1. Rate of Reactions {Chemical Kinetics}

2. Chemical Kinetics Chemical kinetics is the branch of chemistry concerned with the rates of chemical reactions A rate is a change in a measurable quantity over time Different chemical reactions proceed from reactants to products at different rates (e.g. combustion of propane and oxidation of silver)

3. Reaction Rate The reaction rate of a chemical reaction is the change in concentration of a reactant or product per unit time Δc is measured in mol/L Δt is measured in s r is measured in mol/(L  s)

4. Fast or Slow Reactions Extremely slow reactions –Iron rusting –Limestone weathering Extremely fast reactions –Explosion

5. Measuring Rate of Reactions Some rate of reactions have detectable change with respect to time Changes that are observable like –When a volume of gas is given off –When this is a change in mass during the reaction –When there are temperature changes –When there are colour changes –When a precipitate forms –When there are pH changes

6. Speed= = = 300km/h = = 10km/h

7. According to Collision Theory what factors determine whether a reaction happens or not? According to Thermochemistry, what two factors determine whether a reaction is favoured?

8. Collision theory Theory that explains how factors affect rate of reaction Before reaction can occur, particles must first collide Particles collide so that bonds are broken and new bonds can form

9. Collision theory Collision of particles must be effective collisions to produce result Effective collisions are collisions which – produce enough energy to overcome energy of activation – correct orientation

10. Orientation

11. Collision theory Activation energy is the minimum energy the colliding particles must overcome so that reaction occur In order for particles to overcome the activation energy, several factors are involved

12. So lets say collision theory is determining that a reaction can happen…. what factors determine how fast it will happen? Mr McCormack is cooking a roast dinner but his over-sized potatoes are not boiling fast enough….what measures can he take to increase the rate of reaction??? How factors affect rate of reaction according to collision theory

13. Size of reactant for solid reactant Temperature of reactant mixture Concentration of solution reactant Presence of catalyst Pressure on the reaction

14. Concentration Rate of reaction increases with increasing concentration Higher concentration, more reacting particles are present Greater probability of an effective collision Faster rate of reaction

15. Concentration – Same no. of moles Amt of product formed Time/s Higher concentration Lower concentration

16. Concentration – Different no. of moles Amt of product formed Time/s Higher concentration Lower concentration

17. Temperature Rate of reaction increases with increasing temperature High temperature, particles have greater heat energy Particles move faster with greater kinetic energy Leading to more collisions between particles Increased probability of effective collision Reactions take place faster Speed of reaction doubles when the temperature rises by 10  C

18. Temperature Amt of product formed Time Higher temperature Lower temperature

19. Catalyst Presence of catalyst increases rate of reaction (Presence of inhibitors decreases rate of reaction) Catalysts lower activation energy of reactants Aids the formation of unstable intermediate products Increases probability of formation of products Faster rate of reaction

20. POTENTIAL ENERGY DIAGRAM REACTION COORDINATE POTETIALENERGYPOTETIALENERGY REACTANTS PRODUCTS A + B AB ACTIVATION ENERGY WITHOUT CATALYST LOWER ACTIVATON ENERGY PATH WITH CATALYST

21. Catalyst Energy Time/s Use of catalyst Absence of catalyst EaEa EaEa

22. Catalyst Amt of product formed Time/s Use of catalyst Absence of catalyst

23. Catalyst Definition: A substance which increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy but remains unchanged at the end of the reaction

24. Pressure Rate of reaction increases with increasing pressure Higher pressures, reacting particles are closer together Increasing concentration per unit volume Greater probability of an effective collision Faster rate of reaction

25. Pressure Amt of product formed Time/s Higher pressure Lower pressure

26. Particle Size Rate of reaction increases when particle size decreases Smaller particles has greater surface area than larger particles of the same mass Greater surface area for collision by another reacting particle Greater probability of an effective collision Faster rate of reaction

27. HOW PARTICLE SIZE AFFECTS CHEMICAL REACTION RATE!

28. Particle size Amt of product formed Time/s Smaller particle size Larger particle size

29. A reaction is fast, the time taken for the reaction is short. A reaction is slow, the time taken for the reaction is long. The rate of reaction depends to the speed of reaction. If a reaction is fast, its rate of reaction is high. If a reaction is slow, its rate of reaction is low. The rate of reaction is inversely proportional with time. Rate ά 1 time taken for reaction Rate of reaction = change in quantity of product / reactant time taken For gas product, rate of reaction = volume of gas time From a graph, the average rate of reaction = the gradient

30. HOW TO DETERMINE RATE OF REACTION FROM GRAPH AVERAGE RATE OF REACTION INSTANTANEOUS RATE OF REATION

31. AVERAGE RATE OF REACTION For the whole exp From X min to Y min On the X min (2 nd ) ( from 2 nd to 1 st ) For first x (3) min ( from 0 to 3rd )

32. INSTANTANEOUS RATE OF REACTION ( rate of reaction at that time) Draw tangent to the graph Y X Rate = Y/X

33. b) Example from the graph, determine: i) The rate of reaction at 120 s Instantaneous rate of reaction = Draw tangent to the graph

34. = 56 – 20 = 0.176 cm 3 s -1 222-18

35. Changes to the graph

36. Changes to the CURVE part of graph II I III Volume of gas Time Use positive catalyst Increase temperature Increase total surface area Use negative catalyst Decrease temperature Decrease TSA

37. Changes to the graph

38. d) Curve I represents the result of the experiment using excess zinc powder and 50cm 3 of 1.0 moldm -3 dilute hydrochloric acid II I III Volume of gas/ cm 3 Time/s Use positive catalyst Increase temperature of reactant Lower concentration of hydrochloric acid

39. (h) Time (seconds)0306090120150180210240 Burette reading (cm 3 )49.533.523.516.010.55.02.0 Volume CO 2 (cm 3 ) 0 162633.53944.547.5 Time(second)0306090120 Burette reading(cm 3 ) x 49.5 y 33.5 z 23.5 Total volume of gas(cm 3 ) x-x 0.00 x-y 16.00 x-z 26.00

40. Volume of CO 2, cm 3 Time, s Connect the point without using ruler! Not all the point is connected

41. Volume of CO 2 cm 3 Time s Cannot make this graph Straight line It’s must be smooth graph

42. Average Rate Of reaction The average rate of reaction in the first 90 seconds. = The total volume of gas released in the first 90 seconds Time taken (i) Time (seconds)0306090120150180210240 Burette reading (cm 3 )49.533.523.516.010.55.02.0 Volume CO 2 (cm 3 ) 0 162633.53944.547.5 = 33.5÷90=0.372 cm 3 s -1 unit

43. i(ii) Time (seconds)0306090120150180210240 Burette reading (cm 3 )49.533.523.516.010.55.02.0 Volume CO 2 (cm 3 ) 0 162633.53944.547.5 47.5÷180= 0.264 cm 3 s -1 The average rate of reaction in the whole experiment. = The total volume of gas released in the whole experiment Time taken =

44. Total volume of Hydrogen gas/cm 3 Time (second) Analysis of Data t p q Rate of reaction at t second = gradient AB = p/q cm 3 s -1 A B Tangent Cannot take directly at x Tangent is a line that touch just 1 point of graph in order to calculate gradient

45. Tangent Only touch 1 point of curve Cannot touch more than 2 points because each of point has a different gradient

46. tangent α

47. Total Volume of CO 2 (cm 3 ) Time (second) Analysis of data A B C D E F t1t1 t3t3 t2t2 Rate of reaction at t 1 = gradient AB Rate of reaction at t 2 = gradient CD Rate of reaction at t 3 = gradient EF Each of point has a different gradient!

48. Two methods to calculate tangent: Total volume of Hydrogen gas/cm 3 Time (second) A B Tangent number of small boxes × value of 1 small unit box Y X

49. Total volume of Hydrogen gas/cm 3 Time (second) A B Tangent Gradient of graph: m = ΔY ΔXΔX x1x2 y2 y1 m = Y2-y1 X2-x1 First Method

50. Total volume of Hydrogen gas/cm 3 Time (second) Analysis of Data t p q Rate of reaction at t second = gradient AB = p/q cm 3 s -1 A B Tangent

51. Total Volume of CO 2 (cm 3 ) Time (second) Analysis of data A B C D E F t1t1 t3t3 t2t2 Rate of reaction at t 1 = gradient AB Rate of reaction at t 2 = gradient CD Rate of reaction at t 3 = gradient EF

52. The Rate Law: Reactant Concentration and Rate The rate of a chemical reaction depends on several factors …. Relating Reactant Concentrations and Rate Consider the general reaction below. aA + bB cC + dD This reaction occurs at constant temperature

53. The reactant formulas are represented by A and B The stoichiometric coefficients are represented by a and b In this section we will look at reaction rates that are not affected by concentrations of products In general, the rate of a reaction increases when the concentrations of reactants increases. The dependence of of the rate of a reaction on the con- centration is given by: OR Rate=k[A] m

54. This relationship can be expressed in a general equation called the rate law equation For any reaction, the rate law equation expresses the relationship between the concentrations of the reactants and the rate of the equation.

55. The letter k represents the proportionality constant called the rate constant There is a different rate constant for each reaction at any given temperature The exponents m and n must be determined by exp- eriment. The do not necessarily correspond to the coefficients of their reactants They are usually 1 or 2, but values of 0, 3 even fractions can occur

56. HOW CONCENTRATION EFFECTS REACTION RATES (REACTION ORDER) REACTIONS WITH RATE EQUATIONS HAVING n = 0 ARE ZERO ORDER REACTIONS. THOSE WITH n = 1 ARE FIRST ORDER AND THOSE WITH n = 2 ARE SECOND ORDER. IN ZERO ORDER REACTIONS, CHANGING THE CONCENTRATION OF THE REACTANT HAS NO EFFECT ON THE RATE. IN FIRST ORDER REACTIONS, RATE CHANGES ONE FOR ONE WITH CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING CONCENTRATION DOUBLES THE RATE.

57. HOW CONCENTRATION EFFECTS REACTION RATES (REACTON ORDER) IN SECOND ORDER REACTIONS, RATE CHANGES RELATIVE TO THE SQUARE OF THE CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING THE CONCENTRATION OF THE REACTANT RESULTS IN THE RATE INCREASING 4 TIMES. THIS KNOWLEDGE OF HOW RATE CHANGES WITH CONCENTRATION DEPENDING ON THE ORDER LETS US FIND REACTION ORDERS BY AN EXPERIMENTAL PROCESS CALLED “METHOD OF INITIAL RATES” REACTION ORDERS MUST BE DETERMINED EXPERIMENTALLY. THEY CAN NEVER BE DETERMINED FROM THE CHEMICAL EQUATION.

58. RXN RATE? FORWARD OR REVERSE RXN? RXN RATE? FORWARD OR REVERSE? RATE = 0 RXN RATE? FORWARD OR REVERSE? REACTION RATES TIME MOLESAMOLESA MOLESAMOLESA MOLESAMOLESA MOLESAMOLESA A  B + C FORWARD RXN RATE = CONSTANT FORWARD RXN RATE = VARIABLE REVERSE RXN RATE = VARIABLE SLOPE OF A TANGENT LINE TO AN AMOUNT VS. TIME GRAPH = RATE GRAPH 1GRAPH 2 GRAPH 3GRAPH 4

59. HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES) USING THE METHOD OF INITIAL RATES REQUIRES THAT A REACTION BE RUN AT SERIES OF DIFFERENT STARTING CONCENTRATIONS AND THE RATE BE DETERMINED FOR EACH. GIVEN THE FOLLOWING DATA FOR THE REACTION A  B + C (TABLE 1) EXPT [A]RATE (M/SEC) 1 1 x 10 -3 4 x 10 -1 2 2 x 10 -3 8 x 10 -1 3 4 x 10 -3 16 x 10 -1 AS CONCENTRATION OF A DOUBLES, RATE DOUBLES. THE REACTION IS FIRST ORDER IN REACTANT A RATE = k[A] 1 OR RATE = k[A]

60. HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES) FOR THE REACTION: A + B  C + D (TABLE 2) 1 1 x 10 -3 1 x 10 –3 4 x 10 -1 2 2 x 10 -3 1 x 10 -3 8 x 10 -1 3 1 x 10 -3 2 x 10 -3 16 x 10 –1 USING EXPT 1 AND 2, [A] DOUBLES AND [B] IS CONSTANT. THE DOUBLING OF THE RATE IS THEREFORE CAUSED BY REACTANT A AND THE ORDER WITH RESPECT TO A IS FIRST. USING EXPT 1 AND 3, [A] IS CONSTANT AND [B] IS DOUBLED. THE FOUR TIMES RATE INCREASE IS THEREFORE CAUSED BY REACTANT B AND THE ORDER WITH RESPECT TO B IS SECOND. RATE = k[A] 1 [B] 2 OR RATE = k[A][B] 2

61. HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS) FROM INITIAL RATES DATA TABLE 1, RATE = k [A], k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 1 [A] = 1 x 10 –3 M, RATE = 4 x 10 –1 M/SEC 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) k = 1 x 10 2 SEC -1 OR 1 x 10 2 / SEC

62. HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS) FROM INITIAL RATES DATA TABLE 2, RATE = k [A] x [B] 2, k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 2 [A] = 1 x 10 –3 M, [B] = 1 x 10 –3 M RATE = 4 x 10 –1 M/SEC 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) x (1 x 10 –3 M ) 2 k = 4 x 10 5 M -1 SEC -1 OR 4 x 10 5 / M x SEC

63. Let us see how the initial rates method works 2N 2 O 3(g) 2NO 2(g) + O 2(g) The general rate law equation for this reaction is: Rate = k [N 2 O 3 ] m To determine the value of m, a chemist performs three experiments. A different initial concentration of [N 2 O3 5 ] 0 Is used for each experiment. The 0 represents t=0

64. ExperimentInitial [N 2 O 3 ] 0 (mol/L)Initial rate (mol/(L.s)) 10.0104.8 x 10 -6 20.0209.6 x 10 -6 30.0301.5 x 10 -5 Value of m can be determined with at least two different methods by inspection rate law equation

65. 1. By inspection  When the [N 2 O 5 ] is doubled expts 1 and 2 doubles  the rate also doubles  when [N 2 O 5 ] is tripled, (expts 1 and 3) the rate triples this indicates a first-order relationship as follows Rate = k [N 2 O 5 ] 1

66. 2. Compare rate law equation using ratios this method very useful when relation between conc and rate are not immediately obvious from data Write the rate expressions for expts 1 and 2 as follows: Rate 1 = k [0.010] m = 4.8 x 10 -6 mol/(L.s) Rate 2 = k [0.020] m = 9.6 x 10 -6 mol/(L.s)

67. Create a ratio to compare the two rates Rate 1 = k(0.010 mol/L) m = 4.8 x 10 -6 mol/L.s Rate 2 k(0.020 mol/L) m 9.6 x 10 -6 mol/L.s Since k is a constant at constant temp you can cancel out k(0.010 mol/L) m = 4.8 x 10 -6 mol/L.s k(0.020 mol/L) m 9.6 x 10 -6 mol/L.s (0.5) m = 0.5 m = 1 (by inspection)

68. Determining the Rate Constant Once you know the rate law equation for a reaction, you can calculate the rate constant using results from any of the experiments Rate = k [N 2 O 5 ] 1 You can use data from any of the three experiments to calculate k 4.8 x 10 -6 mol/(L.s) = k(0.010 mol/L)

69. k = 4.8 x 10 -6 mol/ (L.s) 0.010 mol/L = 4.8 x 10 -4 s -1

70. TEMPERATURE & REACTION RATE AT ANY TEMPERATURE THE MOLECULES IN A SYSTEM HAVE A DISTRIBUTION OF KINETIC ENERGIES (SIMILAR TO THE DISTRIBUTION OF THE SPEEDS OF CARS ON A HIGHWAY). AS THE TEMPERATURE INCREASES, THE AVERAGE KINETIC ENERGY OF THE MOLECULES INCREASE (THEY MOVE FASTER AT HIGHER TEMPERATURES) AND THEREFORE COLLIDE WITH EACHOTHER MORE FREQUENTLY AND HIT HARDER WHEN THEY DO COLLIDE. AS A RESULT, REACTION RATE INCREASES WITH TEMPERATURE.

71. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES LOW ENERGY MOLECULES HIGH ENERGY MOLECULES AVERGE ENERGY MOLECULES

72. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 TEMPERATURE = T 2 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES TEMP 2 > TEMP 1 AT HIGHER TEMPERATURES, MOLECULES HAVE HIGHER ENERGIES ON AVERAGE

73. TEMPERATURE & REACTION RATE IN ORDER TO REACT, MOLECULES MUST COLLIDE WITH SUFFICIENT ENERGY. THIS MININIUM ENERGY FOR REACTION IS CALLED ACTIVATION ENERGY. AS THE TEMPERATURE OF A SYSTEM IS INCREASED, THE NUMBER OF MOLECULES WITH THE NECESSARY ENERGY FOR REACTION (THE ACTIVATION ENERGY) INCREASES.

74. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 TEMPERATURE = T 2 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES ACTIVATION ENERGY MOLECULES WITH SUFFICIENT ENERGY TO REACT AT T 1 AS TEMPERATURE, REACTION RATE MOLECULES WITH SUFFICIENT ENERGY TO REACT AT T 2

75. TEMPERATURE & REACTION RATE THE ENERGY CHARACTERISTICS OF A CHEMICAL REACTION CAN BE SHOWN ON A POTENTIAL ENERGY DIAGRAM. THIS GRAPH SHOWS THE ENERGY STATE OF THE SYSTEM AS REACTANTS PROCEED THROUGH THE ACTIVATED COMPLEX TO FORM THE PRODUCTS. THE ACTIVATED COMPLEX IS THE INTERMEDIATE STATE (MOLECULAR FORM) WHICH REACTANTS GO THROUGH AS THEY CONVERT INTO THE PRODUCTS. THE ACTIVATION ENERGY IS THE ENERGY REQUIRED TO FORM THE INTERMEDIATE ACTIVATED COMPLEX MOLECULE.

76. POTENTIAL ENERGY DIAGRAM REACTION COORDINATE POTETIALENERGYPOTETIALENERGY REACTANTS PRODUCTS ACTIVATED COMPLEX A + B AB  H ACTIVATION ENERGY EXOTHERMIC  H = (-) ENERGY IS RELEASED

77. The Effect of Surface Area on Reaction Rate

78. Which will cook faster? One 60g whole potato… …or 60g of small potato pieces?

79. The small potato pieces…

80. Why do they cook faster? Because if you compare a 60g potato with 60g of small potato pieces, the small potato pieces have a LARGER SURFACE AREA

81. So what? The larger the surface area, the more collisions that can happen between the boiling water and the potato at the same time This means the potato cooks faster

82. SURFACE AREA & REACTION RATE REACTANT THAT IS INTERIOR CANNOT BE ATTACKED UNTIL THE EXTERIOR REACTANT IS CONSUMED. THE REACTION RATE IS SLOW!

83. SURFACE AREA & REACTION RATE WHEN SURFACE AREA IS INCREASED MORE REACTANTS ARE EXPOSED TO EACHOTHER SIMULTANEOUSLY AND THE REACTION IS RAPID.

84. REACTION MECHANISMS ALL REACTIONS, NO MATTER HOW SIMPLE FOLLOW A PARTICULAR REACTION PATHWAY CALLED A MECHANISM. IT IS A SERIES OF STEPS WHICH LEAD TO THE FORMATION OF PRODUCTS. DURING THE PROCESS OFTEN MOLECULES CALLED INTERMEDIATES ARE FORMED AND SUBSEQUENTLY CONSUMED. THE SLOWEST STEP IN THE SERIES OF STEPS THAT MAKE UP THE MECHANISM IS CALLED THE RATE DETERMINING STEP. THE SPEED OF THE RATE DETERMINING STEP DEPENDS ON THE COMPLEXITY OF THE STEP (NUMBER OF MOLECULES INVOLVED CALLED THE MOLECULARITY) AND THE BOND STRENGTHS OF THE REACTING COMPONENTS

85. NO NO 2 O2O2 REACTION MECHANISM FOR; 2 NO + O 2  2 NO 2 STEP I 2 NO  N 2 O 2 (fast) STEP II N 2 O 2 + O 2  2 NO 2 (slow) N 2 O 2 INTERMEDIATE

86. REACTION MECHANISMS UNLIKE OVERALL REACTIONS, THE REACTON ORDERS FOR THE STEPS IN A MECHANISM ARE BASED ON THE NUMBER OF MOLECULES REACTING IN THAT STEP. IN THE REACTION MECHANISM FOR: 2 NO + O 2  2 NO 2 STEP I 2 NO  N 2 O 2 (fast) STEP II N 2 O 2 + O 2  2 NO 2 (slow) SINCE THERE IS ONLY ONE N 2 O 2 AND ONE O 2 IN THE RATE DETERMING STEP (THE SLOW STEP), THE RATE EQUATION FOR THE REACTION IS: RATE = k [N 2 O 2 ] x [O 2 ] THE REACTION IS FIRST ORDER IN BOTH REACTANTS