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 → N 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) over time decreases
2. [HCl] over time decreases
3. [CuCl2] over time increases
4. Volume of CO2 over time increases
5. Mass of a open beaker over time decreases
6. Pressure of a closed beaker over time increases
7. Colour of the blue Cu2+ over time increases

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 minutes
with 2.00 M HCl. If the rate of consumption
of HCl is g/min, calculate the amount
of Al remaining.

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

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

15. N H2 → 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. N H2 → 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 g

17. N H2 → 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 g mole NH3

18. N H2 → 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 g mole NH mole

19. N H2 → 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 = g
min g mole NH mole min

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?
N H2 → 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?

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?

37. What happens to the KE, PE, and total energy as the activated complex turns into 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. Rate Ea What is the relationship between Ea and rate? Draw a graph.:
Inverse Ea
Rate

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