Chemical Reactions: Kinetics

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Lecture materials for the Introductory Chemistry course for Forensic Scientists, University of Lincoln, UK. See http://forensicchemistry.lincoln.ac.uk/ for more details.

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Chemical Reactions: Kinetics

  1. 1. Rate of Reaction University of Lincoln presentation This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  2. 2. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  3. 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? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  4. 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 q m ΔH 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  5. 5. How do ignitable liquids burn? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  6. 6. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  7. 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) 67.2 MJ 67.2x106 J time(s) 840000 s 80 Js 1 80 Js 1 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  8. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  9. 9. Factors affecting the rate of a chemical reaction 1. Concentration (hydrogen peroxide demo) 2. Pressure (gases) 3. Temperature (glowstick) 4. Surface area (dust explosion) 5. Catalysis This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  10. 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 A B C D ΔA ΔB ΔC ΔD Rate Δt Δt Δt Δt This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  11. 11. H2O2(l) H2O(l) + ½O2(g) 200 Amount of H 2O2 remaining (x105mol) 150 100 50 0 0 50 100 150 Time (s) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  12. 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? 0.4 Concentration of H2O2 (mol dm-3) 0.35 Concentration = 0.3 mol dm-3 s-1 0.3 Rate = gradient = 0.0068 mol dm-3 s-1 0.25 0.2 Concentration = 0.1 mol dm-3 s-1 0.15 Rate = gradient = 0.0023 mol dm-3 s-1 0.1 0.05 0 0 50 100 150 Time (s) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  13. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  14. 14. Rate plot for the decomposition of hydrogen peroxide 0.008 Rate of Reaction (mol dm -3 s-1) 0.007 0.006 0.005 0.004 0.003 0.002 0.001 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 -3 Hydrogen Peroxide concentration (mol dm ) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  15. 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 1 1 Rateα H2O2 Rate k H2O2 − 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. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  16. 16. Rate of reaction can be measured from the rate that oxygen gas is produced. Inverted burette 50 40 30 Yeast suspension 20 +hydrogen peroxide solution 10 Water 0 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  17. 17. Vary the starting concentration and measure the initial rate 30 [H2O2]= 0.40 mol dm-3 25 Volume O 2/cm2 20 [H2O2]= 0.32 mol dm-3 [H2O2]= 0.24 mol dm-3 15 [H2O2]= 0.16 mol dm-3 10 [H2O2]= 0.08 mol dm-3 5 0 0 50 100 150 200 Time (s) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  18. 18. Initial Rate can be measured 30 Initial 25 gradient 0.51cm3s-1 Volume O2/cm-3 20 15 10 5 0 0 50 100 150 200 Time (s) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  19. 19. Plot Initial Rate as a function of starting concentration 0.5 Rate of reaction (cm2(O2)s-1) 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 -3 Conc of Hydrogen peroxide (mol dm ) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  20. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  21. 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 0.32 0.41 0.4 0.51 Determine the order of reaction with respect to hydrogen peroxide and calculate the value of the rate constant. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  22. 22. General Rate equations n Rate k A 0 Zero order Rate k A k Units of k ? First order 1 Units of k ? Rate k A kA Second order 2 Units of k ? Rate k A This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  23. 23. Further Analysis of data • Logarithms can be very useful • Plot of log rate as a function of log concentration (p439 Housecroft) n Rate k A n n logRate logk A logk log A logk nlog A 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  24. 24. Half-life Time taken for the concentration of reactant A at time t, [A]t to fall to half its value. At 1 ln kt ln kt A0 2 0.693 kt 0.693 t k A constant half-life for a first order reaction Progress curve and measure t½ at several different points. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  25. 25. Constant half-life 200 Amount of H 2O2 remaining (x10 5mol) Going from 200 x 10-5 150 mol to 100 x 10-5 mol takes 27s 100 Going from 100 x 10-5 mol to 50 x 10-5 mol takes 27s 50 Going from 50 x 10-5 mol 27s 27s 26s to 25 x 10-5 mol takes 26s 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 Time (s) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  26. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  27. 27. Reactions with more than one reactant A + B → products n m Rate k A B e.g. C12H22O11 + H2O → C6H12O6 + C6H12O6 sucrose glucose fructose Rate k C12H22 O11 H2O First order with respect to each reactant Second order reaction (sum of orders in rate equation) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  28. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  29. 29. An Example Reaction peroxodisulfate (VI) and iodide ions 2 2 S2O8 2I 2SO4 I2 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. 2 1 1 Rate k S2O8 I 3. k is determined by rearranging the rate equation. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  30. 30. Iodine clock data from experiment This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  31. 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? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  32. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  33. 33. Only those molecules with sufficient energy can Number of molecules with kinetic react Activation enthalpy Ea =50kJ mol-1 energy E Kinetic energy (E) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  34. 34. Increasing Temperature increases Rate of Reaction Number of molecules 300 K with energy greater Number of molecules with kinetic than 50kJ mol-1 at 300 K 310 K Number of molecules energy E with energy greater than 50kJ mol-1 at 310 K Activation enthalpy Ea =50kJ mol-1 Kinetic energy (E) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  35. 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  36. 36. Combining Activation Energy and enthalpy Both can be shown on an Exothermic reaction enthalpy level diagram enthalpy Positive Add Ea to the A+B diagram ΔH Negative enthalpy C+D Draw a diagram for an endothermic reaction Reaction coordinate This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  37. 37. Rate equations and Temperature n m Rate k A B Ea The Arrhenius equation k Ae RT 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? EA It might be easier to do this ln k ln A RT increase temperature increase k increase rate decrease Ea increase k increase rate This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  38. 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 54 8.7 x 106 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? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  39. 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 Ea EA k Ae RT lnk lnA RT Plot lnk against 1/T: gradient = -EA/R; intercept = lnA This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  40. 40. Calculating Ea 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  41. 41. How do we explain catalysis? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  42. 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? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  43. 43. A catalyst provides an alternative path for the reaction with a lower activation enthalpy Uncatalysed Activation enthalpy of reaction uncatalysed reaction Enthalpy Catalysed Activation enthalpy of reaction catalysed reaction Reactants Products Progress of reaction This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  44. 44. Acknowledgements • JISC • HEA • Centre for Educational Research and Development • School of natural and applied sciences • School of Journalism • SirenFM • http://tango.freedesktop.org This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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