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Kinetics Lesson 7 Catalysts & Review

Kinetics Lesson 7 Catalysts & Review. Commercial Catalysts H 2 SO 4 Many Organic Reactions Pt Surface Catalyst for Diatomic Gases Pd Catalytic Converter Automobile Convert CO → CO 2 Rh Catalytic Converter Automobile Convert NO 2 → N 2 + O 2 Biological Catalysts

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Kinetics Lesson 7 Catalysts & Review

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  1. Kinetics Lesson 7 Catalysts & Review

  2. Commercial Catalysts H2SO4 Many Organic Reactions Pt Surface Catalyst for Diatomic Gases Pd Catalytic Converter Automobile Convert CO → CO2 Rh Catalytic Converter Automobile Convert NO2 → N2 + O2 Biological Catalysts Enzymes Biological Catalysts Amylase pepsin

  3. CuCO3(s) + 2HCl(aq) CuCl2(aq) + CO2(g) + H2O(l) blue Describe seven different ways to monitor the rate of the above reaction. State how each property would change as the reaction proceeds. 1. Mass of CaCO3(s) overtime decreases 2. [HCl] overtime decreases 3. [CuCl2] overtime increases 4. Volume of CO2 over timeincreases 5. Mass of a open beaker over timedecreases 6. Pressure of a closed beaker over timeincreases 7. Colour of the blue Cu2+ over timeincreases

  4. CuCO3(s) + 2HCl(aq) CuCl2(aq) + CO2(g) + H2O(l) blue Describe five different ways to increase the rate of the above reaction. 1. Increase the temperature 2. Increase [HCl] 3. Add a catalyst 4. Increase the surface area of CuCO3(s) 5. Agitate We can't change the nature of the reactant because then we wouldn't have the same reaction. Replacing HCl with H2SO4 would be faster but a different reaction.

  5. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170

  6. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min

  7. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl 1 min

  8. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole 1 min 36.5 g

  9. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole x 2 mole Al 1 min 36.5 g 6 mole HCl

  10. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1 min 36.5 g 6 mole HCl1 mole

  11. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl1 mole

  12. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl1 mole Al remaining = 2.00 g - 1.48 g = 0.52 g

  13. 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) 2.00 g of Al is placed in a beaker  and allowed to react for 12.00 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170 12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl1 mole Al remaining = 2.00 g - 1.48 g = 0.52 g Beware subtraction- loss of 1 sig fig!

  14. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.

  15. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min. 2.0 g NH3 min

  16. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min. 2.0 g NH3 x 1 mole min 17.0 g

  17. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min. 2.0 g NH3 x 1 mole x 3 mole H2 min 17.0 g 2 mole NH3

  18. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min. 2.0 g NH3 x 1 mole x 3 mole H2 x 2.0 g min 17.0 g 2 mole NH3 1 mole

  19. N2 + 3H2→ 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min. 2.0 g NH3 x 1 mole x 3 mole H2 x 2.0 g = 0.35 g min 17.0 g 2 mole NH3 1 molemin

  20. Label the Ea forward, Ea reverse, Activated complex, and ∆H. Energy of the Activated complex Ea for Ea rev ∆H

  21. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Activated complex reactants

  22. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Ea= PE activated complex - PE reactants Activated complex reactants

  23. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.

  24. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.

  25. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Ea=

  26. Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Ea= PE activated complex - PE reactants Activated complex reactants

  27. Why does this reaction have more than one step? N2 + 3H2→ 2NH3 More than three reactant particles.

  28. A colllision that is not successful does not have: Favourable geometry Sufficient energy

  29. Define activated Complex Unstable Reaction Intermediate High PE Low KE Bonds forming and breaking

  30. Define Activation Energy The minimum amount of energy required for a successful collision.

  31. Why does gasoline burn faster than wax? Nature of the reactant.

  32. What happens to the KE, PE, and total energy as the activated complex forms?

  33. What happens to the KE, PE, and total energy as the activated complex forms?

  34. What happens to the KE, PE, and total energy as the activated complex forms? PE

  35. What happens to the KE and PE as the activated complex turns into products?

  36. What happens to the KE, PE, and total energy as the activated complex turns into products? Activated complex products

  37. What happens to the KE, PE, and total energy as the activated complex turns into products? Activated complex PE products

  38. What happens to the KE, PE, and total energy as the activated complex turns into products? The PE decreases, the KE increases, and the total energy is constant. Activated complex PE products

  39. If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ?

  40. If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ? 125 kJ 250 kJ

  41. If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ? 125 kJ 250 kJ ΔH = -125 kJ

  42. What is the relationship between Ea and rate? Draw a graph.:

  43. What is the relationship between Ea and rate? Draw a graph.: Inverse

  44. What is the relationship between Ea and rate? Draw a graph.: Inverse Rate Ea

  45. According to the collision theory, how does increasing temperature increase the rate More collisions and harder collisions

  46. According to the collision theory, how does increasing concentration increase the rate More collisions

  47. According to the collision theory, how does a catalyst increase the rate Lowers the Ea and allows low energy collisions to be successful.

  48. What is the formula of the activated complex in the reaction below: CH3OH + HCl → CH3Cl + H2O CH5OCl

  49. Why doesn't an unlit candle burn according to the collision theory? The Ea is too high. A match lights to candle The match provides the Ea The candle continues to burn. Exothermic reaction

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