Rate of Reaction Analysis

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?


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


How do ignitable liquids burn?


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


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)

67.2 MJ 67.2x106 J
time(s) 840000 s
80 Js 1 80 Js 1


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

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


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

ΔA ΔB ΔC ΔD
Rate
Δt Δt Δt Δt


H2O2(l) H2O(l) + ½O2(g)

200
Amount of H 2O2 remaining (x105mol)

150

100

50

0
0 50 100 150
Time (s)


Calculating Rate of Reaction
The gradient of tangent to the curve is the rate of
reaction
What happens to the reaction rate with time?
0.4
Concentration of H2O2 (mol dm-3)

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)


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


Rate plot for the decomposition of hydrogen
peroxide
0.008
Rate of Reaction (mol dm -3 s-1)

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 0.35
-3
Hydrogen Peroxide concentration (mol dm )


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

1 1
Rateα H2O2 Rate k H2O2

− 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.


Rate of reaction can be measured from the
rate that oxygen gas is produced.

Inverted burette
50

40

30

Yeast suspension 20

+hydrogen peroxide
solution 10

Water
0


Vary the starting concentration and
measure the initial rate
30

[H2O2]= 0.40 mol dm-3
25
Volume O 2/cm2

20 [H2O2]= 0.32 mol dm-3

[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)


Initial Rate can be measured
30
Initial
25 gradient
0.51cm3s-1
Volume O2/cm-3

20

15

10

5

0
0 50 100 150 200
Time (s)


Plot Initial Rate as a function of starting
concentration

0.5
Rate of reaction (cm2(O2)s-1)

0.4

0.3

0.2

0.1

0
0 0.1 0.2 0.3 0.4
-3
Conc of Hydrogen peroxide (mol dm )


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


For you to do: Initial rate data
[H2O2]/mol dm-3 Rate/cm3 O2 s-1
0.08 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.


General Rate equations
n
Rate k A

0
Zero order Rate k A k
Units of k ?

First order 1
Units of k ?
Rate k A kA
Second order 2
Units of k ? Rate k A


Further Analysis of data
• Logarithms can be very useful
• Plot of log rate as a function of log
concentration (p439 Housecroft)
n
Rate k A
n n
logRate logk A logk log A logk nlog A
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


Half-life
Time taken for the concentration of
reactant A at time t, [A]t to fall to half its
value.
At 1
ln kt ln kt
A0 2
0.693
kt 0.693 t
k
A constant half-life for a first order reaction
Progress curve and measure t½ at several different points.


Constant half-life

200
Amount of H 2O2 remaining (x10 5mol)

Going from 200 x 10-5
150 mol to 100 x 10-5 mol
takes 27s

100
Going from 100 x 10-5
mol to 50 x 10-5 mol
takes 27s
50
Going from 50 x 10-5 mol
27s 27s 26s to 25 x 10-5 mol takes 26s

0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
Time (s)


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


Reactions with more than one reactant
A + B → products

n m
Rate k A B
e.g. C12H22O11 + H2O → C6H12O6 + C6H12O6
sucrose glucose fructose

Rate k C12H22 O11 H2O
First order with respect to each reactant
Second order reaction (sum of orders in rate equation)


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


An Example Reaction
peroxodisulfate (VI) and iodide ions
2 2
S2O8 2I 2SO4 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.
2 1 1
Rate k S2O8 I
3. k is determined by rearranging the rate equation.


Iodine clock data from
experiment


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?


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


Only those molecules with sufficient energy can
Number of molecules with kinetic react

Activation
enthalpy Ea =50kJ
mol-1
energy E

Kinetic energy (E)


Increasing Temperature increases Rate of
Reaction
Number of molecules
300 K with energy greater
Number of molecules with kinetic

than 50kJ mol-1 at
300 K
310 K

Number of molecules
energy E

with energy greater
than 50kJ mol-1 at
310 K

Activation
enthalpy Ea =50kJ
mol-1

Kinetic energy (E)


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


Combining Activation Energy and
enthalpy
Both can be shown on an Exothermic
reaction
enthalpy level diagram

enthalpy
Positive
Add Ea to the A+B
diagram
ΔH
Negative
enthalpy
C+D
Draw a diagram for an
endothermic reaction
Reaction coordinate


Rate equations and Temperature
n m
Rate k A B
Ea
The Arrhenius equation k Ae 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?
EA
It might be easier to do this ln k ln A
RT
increase temperature increase k increase rate
decrease Ea increase k increase rate


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?


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

Ea EA
k Ae RT lnk lnA
RT

Plot lnk against 1/T: gradient = -EA/R; intercept = lnA


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


How do we explain catalysis?


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?


A catalyst provides an alternative path for the
reaction with a lower activation enthalpy

Uncatalysed Activation enthalpy of
reaction uncatalysed reaction
Enthalpy

Catalysed
Activation enthalpy of
reaction
catalysed reaction
Reactants

Products

Progress of reaction


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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