1. Rate of Reaction University of Lincoln presentation
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3. 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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4. 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
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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5. How do ignitable liquids burn?
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6. 2 litre petrol bomb takes about 10s to 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
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7. 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)
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8. 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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9. Factors affecting the rate of a chemical reaction
Concentration (hydrogen peroxide demo)
Pressure (gases)
Temperature (glowstick)
Surface area (dust explosion)
Catalysis
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10. 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
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11. H2O2(l)
H2O(l) + ½O2(g)
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12. Calculating Rate of Reaction
The gradient of tangent to the curve is the rate of reaction
What happens to the reaction rate with time?
Concentration = 0.3 mol dm-3 s-1
Rate = gradient = mol dm-3 s-1
Concentration = 0.1 mol dm-3 s-1
Rate = gradient = mol dm-3 s-1
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13. 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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15. 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
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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16. Rate of reaction can be measured from the rate that oxygen gas is produced.
Inverted burette
Yeast suspension +hydrogen peroxide solution
Water
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17. Vary the starting concentration and measure the initial rate
[H2O2]= 0.40 mol dm-3
[H2O2]= 0.32 mol dm-3
[H2O2]= 0.24 mol dm-3
[H2O2]= 0.16 mol dm-3
[H2O2]= 0.08 mol dm-3
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18. Initial Rate can be measured
Initial gradient 0.51cm3s-1
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19. Plot Initial Rate as a function of starting concentration
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20. 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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21. 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
0.41
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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22. General Rate equations
Zero order
Units of k ?
First order
Units of k ?
Second order
Units of k ?
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23. Further Analysis of data
Logarithms can be very useful
Plot of log rate as a function of log concentration (p439 Housecroft)
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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24. Half-life
Time taken for the concentration of reactant A at time t, [A]t to fall to half its value.
A constant half-life for a first order reaction
Progress curve and measure t½ at several different points.
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25. Constant half-life 27s 27s 26s
Going from 200 x 10-5 mol to 100 x 10-5 mol takes 27s
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
27s
27s
26s
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26. 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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27. Reactions with more than one reactant
A + B → products
e.g. C12H22O11 + H2O → C6H12O6 + C6H12O6
sucrose glucose fructose
First order with respect to each reactant
Second order reaction (sum of orders in rate equation)
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28. 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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29. An Example Reaction peroxodisulfate (VI) and iodide ions
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.
3. k is determined by rearranging the rate equation.
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30. Iodine clock data from experiment
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31. 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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32. 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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33. Only those molecules with sufficient energy can react
Number of molecules with kinetic energy E
Kinetic energy (E)
Activation enthalpy Ea =50kJ mol-1
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34. Increasing Temperature increases Rate of Reaction
Number of molecules with kinetic energy E
Kinetic energy (E)
Activation enthalpy Ea =50kJ mol-1
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
300 K
310 K
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35. 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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36. Combining Activation Energy and enthalpy
Both can be shown on an enthalpy level diagram
Exothermic reaction
Positive enthalpy
Add Ea to the diagram
A+B ΔH
Negative enthalpy
C+D
Draw a diagram for an endothermic reaction
Reaction coordinate
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37. Rate equations and Temperature
The Arrhenius equation
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 increase k increase rate
decrease Ea increase k increase rate
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38. 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
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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39. 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
Plot lnk against 1/T: gradient = -EA/R; intercept = lnA
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40. Calculating Ea Temperature/K k/dm3 mol-1 s-1 296 2.9 x 10-3 302
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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41. How do we explain catalysis?
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42. 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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43. A catalyst provides an alternative path for the reaction with a lower activation enthalpy
Uncatalysed reaction
Activation enthalpy of uncatalysed reaction
Catalysed reaction
Enthalpy
Activation enthalpy of catalysed reaction
Reactants
Products
Progress of reaction
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44. Acknowledgements JISC HEA
Centre for Educational Research and Development
School of natural and applied sciences
School of Journalism
SirenFM
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