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SOME IMPORTANT GRAPH REPRESENTATION

Use in chemical kinetics :

(1) y = mx + C

(2) y = –mx + C

(3) y = mx





2

(4) y = –mx

Slope = –m = tan

Intercept = zero

(5) y = ex

(6) y = e–x





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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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





27

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





28

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





29

[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





30

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





31

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





32

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.





33

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]





34

[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]





35

= 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


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.





36

i.e. T then k

The graph of ln k vs 1T

is given below


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


i.e. ln A = 2.3

& A = e2.3 = 9.974

i.e. aERT

= 100T

Ea = 100 × R = 100 × 8.314





37

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


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.





38

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.





39

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





40

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.





41

(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





42

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





43

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





44

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





45

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)





46

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





47

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.





48

(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.





49

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]





50

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)





51

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





52

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





53

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





54

(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*]





55

& 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]





56

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]





57

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)





58

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]





59

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





60

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.”





61

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.





62

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





63

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





64

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.





65

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





66

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)





67

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].





68

(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 )





69

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





70

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.





71

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





72

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)





73

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 .





74

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)





75

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)


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)





76

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


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





77


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





78

Ea = 1 42

a aa

E EE

2 2


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)





79

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





80

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)


[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)





81

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)





82

(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).

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