Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture...

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Lecture 6: Diffusion and Reaction kinetics 12-10-2010 Lecture plan: – diffusion thermodynamic forces evolution of concentration distribution reaction rates and methods to determine them reaction mechanism in terms of the elementary reaction steps basic reaction types – problems

Transcript of Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture...

Page 1: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Lecture 6: Diffusion and Reaction kinetics12-10-2010

• Lecture plan:– diffusion

• thermodynamic forces• evolution of concentration distribution

– reaction rates and methods to determine them

– reaction mechanism in terms of the elementary reaction steps

– basic reaction types– problems

Page 2: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Diffusion

• Thermodynamic force

,p T

dw d dxxμμ ∂⎛ ⎞= = ⎜ ⎟∂⎝ ⎠

If the chemical potential depends on position, the maximum nonexpansion work

μ μ μ+ Δw μ= Δ

Comparing with dw Fdx= −

,p T

Fxμ∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

Thermodynamic force

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Diffusion• Thermodynamic force of concentration gradient

0 lnRT aμ μ= +

,

ln

p T

aF RTx

∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠ ,p T

RT cFc x

∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

For ideal solution

~ ~ ~ cJ drift velocity Fx∂∂

Fick’s law of diffusion:

Particles flux:

Page 4: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Diffusion• The Einstein relation

cJ Dx∂

= −∂

s AtcJ scA tΔ

= =Δ

csc Dx∂

= −∂

D c DFsc x RT∂

= − =∂

In the case of charged particle (ion) in electric field we know drift speed vs force relation for ion mobility, so we can deduce diffusion constant

ADzeN Es uERT

= =F

uRTDz

=

For example: for 8 25 10 /m sVμ −= ⋅ 9 21 10 /D m s−= ⋅we find

Einstein relation

Page 5: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Diffusion• The Nernst-Einstein equation 2 2FF= z Dz

RTλ μ=Molar conductivity of ions in the solution

22 2 F( )m z D z D

RTν ν+ + + − − −Λ = +

• The Stokes-Einstein equation

= ezEEf

μ

FRTDzμ

=F

A

A

zekN TzeRT kTDfz fzeN f

= = =

Using Stokes’s law6kTD

aπη=

Frictional force

No charge involved -> applicable to all molecules

Page 6: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Diffusion equation• How concentration distribution evolves with time due to diffusion

2

2

2

2

c JAdt J Adt J Jt Aldt Aldt l

c c c c cJ J D D D D c l Dlx x x x x x

c cDt x

′ ′∂ −= − =

∂′ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞′− = − + = − + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

∂ ∂=

∂ ∂

Diffusion with convectionc JAdt J Adt J Jt Aldt Aldt l

c cJ J cv c v c c l v vlx x

c cvt x

′ ′∂ −= − =

∂⎛ ⎞⎛ ⎞∂ ∂⎛ ⎞′ ′− = − = − + =⎜ ⎟⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠⎝ ⎠

∂ ∂=

∂ ∂due to convection only

2

2

c c cD vt x x∂ ∂ ∂

= −∂ ∂ ∂

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Solution of diffusion equation

• 2nd order differential equation: two boundary condition are required for spatial dependence and single for time dependence

2

2

c c cD vt x x∂ ∂ ∂

= −∂ ∂ ∂

Example 1: one surface of a container (x=0) is coated with N0 molecules at time t=0

2 / 401/ 2( , )

( )x Dtnc x t e

A Dtπ−=

2 / 403/ 2( , )

8( )r Dtnc r t e

Dtπ−=

Example 2: dissolution infinitely small solid containing N0 molecules at time t=0 in 3D solvent

Page 8: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Diffusion probabilities

• Probability to find a particle at a given slab of thickness dx is proportional to the concentration there:

• The mean distance traveled by the particles:

0( ) ( ) /Ap x c x AN dx N=

21/ 2

/ 41/ 2

00 0

( ) 1 2( )

x DtAc x AN dx Dtx xe dxN Dtπ π

∞ ∞− ⎛ ⎞= = = ⎜ ⎟

⎝ ⎠∫ ∫

1/ 22 1/ 2(2 )x Dt=

Diffusion in unstirred solution, D=5x10-10 m2/S

Page 9: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Random Walk

• Apparently diffusion can be modeled as a random walk, where particle is jumping distance λ in a time τ. Direction of the jump is chosen randomly

• One dimensional walk:2 2

1/ 2/ 22 x tp e

tτ λτ

π−⎛ ⎞= ⎜ ⎟

⎝ ⎠Comparing with the solution of diffusion equation

2

2D λ

τ= Einstein-Smoluchowski equation

Connection between microscopic and macroscopic parameters.

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RATES OF CHEMICAL REACTIONS

Page 11: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Rates of chemical reactions2 3A B C D+ ⎯⎯→ +

Instantaneous rate of consumption of a reactant:

[ ] /d R dt−Instantaneous rate of formation of a product:

[ ] /d P dtFrom stoichiometry

[ ] 1 [ ] [ ] 1 [ ]3 2

d D d C d A d Bdt dt dt dt

= = − = −

Rate of the reaction:

1 i

i

dn dvdt dt

ξν

= =

In case of heterogeneous reaction the rate will be defined as mol/m2s

Page 12: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Reaction orderReaction rate is generally dependent on temperature, pressure, concentration of species, the phases where reaction occurs etc. However, an empirical relation called a rate law exists stating that:

[ ] [ ] ...a bv k A B=

rate constant,depends only on T and not on the concentration

The power is generally not equal to the stoichiometric coefficients, has to be determined from the experiment

Order of a reaction:

[ ][ ]v k A B= First order in A, first order in B, overall second order.

v k= Zero order.

1/ 2[ ] [ ]v k A B= Halforder in A, first order in B, overall tree-halves order.

algebraic dependence on the reagent concentration raised to some power

overall reaction order: ....α β+reaction order with respect to species A: α

1M s−⎡ ⎤⋅⎣ ⎦

1 1M s− −⎡ ⎤⋅⎣ ⎦

Page 13: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Measuring the rates of chemical reactions

• Experimental measuring progress of the reaction

– Monitoring pressure in the reaction involving gases

2 5 2 2

0

2 ( ) 4 ( ) ( )1(1 ) 22

3(1 )2

N O g NO g O g

n n n

p p

α α α

α

⎯⎯→ +

= +

– Absorption at particular wavelength (e.g. Br2 below)

2 2( ) ( ) 2 ( )H g Br g HBr g+ ⎯⎯→

– Conductance of the ionic solution

3 3 2 3 3( ) ( ) ( ) ( ) ( ) ( ) ( )CH CCl aq H O aq CH COH aq H aq Cl aq+ −+ ⎯⎯→ + +

Page 14: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Measuring the Rates of chemical reactions

•Flow method •Stopped-flow method (down to 1 ms range)

•Flash photolysis(down to 10 fs = 10-14 s range)

•Chemical quench flow

•Freeze quench method

concentration as function of time

concentration as function of position

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Measuring the rates of chemical reactions

• Determination of the rate lawUsual technique is the isolation method, where all the components except one are present in large amounts (therefore their concentration is constant)

0[ ][ ] [ ]v k A B k A′= =It usually accompanied by the method of initial rates, when several initial concentration of A measured (again assuming that the concentrations are constant)

0 0 0 0[ ] log log log[ ]av k A v k a A′ ′= = +

Page 16: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Measuring the rates of chemical reactions

• Example 22 ( ) ( ) ( ) ( )I g Ar g I g Ar g+ ⎯⎯→ +

A B

2α =1β =

Page 17: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Integrated rate laws• First order reaction.

– Let’s find concentration of reagent A after time t

[ ] [ ]d A k Adt

= −0 0

[ ][ ]

A t

A

d A k dtA

= −∫ ∫

0[ ] [ ] ktA A e−=0

[ ]ln[ ]

A ktA

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Slope -> k

– Half-life – time required for concentration to drop by ½.

0

1/ 2 1/ 20

1 [ ] ln 22ln ln 2[ ]

Akt t

A k

⎛ ⎞⎜ ⎟

= − = =⎜ ⎟⎜ ⎟⎝ ⎠

– Time constant – time required for concentration to drop by 1/e: 1

kτ =

Page 18: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Integrated rate laws• Second-order reactions

2[ ] [ ]d A k Adt

= −0

20

[ ][ ]

A t

A

d A k dtA

= −∫ ∫

0

0

[ ][ ]1 [ ]

AAkt A

=+0

1 1[ ] [ ]

ktA A

− =Second order

First order– Half-life – depends on initial

concentration

00 1/ 2

1/ 2 0 0

[ ]1 1[ ]2 1 [ ] [ ]

AA tkt A k A

= =+

The concentration of the reagent drops faster in the 1st

order reaction than in the 2nd order reaction

Page 19: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Reaction approaching equilibrium• Generally, most kinetics measurements are made far from equilibrium

where reverse reactions are not important. Close to equilibrium the amount of products is significant and reverse reaction should be considered.

[ ]'[ ]

A B v k AB A v k B⎯⎯→ =

⎯⎯→ =

0[ ] [ ] [ ] [ ] [ ]dA k A k B A B Adt

′= − + + =

0[ ] ([ ] [ ])dA k A k A Adt

′= − + −

( )

0[ ] [ ]k k tk keA A

k k

′− +′ +=

′+

0 0

[ ][ ] [ ] , [ ] [ ] ,

[ ]eq

eq eqeq

Bk k kA A B A Kk k k k A k

′= = = =

′ ′ ′+ +

Generally: ...a b

a b

k kKk k

= × ×′ ′

Page 20: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

The temperature dependence of reaction rate

• The rate constants of most reaction increase with the temperature.

• Experimentally for many reactions k follows Arrhenius equation

ln ln aEk ART

= −

A – pre-exponential (frequency) factor,Ea – activation energy

High activation energy means that rate constants depend strongly on the temperature, zero would mean reaction independent on temperature

Page 21: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

The temperature dependence of reaction rate

• Stages of reaction:– Reagents

– Activation complex

– Product

/aE RTk Ae−=

Transition state

•Activation energy is the minimum kinetic energy reactants must have to form the products.•Pre-exponential is rate of collisions•Arrhenius equation gives the rate of successful collisions.

Fraction of collision with required energy

reaction coordinate: e.g. changes of interactomicdistances or angles

Page 22: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Elementary reactions• Most reactions occur in a sequence of steps called elementary reactions.• Molecularity of an elementary reaction is the number of molecules coming together

to react (e.g. uni-molecular, bimolecular)

Uni-molecular: first order in the reactant

[ ] [ ]d AA P k Adt

⎯⎯→ = −

Bimolecular: first order in the reactant

[ ] [ ][ ]d AA B P k A Bdt

+ ⎯⎯→ = −

Proportional to collision rate

2H Br HBr Br+ ⎯⎯→ +

Page 23: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Consecutive elementary reactionsa bk kA I P⎯⎯→ ⎯⎯→

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ]

a

a b

b

d A k Adt

d I k A k Idt

d P k Idt

= −

= −

=

Solution for A should be in a form: 0[ ] [ ] ak tA A e−=

0 0[ ] [ ] [ ] [ ] ( )[ ]a a bk t k t k ta

b ab a

kd I k I k A e I e e Adt k k

− − −+ = = −−

0 0[ ] [ ] [ ] [ ] [ ] 1 [ ]a bk t k t

a b

b a

k e k eA I P A P Ak k

− −⎛ ⎞−+ + = = +⎜ ⎟−⎝ ⎠

Page 24: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Consecutive elementary reactions• The (quasi) steady-state approximation

[ ] 0d Idt

=

Then [ ] [ ] [ ] 0a b

d I k A k Idt

= − =

[ ] [ ] [ ]b ad P k I k Adt

= ≈and

( ) 0[ ] 1 [ ]ak tP e A−≈ −

Page 25: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Consecutive elementary reactions• Pre-equilibria

Then

[ ] [ ] [ ][ ]b bd P k I k K A Bdt

= =and

,a a bk k kA B I P′+ ←⎯⎯→ ⎯⎯→This condition arises when k’a >>kb.

[ ][ ][ ]

a

a

kIKA B k

= =′

Second order form with composite rate constant

k

Page 26: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Unimolecular reactions• As the molecule acquires the energy as a result of

collision why the reaction is still a first order?

• Lindemann-Hinshelwood mechanism*

* 2

** *

** *

[ ] [ ]

[ ] [ ][ ]

[ ] [ ]

a

a

b

ka

ka

kb

d AA A A A k Adt

d AA A A A k A Adt

d AA P k Adt

+ ⎯⎯→ + =

′+ ⎯⎯→ + = −

⎯⎯→ = −

If the last step is rate-limiting the overall reaction will have first order kinetics

Page 27: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Lindemann-Hinshelwood mechanism

*2 * *

2*

2*

[ ] [ ] [ ][ ] [ ]

[ ][ ][ ]

[ ][ ][ ]

a a b

a

b a

b ab

b a

d A k A k A A k Adt

k AAk k A

k k AdP k Adt k k A

′= − −

=′+

= =′+

** 2

** *

** *

[ ] [ ]

[ ] [ ][ ]

[ ] [ ]

a

b

ka

ka

kb

d AA A A A k Adt

d AA A A A k A Adt

d AA P k Adt

+ ⎯⎯→ + =

′+ ⎯⎯→ + = −

⎯⎯→ = −

If the rate of deactivation is much higher that unimolecular decay than:

2[ ] [ ][ ]

b a b a

b a a

k k A k k AdPdt k k A k

= ≈′ ′+

The Lindemann-Hinshelwood mechanism can be tested by reducing the pressure (slowing down the activation step) so the reaction will switch to the second order.

Page 28: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

The activation energy of the composite reaction

Let’s consider Lindemann-Hinshelwood mechanism and apply Arrhenius-like temperature dependence to each rate constant

( )( )( )( ) / ( ) /

{ ( ) ( ) ( )}/( ) /

a a

a a a

a

E a RT E b RTa b E a E a E a RTa b a b

E a RTa aa

A e A ek k A Ak ek AA e

− −′− + −

′−= = =

′ ′′

( ) ( ) ( )

( ) ( ) ( )

a a a

a a a

E a E b E a

E a E b E a

′+ >

′+ <

Overall activation energy can be positive or negative

Page 29: Lecture 6: Diffusion and Reaction kineticshomes.nano.aau.dk/lg/PhysChem2010_files/Physical...Lecture 6: Diffusion and Reaction kinetics 12-10-2010 • Lecture plan: – diffusion •

Problems• P20.35. The diffusion coefficient of particular RNA molecule is 1.0x10-11

m2/s. Estimate time required for a molecule to diffuse 1 um from nucleus tothe cell wall

• E21.11a The second-order rate constant for the reaction

CH3COOC2H5(aq) + OH-(aq) → CH3CO2-(aq) + CH3CH2OH(aq)

is 0.11 dm3 mol-1 s-1. What is the concentration of ester after (a) 10 s, (b)10 min when ethyl acetate is added to sodium hydroxide so that the initialconcentrations are [NaOH] = 0.050 mol dm-3 and [CH3COOC2H5] = 0.100mol dm-3?

• E21.17a The effective rate constant for a gaseous reaction that has aLindemann–Hinshelwood mechanism is 2.50 × 10-4 s-1 at 1.30 kPa and2.10 × 10-5 s-1 at 12 Pa. Calculate the rate constant for the activation stepin the mechanism.