is half order in A, first order in B and three half 32
order overall.
Molecularity of a Reaction : The number of reacting species (atoms, ions or molecule) taking
parting an elementary reaction, must collide simultaneously, in order bring about a chemical reaction
is called molecularity of a reaction.
Relationship between Rate law, order and the rate constant :
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5
kA B
Then, rate of reaction = [ ] [ ]nd A k Adt
The unit of rate or reaction is mol liter–1 sec–1 i.e. mol L–1 s–1.
where M represent mol L–1 or moles per liter & n is order of reaction.
The unit of rate constant (k)
Rate of reaction = k[A]n
unit of rate of reaction = unit of k × [unit of concentration]n
MS–1 = unit of k × [M]n
unit of k = 1
1 n 1n
[MS ] [M S ][M]
i.e., unit of k = 1 n 1 1 n n 1 1M S mol L S
Rate law Order Unit of k
Rate = k Zero MS–1
Rate = k[A] First order S–1
Rate = k[A]2 Second order M–1 S–1
Rate = k[A][B] Second order M–1 S–1
Rate = k[A][B][C] Third order M–2 S–1
Prob. Find the order of the reaction if unit of rate constant or the reaction is (dm3)3/2
mol–3/2 s–1.
Sol. Unit of rate constant = (dm3)3/2 mol–3/2 s–1 (given)
We know that,
Unit of rate constant = M1 – n s–1
For nth order
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i.e. 1 n 1M s = 3 3/ 2 3 / 2 1(dm ) (mol) s
= 3/ 2
13
mol sdm
= 3/ 2
1mol sL
1 L = 1 dm3
& molL
= M
1 n 1M s = 3/ 2 1M s
M1 – n = M–3/1
1 – n = 32
n = 3 512 2
i.e. it is 52
order reaction.
Determing Reaction order :
Using the following data for the reaction, we determine the order of the reaction with respect
to A and B, over all order and rate constant for the reaction
[A] (M) [B] (M) Initial rate (Ms–1)
2.30 × 10–4 3.10 × 10–5 5.25 × 10–4
4.60 × 10–4 6.20 × 10–5 4.20 × 10–3
9.20 × 10–4 6.20 × 10–5 1.70 × 10–2
Sol. A + B Product
rate of reaction = k[A]a [B]b
5.25 × 10–4 = k[2.30 × 10–4]a [3.10 × 10–5]b ...(1)
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4.20 × 10–3 = k[4.60 × 10–4]a [6.20 × 10–5]b ...(2)
1.70 × 10–2 = k[9.20 × 10–4]a [6.20 × 10–5]b ...(3)
Divide equation (2) by equation (3), we get
3
24.20 101.70 10
= 4 a 5 b
4 a 5 bk[4.60 10 ] [6.20 10 ]k[9.20 10 ] [6.20 10 ]
2.47 × 10–1 = (0.5)a
(0.247) = (0.5)a
(0.5 × 0.5) (0.5)a
(0.5)a (0.5)a
or taking log we can find the value of a.
a = 2
Divide equation (1) by equation (2) we get
4
35.25 104.20 10
= 4 a 5 b
4 a 5 bk[2.30 10 ] [3.10 10 ]k[4.60 10 ] [6.20 10 ]
1.25 × 10–1 = [0.5]a [0.5]b = [0.5]2 [0.5]b
= 0.25 [0.5]b
5 × 10–1 = [0.5]b
0.5 = [0.5]b
b = 1
Therefore, the reaction is second order in A and first order in B and third order overall.
rate = k[A]2 [B]
5.2 × 10–4 Ms–1 = k(2.3 × 10–4 M)2 (3.1 × 10–5]M
k = 3.17 × 108 M–2 s–1
i.e. the over all rate law is
rate = (3.17 × 108 M–2 s–1) [A]2 [B]
Integrated Rate law Expression :
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Integrated rate law expression provide the predicted temporal evolution in reactant and product
concentrations for reactions having an assumed order dependence.
(1) Zero-order Reaction : Consider the following elementary reaction
kA P
For zero-order reaction, the rate law is
rate = d[A] d[P]r k[A] kdt dt
k is rate constant.
r = d[A] kdt
–d[A] = k dt
If at t = 0, the initial concentration is [A]0 and the concentration at t = t, is [A], then integration
yields
0
[A]
[A]
d[A] = t t
0t 0
k dt
[A]0 – [A] = kt
This is integrated rate equation for a zero-order reaction in terms of reactant.
d[P]dt
= k
d[P] = k dt
at t = 0, [P] = 0
and at t = t, [P] = [P]
then integration yields
[P] [P]
[P] 0
d[P]
=
t t
t 0
k dt
[P] = kt
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This is integrated rate law equation for a zero-order reaction in terms of product.
i.e. [A]0 – [A] = kt = [P]
Graph representation of zero-order reaction
[A]0 – [A] = kt
[A] = –kt + [A]0
y = mx + c
Graph of reactant vs time.
[P] = kt y = mx
Graph of concentration of product vs time.
01 [A] [A]2
= kt
[A]0 – [A] = 2kt …(i)
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When t = 0 then [P] = 0 and t = t then [P] = [P]
1 d[P]3 dt
= k
1 d[P]3
= kt
[P]
0
1 d[P]3 =
t
0
k dt
1 [P]3
= kt
[P] = 3kt …(ii)
kt = 01 1[P] [A] [A]3 2
…(iii)
Problem. Find the integrated rate law expression for an elementary zero order reaction given
below.
kA 2B P
Sol.
kA 2B P
The rate law of above elementary reaction is given below
d[A]dt
= 1 d[B] d[P] k[A] [B] k2 dt dt
d[A]dt
= k
0
[A]
[A]
d[A] = t t
t 0
k dt
– [[A] – [A]0] = kt
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[A]0 – [A] = kt …(i)
1 d[B]2 dt
= k
0
[B]
[B]
1 d[B]2
= t t
t 0
k dt
01 [B] [B]2
= kt
01 [B] [B]2
= kt …(ii)
d[P]dt
= k
[P]
0
d[P] = t
0
k dt
[P] = kt …(iii)
From equation (i), (ii) & (iii) we get
[A]0 – [A] = 01 [B] [B]2
= [P] = kt
(2) First-order reaction
Consider the following elementary reaction
A P
If the reaction is first order with respect to [A], the rate law expression is
Rate = d[A] d[P] k[A]dt dt
k is rate constant
r = d[A] k[A]dt
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d[A][A]
= k dt
If t = 0, the initial concentration is [A]0 and the concentration at t = t, is [A], then integrating
yields
0
[A]
[A]
d[A][A]
= t
0
k dt
0
[A]
[A]ln A = kt
0
[A]ln[A]
= –kt
[A] = [A]0 e–kt ….(i)
or 0[A]ln[A]
= kt …(ii)
Using this idea, the concentration of product with time for this first-order reaction is :
[P] + [A] = [A]0
[P] = [A]0 – [A]
[P] = [A]0 – [A]0 e–kt
[P] = [A]0 (1 – e–kt) ….(iii)
Graph representation of first order reaction
[A] = [A]0 e–kt
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Plot of concentration vs time.
0[A]ln[A]
= kt
ln [A] = –kt + ln [A]0
Plot of log [A] vs time.
t1/2 i.e. half life time of first order reaction
0[A]ln[A]
= kt
when t = t1/2; then [A] = 0[A]2
0
0
[A]ln [A]2
= kt1/2
ln 2 = kt1/2
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t1/2 = 0.693k
Problem. The half life for the first order decomposition of N2O5 is 2.05 × 105 s. How long will
it take for a sample of N2O5 to decay to 60% of its initial value ?
Sol. We know that, t1/2 = 0.693k
k = 41/ 2
0.693 0.693t 2.05 10 s
The time at which the sample has decayed to 60% of its initial value then
kt = 0[A]ln[A]
(3.38 × 10–5) t = 2.303 log 10060
= 0.5109
T = 1.51 × 104 s
Problem. Find the t3/4 i.e. 34
life time of first order reaction.
kA P
Sol. kA P
Integrated rate law expression is
0[A]ln[A]
= kt
when t = t3/4 than [A] = [A]0 – 34
[A]0s
[A] = 0[A]4
then 0
0
[A]ln [A]4
= kt3/4
ln 4 = kt3/4
t3/4 = ln 4 2 ln 2 1.38k k k
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(3) Second-order reaction : (Type I) Consider the following elementary reaction,
k2A P
If the reaction is second order with respect to [A], the rate law expression is
rate = 21 d[A] d[P]r k[A]2 dt dt
k is rate constant
r = 21 d[A] k[A]2 dt
21 d[A]2 [A]
= k dt
2d[A][A]
= 2k dt
If t = 0, the initial concentration is [A]0 and the concentration at t = t, is [A], then integration yields
0
[A]
2[A]
d[A][A]
= t t
t 0
2k dt
0
1 1[A] [A]
= 2kt
0
1 1[A] [A]
= kt …(i)
The concentration of product with time for second order reaction
[P] = 0[A] [A]2
1[A]
= eff0
1 k t[A]
or [A] = eff
11 k t
[A]
then [A]0 = 0
eff
[A]k t [A] 1
2
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[P] = 0
0
[A] 112 2k [A] 1
…(ii)
t1/2 i.e. Half-life time of second order reaction (type I)
0
1 1[A] [A]
= keff t
when t = t1/2 then [A] = 0 00
[A] [A][A]2 2
0 0
1 1[A] [A]
2
= keff t1/2 = 0
1[A]
t1/2 = eff 0
1k [A]
Second-order reaction (Type II) Second order reactions of type II involves two different reactants A and B, as follows
kA B P
Assuming that the reaction is first order in both A and B, the reaction rate is
r = d[A] d[B] d[P] k[A][B]dt dt dt
If t = 0 then the initial concentration are [A]0& [B]0 and the concentration at t = t, are [A] & [B].
The loss of reactant i.e. the formation of product is equal to [A]0 – [A] = [B]0 – [B] = [P] [B]0 – [A]0 + [A] = [B]
then d[A]dt
= k[A][B]
d[A][A][B]
= k dt
the integration yield
0
[A]
[A]
d[A][A][B]
=
0
[A]t
0 00 [A]
d[A]k dt[B] [A] [A] [A]
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0
[A]
[A]
d[A][A] [A]
=
t
0
k dt
let = [B]0 – [A]
The solution to the integral involving [A] is given by
dxx(C x) = 1 C xln
C x
Using this solution to the integral, the integrated rate law expression becomes
[A]
[A]
1 [A]ln[A]
= kt
0
[A]
[A]
1 [A]ln[A]
= kt
0
0
[A]1 [A]ln ln[A] [A]
= kt
0 0 0 0 0
0
[B] [A] [A] [B] [A] [A]1 ln ln[A] [A]
= kt
0
0
[B]1 [B]ln ln[A] [A]
= kt
0
0
[B][A]1 ln[B] [A]
= kt
0
0 0 0
[A] [B]1 ln[B] [A] [A][B]
= kt
Graph representation of second order reaction of type I
0
1 1[A] [A]
= kt
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1[A]
= 0
1 kt[A]
Y = mx + C
Plot of concentration vs time
(4) nth order reaction where n 2 :
An nth order reaction may be represented as
nA Products
the rate law is,
rate = n1 d[A]r k[A]n dt
where k is rate constant for nth order reaction
1 d[A]n dt
= k[A]n
nd[A][A]
= –nk dt
If at t = 0, the initial concentration is [A]0 and the concentration at t = t, is [A], then
integration yields
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0
[A]
n[A]
d[A][A] =
t
0
nk dt
Let nk = k’
0
[A]
n 1[A]
1(1 n) [A]
= –k’t
0
[A]
n 1[A]
1 1n 1 [A]
= k’t
n 1 n 10
1 1 1n 1 [A] [A]
= k’t …(1)
t1/2 i.e. Half life time of nth order reaction
n 1 n 10
1 1 1n 1 [A] [A]
= k’t
Where t = t1/2 then [A] = 0 00
[A] [A][A]2 2
n 1 n 100
1 1 1n 1 [A][A]
2
= k’t1/2
kt1/2 = n 1
n 10
2 1(n 1) [A]
t1/2 = n 1
n 10
1 2 1k(n 1) [A]
…(2)
i.e. t1/2 n 10
1[A]
…(3)
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Thus we can say that t1/2 of the reaction is inversely proportional to the initial concentration of
reactant, except first order reaction.
So, for a first order reaction (n = 1), t1/2 is independent on [A]0 for a second order reaction
(n = 2), t1/2 is dependent on [A]0
t1/2 0
1[A]
for a nth order reaction
t1/2 n 10
1[A]
Note : For the elementary reaction, the order of reaction is equal to the molecularity of the
reaction.
Problem. Find the rate law for the following reaction.
Sol
Rate law is
(1) d[B]dt
= k1[A]
(2) d[C]dt
= k2[A]
(3) d[A]dt
= k1[A] + k2[A] = (k1 + k2)[A]
Problem. Find the rate law for the following reaction.
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Sol. (1) d[A]dt
= k1[A]
(2) d[B]dt
= k1[A] – k2[B] – k3[B]
= k1[A] – (k2 + k3) [B]
(3) d[C]dt
= k2[B]
(4) d[D]dt
= k3[B]
Consecutive elementary reaction (Series reaction) :
Consider the following series reaction scheme
Ik kA I P
In this, the reactant A decays to four intermediate I, and this intermediate undergoes
subsequent decay resulting in the formation of product P. The above series is elementary first order
reaction.
Then the rate law expression is :
d[A]dt
= –kA[A] …(1)
d[I]dt
= kA[A] – kI[I] …(2)
d[P]dt
= kI[I] …(3)
Let only the reactant A is present at t = 0 such that
[A]0 0, [I]0 = 0, [P]0 = 0
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then the rate law expression is
d[A]dt
= –kA[A]
0
[A]
[A]
d[A][A] =
k
A0
k dt
[A] = Ak t0[A] e …(4)
The expression for [A] is substituted into the rate law of I resulting in
d[I]dt
= A Ik [A] k [I]
= Ak tA 0 Ik [A] e k [I]
Id[I] k [I]dt
= Ak tA 0k [A] e
This differential equation has a standard form and after setting
[I]0 = 0, the solution is
[I] = A Ik t k tA0
I A
k e e [A]k k
The expression for [P] is
[A]0 = [A] + [I] + [P]
[P] = [A]0 – [A] – [I]
So [P] = I Ak t k t
A I0
I A
k e k e 1 [A]k k
b
Case I. Let kA >> kI
i.e. A 2k kfast slowA I P
then kI – kA –kA
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and Ak te 0
[P] = [A]0 – [A] – [I]
= [A]0 – [A]0 A A Ik t k t k tA 0
I A
k [A]e e ek k
[P] = A Ak t k t
A I0
I A
k e k e 1 [A]k k
i.e. [P] = Ak t
A0
A
k e 0 1 [A]k
= Ik tA0
A
k e 1 [A]k
[P] = [A]0 Ik t(1 e )
The rate of formation of product can be determined by slowest step.
[A] = [A]0 Ak te …(1)
[I] = A Ik t k tA 0
I A
k [A] (e e )[k k ]
[I] = Ik tA 0
A
k [A] ( e )k
[I] = Ik t0[A] ( e ) …(2)
[P] = [A]0 Ik t(1 e ) …(3)
The graph representation for case I i.e. when kA >> kI.
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Case II. kI >> kA
A Ik kslow fastA I P
kI – kA kI
[P] = [A]0 Ak t(e )
[A] = Ak t0[A] e
The graph representation of case II i.e. when kI >> kA.
The Steady-State Approximation.
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The steady-state approximation assume that, after an initial induction period, an interval during
which the concentration of intermediate ‘I’ rise from zero, and during the major part of the reaction,
the rates of change of concentration of all reaction intermediate are negligibly small.
d[I]dt
= 0
Problem. Consider the following reaction
A Ik kA I P
assuming that only reactant A is present at t = 0, what is the expected time dependence of [P]
using the steady state approximation ?
Sol. The differential rate expression for this reaction are :
d[A]dt
= –kA[A]
d[I]dt
= kA[A] – kI[I]
d[P]dt
= kI[I]
Applying the steady sate for I we get
d[I]dt
= 0 = kA[A] – kI[I]
A
I
kk
= [I][A]
[I] = A
I
k[A]k
and [A] = Ak t0[A] e
[I] = Ak tA0
I
k [A] ek
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then d[P]dt
= Ak tAI I 0
I
kk [I] k [A] ek
= Ak tA 0k [A] e
[P]
0
d[P] = A
tk t
A 00
k [A] e dt
[P] = Ak ta 0
A
1k [A] (1 e )k
[P] = Ak t0[A] (1 e )
This is expression for [P].
Problem. Using steady state approximation find the rate law for d[P]dt
for the following given
equation
1 2 3k k k1 2A I I P
Sol. d[A]dt
= –k1[A]
1d[I ]dt
= –k2[I1] + k1[A]
2d[I ]dt
= –k3[I2] + k2[I1]
d[P]dt
= k3[I2]
I1 & I2 are intermediate & apply steady state approximation on intermediate, we get
1d[I ]dt
= 0 = –k2[I1] + k1[A]
[I1] = 1
2
k [A]k
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2d[I ]dt
= 0 = –k3[I2] + k2[I1]
[I2] = 1 2 11
2 3 2
k k k[I ] [A]k k k
[I2] = 1
3
k [A]k
and d[P]dt
= k3[I2]
= 13
3
kk [A]k
d[P]dt
= k1[A]
Problem. Using steady state approximation, derive the rate law for the decomposition of
N2O5.
2N2O5(g) 4NO2(g) + O2(g)
On the basis of following mechanism.
N2O5 ak NO2 +NO3
ak '2 2 2 5NO NO N O
bk2 3 2 2NO NO NO O NO
ck2 5 2 2 2NO N O NO NO NO
Sol. The intermediate are NO & NO3.
The rate law are :
d[NO]dt
= kb[NO2[[NO3] – kc [NO][N2O5] = 0
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3d[NO ]dt
= ka[N2O5] – ka’ [NO2][NO3] – kb[NO2][NO3] = 0
2 5d[N O ]dt
= –ka[N2O5] + ka’[NO2][NO3] – kc[NO][N2O5]
and replacing the concentration of intermediate by using the equation above gives
2 5d[N O ]dt
= a b 2 5
a b
2k k [N O ]k ' k '
Parallel Reaction : Parallel reaction are those reaction in which the reactant can form one of
two or more products.
Consider the following parallel reaction in which reactant A can form two products B & C.
The rate law for the reactant and products are :
d[A]dt
= –kB[A] – kc[A] = –(kB + kC)[A] …(1)
d[B]dt
= kB[A] …(2)
d[C]dt
= kC[A] …(3)
Integration of equation (1) with the initial condition [A]0 0 and [B] = 0 = [C] yields
[A] = [A]0 B C(k k )te …(4)
Integration of equation (2), we get
[B]
0
d[B] = B C
t t(k k )t
B B 00 0
[A] k k [A] e
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[B] = B C
t(k k )t
B 0B C 0
ek [A](k k )
[B] = B C(k k )tB 0
B C
k [A] 1 ek k
…(5)
Similarly
[C] = B C(k k )tC 0
B C
k [A] 1 ek k
…(6)
i.e. the ratio of concentration of product is
[B][C]
= B
C
kk
i.e. the product concentration ratio remains constant with time.
The yield, , is defined as the probability that a given product will be formed by decay of the
reactant.
i = i
nn
kk
The quantum yield of product [B] is
B = yield of [B] = B
B C
kk k
The quantum yield of product [C] is
C = yield of [C] = C
B C
kk k
Problem. Find the quantum yield of [B] & [C] in the following reaction
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Sol. 1 d[B]2 dt
= k1[A]
d[B]dt
= 2k1[A] …(1)
d[C]dt
= 2k2[A] …(2)
then B = 1 1
1 2 1 2
2k k2k 2k k k
C = 2 2
1 2 1 2
2k k2k 2k k k
Problem. Find the quantum yield of [B], [C] & [D] in the following reaction.
Sol. 1 d[B]2 dt
= k1[A]
d[B]dt
= 2k1[A] …(1)
d[C]dt
= 2k1[A] …(2)
d[D]dt
= 2k1[A] …(3)
The ratio of formation of product [B], [C] & [D] are
d[B] d[C] d[D]:dt dt dt
= 2 : 1 : 2
B = 1 1
1 1 2 1
2k 2k 22k k 2k 5k 5
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C = 1
1
k 15k 5
D = 1
1
2k 25k 5
Reversible reactions and Equilibrium
Consider the following reaction in which the forward reaction is first order in A, and the back
reaction is first order in B :
A
B
kk
A B
The forward and back rate constant are kA & kB. Then rate law are
d[A]dt
= –kA[A] + kB[B]
d[B]dt
= kA[A] – kB[B]
Only reactant is present at t = 0 and the concentration of reactant and product for t > 0 must be
equal to the initial concentration of reactant.
[A]0 = [A] + [B]
then d[A]dt
= –kA[A] + kB[B]
= –kA[A] + kB([A]0 – [A])
= –[A] (kA + kB) + kB[A]0
0
[A]
A B B 0[A]
d[A][A] (k k ) k [A] =
t
0
dt
dxa bx = 1 ln (a bx)
b
Using this relationship we get
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0
[A]A B B 0 [A]
A B
1 ln [A] (k k ) k [A] tk k
A B B 0
0 A B B 0
[A](k k ) k [A]ln[A] (k k ) k [A]
= –t(kA + kB)
A B B 0
A 0
[A](k k ) k [A]lnk [A] = –(kA + kB)t
A B B 0
A 0
[A](k k ) k [A]k [A] = A B(k k )te
[A] (kA + kB) – kB[A]0 = A B(k k )tA 0k [A] e
[A] = A B(k k )t
B 0 A 0
A B
k [A] k [A] ek k
[A] = A B(k k )t
B A
A B
k k ek k
Then [B] = A B(k k )t
B A0
A B
k k e[A] 1k k
As t , the concentration reach their equilibrium values then
[A]eq = B0
t A B
klim [A] [A]k k
& [B]eq = A 00
A B
k [A][A] [A]k k
It follows that the equilibrium constant of the reaction is
kC = eq A
eq B
[B] k[A] k
i.e. kC = A
B
kk
kC is equilibrium constant in terms of concentration.
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At equilibrium, the forward and reverse rates must be same so,
KA[A]eq = kB[B]eq
Problem. Using the following equation A C
mechanism,
(i) 1
2
kk
A B
(ii) 3kB C
(a) Find rate of reaction ?
(b) Find rate of reaction when (i) is fast.
Sol. (a) From the rate law
d[A]dt
= –k1[A] – k2[B]
d[B]dt
= k1[A] – k2[B] – k3[B]
d[C]dt
= k3[C]
[B] is intermediate then we apply SSA then we get
0 = k1[A] – k2[B] – k3[B]
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[B] = 1
2 3
k [A]k k
then d[C]dt
= 1 3
2 3
k k [A]k k
(b) When (i) is fast then
k1[A] = k2[B]
i.e. 1
2
k [A]k
= [B]
& d[C]dt
= k3[B] = 1 3
2
k k [B]k
Problem. Using the following equation
2NO2 + F2 2NO2F
Mechanism
NO2 + F2 1k NO2F + F (slow)
F + NO2 2k NO2F (fast)
Find the rate of reaction.
Sol. From the rate law
2d[NO F]dt
= k1[NO2][F2] + k2[F][NO2]
d[F]dt
= k1[NO2][F2] – k2[F][NO2]
= 0 (F is intermediate)
[F] = 1
2
kk
[F2]
2d[NO F]dt
= k1[NO2][F2] + 2 1
2
k kk
[F][NO2]
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= 2k1[F2][NO2]
Arrhenius Equation (Expression)
The following empirical relationship between temperature (T), rate constant (k) and activation
energy (Ea) is known as the Arrhenius expression :
K = aE / RTAe
A is constant known as the frequency factor or Arrhenius pre-exponential factor.
A & Ea is temperature independent.
The unit of A is always equal to the unit of rate constant (k)
k = aE / RTAe ….(1)
taking natural log of this equation, we get
ln k = ln A – aERT
…(2)
or log k = log A – aE2.303 RT
…(3)
Problem. Prove that on increasing the activation energy, the rate constant will be decreasing
and on increasing the temperature, the rate constant will be increasing.
Sol. k = aE / RTAe
ln k = ln A – aERT
when Ea increase then the value of aERT
increases and the value of ln k i.e. k will be
decreases.
i.e. Ea then k
When T increase then the value of aERT
decreases and the value of ln k i.e. k will be
increase.
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i.e. T then k
The graph of ln k vs 1T
is given below
ln k = ln A – aERT
ln k = aE 1 ln ART T
y = –m . x + C
Problem. Using the given equation find the value of A & Ea.
ln k = 100 J2.3T
Sol. We know that
k = aE / RTAe
ln k = ln A – aERT
i.e. ln A = 2.3
& A = e2.3 = 9.974
i.e. aERT
= 100T
Ea = 100 × R = 100 × 8.314
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Ea = 831.4 J mol–1
Problem. When temperature is increased then t1/2 of reaction will be
(a) remains constant (b) increased
(c) decreased (d) first increase and then decrease
Sol. We know that
t1/2 = 0.693k
& k = aE / RTAe
t1/2 = aE / RT
0.693Ae
t1/2 aE / RTe
i.e. on increasing T, aE / RTe decrease then we can say that on increasing temperature (T), the
t1/2 of the reaction will decrease.
i.e. T then t1/2
i.e. The correct answer is (C).
Variation of rate constant with temperature
We know that k = aE / RTAe
ln k = ln A – aERT
If k1 and k2 be the value of rate constant at temperature T1 and T2, we can derive
2
1
klnk
= a 2 1
1 2
E T TR T T
or 2
1
klogk
= a 2 1
1 2
E T T2.303 R T T
Temperature Coefficient.
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The ratio of rate constant of a reaction at two different temperature differing by 10 degree is
know as temperature coefficient.
i.e. Temperature coefficient = T 10
T
kk
Standard Temperature coefficient = 35 308
25 298
k kk k
= 2 to 3
Problem. In the reaction mechanism
1 a 3 a1 3
2 a2
k , E k , E2 3k , E
X Y Z P, k k
Find the overall rate constant (koverall) and Activation energy Ea (overall).
Sol. From the above reaction, the ate of formation of product is
d[P]dt
= k3[Z] …(1)
and d[Z]dt
= k1[X][Y] – k2[Z] – k3[Z]
= k1[X][Y] – (k2 + k3)[Z]
0 = k1[X][Y] – k2[Z] [ k2 >> k3]
SSA on intermediate.
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then [Z] = 1
2
k [X][Y]k
then we find, d[P]dt
= k3[Z]
= 1 3
2
k k [X][Y]k
d[P]dt
= koverall [X][Y]
i.e. koverall = 1 3
2
k kk
…(2)
koverall = 1 3
2
k kk
Aoverall . overallERTe
=
1 3
2
E ERT RT
1 3ERT
2
A e A e
A e
i.e. Aoverall = 1 3
2
A AA
and overallERTe
=
1 3 2E E ERT RT RTe
i.e. overallERT
= 1 3 2E E ERT
Eoverall = E1 + E3 – E2
Problem. What is the energy of activation of the reaction if it rate doubles when temperature
is raised .290 to 300 K.
Sol. We know that
2
1
klogk
= a 2 1
1 2
E T T2.303 R T T
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2klogk
= aE 300 2902.303 R 300 290
log 2 = aE 102.303 8.314 300 290
Ea = 50145.617 J
Ea 50.145 kJ
Problem. A plot of log k versus 1T
gave a straight line of which the slope was found to be
–1.2 × 104 K. What is the activation energy of the reaction.
Sol. k = aE / RTAe
log k = aElog A2.303 RT
log k = aE 1 log A2.303 R T
y = m x + C
where m = slope of line
then aE2.303 R
= slope
Ea = –2.303 R (slope)
= –2.303 × 8.314 × (–1.2 × 104 K)
= 1.0 × 105 J mol–1
Fast Reaction.
Fast reactions are studies by following methods
(1) Stopped-Flow technique : For reaction that occur on timescales as short at 1 ms (10–3 s)
(2) Flash photolysis technique : Reaction that can be triggered by light are studied using flash
photolysis.
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(3) Perturbation-relaxation methods : A chemical system initially at equilibrium is perturbed
such that the system is not longer at equilibrium. By following the relaxation of the system back
toward equilibrium, the rate constant for the reaction can be determined.
The temperature perturbation or T-jump are most important type of perturbation.
Problem. Using the T-jump method find out the relaxation time () of the following reaction,
1
1
k '
k 'A B
1
1
k
kA B
Sol. Let a be the total concentration of (A + B) and x the concentration of B at any instant. Then
rate 1 1d[B] dx k (a x) k (x)dt dt
If xe is the equilibrium concentration, then
x = x – xe or x = x + xe
Since d( x)dt = dx
dt, we have
d( x)dt = k1(a – xe – x) – k–1 (xe + x)
at equilibrium, dxdt
= 0 and x = xe. Hence
k1(a – xe) = k–1 xe
then d( x)dt = –(k1 + k–1)x
= –krx
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where kr = k1 + k–1
= relaxation rate constant
then d xdt = –kr dt
0
x
x
d xdt
=
t
r0
k dt
x = rk t0x e
Then reciprocal of kr i.e. 1rk is called relaxation time.
It is represented by
= 1 1r 1 1k (k k )
= 1 1
1k k
Problem. Find the relaxation time for the following reaction.
1
1
kk
A B C
Sol. Let a be the total concentration and x the concentration of B which is equal to the
concentration of C. Then, the rate law is given by
r 21 1
dx k (a x) k xdt
Now x = x – xe
xe = equilibrium concentration of x
d( x)dt = k1(a – xe – x) – k–1 (xe + x)2
= k1(a – xe) – k–1 x – k–1xe2 – 2k–1 xe x – k–1(x)2
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at equilibrium, dxdt
= 0, hence
k1(a – xe) = k–1 xe2
we get d( x)dt = –k1 + x – 2k–1 xe x – k–1(x)2
x is very small than (x)2 is neglected,
d( x)dt = 1 1 e r(k 2k x ) x k x
where kr = k1 + 2k–1 xe
is the relaxation rate constant.
and d( x)dt = –kr x
d xx
= –kr dt
d xx = rk dt
x = rk t0x e
The relaxation time in this case is
= 11 1 e
1 1 e
1(k 2k x )k 2k x
Problem. The relaxation time for the fast reaction 1
1
kk
A B
is 10 µs and the equilibrium
constant is 1.0 × 10–3. Calculate the rate constant for the forward and the reverse reactions.
Sol. = 6 5
1 1
1 10 10 s 10 sk k
K = equilibrium constant = 1.0 × 10–3 = 1
1
kk
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k1 = 1.0 × 10–3 k–1
= 10–5 = 1 1
1k k
= 3 31 1 1
1 110 k k (1 10 ) k
10–5 = 31
110 k
k–1 = 108 s–1
k1 = 105 s–1
The Collision Theory of Bimolecular Gaseous Reaction.
The reaction between two species takes place only when they are in contact i.e. the reactant
species must be collide before they react.
Consider the bimolecular elementary reaction.
A + B P
rate = v = d[A] d[B] k[A][B]dt dt
The rate of reaction to be proportional to the rate of collision i.e. the mean sped of the
molecules, their collision cross-section () and the number of densities of A and B.
Using kinetic theory of gases, the rate of bimolecular collisions per second per cm–3 between
unlike molecule is given by
ZAB = 1/2
2A B AV
8 kTn n (d )
Where nA & nB are number of A and B molecules, dAV is the average collision diameter
defined as A Bd d2 and µ is the reduced mass defined as A B
A B
m mm m
ZAB = collision frequency
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The detailed analysis of the bimolecular collisions leads to the result that the number of
collision per second per cm3 between molecules A and B is given by
rate = ZAB 0E / RTe = no. of collision
where E0 = Energy generated by collision
then the rate of relative collision is given by
Adndt
= 0E / RTABZ e
Ad(N A)dt
= 01/2
E / RT2A B av
8 kTn n (d ) e
Ad[A]N
dt = 0
1/ 2E / RT2
A A AV8 RTN [A] N [B] (d ) e
[A] = A B
A A
n n& [B]N N
d[A]dt
= 01/ 2
E / RT2A AV
8 RTN [B][A] (d ) e
d[A]dt
= 01/ 2
E / RT2A av
8 RTN (d ) T [A][B]e
let M = 1/ 2
2A av
8 RN (d )
then d[A]dt
= 0E / RTM T [A][B]e …(1)
we know that
d[A]dt
= k[A][B]
then k = 0E / RTM T e …(2)
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The collision theory can be generalized by introducing the steric factor, P, into the equatiohn
for the bimolecular rate constant.
Then k = 0E / RTPM T e
Relation between Ea and E0 :
By between equation; k = aE / RTAe
By collision theory, k = 0E / RTPM T e
Taking natural log we get
ln k = aEln ART
…(1)
ln k = ln P + ln M + 0Eln TRT
…(2)
Differentiate both equation (1) and (2) with respect to T we get
d ln kdT
= 1a aE Ed ln A d d0 TdT dT RT R dT
d ln kdT
= a2
ERT
…(1a)
and d ln kdT
= 0Ed ln P d ln M d ln T ddT dT dT dt RT
d ln kdT
= 10E1 d ln T d0 0 T2 dT R dT
d ln kdT
= 02
E12T RT
…(2a)
Comparing equation (1a) & (2a) we get
a2
ERT
= 02
E12T RT
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Ea = 22
02
RT ERT2T RT
Ea = 0RT E2
…(3)
The expression for Arrhenius pre-exponential factor using collision theory
We know that
k = aE / RTAe [by Arrhenius equation]
k = 0E / RTPM T e [by collision theory]
then aE / RTAe = 0E / RTPM T e
0
RT E / RT2Ae
= 0E / RTPM T e
0RT E
2RT RTAe e
= 0E / RTPM T e
Ae–1/2 = PM T
A = 1/ 2PM T e …(1)
We know that
M = 1/ 2
2A AV
8 RN d
then A = 1/ 2
2 1/ 2A AV
8 RP N d T e
…(2)
Activated complex Theory of Bimolecular Reaction or Transition state Theory or Eyring
Equation
The activated complex forms between reactants as they collide. The difference between the
energy of the activated complex and the energy of the reactants is the activation energy, Ea.
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(a) Exothermic reaction
(b) Endothermic reaction According to Eyring the equilibrium is between reactants and the activated complex.
Consider A and B react to form an activated complex that undergoes decay, resulting in
product formation.
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The activated complex represents the system at the transition state.
This complex is stable.
1 2k k#A B (AB) Product
where (AB)# is the activated complex and k1 is the equilibrium constant between reactants and
activated complex.
If (AB)# one of the vibrational degrees of freedom has become a translational degree of
freedom.
From the classical mechanics,
Energy = kBT = A
RTN
kB = Boltzmann constant
from the quantum mechanics,
energy = hv
than hv = A
RTN
v = A
RTh N
The vibrational frequency v is the rate at which the activated complex move across the energy
barrier i.e. the rate constant k2 is identified by v.
Then the reaction is
r = # #2
d[A] k (AB) (AB)dt
k1 = #(AB)
[A][B]
(AB)# = k1[A][B]
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then r = #d[A] (AB)dt
r = v k1[A][B] = A
RTN h
k1 [A][B]
for conventional rate
r = 2d[A] k [A][B]
dt
k2[A][B] = A
RTN h
k1[A][B]
k2 = A
RTN h
k1
i.e. k2 = v × k1
k2 = frequency × k1 …(1)
where k1 = equilibrium constant = keq
Relation between k1 and G# :
k1 = equilibrium constant = #G / RTe
& G# = H# – TS#
where G#, H# & S# are the standard free energy of activation, enthalpy of activation and
entropy of activation.
k2 = keq × frequency = k1× frequency
k2 = #G / RT
A
RTeN h
…(2)
k2 = # #( H T S ) / RT
A
RT eN h
k2 = # #H / RT S / R
A
RT e . eN h
…(3)
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H# = E# + ngRT
k2 = # #gnE / RT S / R
A
RT e . e . eN h
…(4)
and ng = difference is number of moles between transition state and reactant
Relation between E# and Ea
We know that
k2 = # #gnE / RT S / R
A A
RT RTe . e . e .N h N h
On taking log
ln k2 = # #
gA
E S Rn ln ln TRT R N h
On differentiate above equation
d ln kdT
= #E 10 0 0
RT T
or d ln kdT
= #E 1
RT T
and d ln kdT
= a2
ERT
(from arrhanius equation)
then a2
ERT
= #E 1
RT T
Ea = E# + RT …(5)
Relation between E0 & E#
Ea = E# + RT
0RTE2
= E# + RT
E0 = Collision energy
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E0 = E# + RT2
Value of A, using above equations
From Eyring theory and Arrhenius theory we have
rate constant = # #ga nE / RT E / RT S / R
A
RTAe e . e . e .N h
aE / RTAe = #gn(E RT) / RT S / R
A
RTe . e . e .N h
A = #
g1 n S / R
A
RTe .N h
…(6)
Problem. Consider the decomposition of NOCl,
2NOCl(g) 2NO(g) + Cl2(g)
The Arrhenius parameters for this reaction are A = 1.00 × 1013 M–1 s–1 and Ea = 104 kJ mol–1.
Calculate H# and S# for this reaction with T = 300 K.
Sol. We know that
H# = E# + ngRT
where ng = difference in number of moles between activated complex and reactant.
H# = E# + (–1) RT
H# = E# – RT
& Ea = E# + RT
then E# = Ea – RT
H# = E# + RT = Ea – RT – RT = Ea – RT
H# = 104 kJ mol–1 – 2(8.314 J mol–1 K–1) (300 K)
= 99.0 kJ mol–1
We know that
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A = #
g1 n S / R
A
RTe .N h
#S / Re = gn 1AN hA e
RT [ng = –1]
Taking log
#S
R = A
2AN hlne RT
S# = (8.314 J mol–1 K–1)
13 1 1 34 23
2 1 1(1 10 M s ) (6.6 10 J-s) (6.023 10 )ln
e (8.314 Jmol K ) (300 K)
S# = –12.7 J mol–1 K–1
Note :
for unimolecular ng = 0
for bimolecular ng = –1
for trimolecular ng = –2
& H# = E# + ngRT [ Ea = E# + RT]
= Ea – RT + ngRT
Ea = H# + RT – ngRT
for unimolecular Ea = H# + RT
for bimolecular Ea = H# + 2RT
for trimolecular Ea = H# + 3RT
The Pre-equilibrium Approximation
Consider the following reaction
1 3
2
k kk
A B I P
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(i) First, equilibrium between the reactants and the intermediate is maintained during the
course of the reaction.
(ii) The intermediate undergoes decay to form product.
Then the rate law expression is
d[P]dt
= k3[I]
I is in equilibrium with the reactant then
[I][A][B]
= 1C
2
k kk
= equilibrium constant
[I] = kC [A][B]
d[P]dt
= k3kC[A][B]
d[P]dt
= keff [A][B]
keff = k3kC = 3 1
2
k kk
The Lindemann Mechanism
Lindemann mechanism for unimolecular reactions involves two steps. First reactants acquire
sufficient energy to undergo reaction through a bimolecular collision.
A + A 1k A* + A
In this, A* is the activated reactant and undergoes one of two reactions.
A* + A 1k A + A
A* 2k P
Then, rate of product formation is
d[P]dt
= k2[A*]
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& rate of formation of A*
*d[A ]
dt = k1[A]2 – k–1[A][A*] – k2[A*]
Applying the steady-state approximation
*d[A ]
dt = 0 = k1[A]2 – k–1[A][A*] – k2[A*]
[A*] = 2
1
1 2
k [A](k [A] k )
Then d[P]dt
= 2
1 2
1 2
k k [A]k [A] k
It state that the observed order dependence on [A] depends on the relative magnitude of k–1[A]
versus k2. At high reactant concentration, k–1[A] > k2 and
d[P]dt
= 1 2
1
k k [A]k
AP n [A]RT V
i.e. the product formation is first order at high pressure.
At low reactant concentration k2 > k–1[A] and
d[P]dt
= k1[A]2
i.e. at low pressure, the rate of formation product is second order in [A].
Then R = 1 2uni
1 2
d[P] k k [A][A] k [A]dt k [A] k
kuni is the apparent rate constant for the reaction defined as
kuni = 1 2 1 2
21 2 1
k k [A] k kkk [A] k k[A]
and uni
1k
= 1
1 2 1
k 1 1k k k [A]
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when k–1[A] >> k2 i.e. at high concentration
kuni = 1 2
1
k kk
when k–1[A] << k2 i.e. at low concentration
kuni = k1[A]
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CATALYSIS
A catalyst is a substance that participates in chemical reaction by increasing the rate of
reaction, yet the catalyst itself remains intact after the reaction is complete.
The mechanism describing a catalytic process is as follows :
S + C 1
1
kk
SC
SC 2k P + C
where S represents the reactant; C is catalyst and P is the product. The reactant or substrate-
catalyst complex is represented by SC and is an intermediate.
The rate expression for product formation is
d[P]dt
= k2[SC] …(1)
Because SC is an intermediate than apply S.S.A. on the formation of SC.
d[SC]dt
= k1[S][C] – k–1[SC] – k2[SC] = 0
[SC] = 1
1 2 m
k [S][C] [S][C]k k k
…(2)
km = 1 2
1
k kk
= composite constant
then d[P]dt
= 2
m
k [S][C]k
…(3)
The relationship between initial concentration and the concentration of all species present after
the reaction is :
[S]0 = [S] + [SC] + [P]
[C]0 = [C] + [SC]
then [S] = [S]0 – [SC] – [P] …(4)
& [C] = [C]0 – [SC] …(5)
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Substituting these values in equation (2), we get
[SC] = 0 0
m
([S] [SC] [P]) ([C] [SC])k
Km[SC] = ([S]0 – [SC] – [P]) ([C]0 – [SC])
[SC] = 0 0
0 m
[S] [C][S] [C] k
then rate of the reaction becomes
R0 = 2 0 0
0 0 m
k [S] [C]d[P]dt [S] [C] k
Case I. [C]0 << [S]0
i.e. much more substrate is present in comparison to catalyst. Then
R0 = 2 0 0
0 m
k [S] [C][S] k
if km < [S]0
then R0 = 2 0 02 0
0
k [S] [C] k [C][S]
i.e. zero order reaction with respect to substrate.
& 0
1R
= m
2 0 0 2 0
k 1 1k [C] [S] k [C]
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when concentration of substrate [S]0 >> km then reaction rate R0 = k2[C]0 = Rmax i.e. the rate of reaction will reach a limiting value where the rate becomes zero order in
substrate concentration. Case II. [C]0 >> [S]0
R0 = 2 0 0
m 0
k [S] [C]k [C]
i.e. the reaction rate is first order in [S]0, but can be first or zero order in [C]0 depending on the size of [C]0 relative to km.
Michaelis-Menten Enzyme Kinetics. Enzyme are protein molecules that serve as catalysts in a chemical reaction.
The kinetic mechanism of enzyme catalyst can be described using the Michaelis-Menten
mechanism.
E + S 1
1
k
k ES 2k E + P
E is enzyme, S substrate, ES is enzyme-substrate complex and P is product.
The mechanism of above reaction is similar to catalytic mechanism.
rate = 2 0 0
0 0 m
k [S] [E][S] [E] k
But in this mechanism substrate concentration is greater than that of enzyme
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i.e. [S]0 >> [E0]
then rate of formation of product in enzyme catalyst is
R0 = 2 0 0
0 m
k [S] [E][S] k
…(1)
The composite constant km is referred to as the Michaelis constant in enzyme kinetics and the
equation is referred to as the Michaelis-Menten rate law.
When [S]0 >> km, the Michaelis constant can be neglected, resulting new expression for the
rate.
R0 = k2[E]0 = Rmax
The reciprocal equation of equation (1) is the Lineweaver-Burk equation i.e.
R0 = 2 0 0 max 0
0 m 0 m
k [S] [E] R [S][S] k [S] k
0
1R
= m
max max
1 kR R [S]
…(2)
This equation is known as Lineweaver-Burk equation.
The plot of reciprocal of rate is known as Linewearver-Burk plot.
k2 is known as turn over numberof the enzyme.
“The turn over number is the maximum number of substrate molecules per uit time that can be
converted into product.”
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This is Linewearver-Burk plot.
We know that d[P]dt
= 2 0 0
m 0
k [E] [S]k [S]
km = 1 2
1
k kk
Case I. ` [S]0 >> km
R = 2 0 0
0
k [E] [S]d[P]dt [S]
=k2[E]0
i.e. rate is maximum due to all enzyme are present
R = Rmax = k2[E]0
This is zero order w.r.t. substrate.
Case II. If [S]0 = km
R = 2 0 0 2 0 0
0 m 0
k [E] [S] k [E] [S]d[P]dt [S] k 2[S]
R = 2 0 maxk [E] R2 2
Case III. If [S] << km
E = 2 0 0
m
k [E] [S]d[P]dt k
R = max 0
m
R [S]k
This is first order w.r.t. substrate.
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This is graph between initial rate and concentration of substrate.
G.S. Eadie Plot
We know that,
1R
= m
max max 0
1 kR R [S]
Multiplying with R,
RR
= m
max max 0
R R kR R [S]
1 = m
max max 0
R R kR R [S]
Multiplying with max
m
Rk
,
max
m
Rk
= max max m
m max m max 0
R RR R kk R k R [S]
max
m
Rk
= m 0
R Rk [S]
or 0
R[S]
= max
m m
R Rk k
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0
R[S]
= max
m m
RRk k
y = mx + C
Homogeneous and Heterogeneous Catalysis.
A homogeneous catalyst is a catalyst that exist in the same phase as the species involved in the
reaction, and heterogeneous catalysts exist in a different phase. Enzymes surve as an example of a
homogeneous catalyst, they exist in solution and catalyze reactions that occur in solution.
In heterogeneous catalysis reaction, an important step in reactions involving solid catalysis is
the absorption of one or more of the reactants to the solid surface. The particles absorb to the surface
without changing their internal bonding. An equilibrium exists between the free and surface-absorbed
species or adsorbate and surface adsorption and deadsorption can be obtained.
A critical parameter in evaluating surface adsorption is the fractional coverage, , defined as
= Number of adsorption sites occupied
Total numbe rof adsorption site
The variation of with pressure at fixed temperature is called adsorption isotherm.
(a) The Langmuir Isotherm
The simplest kinetic model describing the adsorption process is known as the Langmuir model,
where adsorption is described by the following mechanism
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R(g) + M(surface) a
d
k
k RM (surface)
R is reagent, M is an unoccupied absorption site of catalyst and RM is an occupied adsorption
site. ka and kd is the rate constant for adsorption and deadsorption.
Three approximations are employed in the Langmuir model :
(1) Adsorption is complete once monolayer coverage has been reached.
(2) All adsorption site are equivalent and the surface is uniform
(3) Adsorption and deadsorption are uncooperative processes. The occupancy state of the
adsorption site will not affect the probability of adsorption or deadsorption for adjacent site.
The rate of change in will depends on the rate constant for adsorption ka, reagent pressure P
and the number of vacant site which is equal to the total number of adsorption sites, N, times the
fraction of sites that are open (1 – )
ddt
= ka PN(1 – )
The change in due to deadsorption is
ddt
= –kd N
At equilibrium, the change in with time is zero i.e.
(ka PN + kdN) = kaPN
=
a
a d
a da d
d d
kP
k P kk kk P k Pk k
= kP
kP 1
where k is the equilibrium constant defined as a
d
kk
.
This equation is the equation for the Langmuir isotherm.
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In many instances adsorption is accompanied by dissociation of the adsorbate, a process that is
described by the following mechanism :
R2(g) + 2M(surface) a
d
k
k 2RM (surface)
ddt
= kaP{N(1 – )}2
& ddt
= –kd (N)2
The condition for no net change leads to the isotherm.
= 1/ 2
1/2(kP)
1 (kP)
i.e. the surface coverage now depends more weakly on pressure than for non-dissociative
adsorption.
(b) The BET isotherm.
If initial adsorbed layer can be act as a substance for further adsorption, then, instead of the
isotherm leveling off the some saturated value at high pressure, it can be expected to rise indefinitely.
The most widely used isotherm dealing with multiplayer adsorption was derived by Brunauer, Emmett
and Teller, and is called the BET isotherm.
mono
VV
= CZ
(1 Z) {1 (1 C)Z} wth Z = 0
PP
P° = vapour pressure above a layer of absorbate
Vmono = volume of monolayer coverage.
C = constant
The Langmuir-Hinshelwood mechanism for adsorption and catalysis.
(1) Unimolecular surface Reaction.
A is reactant and S is the vacant site on surface.
If r is the rate of the reaction, then according to the Langnuir-H. hypothesis,
r
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r = k2
Apply S.S.A. for the formation of [AS].
d[AS]
dt = k1[A][S] – k1[AS] – k2[S] = 0
[AS] = 1
1 2
k [A][S]k k
If Cs is the total concentration of active site on surface, then the concentration [S] of the vacant
sites on the surface is equal to the product of Cs and (1 – ). Thus
[S] = Cs (1 – )
Also, the concentration of AS on the surface is,
[AS] = Cs
then Cs = 1 1 s
1 2 1 2
k [A][S] k [A] C (1 )k k k k
or 1
= 1 2
1
k kk [A]
or 1
1
= 1 2
1
k kk [A]
or 1
= 1 2 1 1 2
1 1
k k k [A] k k1
k [A] k [A]
= 1
1 1 2
k [A]k [A] k k
…(1)
thus, r = 1 2
1 1 2
k k [A]k [A] k k
...(2)
The concentration is expressed in terms of partial pressure
Then r = 1 2 A
1 A 1 2
k k Pk P k k
…(3)
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or 1r
= 1 2
2 1 2 A
1 k kk k k P
Two limiting cases
Case I. k2 > k1PA + k–1
r = 2 1 A
2
k k Pk
r = k1PA it is first order w.r.t. A
Case II. k2 << k1PA + k–1
r = 2 1 A
1 A 1
k k Pk P k
r =
12 A
1
1 A 1
kk P
kk P k
1
1
kk
= keq
R = 2 eq A
eq A
k k Pk P 1
Two situations arise depending upon the pressure :
(1) at low pressure 0 and keq PA << 1 so that
r = k2keqPA
It is first order w.r.t. PA or [A].
(a) at high pressure; 1 and keqPA >> 1 so that
r = k2
it is zero order with respect to PA or [A].
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(2) Bimolecular surface reaction A + B P
rate = r = A Bk
A = A A
A A B B
k P1 k P k P
B = B B
A A B B
k P1 k P k P
then it follow the rate law
r = A B A B2
A A B B
k k k P P(1 k P k P )
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PHOTOCHEMISTRY Photochemistry process involve the initiation of a chemical reaction through the absorption of
a photon by an atom or molecule.
When a molecule absorbs a photon of light, the energy is photon is transferred to the molecule.
The energy of a photon is given by the Planck equation :
E = hv = hv
h = Planck constant
= 6.626 × 10–34 J-s
c = speed of light in vacuum
= 3 × 108 ms–1
v = frequency of light
and = wave length of light
The phenomena of photochemistry of photochemistry as best explained by Jablonski diagram.
So, S1, S2, T1 & T2 are electronic level.
S singlet
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T triplet
Loss of excess electronic energy through the emission of a photon is known as rediative
transition.
The process by which photons are emitted in radiative transition between S1 and S0 is known
as fluorescence.
The process by which photons are emitted in radiative transition between T1 and S0 is known
as phosphorescence.
The life time for phosphorescence is longer (10–6 s) than fluorescence (10–9 s)
Photo physical reactions are corresponding rate expression
Process Reaction Rate
Absorption/excitation S0 + hv S1 ka[S0]
Fluorescence S1 S0 + hv kf[S1]
Internal conversion S1 S0 kic[S1]
Intersystem crossing S1 T1 sisc 1k [S ]
Phosphoresence T1 S0 + hv kp[T1]
Intersystem crossing T1 S0 sisc 1k [T ]
Quantum yield = Number of events
number of photons absorbed
= abs
rate of process rintensity of light absorbed I
The Bear Lambert Law.
When a beam of monochromatic radiation of a suitable frequency passes through a solution, it
is absorbed by the solution.
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I0 = intensity of incident light
I = intensity of transmitted light
and Ia = Intensity of the light absorbed = I0 – I
Absorbance of solution, A = 0Ilog C
I
A = 0Ilog C
I …(1)
where A = absorbance
C = concentration of solution
= molar extinction coefficient
or molar absorption coefficient
(unit = concentration length–1)
l = path length
Transmittance; T = 0
II
T = 0
II
…(2)
The absorbance of a solution is additive whereas the transmittance is multiplicative.
Problem. A monochromatic light is incident on solution of 0.05 molar concentration of an
absorbing substance. The intensity of the radiation is reduced to one-fourth of the initial value after
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passing through 10 cm length of the solution. Calculate the molar extinction coefficient of he
substance.
Sol. From bear Lambert law
0
Ilog
I = C
0
II
= 14
= 0.25 = 25%
i.e. 0II
= 1
0.25 = 4
log 4 = Cl = × 1 10 cm × 0.05 3moldm
2 × 0.3010 = 0.5
= 1.204 dm3 mol–1 cm–1
Problem. A substance when dissolved in water at 10–3 M concentration absorbs 10 percent of
an incident radiation in a path of 1 cm length. What should be the concentration of the solution in
order to absorb go per cent of the same solution.
Sol. 10% absorbed then 90% transmitted
then T = 0
II
= 90%
log 0II
= Cl
1
log90%
= 100
log90
= × 10–3 × 1 …(1)
90% absorbed then 10% transmitted
then T = 0
II
= 10% = 10100
log 0II
= log 10010
= Cl = × C × 1 …(2)
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equation (1)equation (2)
= 3
100log 10 190
100 C 1log10
100log
90log 10
= 310
C
0.04575 = 310
C
C = 0.0218 mol dm–3.
Objective Questions asked in previous years of Gate and GRF examination.
Problem. In carbon-dating application of radio isotopes, 14C emits (JRF – June 2012)
(1) -particle (2) -particle
(3) --particle (4) positron
Sol. 14 14 06 7 1C N -particle ( e )
Correct answer is (1)
Problem. With increase in temperature, the Gibb’s free energy for the adsorption of a gas on a
solid surface. (JRF – June 2012)
(1) becomes more positive from a positive value
(2) becomes more negative from a positive
(3) becomes more positive from a negative value
(4) becomes more negative from a negative value
Sol. From Langmuir isotherm, the fractional coverage is
= kP
kP 1
P T
Then higher the pressure or temperature lower the value of fractional coverage .
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R(g) + M(surface) RM (surface)
if decreases then the formation of RM decreases.
i.e. rate of formation of RM decreases.
This indicate that Gibbs from energy of adsorption become positive.
So, increase the temperature, the Gibbs free energy of adsorption of a gas on a solid surface
become more positive from a negative value.
The correct answer is (3).
Problem. One of the assumption made in the conventional activated complex theory is :
(JRF – June 2012)
(1) equilibrium is maintained between the reactants and the activated
(2) equilibrium is maintained between the reactants and the product
(3) equilibrium is maintained between the products and the activated complex
(4) equilibrium is maintained between the reactants, the activated complex and the products
Sol. According to the Eyring the equilibrium is maintained between reactants and the
activated complex
A + B 1k (AB)# 2k Products
(AB)# is the activated complex.
The correct answer is (1)
Problem. For a reaction, the rate constant k at 27°C was found to be
k = 5.4 × 1011 e–50
The activation energy of the reaction is (JRF – June 2012)
(1) 50 J mol–1 (2) 415 J mol–1
(3) 15000 J mol–1 (4) 125000 J mol–1
Sol. From Arrhenius equation
k = aE / RTAe …(1)
& k = 5.4 × 1011 e–50 …(2)
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From equation (1) & (2) we get
aE / RTAe = e–50
aERT
= 50
Ea = RT × 50 = 50 × 8.314 J K–1 mol–1 × 300 K
Ea = 124710 J mol–1 125000 J mol–1
The correct answer is (4)
Problem. The carbon-14 activity of an old wood sample is found to be 14.2 disintegration
min–1 g–1. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is 15.3
disintegration min–1 g–1 (t1/2 carbon =14) = 5730 year), is (JRF – June 2012)
(1) 5000 year (2) 4000 year
(3) 877 year (4) 617 year
Sol. Activity = AA
dNkN
dt
k = rate constant
and NA = no. of atom
Activity of old wood = k Nold = 14.2 …(1)
Activity of new wood = k Nnew = 15.3 …(2)
From equation (1) & (2) we get
old
new
k Nk N
= 14.215.3
or new
old
NN
= 15.314.2
…(3)
We know that
k × t1/2 = 0.693
k = 1/ 2
0.633 0.693t 5730
…(4)
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We know that
k t = 2.303 log 0
t
NN
i.e. t = new
old
2.303 Nlog
k N
= 2.303 15.2
log0.693 14.25730
= 5730 2.303 15.3
log0.693 14.2
T = 617 year
The correct answer is (4).
Problem. Using cuvettes of 0.5 cm path length, a 10–4 M solution of a chromphone shows
50% transmittance at certain wave length. The molar extinction coefficient of the chromphre at this
wave length is (log 2 = 0.3010) (JRF – June 2012)
(1) 1500 M–1 cm–1 (2) 3010 M–1 cm–1
(3) 5000 M–1 cm–1 (4) 6020 M–1 cm–1
Sol. Transmittance = T = 0
I 5050%
I 100
Absorbance = A = Cl = log 0II
Cl = log 10050
= log 2 = 0.3010
= 40.3010 0.3010
C 10 M 0.5 cml
= 4 1 1
40.3010 0.3010 10 M cm
0.510 M 0.5 cm
= 6020 M–1 cm–1
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The correct answer is (4).
Problem. The rate law for one of the mechanisms of the pyrolysis of CH3CHO at 520°C and
0.2 bar is :
Rate = 1/ 2
3/212 3
4
kk [CH CHO]2k
The overall activation energy Ea in terms of the rate law is : (JRF – June 2012)
(1) Ea(2) + Ea(1) + 2Ea(4) (2) Ea(2) + 12
Ea(1) – Ea(4)
(3) Ea(2) + 12
Ea(1) – 12
Ea(4) (4) Ea(2) – 12
Ea(1) + 12
Ea(4)
Sol. Rate = 1/ 2
3/212 3
4
kk [CH CHO]2k
= koverall [CH3CHO]3/2
i.e. koverall = 1/ 2
12
4
kk2k
Aoverall aE / RTe = a
a4
1/2E / RT
1E / RT
4
A e
2A e
i.e. Aoverall = 1/ 2
12
4
AA2A
and aE / RTe = a1
a2a4
1/ 2E / RTE / RT
E / RTeee
= a a a2 1 4E / RT E / 2RT E / 2RTe e e
or aERT
= 2 1 4a a aE E ERT 2RT 2RT
–Ea = 1 42
a aa
E EE
2 2
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Ea = 1 42
a aa
E EE
2 2
The correct answer is (3).
Problem. In the Michaelis-Menten mechanism of enzyme kinetics, the expression obtained as
0
V[E] [S]
= 1.4 × 102 – 4
0
10 V[E]
The value of k3 and k(Michaelis constant, mol L–1) are (JRF – June 2012)
(1) 1.4 × 1012, 104 (2) 1.4 × 108, 104
(3) 1.4 × 108, 10–4 (4) 1.4 × 1012, 10–4
Sol. We know that Michaelis Menten equation is:
rate = V = 3 0 0
0 m
k [S] [E][S] k
…(1)
1V
= 0 m
3 0 0
[S] kk [S] [E]
1V
= 0 m
3 0 0 3 0 0
[S] kk [S] [E] k [S] [E]
1V
= m
3 0 3 0 0
1 kk [E] k [S] [E]
Multiply this equation by 3
m
k Vk
we get
3
m
k Vk V
= 3 3 m
m 3 0 m 3 0 0
k V k V1 kk k [E] k k [S] [E]
3
m
kk
= m 0 0 0
V Vk [E] [S] [E]
or 0 0
V[S] [E]
= 3
m m 0
k Vk k [E]
…(2)
and 0 0
V[E] [S]
= 4
12
0
10 V1.4 10[E]
…(3)
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Comparing equation (2) & (3) we get
m
1k
= 104
i.e. km = 10–4
and 3
m
kk
= 1.4 × 1012
k3 = 1.4 × 1012 × 10–4
= 1.4 × 108
i.e. The correct answer is (3).
Problem. The Langunier adsorption isotherm is given by kP1 kP
, where P is the pressure
of the adsorbate gas. The Langmuir adsorption isotherm for a diatomic gas A2 undergoing dissociative
adsorption is (JRF Dec.2011)
(1) kP1 kP
(2) 2kP1 2kP
(3) 2
2(kP)
1 (kP)
(4)
1/ 2
1/ 2(kP)
1 (kP)
Sol. R(g) + M(surface) a
d
k
k RM (surface)
then = kP
1 kP
if R2(g) + 2M (surface) a
d
k
k 2M (surface)
= 1/ 2
1/ 2(kP)
1 (kP)
i.e. the correct answer is (4).
Problem. The overall rate of following complex reaction
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1k22A A (fast equilibrium)
A + B 2k C (fast equilibrium)
A2 + C 3k P + 2A (slow)
The steady state approximate would be (JRF Dec.2011)
(1) k1k2k3[A]3[B] (2) k1k2k3[A][B]3
(3) k1k2k3[A][B]2 (4) k1k2k3[A][B]
Sol. 1k22A A (fast equilibrium)
then k1 = 22
[A ][A]
…(1)
A + B 2k C (fast equilibrium)
then k2 = [C]
[A][B] …(2)
A2 + C 3k P + 2A (slow)
The rate of formation of product P is
d[P]dt
= k3[A][C] …(3)
From equation (1) & (2) we get
[A2] = k1[A]2
& [C] = k2[A][B]
then d[P]dt
= k3k1[A]2 k2[A][B]
d[P]dt
= k1k2k3[A]3[B]
i.e. the correct answer is (1)
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Problem. The species 19Ne and 14C emit a position and -particle respectively. The resulting
species formed are respectively (JRF June 2011)
(1) 19Na and 14B (2) 19F and 14N
(3) 19Na and 14N (4) 19F and 14B
Sol. 19 19 010 9 1Ne F e
and 14 14 06 7 1C N e
i.e. the correct answer is (b).
Problem. The half life of a zero order reaction (A P) is given by (k = rate constant)
(JRF June 2011)
(1) 01/ 2
[A]t
2k (2) 1/ 2
2.303t
k
(3) 01/ 2
[A]t
k (4) 1/ 2
0
1t
k[A]
Sol. kA P
d[A]
dt = k[A]0 = k
if t = t1/2 then [A] = 0[A]2
Thus [A]0 – 0[A]2
= k t1/2
0[A]2
= k t1/2
t1/2 = 0[A]2k
i.e. the correct answer is (1).
Problem. The concentration of a reactant undergoing decomposition was 0.1, 0.08 and 0.067
mol L–1 after 1.0, 2.0 and 3.0 hr respectively. The order of the reaction is (JRF Dec. 2011)
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(1) 0 (2) 1
(3) 2 (4) 3
Sol. If kA P
d[A]
dt = n
0k[A]
0
2 1
{[A] [A] }t t
= n0k[A]
0
2 1
[A] [A]t t
= n0k[A]
[A]0 = concentration at t1
and [A] = concentration at t2
0
2 1
[A] [A]t t
= n0k[A]
0.1 0.08
2 1
= k[0.1]n = 0.02
k[0.1]n = 0.02 …(1)
0.08 0.067
3 2
= k[0.08]n = 0.013
k[0.08]n = 0.013 …(2) Equation (1) divide by equation (2) we get
n
nk[0.1]
k[0.08] =
0.020.013
n0.1
0.08
= 1.5385
[1.25]n = 1.5385 [1.25]n = 1.25 × 1.25 = [1.25]2 n = 2 i.e. second order reaction. i.e. the correct answer is (3).