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Chemical Kinetics SCH4U: Grade 12 Chemistry. Unit Mind Map.

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Presentation on theme: "Chemical Kinetics SCH4U: Grade 12 Chemistry. Unit Mind Map."— Presentation transcript:

1 Chemical Kinetics SCH4U: Grade 12 Chemistry

2 Unit Mind Map

3 Chemical Kinetics Chemical kinetics? Why do we study this? Who makes use of chemical kinetics?

4 Chemical Kinetics Chemical kinetics  studying how we can make reactions go faster or slower Why do we study this? Who makes use of chemical kinetics?

5 Chemical Kinetics Chemical kinetics  studying how we can make reactions go faster or slower Why should we study this? Economics effects  Materials are expensive Developing pharmaceutical drugs Who makes use of chemical kinetics?

6 Chemical Kinetics Chemical kinetics  studying how we can make reactions go faster or slower Why should we study this? Economics effects  Materials are expensive Developing pharmaceutical drugs Who makes use of chemical kinetics? Biologists  metabolic rxns, food digestion, bone regeneration Automobile engineers  rate of rusting, decrease pollutants Agriculture  slow down food ripening

7 Reaction Rates What is a chemical reaction rate? Possible formula for measuring rate?

8 Reaction Rates What is a chemical reaction rate? measure of how fast reactants are used up or how fast products are produced Possible formula for measuring rate?

9 Reaction Rates What is the reaction rate? measure of how fast reactants are used up or how fast products are produced Possible formula for measuring rate? Rate = Δ Concentration/Δ Time

10 Sample problem N 2 (g) + 3H 2 (g)  2NH 3 (g) What is the average rate of production of ammonia for the system if the concentration is 3.5mol/L after 1.0min and 6.2mol/L after 4.0 min?

11 Sample problem N 2 (g) + 3H 2 (g)  2NH 3 (g) What is the average rate of production of ammonia for the system if the concentration is 3.5mol/L after 1.0min and 6.2mol/L after 4.0 min? ANSWER: 0.9mol/(L*min)

12 Measuring reaction rates Graphically Average rate of reaction  take the slope of the secant of the line Instantaneous rate of reaction  take the slope of the tangent of the line

13 Measuring reaction rates Graphically Average rate of reaction  take the slope of the secant of the line 1.Draw a secant line between two points 2.Calculate the slope Slope = Rise/Run = Δconcentration/ Δ time

14 Measuring reaction rates Graphically Instantaneous rate of reaction  take the slope of the tangent of the line 1.Draw a tangent line to the graph 2.Calculate the slope of the tangent line Slope = Rise/Run = Δconcentration/ Δ time

15 Measuring reaction rates What are some factors that we can measure experimentally to determine the rate of reaction ?

16 Measuring reaction rates Production of a gas Production of ions Changes in colour

17 Reaction rates and stoichiometry CO (g) + NO 2(g)  CO 2(g) + NO (g) In this reaction, the ratio of CO to NO is 1:1 Therefore, the disappearance of CO is the same as the production of NO Rate = - Δ [CO]/Δt = + Δ[NO]/Δt

18 Reaction rates and stoichiometry Suppose the ratio is NOT 1:1? Example, H 2 (g) + I 2 (g)  2HI (g) 2 mols of HI are produced for every 1 mol H 2 used Rate = The rate at which H 2 is used up is only half of which HI is produced

19 Sample Problem IO 3 - (aq) + 5I - (aq) + 6H + (aq)  3I 2 (aq) + 3H 2 O (l) The rate of consumption of iodate ions (IO 3 - ) is determined experimentally to be 3.0 x mol/(L*s). What are the rates of reaction or the other reactants and products in this reaction? Complete p.364 #3, 4 Homework: practice problem package

20 Collision Theory Why do reactions occur the way that they do? VS

21 Collision Theory Collision theory  reactions can only occur if: 1.There is a collision between molecules

22 Collision Theory Collision theory  reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way

23 Collision Theory Collision theory  reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way 3. enough energy is provided to break the chemical bonds that hold molecules together (Activation energy)

24 Activation Energy Activation Energy  minimum potential energy the system needs to overcome for the molecules to react

25 Factors That Affect Rate of Reaction What are some of the factors that may speed up or slow down a chemical reaction? P.392

26 Factors That Affect Rate of Reaction What are some of the factors that may speed up or slow down a chemical reaction? Chemical nature of the reactant Concentration of reactants Surface area of reactants Temperature Catalysts Get into groups, brainstorm and then explain these factors to the rest of class in a creative way, making use of collision theory

27 Lesson 2 Review collision theory Rate Laws Group Practice problems Individual practice problems homework

28 Unit Mind Map

29 Collision theory

30 Collision Theory Collision theory  reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way 3. enough energy is provided to break the chemical bonds that hold molecules together (Activation energy)

31 Factors That Affect Rate of Reaction

32 Chemical nature of the reactant Concentration of reactants Surface area of reactants Temperature Catalysts

33 The Rate Law Mathematical relationship between reaction rate and factors that affect it Determined empirically (experimentally) Rate = k[X] m [Y] n e.g. 2NO 2 + F 2  2NO 2 F The above reaction is 1 st order with respect to NO 2 and 2 nd order with respect to F 2. What is the rate law equation?

34 The Rate Law Answer: e.g. 2NO 2 + F 2  2NO 2 F rate = k[NO 2 ] 1 [F 2 ] 2

35 Steps to solve rate law problems eg. 2BrO 3 - (aq) + 5HSO 3 - (aq)  Br 2 (g) + 5SO 4 2- (aq) + H 2 O (l) + 3H + (aq) Rate = k [BrO 3 - ] m [HSO 3 - ] n

36 Steps to solve rate law problems eg. 2BrO 3 - (aq) + 5HSO 3 - (aq)  Br 2 (g) + 5SO 4 2- (aq) + H 2 O (l) + 3H + (aq) 1.Write out rate equation: rate = k [BrO 3 - ] m [HSO 3 - ] n 2.pick a trial where [BrO 3 - ] changes but [HSO 3 - ] stays constant= trial 1 and trial 2 3.Using a ratio, determine the relationship between change in concentration and change in rate = (trial 1/trial2) = (4.0/2.0) = (1.6/0.8) = 2 4. Using the following chart, determine the rate order of [BrO 3 - ]

37 Steps to solve rate law problems 5. If the concentration is doubled (2.0 to 4.0), the rate also doubles (0.80 to 1.60), therefore the rate order is 1 m=1

38 Steps to solve rate law problems 6. To find the rate order of the other reactant, follow the same steps but pick a trial where [HSO 3 - ] changes but [BrO 3 - ] stays constant= trial 2 and trial 3 = (6.0/3.0) = (0.8/0.2) = 2 = 4 7. Use the following table to determine the rate order

39 Steps to solve rate law problems 8. As the concentration doubles the rate was multiplied by 4, therefore rate order of HSO 3 - = 2 (n=2) 9. Plug values into rate equation rate = k [BrO 3 - ] m [HSO 3 - ] n and solve for k.

40 Example 1 From the data collected above, determine the rate law of the following equation: aA + bB  products Rate = k[A] m [B] n

41 Example 1 aA + bB  products Rate = k[A] m [B] n m = 0; n = 1; k = 0.6s -1 Rate = 0.6s -1 [B] 1

42 Example 2 From the data collected above, determine the rate law of the following equation: 2NO 2 + F 2  2NO 2 F Rate = k[A] m [B] n

43 Example 2 2NO 2 + F 2  2NO 2 F Rate = k[A] m [B] n m = 2, n = 0, k = 4 x M -1 s -1 Rate =4 x M -1 s -1 [NO 2 ] 2

44 Team problem Get into groups of 3 and obtain problem clues a) Determine the order with respect to each reactant b) Determine the overall order of reaction c) Write the rate expression for the reaction. d) Find the value of the rate constant, k.

45 Practice Problem From the data collected above, determine the rate law for the following equation: IO 3 - (aq) + 5I - (aq) + 6H + (aq)  3I 2 (aq) + 3H 2 O (l)

46 Practice Problem From the data collected above, determine the rate law for the following equation: IO 3 - (aq) + 5I - (aq) + 6H + (aq)  3I 2 (aq) + 3H 2 O (l) m = 1; n = 1; o = 2; k = 5M -3 s -1 Rate = 5M -3 s -1 [IO 3 - ] 1 [I - ] 1 [H + ] 2

47 Homework P. 377 practice # 1, 2, 3, 6, P. 415 # 15 a-e

48 Lesson 3 Reaction mechanisms: Ping pong activity Assembly line Elephant toothpaste demonstration Practice problems

49 Unit Mind Map

50 Reaction mechanisms Rate Laws  determined experimentally Equation: Reactants products This actually occurs in a series of steps called elementary steps Analogy: Cooking takes place in several steps

51 Reaction mechanisms What are the chances of the following reaction occurring in one step? Hint: collision theory 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) actually occurs in 3 separate steps

52 Reaction mechanisms 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) actually occurs in 3 separate steps: HBr (g) + O 2 (g)  HOOBr (g) (slow) HOOBr (g) + HBr (g)  2HOBr (g) (fast) 2HOBr (g) + 2HBr (g)  2H 2 O (g) + 2Br 2 (g) (fast) 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2

53 Reaction mechanisms 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) actually occurs in 3 separate steps: HBr (g) + O 2 (g)  HOOBr (g) (slow) HOOBr (g) + HBr (g)  2HOBr (g) (fast) 2HOBr (g) + 2HBr (g)  2H 2 O (g) + 2Br 2 (g) (fast) 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 Rate law: rate = k[HBr 2 ][O 2 ]

54 Reaction mechanisms: working backwards Overall equation: 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) Rate law: rate = k[HBr] 1 [O 2 ] 1 Devise a proposed mechanism for this reaction

55 Reaction mechanisms: working backwards Overall equation: 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) Rate law: rate = k[HBr][O 2 ] Devise a proposed mechanisms for this reaction Step 1. using the rate law, write out the rate- determining (slow) step Step 2. use the overall equation to determine what needs to be added to achieve the overall equation Step 3. cross out intermediates and add up what is left to produce the overall equation

56 Reaction mechanisms: working backwards 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2 (g) actually occurs in 3 separate steps: HBr (g) + O 2 (g)  HOOBr (g) (slow) HOOBr (g) + HBr (g)  2HOBr (g) (fast) 2HOBr (g) + 2HBr (g)  2H 2 O (g) + 2Br 2 (g) (fast) 4HBr (g) + O 2 (g)  2H 2 O (g) + 2Br 2

57 Potential Energy Diagram

58 Elephant toothpaste demonstration

59 Elephant toothpaste demonstration reaction: 2H 2 O 2 O 2 + 2H 2 O Rate = k[H 2 O 2 ] 1 [I - ] 1

60 Elephant toothpaste demonstration Elephant toothpaste demonstration reaction: 2H 2 O 2 O 2 + 2H 2 O Rate = k[H 2 O 2 ] 1 [I - ] 1 Proposed Reaction Mechanism: H 2 O 2 + I -  IO - + H 2 O(Slow) H 2 O 2 + IO -  I - + H 2 O + O 2 (Fast) 2H 2 O 2  O 2 + 2H 2 O

61 Elephant toothpaste demonstration Possible energy potential diagram?

62 Practice Problem Propose a possible mechanism for the following reaction: 2N 2 O 5 (g)  2N 2 O 4 (g) + O 2 (g) r = k[N 2 O 5 ] 1

63 Practice Problem Propose a possible mechanism for the following reaction: 2N 2 O 5 (g)  2N 2 O 4 (g) + O 2 (g) r = k[N 2 O 5 ] 1 Possible mechanism: N 2 O 5  N 2 O 4 + O(Slow) O + N 2 O 5  N 2 O 4 + O 2 (Fast) 2N 2 O 5  2N 2 O 4 + O 2

64 Practice Problem Propose a possible mechanism for the following reaction: 2NO 2 + F 2  2NO 2 F r = k[NO 2 ] 1 [F 2 ] 1

65 Practice Problem Propose a possible mechanism for the following reaction: 2NO 2 + F 2  2NO 2 F r = k[NO 2 ] 1 [F 2 ] 1 Possible mechanism: NO 2 + F 2  NO 2 F + F(Slow) F + NO 2  NO 2 F(Fast) 2NO 2 + F 2  2NO 2 F

66 Homework p. 390 #2 p. 391 # 1, 2, 3 Also, complete worksheet.


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