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7/27/2019 CHEM1020 Kinetics lecture notes
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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]
2013 CHEM1020 Chemistry for Science & Engineering Module 3.157
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?
2013 CHEM1020 Chemistry for Science & Engineering Module 3.158
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).
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.159
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.160
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.161
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.162
Factors that affect the rate of reaction
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.163
Factors that affect the rate of reaction
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.164
Factors that affect the rate of reaction
Temperature#
5H2O2(aq) + 2KMnO4(aq) + 3H2SO4(aq) !5O2(g) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)
*
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.165
Factors that affect the rate of reaction
Concentration #
*
Mg(s) + 2HCl(aq) !MgCl2(aq) + H2(g)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.166
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.167
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.168
This Rate describes the rate of disappearanceof C4H9Cl.
The reaction of butyl chloride in water
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.169
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.170
Example Calculation *
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.171
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.
The reaction of butyl chloride in water
2013 CHEM1020 Chemistry for Science & Engineering Module 3.172
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)
The reaction of butyl chloride in water
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.173
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.174
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.175
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.176
Rate Laws (Differential) Order of Reactions Method of Initial Rates
Lecture 9 (Module 3)Rate Laws & Order of Reaction
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.177
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.178
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.179
p 636
Consider the reaction: 2HI(g) !H2 (g) + I2(g)
The Rate Law
Rate of reaction = k[HI]n
rate constant/coefficient
order
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.180
Write the rate equations for the following
reactions:
*
2013 CHEM1020 Chemistry for Science & Engineering Module 3.181
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.182
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.183
12_1575
Time Time
2NO2(g) !2NO(g) + O2(g)
The decomposition of NO2. *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.184
_
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)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.185
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.186
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)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.187
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.188
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
! "#!"" "#!"" $#" & !"'(
) "#!"" "#)"" $#" & !"'(
* "#)"" "#!"" !+#" & !"'(
Determine:
(a) the rate law, and(b) the magnitude of the rate constant.
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.189
Solution
General form of rate law:
(a) By observation, doubling [A] !rate increases by 4,hence n = 2:
(b) changing [B] has no effect, hence m = 0.
Rate = !d A[ ]
dt
=k[A]n[B]
m
Rate == k[A]2[B]
0=k[A]
2
So reaction is second order in A, zero order in B andsecond order overall.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.190
(b) Select any experiment and substituteconcentrations to calculate k
Note, since
then
Kinetics _2.16
113
2
15
2
2
sM104.0
M)(0.100
sM104.0
[A]
rate,[A]rate
!!!
!!
"=
"==#= kk
1
LmolM
!
=
M!1
s!1
= mol L!1( )
!1
s!1
= mol!1
L s!1
Example
2013 CHEM1020 Chemistry for Science & Engineering Module 3.191
Units of rate constant,k
Rate =k
[A]
n
The rate has units of mol L-1s-1
The rate constant, k, will then have theunits: mol L-1s-1 x (mol-1L)n
For example: if n= 0 khas units mol L-1s-1
if n= 1 khas units s-1
if n= 2 khas units mol-1L s-1
mol L-1s-1 = (units of k)$(mol L-1)n
(units of k)= mol L-1s-1$(mol L-1)-n
= mol L-1s-1$(mol-1L)n
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.192
!"#$%&'$() +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.
*
2013 CHEM1020 Chemistry for Science & Engineering Module 3.193
Answer *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.194
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.195
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.196
Zero, 1stand 2ndorder reactions Integrated rate laws Half-lives
Lecture 10 (Module 3)Zero, 1st& 2ndOrder Reactions
2013 CHEM1020 Chemistry for Science & Engineering Module 3.197
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.198
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.199
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.200
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.201
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.202
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.203
Practice Problem *
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.204
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"
2013 CHEM1020 Chemistry for Science & Engineering Module 3.205
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.206
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.207
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.208
Zero Order Reaction
A plot of [A] versus tgives a straight line.
,-."
!
,-.
/0123 4 '#
0
[A] = -kt+ [A]0
2013 CHEM1020 Chemistry for Science & Engineering Module 3.209
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.210
Compare the orders of reaction (mathematically)
1storder reaction
Zero order reaction
2ndorder reaction
Imagessource
d:
Purdue
Un
ivers
ity
2013 CHEM1020 Chemistry for Science & Engineering Module 3.211
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.212
Decomposition of nitrogen dioxide *
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.213
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.214
Half lives
Radioactive decay
Lecture 11 (Module 3)Radioactive Decay and Half Lives
2013 CHEM1020 Chemistry for Science & Engineering Module 3.215
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 +!=
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.216
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 ]
2013 CHEM1020 Chemistry for Science & Engineering Module 3.217
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.218
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.219
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.220
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 *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.221
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.222
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 *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.223
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.224
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
+!+
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.225
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.226
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.227
*
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.228
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.229
a) what is the concentration after 1 year?
decays by hydrolysis inlake water at 12C
C0= 5.0 x 10-7g cm-3
pesticide decay productsk= 1.45 yr -1
A pesticide decays with 1storder kinetics*
2013 CHEM1020 Chemistry for Science & Engineering Module 3.230
decays by hydrolysis inlake water at 12C
C0= 5.0 x 10-7g cm-3
pesticide decay productsk= 1.45 yr -1
A pesticide decays with 1storder kinetics*
b) how long to drop to a safe level of 3.0 x 10 -7g cm-3?
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.231
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 =!="
2013 CHEM1020 Chemistry for Science & Engineering Module 3.232
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.233
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
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.234
Practice Problem *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.235
Summary
2013 CHEM1020 Chemistry for Science & Engineering Module 3.236
Elementary reactions Arrhenius Equation Rate determining step Activation Energy
Lecture 12 (Module 3)Reaction Mechanisms
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.237
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.238
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.239
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.240
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.241
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:
2013 CHEM1020 Chemistry for Science & Engineering Module 3.242
The molecules must have the right orientation.
NO2Cl + Cl&!NO2+ Cl2
p 652
Nitrylchloride
Requirements for Two Molecules to React:
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.243
The molecules must have sufficient energy.
The molecules must collidewith sufficient kinetic energyto overcome the activationenergy, Ea.
Requirements for Two Molecules to React:
2013 CHEM1020 Chemistry for Science & Engineering Module 3.244
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.245
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.246
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.247
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.
!"#
$%&'
( RTE
a
e
k = A exp! Ea /RT( )
2013 CHEM1020 Chemistry for Science & Engineering Module 3.248
RTEAk a
!=lnln
Arrhenius Equation
Using the Arrhenius equation
Take the natural log of both sides:
gives linear equation of form y= mx+ c whereT
xky 1,ln ==
y-intercept:
Slope:!
Ea
R
lnA
Ea= activation energyA = pre-exponential or frequency fact or
If kis measured at severaltemperatures we can determine Eafrom a plot of ln kversus 1/T.
k = A exp! Ea /RT( )
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.249
Using the Arrhenius equation Decomposition acetaldehydeCH3CHO !CH4+ CO
T (K) 700 730 760 790 810 840 910 1000
k (L mol-1
s-1
) 0.011 0.035 0.105 0.343 0.789 2.17 20.0 145
1/T (K-1
) x103 1.43 1.37 1.32 1.27 1.23 1.19 1.10 1.00
ln k (L mol-1s-1) -4.51 -3.35 -2.25 -1.07 -0.24 0.77 3.00 4.98
Raw data kversus T
Transformed dataln k versus 1/T
RT
EAk a!=lnlnStep 1. Plot ln k versus 1/T
Slope = -Ea/R = -2.27 x 104K
Intercept = ln A = 27.0
Ea= 2.27 x 104K x 8.3145 J K-1mol-1= 188 kJ mol-1
A = 1.1x1012L mol-1s-1
slope = -Ea/RStep 2. Linear regression least squares fit
2013 CHEM1020 Chemistry for Science & Engineering Module 3.250
Can also use values of kat just two temperatures:
Subtract first equation from second:
Using the Arrhenius equation
ART
Ek
ART
Ek
a
a
lnln
lnln
2
2
1
1
+!
=
+!
=
ART
EA
RT
Ekk aa lnlnlnln
12
12 !
++!
=!
Temperature 1
Temperature 2
2
1 2 1
1 1ln
ak E
k R T T
! "#$ = #% &
' (
2013 CHEM1020 Chemistry for Science & Engineering Module 3.251
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.252
Transition State
Reactions go from reactants to products via a transition state,or activated complex.
(at the transition state, 50% chance offorming product)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.253
A plot of potential energy against reaction progress is knownas a reaction profile.
Multi-step Reactions.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.254
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.255
Practice Problem *
2013 CHEM1020 Chemistry for Science & Engineering Module 3.256
Ea1> Ea2: Reaction 1 is rate limiting
Multi-step Reactions: Rate determining step
2013 CHEM1020 Chemistry for Science & Engineering Module 3.257
Multi-step Reactions: Rate determining step
Ea1< Ea2: Reaction 2 is rate limiting
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.258
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)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.259
Reaction Mechanisms
From experiment, the rate is known to be rate = k[NO2]2
NO2(g) + CO(g) !NO(g) + CO2(g)
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.260
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) + CO(g) !NO(g) + CO2(g)
NO2(g) + CO(g) !NO(g) + CO2(g)
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.261
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$10-5 s-1at 430oC.
a) What is the activation energy in kJ mol -1?b) What is the value of the pre-exponential factor,A, for thisreaction?c) What is the rate constant for the reaction at 350oC?
k = A exp(-Ea/RT)
*
lnk2
k1
=
!Ea
R
1
T2
!1
T1
"
#$%
&'
2013 CHEM1020 Chemistry for Science & Engineering Module 3.262
Example Q
a) What is the activation energy ?
b) What is the pre-exponential factorA?
c) What is the rate constant at 350oC?
*
2013 CHEM1020 Chemistry for Science & Engineering Module 3.263
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.264
Catalysis Intermediates
Lecture 13 (Module 3)Part A: Catalysis
2013 CHEM1020 Chemistry for Science & Engineering Module 3.265
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)
2013 CHEM1020 Chemistry for Science & Engineering Module 3.266
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.267
Energy Profile
Reactions go from reactants toproducts via a transition state, oractivated complex.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.268
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.269
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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2013 CHEM1020 Chemistry for Science & Engineering Module 3.270
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:
2013 CHEM1020 Chemistry for Science & Engineering Module 3.271
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.272
Decomposition of H2O2 *
H2O2_MnO2_catalyst.mov
H2O2(aq) !H2O(l) + %O2(g)
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.273
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
2013 CHEM1020 Chemistry for Science & Engineering Module 3.274
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.
2013 CHEM1020 Chemistry for Science & Engineering Module 3.275
Summary
Catalystsspeed up chemical reactions without themselves
being consumed.
Proteins which act as catalysts are called enzymes.
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2013 CHEM1020 Chemistry for Science & Engineering Module 3.276
Lecture 13 (Module 3)Part B: Exam Revision