1. RATES OF REACTION CONTENTS Prior knowledge Collision Theory Methods for increasing rate Surface area Temperature Catalysts Light Pressure Concentration Check list

2. Before you start it would be helpful to… know how the energy changes during a chemical reaction know the basic ideas of Kinetic Theory know the importance of catalysts in industrial chemistry RATES OF REACTION Rate of reaction = Change in concentration / time for change to happen = Change In reactants / time = Change in products / time

3. MEASURING RATES OF REACTION Measure volume of a gas when gas is produced In some reactions it is easy to see how the rate changes; eg in a reaction that produces bubbles of gas: Mg(s) + 2HCl(aq) -7 MgCliaq) + H2(g) you can vary the reaction conditions - eg the concentration of the acid, the temperature or the surface area of the magnesium - and see what effect this has on the rate by simply counting the number of bubbles at regular time intervals, eg every 15 s. For more accurate results you can measure the volume of gas given off at regular time intervals throughout the reaction, using the apparatus shown in fig. 2.6.5(a). You can then plot a graph of the total volume of gas produced against time (see fig. 2.6.5(b)).

6. In fig. 2.6.5(b) you can see that the graph is steepest at the start of the reaction and the steepness (rate) reduces throughout the reaction. If you draw a tangent to the curve at the start of the reaction (as shown by the dotted line) you can work out the gradient of the line. This is the initial rate of reaction. The reaction rate changes throughout the reaction, as does the concentration of reactants and products, so it is most accurate for comparing the initial rates of different reactions. You could study the decomposition of hydrogen peroxide using the same apparatus.

7. Coloriemetry  Use when reactants or products are colored Colored compounds example : Iodine Conductometry  Use when changes in ions present is solution

8. Titration

9. Measuring the change in mass of the reaction mixture Another way of monitoring the change in concentration of a gaseous product is to measure the mass of the total reaction mixture as the reaction progresses, and see how quickly it decreases. For example, the rate of the reaction between calcium carbonate and dilute hydrochloric acid: CaC03(s) + 2HCl(aq) 7 CaC12(aq) + H20(1) + C02(g) can be monitored by measuring the mass of the apparatus every 15 seconds. The mass decreases as carbon dioxide escapes from the solution. This method can be used to investigate the effect of changing the temperature, the concentration of the acid or the particle size of the calcium carbonate.

10. Important graphs

13. Reactions are fastest at the start and get slower as the reactants concentration drops. In a reaction such as A + 2B ——> C the concentrations might change as shown RATE CHANGE DURING A REACTION Reactants (A and B) Concentration decreases with time Product (C) Concentration increases with time the steeper the curve the faster the rate of the reaction reactions start off quickly because of the greater likelihood of collisions reactions slow down with time as there are fewer reactants to collide TIME CONCENTRATION B A C

14. Experimental Investigation the variation in concentration of a reactant or product is followed with time the method depends on the reaction type and the properties of reactants/products e.g.Extracting a sample from the reaction mixture and analysing it by titration. - this is often used if an acid is one of the reactants or products Using a colorimeter or UV / visible spectrophotometer. Measuring the volume of gas evolved. Measuring the change in conductivity. More details of these and other methods can be found in suitable text-books. MEASURING THE RATE

15. RATEHow much concentration changes with time. It is the equivalent of velocity. MEASURING THE RATE y CONCENTRATION gradient = y x x TIME the rate of change of concentration is found from the slope (gradient) of the curve the slope at the start of the reaction will give the INITIAL RATE the slope gets less (showing the rate is slowing down) as the reaction proceeds THE SLOPE OF THE GRADIENT OF THE CURVE GETS LESS AS THE REACTION SLOWS DOWN WITH TIME

16. CHEMICAL KINETICS Introduction Chemical kinetics is concerned with the dynamics of chemical reactions such as the way reactions take place and the rate (speed) of the process. One can look at the QUALITATIVE and the QUANTITATIVE aspects of how the rate (speed) of a reaction can be changed. Chemical kinetics plays an important part in industrial chemistry because the time taken for a reaction to take place and the energy required are of great economic importance. The kinetic aspect of chemistry is often at odds with the thermodynamic side when considering the best conditions for industrial production. The concepts met in this topic can be applied throughout the theoretical and practical aspects of chemistry. COLLISION THEORY The basis of the study is COLLISION THEORY...

17. COLLISION THEORY Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy – ACTIVATION ENERGY plus particles must approach each other in a certain relative way - STERIC EFFECT

18. COLLISION THEORY Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy – ACTIVATION ENERGY plus particles must approach each other in a certain relative way - STERIC EFFECT According to collision theory, to increase the rate of reaction you need... more frequent collisionsincrease particle speedor have more particles present more successful collisionsgive particles more energyor lower the activation energy

19. INCREASING THE RATE INCREASE THE SURFACE AREA OF SOLIDS INCREASE TEMPERATURE SHINE LIGHT ADD A CATALYST INCREASE THE PRESSURE OF ANY GASES INCREASE THE CONCENTRATION OF REACTANTS INCREASE THE SURFACE AREA OF SOLIDS INCREASE TEMPERATURE SHINE LIGHT ADD A CATALYST INCREASE THE PRESSURE OF ANY GASES INCREASE THE CONCENTRATION OF REACTANTS The following methods may be used to increase the rate of a chemical reaction

20. INCREASING SURFACE AREA Increases chances of a collision - more particles are exposed Powdered solids react quicker than larger lumps Catalysts (e.g. in catalytic converters) are finely divided for this reason + In many organic reactions there are two liquid layers, one aqueous, the other non-aqueous. Shaking the mixture increases the reaction rate as an emulsion is often formed and the area of the boundary layers is increased giving more collisions. CUT THE SHAPE INTO SMALLER PIECES 1 3 1 SURFACE AREA 9+9+3+3+3+3 = 30 sq units SURFACE AREA 9 x (1+1+1+1+1+1) = 54 sq units 1 1 1 3

21. Effectincreasing the temperature increases the rate of a reaction particles get more energy so can overcome the energy barrier particle speeds also increase so collisions are more frequent INCREASING TEMPERATURE ENERGY CHANGES DURING A REACTION As a reaction takes place the enthalpy of the system rises to a maximum, then falls A minimum amount of energy is required to overcome the ACTIVATION ENERGY (E a ). Only those reactants with energy equal to, or greater than, this value will react. If more energy is given to the reactants then they are more likely to react. Typical energy profile diagram for an exothermic reaction

22. INCREASING TEMPERATURE According to KINETIC THEORY, all particles must have energy; the greater their temperature, the more energy they possess. The greater their KINETIC ENERGY the faster they travel. ZARTMANN heated tin in an oven and directed the gaseous atoms at a rotating disc with a slit in it. Any atoms which went through the slit hit the second disc and solidified on it. Zartmann found that the deposit was spread out and was not the same thickness throughout. This proved that there was a spread of velocities and the distribution was uneven. ZARTMANN’S EXPERIMENT

23. Experiments showed that, due to the many collisions taking place between molecules, there is a spread of molecular energies and velocities. no particles have zero energy/velocity some have very low and some have very high energies/velocities most have intermediate velocities. INCREASING TEMPERATURE MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY

24. Increasing the temperature alters the distribution get a shift to higher energies/velocities curve gets broader and flatter due to the greater spread of values area under curve stays constant - corresponds to the total number of particles T1T1 T2T2 TEMPERATURE T 2 > T 1 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY

25. Decreasing the temperature alters the distribution get a shift to lower energies/velocities curve gets narrower and more pointed due to the smaller spread of values area under curve stays constant T1T1 T3T3 TEMPERATURE T 1 > T 3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY

26. REVIEW no particles have zero energy/velocity some particles have very low and some have very high energies/velocities most have intermediate velocities as the temperature increases the curves flatten, broaden and shift to higher energies T1T1 T2T2 T3T3 TEMPERATURE T 2 > T 1 > T 3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY

27. EaEa ACTIVATION ENERGY - E a The Activation Energy is the minimum energy required for a reaction to take place The area under the curve beyond E a corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER INCREASING TEMPERATURE KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY

28. Explanation increasing the temperature gives more particles an energy greater than E a more reactants are able to overcome the energy barrier and form products a small rise in temperature can lead to a large increase in rate T1T1 T2T2 TEMPERATURE T 2 > T 1 EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE KINETIC ENERGY FRACTION OF MOLECUES HAVING A GIVEN KINETIC ENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER

29. INCREASING TEMPERATURE  If Temperature is increased, Rate increases because large proportion of molecules have high energies (grater than activation energy)  So more successful collisions Temperature Rate

30. TEMPERATURE temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. REACTION TYPE  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT Predict the effect of a temperature increase on the equilibrium position of... H 2 (g) + CO 2 (g) CO(g) + H 2 O(g)  = + 40 kJ mol -1 moves to the RHS 2SO 2 (g) + O 2 (g) 2SO 3 (g)  = - ive moves to the LHS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

31. CH 4 + H 2 O CO + 3H 2 Reactants Products  negative Reactants Products  positive INCREASING TEMPERATURE Yield (If temperature is increasing) Example  positive  Yeild increases as temperature increases and if reaction is endothermic Less yield More yield

32. Energy profile diagrams Exothermic

33. Energy profile diagrams Endothermic Eenrgy Progress of reaction reactants products

34. Catalysts provide an alternative reaction pathway with a lower Activation Energy (E a ) Decreasing the Activation Energy means thet more particles will have energy equal to or grater than activation energy Catalysts remain chemically unchanged at the end of the reaction. ADDING A CATALYST WITHOUT A CATALYSTWITH A CATALYST

35. Reaction profiles for a reaction a) without a catalyst and b) with a catalyst. The dip in the curve of the pathway with a catalyst shows where an unstable intermediate forms.

36. The area under the curve beyond E a corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. If a catalyst is added, the Activation Energy is lowered - E a will move to the left. MOLECULAR ENERGY EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER ADDING A CATALYST NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

37. The area under the curve beyond E a corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. Lowering the Activation Energy, E a, results in a greater area under the curve after E a showing that more molecules have energies in excess of the Activation Energy EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY ADDING A CATALYST EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

38. work by providing an alternative reaction pathway with a lower Activation Energy using catalysts avoids the need to supply extra heat - safer and cheaper catalysts remain chemically unchanged at the end of the reaction. Types Homogeneous Catalysts Heterogeneous Catalysts same phase as reactantsdifferent phase to reactants e.g. CFC’s and ozone e.g. Fe in Haber process CATALYSTS - A REVIEW

39. work by providing an alternative reaction pathway with a lower Activation Energy using catalysts avoids the need to supply extra heat - safer and cheaper catalysts remain chemically unchanged at the end of the reaction. Types Homogeneous Catalysts Heterogeneous Catalysts same phase as reactantsdifferent phase to reactants e.g. CFC’s and ozone e.g. Fe in Haber process CATALYSTS DO NOT AFFECT THE POSITION OF ANY EQUILIBRIUM but they do affect the rate at which equilibrium is attained a lot is spent on research into more effective catalysts - the savings can be dramatic catalysts need to be changed regularly as they get ‘poisoned’ by other chemicals catalysts are used in a finely divided state to increase the surface area CATALYSTS - A REVIEW

40. Catalysts are widely used in industry because they… CATALYSTS - WHY USE THEM?

41. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT CATALYSTS - WHY USE THEM?

42. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT enable different reactions to be usedBETTER ATOM ECONOMY REDUCE WASTE CATALYSTS - WHY USE THEM?

43. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT enable different reactions to be usedBETTER ATOM ECONOMY REDUCE WASTE are often enzymesGENERATE SPECIFIC PRODUCTS OPERATE EFFECTIVELY AT ROOM TEMPERATURES CATALYSTS - WHY USE THEM?

44. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT enable different reactions to be usedBETTER ATOM ECONOMY REDUCE WASTE are often enzymesGENERATE SPECIFIC PRODUCTS OPERATE EFFECTIVELY AT ROOM TEMPERATURES have great economic importance in the industrial production ofPOLY(ETHENE) SULPHURIC ACID AMMONIA ETHANOL CATALYSTS - WHY USE THEM?

45. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT enable different reactions to be usedBETTER ATOM ECONOMY REDUCE WASTE are often enzymesGENERATE SPECIFIC PRODUCTS OPERATE EFFECTIVELY AT ROOM TEMPERATURES have great economic importance in the industrial production ofPOLY(ETHENE) SULPHURIC ACID AMMONIA ETHANOL can reduce pollutionCATALYTIC CONVERTERS CATALYSTS - WHY USE THEM?

46. Catalysts are widely used in industry because they… allow reactions to take place at lower temperaturesSAVE ENERGY (lower E a ) REDUCE CO 2 OUTPUT enable different reactions to be usedBETTER ATOM ECONOMY REDUCE WASTE are often enzymesGENERATE SPECIFIC PRODUCTS OPERATE EFFECTIVELY AT ROOM TEMPERATURES have great economic importance in the industrial production ofPOLY(ETHENE) SULPHURIC ACID AMMONIA ETHANOL can reduce pollutionCATALYTIC CONVERTERS CATALYSTS - WHY USE THEM?

47. increasing the pressure forces gas particles closer together  If Pressure is increased, Rate increases because more particles per unit volume / grater concentration  So more frequent collessions / successful collessions takes place  The more frequent the collisions, the greater the chance of a reaction INCREASING THE PRESSURE Pressure Rate

48. PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM Predict the effect of an increase of pressure on the equilibrium position of.. 2SO 2 (g) + O 2 (g) 2SO 3 (g) MOVES TO RHS :- fewer gaseous molecules H 2 (g) + CO 2 (g) CO(g) + H 2 O(g) NO CHANGE:- equal numbers on both sides FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

49. N 2 + 3H 2 2NH 3 4 Total Moles : 2  If pressure increases NH 3 yield will be more as more moles are in right hand side so forward reaction takes place Few moles More moles Low pressure High pressure Yield

50. Increasing concentration = more frequent collisions = increased rate of reaction INCREASING CONCENTRATION However, increasing the concentration of some reactants can have a greater effect than increasing others Low concentration = fewer collisionsHigher concentration = more collisions

51. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM Predict the effect of increasing the concentration of O 2 on the equilibrium position 2SO 2 (g) + O 2 (g) 2SO 3 (g) EQUILIBRIUM MOVES TO RHS SUMMARY REACTANTS PRODUCTS INCREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE RIGHT THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM DECREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE RIGHT Predict the effect of decreasing the concentration of SO 3 on the equilibrium position EQUILIBRIUM MOVES TO RHS

52. 11.3 The Arrhenius Equation

53. dependence of the rate constant temperature Arrhenius equationThe dependence of the rate constant of a reaction on temperature can be expressed by the following equation known as Arrhenius equation. E a = activation energy k = rate constant R = gas constant T = absolute temperature e = the base of the natural logarithm scale A = quantity represents the collision frequency (frequency factor) k=A

54. The relation ship between the rate constant, k and temperature can be seen in the k vs T graph: k = A e -E a ∕ RT 1/T (K -1 )

55. By taking the natural logarithm of both sides. ln k = ln Ae -Ea/RT ln k = ln A – Ea/RT The equation can take the form of a linear equation ln k = ( ) ( ) + ln A A plot of ln k versus gives a straight line Slope m = and intercept c with the y axis is ln A. y m x c

57. Graph Representation Of The Arrhenius Equation Plotting a ln k vs 1 / T graph would show a clearer relationship between k (Rate constant) and temperature Where, E a = Activation Energy R = 8.314 Jmol -1 K -1 T = Absolute Temp A = Collision freq. factor

58. Example 1 k (1/M s)T (K) 3.52 x 10 -7 283 3.02 x 10 -5 356 2.19 x 10 -4 393 1.16 x 10 -3 427 3.95 x 10 -2 508 The rate constants for decomposition of acetaldehyde 2HI (g) H 2(g) + I 2(g) were measured at five different temperatures.The data are shown below. Plot ln k versus 1/T, and determine activation energy (in kJ/mole) the activation energy (in kJ/mole) for the reaction.

59. Solution We need to plot ln k on the y-axis versus 1/T on the x-axis. From the given data we obtain ln k1/T (K -1 )  14.860 1.80 x 10 -3  10.408 1.59 x 10 -3  8.426 1.50 x 10 -3  6.759 1.43 x 10 -3  3.231 1.28 x 10 -3

60. -16 -14 -10 -8 -6 -4 -2 1.41.6 1.8 x 10 -3 1/T(K -1 ) -12 yy xx Plot of ln k versus 1/T. The slope of the line is calculated from two pairs of coordinates. ln k

61. Slope = (-14.0) – (-3.9) = -2.24 x 104 K (1.75– 1.3) x 10 -3 K -1 Finally, calculate the activation energy from the slope: The slope, m = –  E a /R. E a = – R (m) = (–8.314 J K -1 mol -1 ) (–2.24 x 104 K) = 1.9 x 105 J mol -1 = 190 kJ mol -1

62. constants k 1 k 2 temperatures T 1 T 2An equation relating the rate constants k 1 and k 2 at temperatures T 1 and T 2 can be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known. ln k 1 = ln A – Ea/RT 1 ln k 2 = ln A – Ea/RT 2 Subtracting ln k 2 from ln k 1 gives ln k 1 – ln k 2 = Ea/R (1/T 2 – 1/T 1 )

63. Example 2 The rate constant of a first-order reaction is 3.46 x 10 - 2 s -1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mole? Solution Given k 1 = 3.46 x 10 -2 s -1 k 2 = ? T 1 = 298 K T 2 = 350 K Subtituting in equation ln =

64. ln 3.46 x 10-2 = 50.2 x 103 J/mol 298K – 350K k 2 8.314 J/K mol (298K)(350K) Solving the equation gives ln 3.46 x 10 -2 = –3.01 k2k2 3.46 x 10 -2 = 0.0493 k 2 k 2 = 0.702 s  1

65. DATA: SOLUTION: Exercise 1: the Activation energy The decomposition of hydrogen iodide, has rate constants of 9.51 x 10 -9 L mol -1 s -1 at 500 K and 1.10x10 -5 L mol -1 s -1 at 600 K. Find E a. E a = 1.76 x 10 5 J/mol = 176 kJ/mol 2 HI (g)  H 2 (g) + I 2 (g) k 1 = 9.51 x 10 -9 L mol -1 s -1 T 1 = 500K k 2 = 1.10 x 10 -5 L mol -1 s -1 T 1 = 600K

66. changeEffect of the rate of reaction Effect on the rate constant Increasing [ ]IncreasesNo effect Decreases [ ]DecreasesNo effect Increases PIncreasesNo effect Decreases PDecreasesNo effect Increasing TIncreases Decreasing TDecreases Addition of catalystIncreases

67. REVISION CHECK What should you be able to do? Recall and understand the statements in Collision Theory Know six ways to increase the rate of reaction Explain qualitatively how each way increases the rate of reaction Understand how the Distribution of Molecular Energies is used to explain rate increase Understand how the importance of Activation Energy Recall and understand how a catalyst works by altering the Activation Energy Explain how the rate changes during a chemical reaction CAN YOU DO ALL OF THESE? YES NO

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