Some concepts in chemical kinetics

P. K. DasDepartment of Inorganic and Physical Chemistry

Indian Institute of ScienceIndian Institute of ScienceBangalore 560012

pkdas@ipc.iisc.ernet.in

Academy workshop, Fakir Chand College, West Bengal, January 20-22, 2016

(g) OH 2 (g) CO (g) O 2 (g) H C 2224 ++ →

Why do reactions occur?

Take the example of burning of methane

e.g., burning of cooking gasO H C H O H O H O

bond breakings O = O bond making bond breakings supply energyO = O

C

HH H

HO = C = O

HH

O

HH

O

bond making release energy

Progress of the reaction

E

Although many reactions are exothermic only a very few are instantaneousand begin without help.

Example ; ∆H = -57 Kcal / mol

However , they do not react .

PbCl2 + K2 CrO4 PbCrO4 + 2 KCl ; ∆H = -ve

OH O H 222 2

1+

PbCl2 + K2 CrO4 PbCrO4 + 2 KCl ; ∆H = -ve

But it reacts instantaneously.

Therefore, we conclude that some reactions in spite of similar energetics arehaving different rates of reaction.

Rate can be measure by looking at the appearance of a product or thedisappearance of a reagent.

dtHdrate

dtHCHOd 2][][

−==time

1rate ∝

products

rate is faster

slowing down

reactants

complete

time

conc

entr

atio

n

Rates are empirically related to concentration.

For example, a much studied reaction

H2 (g) +I2 (g) 2HI (g) ; ∆H = -2 kcal/mol

rate const

Rate = k C1H2 C1

I2

Rate increases if one of the reactants are increase d. But this reaction do notproceed at room temperature. The order is 1. w .r .t . both reactants. Overall 2.Order of a reaction are experimental quantities

In the previous example of H 2 + I2 reaction the initial step is the dissociation ofthe I 2 , molecule

I2 + hv vis I2 *

I2 * 2 I

2I + H 2 2 HI

2 I + M I2 + M

You cannot tell about the order by looking at the s toichiometry (numberof molecules that gives a balanced equation).

Take a similar reaction ,

H2 + Br2 2HBr ∆H = -24.6 kcal/mol

It also has an activation energy but the

Which is very much different from the H (g) +I (g)

]][[]

1

2

21

22

BrHBrk

][Br[HrateI

1

−

+=

Which is very much different from the H2 (g) +I2 (g)

The mechanism for this reaction is then,hν

Br2 2Br InitiationBr + H2 HBr + H PropagationH + Br2 HBr + BrH + H Br H2 + Br Inhibition2 Br + M Br 2 + M Termination

This is free radical chain mechanism consists of 5 elementary steps .Activation energy for certain reaction can be very high and to carry out these reaction is real time scales, need to lower t he activation energy.

The catalyst increases the rate of a chemical react ion.

N2 (g) + 3H2 (g) 2 NH3(g) ;2 CO (g) + O2 (g) 2CO2 (g)hydrocarbon (g) + O 2 (g) CO2 (g) + H2O(g)2 NO (g) N2 (g) + O2 (g)

Fe Haber’s process

2 NO (g) N2 (g) + O2 (g)2 NO (g) + 2 CO(g) N2 (g) + 2 CO2 (g)

Catalytic converter Rh and Pt catalysts honeycomb su rface brings out theimportance of surface on the rate of a chemical rea ction.All enzymes are catalysts for a specific reaction.Urease hydrolyses urea to NH 3, insulin regulates glucose uptake by cells insulin produced in the pancreas goes to the blood and help glucose uptake by the cell. If the rate of any of these ste ps one attend one ends up with more or less sugar in the blood.

Branched chain reaction is another variety of a cha in reaction with adifferent consequence.

2H2 + O2 = 2H2 O ; ∆H = - 54 kcal /mol

This is harmless as such but if there is a initiati on by a spark, flame, etc., itexplodes (generates enormous amount of heat very fa st).

hνH2 2 H InitiationH2 2 H InitiationO2 2 OH + O2 O H + O chain branchingsO + H2 O H + H two radicals from oneO H + H2 H2 O + H PropagationH + wall ½ H 2 TerminationH + O2 + M H O2 + M

If the free radicals H , OH, O are produced at low pressure, they react tothe wall and get destroyed. The reaction goes smoot hly. At higher

pressure,the conc. of the radicals grows tremendously and in creases the rate of thereaction very much. The system explodes.

I(Slow reaction)

I I(Chain-reaction

explosion)

I I I(Atmospheric

flame)

I V(Thermal

explosion)

Tota

l rat

e

lower explosion limits

explosion) explosion)

P1 P2 P3

P1 First chain reaction explosion limitP2 Second P3 Third, thermal explosion limit

BIA →→

( )0

dt

Id=

Steady - state approximation mainly for simplificati on

at any time t, i.e., [I] << [A] or [B]

This is qualitatively different from the equilibriu m approximation which system at when a reaction system reaches equilibriu m, the forward and reverse reactions occur at the same rate.

BAi.e., →

[ ] [ ]

[ ] [ ]

[ ][ ] rk

fk

eB

eAeqk

and

0eBrkeAf

k

0dt

Bd

dt

Ad

mequilibriuat

BAi.e.,

==

=+−

=−

=

→

equilibrium rate constants

Tk

*EAln lnk

T*E

eAk

B

B k

−=

−

=

Most interesting will be T > dependence rate normal ly increases with increasing temperature.

Arrhenius empirical

Plotting lnk vs.1/T one gets a straigh tline. The s lope gives E*The activation energy E* can be related to enthalpy of a reaction

Collision theory of reaction rates: A+B → C + D

A = Frequency factor, kB = Boltzmann constant, E* = Activation energy

Bn

An

AB2µ3

2

2Bσ

Aσ

πAB

Z TkB

=

Collision theory of reaction rates: A+B → C + D

dtB

dn

dtA

dndtAB]/ d −=−=≡ [

Tk*E

eZdtdn B

ABA

−=−

But only those collisions are reactive for which E > E*

Number of collisions/c.c. s. If reaction occurred with every collision

Rate

BA

A nkndt

dn =−⇒From empirical law

( )( )

TαA

TB

k*E

e

Bm

Am/

Bm

A2m

3kT

4Bσ

Aσ

πk

−

+

+=

Comparing

For simple reactions it is okay but it gives an ove restimate of rate

constants for complex reactions. In order to reduce the calculated rate

1P

Tk*E

PAek B

≤

−=

constant and bring agreement to expt. People introd uced fudge factors

that are less than one. One such factor is steric fa ctor, P

300 K350 K400K450 K

u /m.s -1

F (

u)

The Distribution of molecular speeds is given by the

Maxwell-Boltzmann Distribution

dueuTπk2

mπ4F(u)du Tk2

2muB2

23

B

−

=

In the above figure the distribution of molecular speed, F (u ) is plottedagainst u (molecular speed) and it shows that as the temperat ureincreases, more molecules are likely to be found with higher values of u.

Bk = Boltzmann constant

This equation gives the probability distribution of a molecule having a speed between u and u + du.

Ene

rgy Ea

Activation Energy (40 kcal /mol)

H2 + I2exothermicity 2 kcal/mol

2HI

Progress of reaction

H2 + I2

Transition State Theory

A + B [#] C + D

T.S.

k = A exp( -E a / RT )

kBT×

Q#

= exp (-E/kBT )k

Arrhenius,1889

h×

Q= exp (-E/kBT )kTS

QAQB

Let us now look at a typical potential energy surface representing a system of particles. It will contain 3N-6coordinates or dimensions for N-atom system. In a A + BC → AB + C triatomic reaction, if the geometry is maintained collinear throughout the reaction then we will have a potential function which will depend on two instead of three coordinates. We can choose the A-B and B-C bond lengths as those two coordinates and represent the surface as the following.

basically frequency and at RT it is 6x10 –12 s -1

Polanyi, Eyring,1935

A schematic 2-D Contour diagram for A + BC � AB + CIt is actually a 2-D representation of the above PES

A schematic 3-D PES for A + BC � AB +C The light blue line is the reaction coordinate and the maximum energy point in that coordinate isthe transition state.

It is actually a 2-D representation of the above PES where energy axis is omitted.The dashed line is reaction coordinate.ν1 corresponds to thesymmetric stretch and ν2 corresponds to the asymmetric stretch of[A—B--C].

A schematic 1-D representation forA + BC � AB + CIt is a section of the PES [Fig.-1a] through the reaction coordinate [dashed line]. The highest energy point on this curve is the transition state.

Reagents

approach

Product

separation

Reagents

approachProduct

separation

Attractive Repulsive

A B C

A B C

A B

A B

A B

C

C

C

A

A

A

B

C

B

B

B

B

C

C

C

C

A

A

σ

ρ

σ = sym. stretch

ρ =asym. stretch

Transition State TheoryTake the reaction A + B C+D

Treat it in one dimension :A + B [ * ] C + D

Assume chemical equilibrium for the first step:

[ * ] Q*[A] [B] Q Q

=[A] [B] QA QB

This is the equilibrium approximation in the TSTQ’s are partition functions per unit volume

QA = h –3 N d ΓA exp(- HA/ kBT )

=

A ∫

HA is the Hamiltonian of isolated A in the vicinity of the T.S. the term containing h is introduced as an adhoccorrection to the classical partition function.

In the vicinity of the TS

H = H# + p2 /2m*

Where p is conjugate momentum to q .

H# is independent of p and q and is dependent on the other ( 3N-1) co-ordinates.

Q* = Q# (δ / h) ( 2πm*kBT )1/2

T.S.

R

P

-δ/2 +δ/2O

P.E.

Reaxn. coordinate

For the passage on the top of the T.S.

[ ] ( )∫∞

0

*/* mpdtdp ( )TBkmp *2/exp2

−

( )TBkmpdp *2/2

exp −∫+∞

∞−δ

[ ] TdtB

k*[ ]B

=

( ) 2/1*2 T

Bkmπδ

Assumed :

1. * is in equilibrium with reactants or products.

2. Equal apriori probability at the T.S.

Therefore, kf = (kBT / h ) Q# /QA QB

Is there a Quantum picture of the TST

1. Tunneling?

2. Zero-point energy?

H + H2 reaction

−

∆=RT

E

R

S

h

Tkek aB expexp

#2 o

1000/T

This shows that tunneling is important

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