Lecture as 1c

8/12/2019 Lecture as 1c

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CHEM 2410 (2010)

A/Prof. Erica Wanlessroom 321, ph. 4921 8846, [email protected]


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Chemical Kinetics

The study ofreaction rates

Rate of reactant consumption& product formation

Response to changing

conditions (T, P)

Understanding reaction

mechanisms

Prediction of equilibrium

Catalysis ($$)

Useful in experimental design,

optimisation & understanding

Why important?


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Outline

Experimental methods

CHEM 1020 revision: rate laws, reactionorder, t1/2, Arrhenius equation

Reaction rate theories Steady state approximation

Chain reactions: Explosions & Polymerisation

Catalysis: heterogeneous, homogeneous &enzymatic


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Reference Materials Skills

Atkins & De Paula Elements of

Physical Chemistry 3-5th

Edition

Atkins & De Paula Physical

Chemistry 6-9th

edition J.L. Latham & A.E. Burgess,

Elementary Reaction Kinetics,

3rd edition, Butterworths

P. Monk Maths for Chemistry

Commonsense

Straight line graphs

Some maths

Do lots of example

questions - look forcommonalities

Use the tutorial times

wisely & bring laptop

or graph paper


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Empirical Kinetics Establish stoichiometry

Identify any side reactions Basic data

[reactant] & [product]

T Method

monitor concentration

vs time

depends on rate (fast,

medium, slow)

E.g. Stopped Flow Technique

Uses spectrophotometry

Relies on Beer-Lambert law

Absorbance concentrationA = l c

where

= molar absorption coefficientl = path length, c = [M]

2 reagents driven quickly into

mixing chamber, then concentration

measured vs time (1 ms resolution)


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Monitoring Reaction Progress

Only requirement is that a

physical property changesas the reaction proceeds

Sometimes a basic physical

measurement suffices

pressure - if # molecules

changes

conductivity - if # ions

changes Often a chemically specific

method is required

Methods to monitor concentration

Spectrophotometry expt B1 piperidine

pH

expt B2 - nitroethane polarimetry

expt B3 - sucrose

GC, MS, NMR

etc.


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Reaction Rate

Change in concentration of a reactant or product

per unit time:

Generally dependent on composition & T

Rate =concentration(t2 ) concentration(t1)

t2 t1

Rate = (concentration)(time)

= dcdt


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Instantaneous Rate

Reaction rates vary with time (generally get slower)

Therefore can only consider the rate at a given instant

Instantaneous rate = slope of concentration vs time graph ie. tangent at time, t

units concentration/time eg. (mol.L-1)/s = mol.L-1.s-1

negative for reactants, positive for products


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Reaction Stoichiometry

Note! the rates of reactant consumption and product formation vary

for the different species in a reaction.

Example: the oxidation of ethene

C2H4 + 3O2 2CO2 + 2H2O

or generally for substance J with stoichiometric coefficient vJ, the

overall reaction rate v is given by,

reactants vJ0and the overall rate, v >0

So, always name the species that the rate measurement refers to!

1

2

d[CO2 ]

dt

=1

2

d[H2O]

dt

= d[C2H4 ]

dt

= 1

3

d[O2 ]

dt

v = 1

vj

d[J]dt


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Concentration Dependence

Mg(s) + 2H+(aq) H2(g) + Mg2+(aq)

How could we monitor

this reaction?


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Rate Laws & Rate ConstantsEmpirical observation: rate is often proportional to molar concentration of

the reactants raised to a power

eg. v = k.[A]n (Rate Law)

where n is called the reaction orderand, k is the rate constant

independent of concentration, but dependent on T

units of k must fit the rate law equation (they vary!)

eg. For n=2v (M/s) = k.[A (M)]2

1111

22

1 ==== smolLsMsMM

s

M

ofunitsso

A

vk ...,

][


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Reaction Order

The power to which the

concentration is raised in the

rate law

eg. v = k[A]2[B]1

the reaction issecondorder in A &

firstorder in B

Overall order = sum of

individual species orders

eg. 2+1 = 3 for the rate law above

Order doesnt have to be integral,

v = k[A]1/2[B]

More complex rate laws have no

defined order,eg. H2(g) + Br2(g) 2HBr(g)

which can only be simplified if

[Br2] >> k[HBr]

then, v = k[H2

][Br2

]1/2

v =k[H2 ][Br2 ]

3 / 2

[Br2 ] + k'[ HBr]


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Rate Laws

Formal expression of the rate as a function of the

concentration of the species involved

Determined experimentally

in general cant be inferred from the chemical equation

Applications

if kis also known, the rate can be calculated for given

initial concentrations

predict concentrations at a later time

any proposed mechanism must agree


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Determining the Rate Law: A+B P

Isolation Method: all except one reactant in large excess (constant concentration)

eg. v = k[A][B]2

but B in excess, [B] [B]0 at all timesso, v = k[A][B]0

2 = k[A]1 where k = k[B]02

This would appear experimentally to be a first order reaction:

v [A]1

but is technically known asPseudo-first orderbehaviour as it is

not the whole story..

The dependence of the rate on each reactant can be found in asimilar way and then combined into the full rate law

e.g. B1 Piperidine experiment


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Determining the Rate Law AP

Method of Initial Rates: the initial rate ismeasured for a series of reactantconcentrations

eg. v = k[A]n

then, v0 = k[A]0n , or

logv0 = log k + n log [A]0(linear graph form: y = intercept +slope.x)

logv0 log k = n log [A]0

So, gradient of graph = nUsing...

log x.y = log x + log y

log xb = b log x


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e.g. Initial Rates

[A]0 v0(/10-3 mol.L-1) (/10-7 mol.L-1.s-1)

5.0 3.6

8.2 9.6

17 4130 130

______________

n = slope = 2 (rxn order) so v = k[A]2

log10k = intercept so,

k=

y = 2.0005x - 0.8444

R2= 1

0.5

1

1.5

2

2.5

0.5 1 1.5log [A]0

logv0


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e.g. Initial Rates (2)

[CH3CHO]0 Initial rate of

CH3CHO consumption

(mol/L) (mol.L-1.s-1 )

0.10 0.020

0.20 0.081

0.30 0.182

0.40 0.318

Derive rate law and k for

CH3CHO(g) CH4(g) + CO(g)

Here the rate goes up by

____ when initialconcentration doubles, thus

n = ____

Rate of rxn =

Now determine k


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Types of Rate LawsDifferential Rate Law: how rate depends on concentration

Derived from observations What weve been working with so far.

Disadvantage

rate is not an observable, it is a derived parameter

Integrated Rate Law: how concentration depends on time

Advantage

concentration & time are both experimental observables Principal uses

predicting concentration at time, t

determining k & reaction order


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Zero Order Reaction A P

[A]t = -kt + [A]0

[P]t = kt

Time

[Re

actant]

integrated rate law

Time

Rea

ctionrate

differential rate law

kdt

Pd

kAkdt

Ad

==

===

][

][][

v

v 0

k units = rate units

e.g. M.s-1 = mol.L-1s-1

Rate is independent of time until reactant is completely consumed

Product concentration will increase linearly with time until reactant runs out

Using...k.dt= kt+c


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First Order Reaction A

PConsumption of A,

Integration,

integrated rate law

v =

d[A]

dt= k[A]

d[A]

[A]= kdt

1

[A]d[A]

[A]0

[A] = kdt0

t

ln[A]{ }[A]0[A]t = ln[A]t ln[A]0 = kt

[A]t = [A]0 ekt

differential rate law

Rate dependent

on reactant

concentration

k units = time-1, e.g. s-1Using...

(1/x).dx = ln x


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Exponential DecayProduct concentration will

increase exponentially with time

[A]t= [A]

0e-kt

or,

ln [A]t = ln [A]0 kty = intercept + slope.x (linear form)

Use this form to determine k, graphically:


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ExampleDecomposition of AzomethaneCH3N2CH3 CH3CH3 + N2

k = - slope of ln[A] vs t

Gas phase rxn so [A] ~ p

Rewrite ln p = ln p0 kt as,

ln p - ln p0 = kt

ln (p/p0) = kt k =

Using...

log x/y = log x - log y


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Second Order Behaviour

Although the initial decay of a second order reaction may be

rapid, later the concentration approaches zero more slowly

than in a first order reaction with the same initial rate


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Defining feature:

1/[A] vs t

is linear with slope = + kunits = concentration-1.time-1, e.g. M-1.s-1 = L.mol-1.s-1

Second type of second order reaction: A+B P

(differential)

yields,

(integrated)

v =d[A]

dt= k[A][B]

kt =1

[B]0 [A]0

ln[B]/[B]0

[A]/[A]0


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eg. C2H5I + C6H5N(CH3)2 I-

Method: titratethe liberated I-

with AgNO3

Confirmedsecondorder reactionin [iodoethane]

Slope = k=

y = 0.001x + 0.6252

R2= 1

0.5

1

1.5

2

2.5

3

0 500 1000 1500 2000

t (s)

1

/[C2H5I](m3

/mol)


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Summary of Rate Laws

Differential Rate Law:rate & concentration

Zero Order

v = d[A]/dt = -k First Order

v = d[A]/dt = -k[A]

Second Order

v = d[A]/dt = -k[A]2

Integrated Rate Law:concentration & time

Zero Order

[A]t = -kt + [A]0 First Order

[A]t = [A]0 e-kt

ln [A]t= ln [A]

0- kt

Second Order1

[A]t=

1

[A]0+ kt


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Half Life t1/2

Time taken forthe [reactant]

to fall to half

of the initialvalue

[A]0 1/2[A]0


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First Order Half Life

First Order: ln [A]t = ln [A]0 - kt

for [A]t = 1/2[A]0

independent of

initial concentration

ln

[A]0

2 = ln[A]0 kt1/2

ln[A]02[A]0

= ln1

2= ln2 = kt1/2

t1/2 =ln2

kUsing...

log x/y = log x - log y


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Example: Decomposition of N2O52N2O5(g) 4NO2(g) + O2(g)

v = k[N2O5]

k = 6.7610-5 s-1


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eg. Kinetics of Bacterial Growth

Log phase dn/dt = kn or n=n0ekt

Generation doubling time half life (ln2/k)


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kt1/2 = 112

[A]0

1[A]0

= 2[A]0

1[A]0

= 1[A]0

t1/2 = 1k[A]0

Second Order Half Life

Second Order A P:

for [A]t = 1/2[A]0

dependent on initial concentration

1

[A]t

1

[A]0= kt


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Approaching Equilibrium

For a first order reaction A B,the rate of the reverse reaction

must be considered when closeto equilibrium

A B v = kf[A]B A v = kb[B]

If [B]0 = 0, then [A] + [B] = [A]0

d[A]

dt= kf[A] + kb[B]

d[A]

dt= k

f[A] + k

b([A]

0 [A])

here kf= 2kb


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Example: Photochromic lenses Fully reversible reaction with well

defined Keq

Thin layer of borosilicate glasscontaining silver halide crystals

Silver is reduced by incident UVradiation

Ag+ + e- + h Ag(s) kf

The silver (colour) is a very long

lived excited state because it is asolid state (viscous) reaction

When UV energy is removed (dark)the reverse reaction (kb) is favoured


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Temperature DependenceEmpirical observation

reaction rate increases with T

Arrhenius noted the following dependency

ln k vs 1/Tlinear with a characteristic slope

The Arrhenius Equation

pre-exponential factor, A units same as k

activation energy, Ea units e.g. kJ/mol

ln k= ln A Ea

RTk= AeEa / RT


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Need to account for vast range of observed

rates and also the Arrhenius system parameters

A & Ea must contain the constraining parametersin any theory

Basic molecular theories

Collision Theory

Activated Complex Theory

RTEaAek/=

Reaction Rate Theories


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The Collision Theory Applies to simple gas phase reactions

Key Idea molecules must collide in order to react

However, only a small fraction of collisions

produces a reaction. Why?


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Factors Affecting Rate O3+NO

O2 + NO2

1.Energy of collision

2. Orientation of collision

3.Successful collision

Th Th A B AB


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The Theory: A + B AB

Collision frequency [A][B]

Fraction of molecules,f

with kinetic E > Ea is

given by the Maxwell-

Boltzmann equation,

f =0 when T=0

f =1 when T is infinite

RTEae/=f

Kinetic E = .mass.speed2

A minimum speed must be overcome


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Now rate = # of activated collisions per unit time

v [A][B]f = [A][B]e-Ea/RT

k e-Ea/RT, the Arrhenius form

The pre-exponential factor, A must involve the

collision frequency which can be calculated from the

kinetic theory of gases

where mA & mB are the molar masses of A & B, and

is the collision cross section.

BA

BA

mm

mmkTA

)( +=

8


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Collision cross-section, AB for thetarget area that molecule B presents

to a colliding molecule of A,AB = d2where d = 0.5(dA + dB)

The collision between A & B occurswhen centre of mass of A is withinAB of centre of mass of B

Modification to account for probabilityof correct orientation (steric fudge!)

AB* = P ABP is usually


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Comparison with ExperimentReaction A

Lmol-1

s-1

Expt.

A

Lmol-1

s-1

Theor.

Ea

kJmol-1

P

2NOCl

2NO + 2Cl9.4x109 5.9x1010 102 0.16

2ClOO2+ Cl2

6.3x107 2.5x1010 0 2.5x10-3

H2 + C2H4

C2H6

1.24x106 7.3x1011 180 1.7x10-6

K + Br2

KBr + Br1.0x1012 2.1x1011 0 4.8

(Aexpt = Atheor. x P)


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Activated Complex Theory

Applies to gas &

solution phase reactions

More sophisticated

steric factor appears

automatically

Key Idea

reactions proceed via an

activated complex or

transition state near the

potential energy

maximum


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The Activated Complex, C

Or the transition state

Definite composition & loose structureNota reaction intermediate

Short-lived (10-12 s)

Think equilibrium

A + B C P

Some molecules entering the transition state may revertto reactants

e.g. F- + CH3Cl FCH3Cl FCH3 + Cl-


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NO + Cl2 NOCl2 NOCl + Cl

Ea(forward) = 85 kJ/mol

E a (reverse) = 2 kJ/mol

NO + Cl2

Reactants

NOCl2

NOCl + Cl

Products

H = 83 kJ/mol

Progress of reaction

Energypermol

++

Endothermic example


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C

EquilibriumA + B C P

Define K,

Rate of formation of P:

where m relates to C

bond energies

K =[C ]

[A][B]

d[P]

dt

= m[C ]

d[P]

dt= k[A][B]

Overall reaction rate is

given by,

Therefore,k=mK

But how to get K?

Th d i A h


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Thermodynamic Approach

Recall, G = H - TS = -RT lnKeq

thus, G = H - TS = -RT ln K

so,ln K =

TS H

RT

K = exp S

R

exp

H

RT

k= mexp S

R

exp H

RT

AeEa /RT

Using...

ex+y=ex.ey

A highly ordered C corresponds to S

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