1. Section 1 Controlling the rate Collision Theory

2. LI To learn about the collision theory (a) S.C. By the end of this lesson you should be able to Using the “Collision Theory” explain the effects of concentration, particle size, temperature, and collision geometry on reaction rates. Calculate the average reaction of a reaction. State that where there is a fixed endpoint in a reaction rate = 1/ t (s -1 )

3. (A) Collision Theory The collision theory states that for a chemical reaction to occur, reactant molecules must (i) collide with enough energy (ii) collide in the correct orientation.

4. Increasing the surface area or decreasing the particle size- the smaller the particles, the greater the surface area, the greater the chance of successful collisions and so increases the rate. Effect of Surface Area Workbook activity

5. Calculating the average rate of reaction The rate of a chemical reaction may be expressed in terms of the changes in concentration or volume or mass over a period of time. This is called the average rate of reaction. Experiment Method A - change in volume Method B – change in mass   in quantity in time Average reaction rate =

6. What to do You are going to follow the rate of the reaction by: A Measuring the volume of gases produced over time B Measuring the loss of mass over time

7. Measuring rate of reaction Two common ways: 1) Measure how fast the products are formed 2) Measure how fast the reactants are used up

8. How can we follow the reaction? If we use a container fitted with a delivery tube we could measure the amount of gas produced. How?

9. What to do - Method A Measure 25 cm 3 of 2 mol l -1 HCl into a conical flask fitted with a stopper and a delivery tube Set up an inverted measuring cylinder of water to collect the gas Add 2g marble chips to the acid Measure the volume of gas every 10 seconds Repeat with 2g crushed marble chips

10. What to do – Method A Record your results in the table. Plot a graph of volume vs time using the same axes for both sets of data rate = change in volume ( the unit is cm 3 s -1 ) time interval Calculate the rate for the 1 st and 2 nd 25 seconds for each set of results

11. What to do –Method B Weigh out 2 g marble chips Measure 25 cm 3 1 mol l -1 HCl into a conical flask Place on balance and zero it. Now add the marble chips to the acid and take mass readings every 10 seconds Repeat using crushed chips

12. What to do – Method B Record your results in a the table. Plot a graph of volume vs time using the same axes for both sets of data rate = change in mass ( the unit is gs -1 ) time interval Calculate the rate for the 1 st and 2 nd 25 seconds for each set of results Swap results

13. time (s)01020304050607080 volume (cm 3 ) C volume (cm 3 ) P Method A results Plot the results on a graph with time on the x axis and volume on the y. Use the same set of axes for both sets of results. 0 14 26 36 44 50 50 50 50 0 22 40 48 50 50 50 50 50 Sample results

14. Rate over 1 st 25 seconds (cm 3 s -1 ) rate over 2 nd 25seconds (cm 3 s -1 ) Whole chips (C) 32-0 25-0 =1.3 50-32 50-25 =0.72 Ground chips (G) 45-0 25-0 =1.8 50-45 50-25 =0.2 Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula rate = change in volume = _____________ cm 3 s -1 time interval Time (s) Volume of gas cm 3

15. Method B results Plot the results on a graph with time on the x axis and mass on the y. Use the same set of axes for both sets of results. time (s)020406080100120140160 Mass (g) C mass(g) G 2.0 1.4 1.0 0.7 0.5 0.35 0.3 0.25 0.25 2.0 0.7 0.4 0.3 0.25 0.25 0.25 0.25 0.25 Sample results

16. Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula rate = change in mass The answer will have the units g s -1 time interval Rate over 1 st 25 seconds (g s -1 ) rate over 2 nd 25seconds (g s -1 ) Whole chips (C) 0.8-2 25-0 =0.05 0.35 -0.8 50-25 =0.018 Ground chips (G) 0.3-2 25-0 =0.068 0.25-0.3 50-25 =1x10 -3 Loss in mass (g) Time (s)

17. Increasing concentration- increasing the number of particles increases the number of collisions and so increases the rate. Effect of Concentration 2 mol/l 5 mol/l Workbook activity

18. Using the graph below calculate the average rate between: (i)0-20 s(ii) 10-20 s (iii) 40-50 s Average rate = change in concentration change in time The answer will have the units moll -1 s -1 Workbook activity

19. You will carry out the reaction using a series of dilutions of the potassium iodide solution. This will be diluted by replacing some of the volume with water. Effect of concentration – the chemical clock challenge Your aim is to find out how changing the concentration of potassium iodide affects the rate of reaction.

20. In the earlier experiments you measured the average rate of reaction over a period of time. Sometimes it is easier to make comparisons by calculating the relative rate of a reaction using the formula Relative Rate = 1/ t (s -1 ) time = 1/ rate (s) This allows a comparison of the rate under different conditions to be compared. Effect of concentration –the chemical clock challenge Experiment

21. 2I - (aq) + H 2 O 2 (aq) + 2H + (aq)  2H 2 O (l) + I 2 (aq) The reaction mixture stays colourless as the iodine molecules are converted back to iodide molecules by the thiosulphate ions. Once all the thiosulphate ions have been used, a blue black colour appears suddenly as iodine reacts with starch. t being a measure of how long it takes for the blue/black colour to form. (when excess I 2 forms) Relative Rate = 1 t Units s -1 + 2S 2 O 3 2- (aa)  2I - (aq) + S 4 O 6 2- (ag) I 2 (aq) Effect of concentration –the chemical clock challenge

22. 1) Using syringes measure out 10cm 3 sulphuric acid 0.1moll -1 10cm 3 sodium thiosulphate 0.005moll -1 1cm 3 starch solution 25cm 3 potassium iodide solution 0.1mol l -1 Into a dry 100cm 3 beaker 2) Measure out 5cm 3 of hydrogen peroxide 0.1moll -1 into a syringe. Add it to the mixture as quickly as possible and start the timer. 3) Stop the clock when the mixture suddenly turns dark blue. 4) Repeat, using 20 cm 3 of potassium iodide solution and 5cm 3 of water. Effect of concentration –the chemical clock challenge

23. Volume of water (cm 3 ) Volume of 0.5 mol l -1 KI (aq) (cm 3 ) Relative conc KI Time (s) Rate (1/t) (s -1 ) 0.025.01 5.020.00.8 10.015.00.6 15.010.00.4 20.05.00.2 Effect of concentration –the chemical clock challenge RESULTS - Plot a graph showing the relative concentration of potassium iodide x axis and the rate of reaction (1/t) on the y axis. Workbook activity

24. Your challenge is to create a series of solutions that will change colour in time to music http://www.youtube.com/watch?v=rSAa iYKF0cshttp://www.youtube.com/watch?v=rSAa iYKF0cs(Daniel Radcliffe) Effect of concentration –the chemical clock challenge

25. Listen to the song and identify points where you want to have a colour change come in Time them accurately. Allocate times to each group. Look at your results and check that these are times you can achieve Calculate the rate that each time requires (rate = 1/t) Read off the required concentration from your graph Effect of concentration –the chemical clock challenge

26. Use the relative concentration to help you work out the volume of water and KI(aq) needed to make up 100 cm 3 of the required concentration. Good luck! Effect of concentration –the chemical clock challenge

27. How does changing the temperature affect the rate of reaction between oxalic acid and potassium permanganate? 5(COOH) 2 (aq) + 6H + (aq) + 2MnO 4 2- (aq)  2Mn 2+ (aq) + 10 CO 2 (aq) 8H 2 O (l) What colour change takes place? purple to colourless This reaction is self indicating. No indicator is needed. Effect of Temperature

28. Temperature ( o C) Time (s)Relative rate 1/t (s -1 ) 40 50 60 70 Plot a graph of 1/time on the vertical (y) axis and average temperature on the horizontal (x) axis. Work out the rise in temperature required to double the rate of reaction. Workbook activity

29. LI To learn about reaction profiles (b) S.C. By the end of this lesson you should be able to Describe the term enthalpy Explain the terms activated complex and activation energy Use potential energy diagrams to identify whether a reaction is exothermic or endothermic Use potential energy diagrams to calculate the enthalpy change for a reaction Use potential energy diagrams to calculate the activation energy for a reaction Explain why it is essential for chemists to predict the quantity of heat absorbed or released in an industrial process

30. (B) Reaction Profiles Every substance contains stored energy known as enthalpy (H). Reaction profiles (potential energy diagrams) can be used to show the energy pathway for a chemical reaction. The profile below shows:

31. A – activation energy for the forward reaction B – activation energy for the reverse reaction C – enthalpy change

32. ENTHALPY  H = H products- H reactants Has the symbol H The  H can be positive or negative The units for  H kJmol -1

33. Activated Complex reactants activated complex products A highly energetic and unstable arrangement of atoms formed between reactants and products. It is an intermediate in the reaction and only exists for a short period of time.

34. Exothermic Reactions ΔH = ΔH is always negative for exothermic reactions. Exothermic reactions include combustion, neutralisation and respiration. Workbook activity

35. Endothermic Reactions ΔH = ΔH is always positive. Endothermic reactions include photosynthesis which takes in energy in the form of light. Workbook activity

36. ReactionTemp before mixing/ o C Temp after mixing/ o C Endothermic or exothermic 10cm 3 NaOH + 10cm 3 HCl 10cm 3 NaHCO 3 + 4 spatulas citric acid 10cm 3 CuSO 4 + spatula of Zn powder 10cm 3 H 2 SO 4 + Mg ribbon Workbook activity

37. For industrial processes it is essential that chemists can predict the quantity of heat taken in or given out as this will influence the design of the process. Workbook questions Runaway reactions such as those causing the disasters in Bhopal and Seveso occur when the rate at which a chemical reaction releases energy exceeds the capabilities of the plant to remove heat. http://news.bbc.co.uk/1/hi/world/south_asia/8392093.stm

38. ACTIVATION ENERGY AND THE REACTION PATHWAY Activation energy (E A ) The activation energy (E A ) is the minimum kinetic energy required by colliding molecules for a reaction to occur. The activation energy in the above graph is 40 kJmol -1. (Start to peak) Workbook activity

39. COLLISION GEOMETRY Favourable geometry Unfavourable geometry

40. LI To learn about catalysts (c) S.C. By the end of this lesson you should be able to name the two different types of catalysts describe how catalysts affect the activation energy of a chemical reaction draw potential energy diagrams to show the effect of a catalyst on the activation energy

41. (C) Catalysts Heterogeneous - example Homogeneous - example Enzymes are biological catalysts, and are protein molecules that work by homogeneous catalysis. When the catalyst and reactants are in different states When the catalyst and reactants are in the same state Workbook activity

42. How heterogeneous catalysts work A catalyst increases the rate of reaction and can be recovered at the end. The catalytic mechanism involves the reactant particles being adsorbed onto the surface of the catalyst. Scholar animation

43. For an explanation of what happens click on the numbers in turn, starting with  How a heterogenous catalyst works

44. Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. How a heterogenous catalyst works

45. Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. Reaction (STEPS 2 and 3) Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur. This increases the chances of favourable collisions taking place. How a heterogenous catalyst works

46. Des orption (STEP 4) There is a re-arrangement of electrons and the products are then released from the active sites Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. Reaction (STEPS 2 and 3) Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur. This increases the chances of favourable collisions taking place. How a heterogenous catalyst works

47. Catalysts at Work When the surface activity of a catalyst has been reduced, i.e. poisoned, the catalyst will stop working. However, catalysts can be regenerated. e.g. burning the carbon (soot) off the catalyst used during cracking in the petrochemical industry. Practical on homogeneous/heterogeneous catalysis Workbook questions

48. Potential energy graphs and catalysts Catalysts lower the activation energy needed for a successful collision. P.E. Reaction path 1 3 2 Activation energy E A without a catalyst Activation energy E A with a catalyst A catalyst does not affect the potential energies of reactants and products ie. start point and end point the same.

49. Potential energy graphs and catalysts Workbook activity

50. LI To learn about Temperature and kinetic energy (d) S.C. By the end of this lesson you should be able to give a definition for temperature use energy distribution diagrams to explain the effect of temperature and catalyst on reaction rate

51. (D) Temperature and Kinetic Energy Workbook activity distribution of the kinetic energy of particles

52. No of collisions with a given K.E. Kinetic energy EAEA Shaded area represents the no.of successful collisions - those with K.E greater than the E A Temperature is a measure of the average kinetic energy of the particles in a chemical. Workbook activity

53. No of molecules Kinetic energy As the temperature increases the particles gain kinetic energy, so more particles have the required E A so there are more successful collision. index EAEA T2T2 T1T1 Workbook activity

54. No. of collisions with K.E. more than activation energy - successful collisions Catalysed reaction E A is reduced - shaded area increases so no. of successful collisions increases. No of collisions with a given K.E. Kinetic energy EAEA EAEA Un-catalysed reaction Catalysts lower the E A Workbook activity

55. Temperature and energy