Rate of Reaction
University of Lincoln
presentation
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Why the difference?
Is it the enthalpy change (Heat of combustion) ?
Paraffin wax 42 MJ kg-1
Petrol 45 MJ kg-1
Is it the temperature?
Yellow/white – 1300oC
Pale orange/yellow – 1100oC
What is it then?
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Approximately how long will a 2 litre
pool of petrol burn for?
Important values: Petrol density = 0.8 kg litre-1
Heat of combustion is 45 MJ kg-1
2 litres of petrol has a mass of 1.6 kg (from the density)
Total energy available from 1.6 kg petrol
= 1.6 kg x 45 MJ kg-1 = 72 MJ
q  m  ΔH
2 litre petrol pool is a 1 MW fire (this is a measured value)
1 MW = 1 MJ s-1 so at this rate it would take 72 s
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How do ignitable liquids burn?
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2 litre petrol bomb takes about 10s to burn.
What is the rate of heat release?
72 MJ in 10 s = 7.2 MW
2 litre petrol fully evaporated takes about 1 s
to burn. What is the rate of heat release?
72 MJ in 1s = 72 MW
Conclusion: Same total energy available but
released at a faster rate
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How long to burn a 1.6 kg candle?
• 1.6 kg paraffin wax at 42 MJ kg-1 can release
67.2 MJ
• Candle flame has a heat release rate of
80 W (80 Js-1)
time(s) 
67.2 MJ
80 Js
1

6
67.2x10
80 Js
1
J
 840000 s
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A candle bomb?
• NASA are researching the paraffin rocket!!
• How can this work?
• Increase rate of combustion
– Increase concentration of the oxidant; use
100% oxygen
– Paraffin as small liquid droplets
• Study of the rates of reaction - kinetics
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Factors affecting the rate of a chemical
reaction
1.
Concentration (hydrogen peroxide demo)
2. Pressure (gases)
3. Temperature (glowstick)
4. Surface area (dust explosion)
5. Catalysis
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Measuring Reaction Rate
• Use a characteristic of the products or the reactants that
can be used as a measure of amount.
– Volume of gas
– Change in mass
– Absorption of light
• rate of decrease of reactant or rate of increase of a
product
A B C D
Rate  
ΔA
Δt

ΔB
Δt

ΔC
Δt

ΔD
Δt
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H2O2(l)
H2O(l) + ½O2(g)
Amount of H 2O2 remaining (x105mol)
200
150
100
50
0
0
50
100
150
Time (s)
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Calculating Rate of Reaction
The gradient of tangent to the curve is the rate of
reaction
Concentration of H2O2 (mol dm-3)
What happens to the reaction rate with time?
0.4
0.35
Concentration = 0.3 mol dm-3 s-1
0.3
Rate = gradient = 0.0068 mol dm-3 s-1
0.25
0.2
Concentration = 0.1 mol dm-3 s-1
0.15
Rate = gradient = 0.0023 mol dm-3 s-1
0.1
0.05
0
0
50
100
150
Time (s)
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A Mathematical Relationship
• Select two other points on the curve and
calculate the rate of reaction at that
concentration of H2O2
• Plot a graph of Rate of Reaction as a
function of H2O2 concentration
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-3 -1
Rate of Reaction (mol dm s )
0.008
Rate plot for the decomposition of hydrogen
peroxide
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0
0
0.05
0.1
0.15
0.2
0.25
0.3
-3
Hydrogen Peroxide concentration (mol dm )
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0.35
What does the graph show?
•
•
•
Graph is a straight line through the origin
The two variables have a linear mathematical relationship
We can say: Rate of Reaction is directly proportional to
Hydrogen Peroxide concentration
Rate α H 2 O 2 
Rate  k H 2 O 2 
1
1
− Easy to predict what happens to reaction when [H2O2] is changed
− [H2O2] x2
Rate x 2
− First Order with respect to H2O2
− k is the rate constant; first order reaction has units of s-1 when the
rate of reaction is measured in mol dm-3 s-1. Show this by rearranging
the rate equation and why are the units of rate important.
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Rate of reaction can be measured from the
rate that oxygen gas is produced.
50
Inverted burette
40
30
20
10
Yeast suspension
+hydrogen peroxide
solution
Water
0
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Vary the starting concentration and
measure the initial rate
30
[H2O2]= 0.40 mol dm-3
Volume O 2/cm2
25
[H2O2]= 0.32 mol dm-3
20
[H2O2]= 0.24 mol dm-3
15
[H2O2]= 0.16 mol dm-3
10
[H2O2]= 0.08 mol dm-3
5
0
0
50
100
150
200
Time (s)
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Initial Rate can be measured
30
Initial
gradient
0.51cm3s-1
Volume O2/cm-3
25
20
15
10
5
0
0
50
100
150
200
Time (s)
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Rate of reaction (cm2(O2)s-1)
Plot Initial Rate as a function of starting
concentration
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
-3
0.4
Conc of Hydrogen peroxide (mol dm )
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Summary
– Decomposition of H2O2 can be followed by
measuring the decrease in H2O2 concentration
or the volume of O2 evolved.
– Rate of reaction can be calculated from the
progress curve at different times or initial rate
measurements.
– Plots of rate as a function of reagent
concentration can be used to determine the
mathematical relationship
– Order of reaction can be determined
– Rate equation can be written
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For you to do: Initial rate data
[H2O2]/mol dm-3
0.08
Rate/cm3 O2 s-1
0.1
0.16
0.215
0.24
0.32
0.32
0.41
0.4
0.51
Determine the order of reaction with respect to
hydrogen peroxide and calculate the value of the rate
constant.
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General Rate equations
Rate  k A 
n
Zero order
Units of k ?
First order
Units of k ?
Second order
Units of k ?
Rate  k A   k
0
Rate  k A   k A 
1
Rate  k A 
2
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Further Analysis of data
• Logarithms can be very useful
• Plot of log rate as a function of log
concentration (p439 Housecroft)
Rate  k A 
n
logRate
 logk A   logk  log A   logk  nlog A 
n
n
Gradient is n; Intercept is log k
Use this method on the initial rate data in slide 21 to
determine order and the value of k
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Half-life
Time taken for the concentration of
reactant A at time t, [A]t to fall to half its
value.
A t
ln
A 0
  kt
 kt   0.693
ln
t
1
2
  kt
0.693
k
A constant half-life for a first order reaction
Progress curve and measure t½ at several different points.
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Constant half-life
Amount of H 2O2 remaining (x10 5mol)
200
Going from 200 x 10-5
mol to 100 x 10-5 mol
takes 27s
150
100
Going from 100 x 10-5
mol to 50 x 10-5 mol
takes 27s
Going from 50 x 10-5 mol
to 25 x 10-5 mol takes 26s
26s
50
27s
27s
0
0
10
20
30
40
50
60
70
80
90
100 110 120 130 140 150 160 170 180
Time (s)
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Using Luminol to detect blood stains
Exponential decay curve
First order write rate equation
Calculate half-life and why is it important
Video clip or demo
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Reactions with more than one reactant
A
+
B
products
→
Rate  k A  B 
n
e.g.
C12H22O11 + H2O
sucrose
→
m
C6H12O6 + C6H12O6
glucose
fructose
Rate  k C 12 H 22 O 11  H 2 O 
First order with respect to each reactant
Second order reaction (sum of orders in rate equation)
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Determining order and rate equations
Difficulties with more than one reactant?
Experimental Design
Principle
Vary one concentration and keep other(s) constant while
measuring rate.
•Initial rate method
•Isolation method
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An Example Reaction
peroxodisulfate (VI) and iodide ions
S2O 8
2
 2I

2
 2SO
4
 I2
Task: Determine the rate equation and a value for k
Design the experiment
1.
initial rate method (vary each concentration)
2.
Plot a graph of log rate as a function of log initial
concentration for each reactant. Gradient of each line is
order of reaction for each reactant.

Rate  k S 2 O 8
3.
2
 I 
1

1
k is determined by rearranging the rate equation.
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Iodine clock data from
experiment
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Collision theory
• Molecules have to collide if they are to react –
increasing frequency of collisions?
• Increasing concentration increases the frequency
of collisions
• Increasing pressure increases frequency of
collisions
• Increasing temperature increases frequency of
collision
• But not just about rate of collisions – how do we
explain slow reactions?
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Activation Energy (Enthalpy)
Ea
• Energy of the collision must be above a
certain value for reactants to react
• Why? Energy is needed to break bonds
(remember bond enthalpies)
• This then creates reactive species to make
new bonds
• The minimum energy required for a
collision to result in chemical reaction is Ea
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Number of molecules with kinetic
energy E
Only those molecules with sufficient energy can
react
Activation
enthalpy Ea =50kJ
mol-1
Kinetic energy (E)
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Number of molecules with kinetic
energy E
Increasing Temperature increases Rate of
Reaction
300 K
310 K
Number of molecules
with energy greater
than 50kJ mol-1 at
300 K
Number of molecules
with energy greater
than 50kJ mol-1 at
310 K
Activation
enthalpy Ea =50kJ
mol-1
Kinetic energy (E)
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Back to petrol
• Petrol vapour reacts with oxygen (air)
• But not spontaneous at room temperature
• Needs ignition. What does ignition do?
– Provides energy to break bonds (endothermic)
– Creates reactive species (free radicals)
– Self-sustaining (can remove ignition source and it
carries on). Why????
– Energy released from the reaction breaks more bonds
and the reaction continues
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Combining Activation Energy and
enthalpy
Both can be shown on an
enthalpy level diagram
Positive
enthalpy
Add Ea to the
diagram
Exothermic
reaction
A+B
Negative
enthalpy
ΔH
Draw a diagram for an
endothermic reaction
C+D
Reaction coordinate
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Rate equations and Temperature
Rate  k  A   B 
n
The Arrhenius equation
m
k  Ae

Ea
RT
k is the rate constant; A is the pre-exponential factor; Ea is the activation
energy; R is the molar gas constant (8.314 J mol-1 K-1); T is the absolute
temperature (Kelvin).
How does it work?
It might be easier to do this
increase temperature
decrease Ea
increase k
increase k
ln k  ln A 
EA
RT
increase rate
increase rate
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The well known ‘rule of thumb’
• Reaction rate doubles if
temperature is increased by 10 oC
Temp/K
Ea/kJ mol-1
A/L mol-1 s-1
k/L mol-1 s-1
313
54
8.7 x 106
8.5 x 10-3
323
54
8.7 x 106
1.6 x 10-2
Check the values of k by calculating them from the Arrhenius equation
using the other values in the table
Calculate k at 333 K. What is happening to the value of k? How will this
affect the rate of this reaction?
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An experiment to determine Ea
• Determine order and rate equation for the reaction
• Measure the rate of reaction at different
temperatures keeping the initial concentrations the
same
• Calculate k at the different temperatures
k  Ae

Ea
RT
lnk  lnA 
EA
RT
Plot lnk against 1/T: gradient = -EA/R; intercept = lnA
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Calculating Ea
Use the data below to calculate a value for the
activation energy for this reaction
Temperature/K
k/dm3 mol-1 s-1
296
2.9 x 10-3
302
4.2 x 10-3
313
8.3 x 10-3
323
1.9 x 10-2
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How do we explain catalysis?
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What are catalysts?
Definition and some examples; reactions and catalysts
Hydrogen peroxide , metals and natural substances
Enzymes
Gases on metal surfaces
What is a different reaction route?
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A catalyst provides an alternative path for the
reaction with a lower activation enthalpy
Enthalpy
Uncatalysed
reaction
Activation enthalpy of
uncatalysed reaction
Catalysed
reaction
Activation enthalpy of
catalysed reaction
Reactants
Products
Progress of reaction
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Acknowledgements
•
•
•
•
•
•
•
JISC
HEA
Centre for Educational Research and Development
School of natural and applied sciences
School of Journalism
SirenFM
http://tango.freedesktop.org
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