1. Kinetics
What are the factors affecting the rates of reactions in making new products?
Why is it important to determine the rates of reactions?
How does our knowledge of rates of reactions influence our ability to make wise decisions for safe products?

2. Rate of Reactions
The speed at which reactants are used up or products are formed.

5. Rates of Reaction Really Quick – Firework Exploding
What is meant by the rate of a chemical reaction.
How to measure the rate of reaction.
What factors affect the rate of reaction.
Rates of Reaction
Really Quick – Firework Exploding
Really Slow - Rusting

6. Why are kinetics important?
In order to control processes.
speed up useful reactions that occur to slowly
slow down reactions that are harmful
Example:
Catalysts are used in our cars to rapidly convert toxic substances into safer substances
Refrigerators are used to slow the process of spoiling in food

7. How do we know a reaction has taken place?
In most cases, there will be observable changes :
the mass change per unit time (e.g. mass of product formed or reactant used up)
the volume of gas given off per unit time during the reaction
the change in temperature
the formation of precipitate
the change in pH
the change in colour

8. Rates of reaction A measure of the “speed” of reactions.
The rate of reaction
is the speed at which reactants are used up or products are formed.
How do we measure the rate of reactions?

9. Measuring Rate
How could you measure the rate of these reactions? Use your definition to help you.

10. How do you measure the rate of reaction?
Amount of product obtained per unit time.
Speed of reaction
= ________________ obtained
time taken
Amount of gas produced per unit time.
= ________________ produced
Mass of product
Volume of gas

11. How do you measure the rate of reaction?
Amount of reactant used up per unit time.
Rate of reaction
= ________________ used up
time taken
Mass of reactant

12. Example : Reaction that produces gases
Reaction of magnesium strip with hydrochloric acid
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)

13. Plot this graph of volume of gas produced against time
Time taken(min) 1 2 3 4 5 6 7 8 9
Vol of Gas porduced(cm3) 17 25 30 34 35

14. Rate of Reaction at 4 minutes = (40-26)/(5.7-2.6)
= cm3/min

16. Region 1 (the 1st minute) What can say about the nature of the curve?
What does this show?
The rate of reaction is the ________.
More reacting particles are present and the number of effective collisions per second is ___________.
The slope of the gradient is
the steepest
fastest
highest

17. Region 2 (1st – 4.5th minute) What does this show?
The rate of reaction is ________ than in region 1.
Fewer reacting particles are present and so the number of effective collisions per second will be ________.
The gradient of the slope is
less steep
slower
lesser

18. Region 3 (after 4.5th minute)
What does this show?
The reaction has ________
At least one of reactants has been _______
Hence no further ________ can occur between the two reactants.
The graph is horizontal.
There is no gradient.
stopped
used up
collision

19. Amount of reactants used up against time
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
Time taken(s) 10 20 30 40 50 60 70 80 90
Loss in Mass(g)
0.000
0.056
0.098
0.129
0.158
0.165

20. Mass loss(g) against time(min)
What can you infer from the graph?
The graph shows a decreasing curve
As the time increases, the mass of calcium carbonate decreases as it is gradually used up in the reaction.
Rate of loss of mass at time, t
= g min-1
Mass of CaCO3(g) A
0.165 B C
Time(min)
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)

21. Concentration of reactants against time
Catalysed decomposition of hydrogen peroxide
H2O2(l)  O2(g) + H2O(l)
Time taken(min)
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
Conc of H2O2(moldm-3)
0.159
0.129
0.101
0.080
0.063
0.050
0.040
0.035
0.025
0.020

22. Concentration(moldm-3) against time(min)
What can you infer from the graph?
Conc of H2O2(moldm-3)
The graph shows a decreasing curve
As the time increases, the concentration of hydrogen peroxide decreases as it decomposes.
Rate of decompostion at time, t
= mol dm-3 min-3 P
0.060 Q R
Time(min)
H2O2(l)  O2(g) + H2O(l)

23. Concentration(moldm-3) against time(min)
Gradient of tangents, Rate
Conc of H2O2(moldm-3)
Conc of H2O2
(moldm-3)
0.16
0.12
0.08
0.04
Rate
(moldm-3min -1)
0.076
0.049
0.033
0.019
[A]0
[A]1
[A]2
[A]3
[A]4
Time(min)
[A]4
[A]3
[A]2
[A]1
[A]0
H2O2(l)  O2(g) + H2O(l)
Conc of H2O2(moldm-3)

24. Rate of Decomposition against Concentration
Draw the line of best fit
The rate of reaction is proportional to the concentration of
dinitrogen pentoxide

25. Not all collisions are successful in producing a reaction.
Only a small fraction will
result in a reaction =
EFFECTIVE COLLISIONS
All colliding particles need a minimum amount of energy for a reaction to occur =
ACTIVATION ENERGY

26. Collision Theory How do reactions occur at the molecular level?
Molecules collide with each other
Form activated complex
reactants molecules
Must have sufficient energy
Energy ≥ Activation energy (Ea - energy to break bonds)
Right orientation (Animation)

27. Concentration If the concentration of the reactants
is increased, there are more particles
present in the same volume so that
they are closer together.
More likely to meet and react.
Higher concentration
Number of collisions increased
In particular, doubling the conc of one of the reactants will double the rate of reaction.
This explain why the greatest rate of reaction occurs as soon as the reactants are mixed, when they are at the highest conc. As the cnc of the chemicals decrease and the rate of reaction decreases as there are fewer collisions.
Higher chance of successful collisions
in a given amount of time
 rate will increase

28. Pressure
With __________ reactions, increasing pressure forces the particles to be closer together. This will increase the ______________ of the reacting particles. Hence, the rate will increase.
gaseous
concentration

29. Particle Size smaller Large lump of solid has _________surface
area. Reaction can occur at the
outside surface.
Smaller lumps of solid have a
________surface area at which
the reaction can happen. The rate
is much faster.
Increased surface area
Number of collisions increased
Animation
larger

30. Temperature
At a higher temperature, there is a higher rate of reaction as the particles are moving around with greater _____________ and ________ more frequently with energy greater than the activation energy.
kinetic energy
collide

31. Temperature Increased number of collisions per unit time
More molecules have enough activation energy
Maxwell-Boltzmann energy distribution curve
Increased temperature, distribution flattens out
More molecules have Ea
Does activation energy change with temperature?
Number of
molecules
At higher temp, the peak of the curve moves to the right so the most likely value of KE for the molecules increase.
The curve flattens so the total area under it and therefore the total number of molecules remain constant.
The area under the curve to the right of the activation energy Ea increases. This means at higher temp, a greater % of molecules have energies = or in excess of the Ea
Kinetic energy

32. Light The rates of some reactions are increased by exposure to light.
Example, photosensitive chemicals undergo partial decomposition in the presence of sunlight.
2AgX(s)  2Ag(s) + X2(g), X represents a halogen
2AgNO3(s)  2Ag(s) + 2NO2(g) + O2(g)
2H2O2(l)  O2(g) + 2H2O(l)
4HNO3(aq)  2H2O(l) + 4NO2(g) + O2(g)

33. Mixtures of hydrogen and bromine, or methane and chlorine react rapidly only in the presence of light.
H2(g) + Br2(g)  2HBr(g)
CH4 (g) + Cl2(g)  CH3Cl (g) + HCl(g)

34. Catalysts
Substance that speeds up the rate of reaction but without being used up (remain chemically unchanged) .
Hold molecules in the right _____________ for more successful collision per unit time.
The activation energy is ________ by the catalyst.
orientation
lowered

35. Importance of catalysts
Decomposition of hydrogen peroxide where manganese (IV) oxide is used to speed up the reaction.
H2O2  H2O + O2
Manufacture of sulfur trioxide where vanadium(V)oxide is used to make sulfuric acid (Contact Process)
2SO2(g) + O2(g)  2SO3(g)
Manufacture of ammonia (Haber Process)
at a high pressure, 250 atmospheres
at medium temperature, 4500C
with the help of porous iron catalyst by reducing magnetite Fe3O4
N2(g) + 3H2 (g) NH3(g)

36. Catalysts hold molecules in right orientation
Homogeneous catalyst (same physical state as the reactants - often both in solution)
Catalysis by Co2+
Heterogeneous catalyst (different physical state from the reactants - often the reactants are gases and catalyst is a solid)

37. Biological catalyst - Enzymes
Enzymes are present in all living organisms. E.g. Catalase found in blood, potato peeling can decompose hydrogen peroxide, H2O2.
They are made of proteins
Highly specific in their actions – act on 1 substance only. E.g. Amylase breaks down starch. Pepsin breaks down proteins.
Can be made inactive by heating
Works best between 35 – 400C.

38. The rate expression Rate α concentration of solution A Rate α [A]
Rate = k [A]
Rate expression is experimentally determined.
xA + yB  C + D
Rate = k[A]m[B]n
Overall order = m + n

39. Example 1 Consider the reaction 2A  B Determine the
order with respect to A
the rate of reaction
value of rate of reaction (with units)
the rate of reaction when [A]=1.3moldm-3
Expt
[A] /moldm-3
Rate / /moldm-3 1
1.0
0.60 2
2.0
1.2 3
5.0
3.0

40. Example 2 Consider the reaction 3A + B  C + D Determine the
order with respect to A (b) order with respect to B
the overall rate of reaction (d) the rate expression
(e) The value of the rate constant (with units)
the rate of reaction when [A]=1.60moldm-3 and [B]=0.30moldm-3
Expt
[A] /moldm-3
[B] /moldm-3
Rate / /moldm-3 1
0.10
0.50 2
0.30
4.50 3
0.20

41. Example 3 Given that 2P + Q  R + S and the data determine
the order with respect to P
the order with respect to Q
the overall rate of reaction
the rate expression
the value of the rate constant (with units)
Expt
[P]/moldm-3
[Q]/ moldm-3
Rate/moldm-3 s-1 1
1.20
2.00
5.00 x 10-3 2
2.40
1.00 x 10-2 3
6.00
8.00
0.100