chem1020 kinetics lecture notes

7/27/2019 CHEM1020 Kinetics lecture notes

1/41

2013 CHEM1020 Chemistry for Science & Engineering Module 3.156

Lecture 8 (Module 3)Chemical Kinetics: An introduction

Blackman, et al., Chapter 15

Nature.com

Dr Gwen [email protected]


How fast a reaction occurs (rate) & factorswhich affect speed

Information about the feasibility of a chemicalreaction

Information about the activation energyrequired for a spontaneous reaction

Information about the series of elementarysteps in a reaction (reaction mechanisms)

Why Chemical Kinetics?


Spontaneity is the inherent tendency of the reaction to

occur, notthe rate at which the reaction will occur.

Spontaneity

Spontaneous does not mean fast!

The reaction of H2and O2gases is spontaneous:

2H2(g) + O2(g) ! 2H2O(l)

"rG!= -237.2 kJ;Kc= 9 x 10

80at 298 K

This reaction is veryslow at room temperature unlessenergy is put into the system (lecture 11).


2/41


Rate of Reaction

Consider any simple reaction:A + B!C

The rate is determined by the rate of disappearance of thereactants, or the rate of appearance of the products.

The concentration of the product, C, will increase with timeuntil it reaches its position of equilibrium (at which point therate will equal zero)

How long it takes a system to reach equilibrium#

many systems dont ever reach equilibrium eg biologicalprocesses.

Rate is expressed as a positive number to indicatemovement in forward direction:

Rate = - d[A]/dt


Reaction Rate

The rate of a reactionis determined by

measuring the changein the concentration ofa reactant or product

with time.

Units = concentration per unit time.For solutions: mol L-1s-1or M s-1.

Rate =!concentration

!time


Factors that affect the rate of reaction

Chemical nature of the reactants Ability of the reactants to come in contact

(collide) with each other (physical state ofreactants, surface area)

Concentrations of the reactants Temperature Rate-accelerating agents: catalysts


3/41



Na(s) + H2O(l)r

p 634Potassium is more reactive than sodium with water

Chemical nature of the reactants

2Na + 2H2O !2Na++ 2OH-+ H2

2K + 2H2O !2K++ 2OH-+ H2

Na(s) + H2O(l)r

Both areGroup 1

alkali metals



Surface area of contactbetween the phasesdetermines the reaction rate

p 634 Ability of the reactants to come in contact. Homogeneous vs heterogeneous reactions Surface area for heterogeneous reactants



Temperature#

5H2O2(aq) + 2KMnO4(aq) + 3H2SO4(aq) !5O2(g) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)

*


4/41



Concentration #

*

Mg(s) + 2HCl(aq) !MgCl2(aq) + H2(g)


http://www.youtube.com/watch?v=1oFXC7coirM

Case Study: The reaction of butyl chloride in water

Reaction rates are experimentally determined throughmacroscopic observations. A classic kinetics reaction is:

C4H9Cl(l) + H2O(l)!C4H9OH(aq) + HCl(aq)

Start with a 0.1000 M solution of butyl chloride in waterand measure [C4H9Cl] at 50 s intervals.


0 200 400 600 800

0.02

0.04

0.06

0.08

0.10

Time / s

Plot [C4H9Cl] as a function of time:

The reaction of butyl chloride in water

Time/s [C4H9Cl]/M

0 0.1000

50 0.0905

10 0.0820

150 0.0741

[C4

H9

Cl]/M

Reaction Rate =change in concentration

change in time


5/41


This Rate describes the rate of disappearanceof C4H9Cl.


NB: The rate of a reaction is alwaysdefined to be positive.

C4H9Cl(l) + H2O(l)!C4H9OH(aq) + HCl(aq)

Reaction Rate = !C

4H

9Cl[ ]

final! C

4H

9Cl[ ]

initial

tfinal ! tinitial

=!" C

4H

9Cl[ ]

"t

Describes the slopeof the previous graph


Example Calculation

A. 1.10 $10-8M s-1

B. 1.90 $10-4M s-1

C. 5.30 $10-2M s-1

D. 2.88 $10-4M s-1

E. 4.12 $104M s-1

Time/s [C4H9Cl]/M

0 0.1000

50 0.0905

10 0.0820

150 0.0741

What is the rate of the reactionbetween 0 and 50 seconds?

Reaction Rate = ! C

4H

9Cl[ ]

final! C

4H

9Cl[ ]

initial

tfinal!

tinitial


Example Calculation *


6/41


The slopeof the tangent to the graph at any point gives is theinstantaneous rate at that time.

0 200 400 600 800

0.02

0.04

0.06

0.08

0.10

Time / s

[C4

H9

Cl]/M

Ave Rate = 1.90$10-4M s-1Ave Rate = 1.70$10-4M s-1

Ave Rate = 1.68$10-4M s-1

Note: The rate is changing(decreasing) with theextent of the reaction.



As we make the interval smaller, the average rate approachesinstantaneous rate. For an infinitely small interval:

The slope of the tangent(derivative of the concentration w.r.t. time)


Reaction Rate =!" C

4H

9Cl[ ]

"t=!

d C4

H9

Cl[ ]dt

Can also define the rate in terms of appearanceof one of theproducts. From the stoichiometry of the reaction:

Reaction Rate = !d C

4H

9Cl[ ]

dt= +

d C4

H9

OH[ ]dt

Reactant Product


Reaction Rate

[A]

[B]

Consider the case:A!2BHere [B] increases by 2 times the rate that [A] disappears.

d[A] = 0.017 mol L -1

d[B] = 0.025 mol L -1

dt= 54.0 s

dt= 74.4 s


7/41


For the general reaction:

aA + bB ! cC + dD

The average rate of appearance of disappearancecan be represented as a normalised rate:

Reaction Rate = !1

a

d A[ ]dt

=!1

b

d B[ ]dt

= +1

c

d C[ ]dt

= +1

d

d D[ ]dt

For the case:A!2B

Reaction Rate = !d A[ ]dt

= +1

2

d B[ ]dt

Note: B is aproduct in this

reaction

p 632


Summary

1. Spontaneous does not mean fast.2. The rate of a reaction is determined by measuring the

change in the concentration of a reactant or product as afunction of time.

3. The rate of a reaction is always positive.4.

The rate of reaction is the rate any species disappears orappears divided by its stoichiometry.


Rate Laws (Differential) Order of Reactions Method of Initial Rates

Lecture 9 (Module 3)Rate Laws & Order of Reaction


8/41


Rate Laws

To predict how the rate of a reaction will change as afunction of time, we need to experimentallydetermine themathematical relationship between the rate of the reactionand the concentrations of reactants.

This mathematical relationship is called a rate law.

Rate laws are normally a simple functions of theconcentrations of the reactants.


For the general reaction:

aA + bB ! cC + dD

The rate law is:Rate = k[A]n[B]m

n is the order in terms of reactant [A] and m is the order interms of reactant [B].

The sum n + m is the overall order of the reaction. The order of a reaction is determined by the mechanism

of the reaction not the stoichiometry.

The Rate Law


p 636

Consider the reaction: 2HI(g) !H2 (g) + I2(g)

The Rate Law

Rate of reaction = k[HI]n

rate constant/coefficient

order


9/41


Write the rate equations for the following

reactions:

*


Rate Laws

Consider the reaction: 2NO2(g) !2NO(g) + O2(g)

Initially only NO2 is present and the rate will be proportional tothe [NO2] raised to some power.

15.3

Rate = k[NO2]n

k= rateconstant

n= the orderof the reactionkand n are determined experimentally

Reaction rates are generally proportional to the product of theconcentrations of each of the reactants raised to some power.


Rate ofproductionof NO

Rate ofdecompositionof NO2

2 xRate ofproductionof O2

= =

Reaction Rates and Stoichiometry

(This is a key reaction in air pollution.)

Decomposition of NO2

2NO2(g) !2NO(g) + O2(g)

Reaction Rate = !1

2

d NO2[ ]

dt=

1

2

d NO[ ]dt

=

d O2[ ]

dt


10/41


12_1575

Time Time

2NO2(g) !2NO(g) + O2(g)

The decomposition of NO2. *


_

0.0003

70s

O2

0.0025

0.005

0.0075

0.0100

0.0006

70s

0.0026

110 s

NO2

NO

50 100 150 200 250 300 350 400

Concentrations(mol/L)

Time (s)

![NO2]

! t

The decomposition of NO2.

To define the rate ofthe reaction wemust consider thestoichiometry of thereactants andproducts.

2NO2(g) !2NO(g) + O2(g)


Rate ofproductionof NO

Rate ofdecompositionof NO2

2NO2(g) !2NO(g) + O2(g)

2 x Rate ofproductionof O2

= =

Reaction Rates and Stoichiometry

Consider the decomposition of NO2

Rate of reactant(NO2)

consumption

Rate of product(NO)

appearance

Rate of product(O2)

appearance

Reaction Rate =!1

2

d NO2[ ]

dt=

1

2

d NO[ ]dt

=

d O2[ ]

dt


11/41


To determine the rate law we need to examine how the ratedepends on the concentration of a reactant.

However, as a reaction proceeds, the reverse reaction mayalso occur. For example in the above reaction, recombinationof NO and O2can also occur

The order of a reactant must be determined under conditionswhere the reverse reaction is unimportant (i.e. when only thereactants are present).

Determining the Rate Law

2NO2(g) !2NO(g) + O2(g)

2NO(g) + O2(g) !2NO2(g)


The Method of Initial Rates

Experimentally, the rate law is determined using the method ofinitial rates.

The initial rate of reaction is the rate just after the reaction hasstarted (just after t = 0). At this point:

a) the concentration of the reactants have changed significantly,b) the reverse reaction is important. If a reaction is first order in particular reactant, then

doubling its concentration will double the initial rate.

If it is second order in a particular reactant, the rate willincrease by factor of 22= 4

If it is zero order in a particular reactant, rate will beindependent of concentration


Example

The data below were collected for the reaction:

A + B!C

!"#$%&'$() +,- .'/0 123

+4- .

'/0 1235(&670 %7)$ /8 9&:7##;

/8 ,.'/0 123:23

! "#!"" "#!"" $#" & !"'(

) "#!"" "#)"" $#" & !"'(

* "#)"" "#!"" !+#" & !"'(

Determine:

(a) the rate law, and(b) the magnitude of the rate constant.


12/41


Solution

General form of rate law:

(a) By observation, doubling [A] !rate increases by 4,hence n = 2:

(b) changing [B] has no effect, hence m = 0.

Rate = !d A[ ]

dt

=k[A]n[B]

m

Rate == k[A]2[B]

0=k[A]

2

So reaction is second order in A, zero order in B andsecond order overall.


(b) Select any experiment and substituteconcentrations to calculate k

Note, since

then

Kinetics _2.16

113

2

15

2

2

sM104.0

M)(0.100

sM104.0

[A]

rate,[A]rate

!!!

!!

"=

"==#= kk

1

LmolM

!

=

M!1

s!1

= mol L!1( )

!1

s!1

= mol!1

L s!1

Example


Units of rate constant,k

Rate =k

[A]

n

The rate has units of mol L-1s-1

The rate constant, k, will then have theunits: mol L-1s-1 x (mol-1L)n

For example: if n= 0 khas units mol L-1s-1

if n= 1 khas units s-1

if n= 2 khas units mol-1L s-1

mol L-1s-1 = (units of k)$(mol L-1)n

(units of k)= mol L-1s-1$(mol L-1)-n

= mol L-1s-1$(mol-1L)n


13/41


!"#$%&'$() +52- .'/0 123

+?=2- .

'/0 1235(&670 %7)$.'/0 123:23

! "#!"" "#"(" "#)

) "#!"" "#")( "#"(

* "#)"" "#"(" *#" & !"'$

Practice Problem

Iodide reacts with thiosulphate to form elemental iodine at

298 K. If the reaction solution contains a tiny amount ofstarch solution, then this I2is oberved as a blue complex.

I-+ S2O82-!products

Determine:

(a) the order of reaction, and (b) the rate constant, k.

*


Answer *


1. n and m are normallyintegers such as 0, 1 or 2 but canbefractions.

2. If the order of a reactant = 0 (zeroth order) the rate of thereaction is independent of the concentration of that reactant.

3. The orders, nand m, and the rate constant, k, depend on themechanism and are determined experimentally.

Rate = k[A]n[B]mnis the order of reactant A

mis the order of reactant B.

Summary


14/41


Summary

1. nand mcannot be inferred from the reaction equation.They depend on the mechanism (and are determinedexperimentally).

2. Reaction rates are generally proportional to the product ofthe concentrations of each of the reactants raised to somepower (the order of a reaction).

3. The order of a reaction depends on the mechanism andmust be determined experimentally.


Zero, 1stand 2ndorder reactions Integrated rate laws Half-lives

Lecture 10 (Module 3)Zero, 1st& 2ndOrder Reactions


A rate law that expresses how a rate depends on concentrationis called a Differential Rate Law, or just the rate law,e.g. rate = k[A]2

A rate law that expresses how the concentration depends ontimeis called an Integrated Rate Law.

Given one rate law, you can determine the other mathematically(i.e. by integration or differentiation).

Types of Rate Laws


15/41


C !products

Differential Rate Law

d[C] = 0.10 mol L-1

d[C] =0.16 mol L-1

dt = 0.38 s

dt = 0.48 s


Consider: A!products

If each molecule of A reacts independently the rate of thereaction will depend linearly on the concentration of [A].That is the reaction will be first order in A:

i.e.

In order to determine the integrated rate law we simplyintegrateboth sides of the differential rate law.

(a differential equation)[A][A]

rate kdt

d=!=

dtkd

!=

[A]

[A]

Integrated Rate Laws: 1st order reaction


or

We now have an expression for how [A] varies with time.

If we integrate this expression from t= 0 (when [A] = [A]0) totime t(when conc. of A is [A]t)

Maths hint

ln[A]t ln[A]0= - (kt- kt0) where t0=0

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

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

1

[A]d[A]

[A]=[A]0

[A]=[A]t! =" k dt

t=0

t=t

!!

!=

=

xkdxk

xdxx ln/1


16/41


First Order Reaction

The integrated rate law is:

A plot of ln[A]versus tgives a straight line, with a slope of -k

y = mx + c

ln[A] = - k t+ ln[A]0

ln[A]0

Time, t

ln[A]

slope= -k

p 645


First Order Reaction

A first-order reaction results in theexponential decay of the reactant.

The larger the rate constant,the more rapid the decay:

klarge= 3ksmall

ln[A]t= - k t + ln[A]0

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


Practice Problem *


17/41


Integrated Rate Laws: Second Order Reaction

Consider: 2A!products

For second order reaction with only one reactant A,

If we again rearrange and integrate both sides:

or

Maths hint

rate =! d[A]

dt=k[A]

2

tkdtkd

t

=!!=

0

2 [A]

1

[A]

1so,

[A]

[A]

0[A]

1

[A]

1+= tk

t

1/ x2dx = !1/ x"


Second Order Reaction

A plot of 1/[A]versus tgives a straight line of slope, k.

1/[A]0

t

1/[A]

slope = k

1

[A]t

=k t +1

[A]0


Rate is independent of the reactant concentrations

Integrated Rate Laws: Zero Order Reaction

(a differential equation)

(rearrange)

In order to determine the integrated rate law wesimply integrateboth sides of the differential rate law.

[ ]kk

dt

dRate ==

!

=0]A[

A

!dA[ ]=kdt

Consider: A!products


18/41


A plot of [A]against tis linear with a slope = -k

Integrated rate law is:

y = mx + c

Integrated Rate Laws: Zero Order Reaction

If we integrate this expression from t= 0 (when [A]= [A]0) to time t(when conc. of A is [A]t)

Maths hint

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

!d A[ ]A[ ]= A[ ]

0

A[ ]= A[ ]t

" = kdtt=0

t= t

" k dx = k x!

! A[ ]A[ ] = A[ ]

0

A[ ] = A[ ]t

= ktt= 0

t= t

A[ ]t! A[ ]

0= 0 ! kt

A[ ]t

= !kt+ A[ ]0

aside


Zero Order Reaction

A plot of [A] versus tgives a straight line.

,-."

!

,-.

/0123 4 '#

0

[A] = -kt+ [A]0


Order

Zero First Second

Rate law Rate = k Rate = k[A] Rate = k[A]2

Integrated rate law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 [A]1

= kt +[A]01

Plot needed to give a straight line [A] versus t ln[A] versus t[A]1

versus tRelationship of rate constant

Slope = -k Slope = -k Slope = kto the slope of straight line

Half-life2k

t1/2 =[A]0

k[A]0t1/2 =

1t1/2

=

ln2

k=

0.693

k

A summary comparison of rate laws and integratedrate laws


19/41


Compare the orders of reaction (mathematically)

1storder reaction

Zero order reaction

2ndorder reaction

Imagessource

d:

Purdue

Un

ivers

ity


Example: The following data were obtained for thegas-phase decomposition of nitrogen dioxide at 300 C

Using the integrated rate equation to determine theorder of a reaction

NO2(g) !NO(g) + 1/2O2(g)

@&'$ . : +A>=- . B

" "#"!""

(" "#""56

!"" "#""+(

)"" "#""$7

*"" "#""*7

Is the reaction first order orsecond order in NO2?

If first order, a plot of ln[NO2] vstime will be linear,if second order, a plot of 1/[NO2]vs time will be linear.


Decomposition of nitrogen dioxide *


20/41


1. A rate law that expresses how a rate depends onconcentration is called a Differential Rate Law, or justthe rate law, e.g. rate = k[A]2

2. A rate law that expresses how the concentration dependson time is called an Integrated Rate Law.

Summary


Half lives

Radioactive decay

Lecture 11 (Module 3)Radioactive Decay and Half Lives


Defined as the time it takes for the

concentration of a reactant, A, todrop to half of its original value,

Since

i.e. or

Half life of a first order reaction

Half-life of a reaction

021 [A][A]

21=

t

21

021

00

[A]

[A]ln,

[A]

[A]ln tktk

t

==

2

12ln tk=

t12

=

ln2

k=

0.693

k

0t ]ln[ln[A] Atk +!=


21/41


Each subsequent half life is the same as the first.

_

[N2O5]0 0.1000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0.0700

0.0800

0.0900

[N2O5]02

[N2O5]04

[N2O5]08

50 150 250 350100 200 300 400

t1/2 t1/2 t1/2

Time (s)

[N2

O5

](mol/L)

2N2O5(aq) 4NO2(aq) + O2(g)

First order

Rate = ! 1

2

d[N2O5 ]

dt=k[N2O5 ]


Half-lives of radioactive isotopes

Da

tasource

d:

Phys

ica

lChem

istry

by

Pau

lMon

k

Isotope Half-Life Source of Radioactive Isotope

12B 0.02 s Unnatural (manmade)

14C 5570 years Natural

40K 1.3 x 109years Natural: 0.011% of all potassium

60Co 10.5 min Unnatural: made for medicinal use

129I 1.6 x 107years Unnatural: fallout from nuclearweapons

238U 4.5 x 108years Natural: 99.27% of all uranium

239Pu 2.4 x 104years Unnatural: by-product of nuclearenergy

A chemical half-life !time required for half the material to have beenconsumed chemically.

A radioactive half-life !time required for half of a radioactive substance todisappear by nuclear disintegration.


Note that for a first order reaction, t1/2 does not depend on initialconcentration. (e.g. radioactive decay).

The half-life of 60Co is 10.5 min. If we start with100 g of 60Co, how much remains after 42 min?

1. Determine how many of the half-lives have occurred duringthe time interval.

2. Successively halve the amount of 60Co, once per half-life.

42 10.5 = 4 half-lives elapse in 42 mins.

100 g 50 g 25 g 12.5 g 6.25 g

1sthalf-life

2ndhalf-life

3rdhalf-life

4thhalf-life


22/41


The rate constant for the first-order transformationof cyclopropane to propene is 5.40 x 10-2h-1.

(a)What is the half life of the reaction?(b)What fraction remains after 18.0 h?(c)What % remains after 51.2 h?

Example calculations


The rate constant for the first-order transformationof cyclopropane to propene is 5.40 x 10-2h-1.

(a)What is the half life of the reaction?

Example calculations *


The rate constant for the first-order transformation ofcyclopropane to propene is 5.40 x 10-2h-1.

(b) What fraction remains after 18.0 h?

Example calculations

From the integrated rate equation

ln x ln y = ln (x/y)

*

0ln[A]ln[A] +!= ktt


23/41


The rate constant for the first-order transformation ofcyclopropane to propene is 5.40 x 10-2h-1.

(c) What % remains after 51.2 h?

Example calculations *


Applications of half lives

Archaeological dating (radioactive decay)

14C used for ages up to 10 x t1/2(~ 50,000 yrs)

Reaction:

Radioactive decay of 14C is a first order process.

6

14 C" 714 N + #

$t1

2

= 5770 yr

Image source: news.bbc.co.uk

Radioactive 14C decay provided evidencethat the Turin Shroud is no older than1290 AD.


14CO, 14CO2

biosphere (water, air, plants)

Due to exchange with the environment the % of 14C in livingorganisms is the same as the atmosphere. After death thisexchange no longer happens.

Therefore, due to radioactive decay the % 14C decreases withtime.

14C is formed in the ionosphere due to the interaction of Nwith cosmic rays.

HCnN 111460

1147

+!+


24/41


Solution: Use the integrated first-order rate equation toevaluate how long ago the plants that formed the charcoaldied.

What is k?

Example: Per gram of carbon, a sample of charcoal containedonly 21.6% of the 14C found in the leaves of a living tree. Given

that t1/2 = 5776 years, how long ago was the charcoal formed?

We are given:

ln(0.216) = -1.196$10-4t

t = 12800 yr

ktc

ct

!=

0

ln

ct

c0

= 0.216

t1/2

=

ln2

k!

0.693

k

1-yryr

4101961

5776

6930 !

"==# ..

k

0ln[A]ln[A] +!= ktt


A small portion of tzis clothing was removed and burntcarefully in pure oxygen. The amount of 14C was found to be50.93% of the amount expected if the naturally occurring fabricprecursors had been freshly picked. How long is it since thecrop of flax was picked? ie what is its age?

Practice Problem


*


25/41


Chemical Decomposition

decays by hydrolysis inlake water at 12C C0= 5.0 x 10

-7g cm-3

a) what is the concentration after 1 year?

b) how long to drop to a safe level of 3.0 x 10 -7g cm-3?

pesticide decay productsk= 1.45 yr -1

A pesticide decays with 1storder kinetics


a) what is the concentration after 1 year?

decays by hydrolysis inlake water at 12C

C0= 5.0 x 10-7g cm-3


A pesticide decays with 1storder kinetics*


decays by hydrolysis inlake water at 12C

C0= 5.0 x 10-7g cm-3


A pesticide decays with 1storder kinetics*

b) how long to drop to a safe level of 3.0 x 10 -7g cm-3?


26/41


Half-life of a second order reaction

i.e.

Integrated rate equation is

Substituting [A] = %[A]0 when t= t%

tk

t

=!

0[A]

1

[A]

1

00 [A]

1

/2[A]

1

21 += tk

t12

=

1

k[A]0

Unlike first order case, half-life doesdepend on initialconcentration.

Hence subsequent half-lives take different times to the first.

21

21

000 [A]

1,

[A]

1

[A]

2tktk =!="


Rate is independent of the concentration of the reactants

Integrated rate law is: [A]t= -k t + [A]0

Note that the half life depends on the initial concentration.

Substitute [A] = %[A]0at t = t%

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

Half life of a Zero Order Reaction

t12

=

[A]0

2k


Decomposition of reaction

2N2O(g)!

N2(g) + O2(g)on a platinum surface.

The reaction is zero order as it is limited by thesurface area of the Pt.

Example of a zero order reaction


27/41


Practice Problem *


Summary


Elementary reactions Arrhenius Equation Rate determining step Activation Energy

Lecture 12 (Module 3)Reaction Mechanisms


28/41


A balanced chemical equation does not tell you how thereaction occurs, i.e.the reaction mechanism

The reaction mechanism consists of a series of simplereactions, known as elementary steps (or elementaryreactions) leading from the reactants to the products.

The rate law depends on the mechanism of the reaction.

The Mechanism of a Reaction


Example

The mechanism of the above reaction is thought toinvolve two elementary reactions.

NO3is an intermediate (an unstable radical). NO3 is produced and used up during the reaction. NO3does notappear in overall equation. k1and k2are the rate constants of the elementary steps.

Rate Law: Rate = k [NO2]2NO2(g) + CO(g) !NO(g) + CO2(g)

NO2(g) + NO2(g) !NO3(g) + NO(g)

NO3(g) + CO(g) !NO2(g) + CO2(g)

k1

k2


The rate law for an elementary reaction is determined by itsmolecularity(the number of molecules involved).

Rate Laws and Elementary Reactions

Elementary Step Molecularity Rate Law

A!products Unimolecular Rate = k[A]

A + A!products(2A!products)

Bimolecular Rate = k[A]2

A + B!products Bimolecular Rate = k[A][B]


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A reaction mechanism must satisfy the followingconditions:

1. The sum of the elementary steps must give theoverall balanced equation

2. The mechanism must agree with the experimentallydetermined rate law

Reaction Mechanisms


The molecules must meet.

Postulate 1:The rate of a reaction will be proportional to the number ofcollisions per unit time.

Postulate 2:

The number of collisions between two molecules will depend

on the productof their concentrations.

For the elementaryreaction A + B !C rate = k[A][B]2ndorder

Requirements for Two Molecules to React:


The molecules must have the right orientation.

NO2Cl + Cl&!NO2+ Cl2

p 652

Nitrylchloride



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The molecules must have sufficient energy.

The molecules must collidewith sufficient kinetic energyto overcome the activationenergy, Ea.



Energy of Activation

Temperature Dependence of Reaction RatesWe know from experience that the rate of most reactions increases withtemperature e.g. eggs cook faster at higher temperature.

Svante August Arrhenius(1859 1927) Swedish, NobelPrize in chemistry 1903.

Arrhenius proposed that the rate of a reactionwas given by the product of 2 terms: atemperature independent term and a term thatincreased exponentially with temperature.

He showed empiricallythat a plot of ln kversus 1/T gave a straight line.


Energy of Activation:Temperature Dependence of Reaction Rates

Arrhenius proposed the rate of a reaction was givenby:

k = A exp! Ea /RT( )

Arrhenius Equation

Pre-exponential or frequencyfactor. Probability that a

collision which could lead toreaction occurs. (temperature

independent)

Probability the collision has enoughenergy to react. Ea= activationenergy. R = universal gas constant(8.314 J mol-1K-1)

Rate constant


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As the temperature increases, the fraction of collisions withsufficient kinetic energy to react increases exponentially.(See L1 Module 1)

p 582

Kinetic energy of a gas


Energy of Collisions

Number of collisionswith E> Ea

=Number ofcollisions

Arrhenius Equation

Use to:1. Predict how the rate constant kmay change with temperature.2. Determine the apparent activation energy.

A- the frequency factor, incorporates the collision frequency and asteric factor. Units of A are the same as k.

!"#

$%&'

( RTE

a

e



RTEAk a

!=lnln

Arrhenius Equation

Using the Arrhenius equation

Take the natural log of both sides:

gives linear equation of form y= mx+ c whereT

xky 1,ln ==

y-intercept:

Slope:!

Ea

R

lnA

Ea= activation energyA = pre-exponential or frequency fact or

If kis measured at severaltemperatures we can determine Eafrom a plot of ln kversus 1/T.



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Using the Arrhenius equation Decomposition acetaldehydeCH3CHO !CH4+ CO

T (K) 700 730 760 790 810 840 910 1000

k (L mol-1

s-1

) 0.011 0.035 0.105 0.343 0.789 2.17 20.0 145

1/T (K-1

) x103 1.43 1.37 1.32 1.27 1.23 1.19 1.10 1.00

ln k (L mol-1s-1) -4.51 -3.35 -2.25 -1.07 -0.24 0.77 3.00 4.98

Raw data kversus T

Transformed dataln k versus 1/T

RT

EAk a!=lnlnStep 1. Plot ln k versus 1/T

Slope = -Ea/R = -2.27 x 104K

Intercept = ln A = 27.0

Ea= 2.27 x 104K x 8.3145 J K-1mol-1= 188 kJ mol-1

A = 1.1x1012L mol-1s-1

slope = -Ea/RStep 2. Linear regression least squares fit


Can also use values of kat just two temperatures:

Subtract first equation from second:

Using the Arrhenius equation

ART

Ek

ART

Ek

a

a

lnln

lnln

2

2

1

1

+!

=

+!

=

ART

EA

RT

Ekk aa lnlnlnln

12

12 !

++!

=!

Temperature 1

Temperature 2

2

1 2 1

1 1ln

ak E

k R T T

! "#$ = #% &

' (


Example: The rearrangement of methyl isonitrile to acetonitrile

was studied at two temperatures, and following values obtainedfor rate constant:Temperature/ C k / s-1

189.7 2.52 x 10-5

198.9 5.25 x 10-5

What is the activation energy of the reaction?

Using the Arrhenius equation *


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Transition State

Reactions go from reactants to products via a transition state,or activated complex.

(at the transition state, 50% chance offorming product)


A plot of potential energy against reaction progress is knownas a reaction profile.

Multi-step Reactions.


Rate Determining StepWhere a mechanism involves a series of elementaryreactions the overall rate of the reaction will be determined

by the rate of the slowest step.This is called the rate determining step

Example: Highway with two tollgates A and B

If gate A processes cars much slower than gate B, the overallrate will depend on rate of progress through gate A. Gate A israte limiting.

If gate B much slower than A, rate only depends on rate ofprogress through gate B. Gate B is rate limiting.


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Practice Problem *


Ea1> Ea2: Reaction 1 is rate limiting

Multi-step Reactions: Rate determining step


Multi-step Reactions: Rate determining step

Ea1< Ea2: Reaction 2 is rate limiting


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Reaction Mechanisms

From experiment, it is known that rate = k[NO2]2

From this we know the rate limiting step in themechanism does not involve the direct interactionof NO2with CO.

NO2(g) + CO(g) !NO(g) + CO2(g)


Reaction Mechanisms

From experiment, the rate is known to be rate = k[NO2]2


The above mechanism would satisfy the rate law if thefirst elementary step was rate limiting.

The mechanism has been proposed to involve two elementaryreactions

slow

fast

NO2(g) + NO2(g) !NO3(g) + NO(g)

NO3(g) + CO(g) !NO2(g) + CO2(g)k2

k1


Reaction Mechanisms

While this mechanism does satisfy rate law: rate = k[NO2]2

it does not provethat it is the true mechanism.

Does it give the right overall equation? "



NO2(g) + NO2(g) + NO3(g) + CO(g) ! NO3(g) + NO(g) + NO2(g) + CO2(g)

slow

fast

NO2(g) + NO2(g) !NO3(g) + NO(g)

NO3(g) + CO(g) !NO2(g) + CO2(g)k2

k1


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Example Q

The conversion of cyclopropane (an anaesthetic), to propenehas a rate constant k= 1.3$10-6s-1 at 400oC and k = 1.1$10-5 s-1at 430oC.

a) What is the activation energy in kJ mol -1?b) What is the value of the pre-exponential factor,A, for thisreaction?c) What is the rate constant for the reaction at 350oC?

k = A exp(-Ea/RT)

*

lnk2

k1

=

!Ea

R

1

T2

!1

T1

"

#$%

&'


Example Q

a) What is the activation energy ?

b) What is the pre-exponential factorA?

c) What is the rate constant at 350oC?

*


Summary

The reaction mechanism consists of a series of elementaryreactionsleading from the reactants to the products.

The overall rate of the reaction will be determined by therate of the slowest step.

To react molecules must meet in the right orientation andcollide with sufficient kinetic energy to overcome theactivation energy, Ea.

The Arrhenius equation can be used to estimate Eaandmodel how the rate constant kmay change with temperature.


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Catalysis Intermediates

Lecture 13 (Module 3)Part A: Catalysis


Catalysis

Catalystsare substances which speed up chemical reactionswithout themselves being consumed. Virtually every chemicalreaction in the human body, the atmosphere and chemicalindustry is affected by catalysts.

Homogeneous catalysts: exist in same phase as the

reactants.Heterogeneous catalysts: exist in a different phase toreactants (e.g. A solid catalyst for gas-phase reaction)


A catalyst:

increases the rate of a reaction is unchanged chemically at the end of the

reaction

is present in same quantity at the end ofreaction

is only required in small amounts does not effect the position of equilibrium


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Energy Profile

Reactions go from reactants toproducts via a transition state, oractivated complex.


Catalysts work by providing an alternative pathway for thereaction, where activation energy barrier is lower:

12_303

!E

Reactants

Products

Catalyzed

pathway

Uncatalyzed

pathway

Reaction progress

Energy

p 668

Catalysis


A catalyst makes a reaction go faster without being

used up in the reaction.

Example: Chlorine catalyzes the decomposition of O3

Cl(g) + O3(g) !ClO(g) + O2(g)

O(g) + ClO(g) !Cl(g) + O2(g)

O3(g) + O(g) ! 2O2(g)

catalyst

intermediate

Catalysis


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Practice Q: Combining concepts

The destruction of the ozone layer goes through a multi-stepreaction, the main one being:

2Cl + 2O3!2ClO + 2O2ClO + ClO !Cl2O2Cl2O2 !Cl + ClO2

ClO2 !Cl + O2

What is the overall reaction?

What are the intermediates?

What are the catalysts?

*

2O3!3O2

catalyst

intermediates:


Catalysis

Ea(uncatalyzed)

Effectivecollisions(uncatalyzed)

Effectivecollisions(catalyzed)

Ea(catalyzed)

(a) (b)

Numberofcollisions

with

agivenenergy

Numberofcollisions

wit

hagivenenergy

Energy Energy

A lower Eameans more molecules will react.

p 668


Decomposition of H2O2 *

H2O2_MnO2_catalyst.mov

H2O2(aq) !H2O(l) + %O2(g)


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Heterogeneous Catalysis

Usually involves a solid surface and gaseousreactants

e.g.

CH2=CH2

CH3-CH3 Catalytic converterPd and Pt particles on ceramic support,promotes conversion of NO to NO2and N2

p 669


Enzymes are Biological Catalysts

Enzymes are large biological moleculeswhich catalyze particular reactions.

For example carbonic anhydrasecatalyses the reaction:

CO2+ H2O ! HCO3-+ H+

This reaction helps remove CO2from cells.Each molecule of carbonic anhydrase catalyses over600,000 reactions per second.


Summary

Catalystsspeed up chemical reactions without themselves

being consumed.

Proteins which act as catalysts are called enzymes.


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Lecture 13 (Module 3)Part B: Exam Revision

chem1020 kinetics lecture notes

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