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Integrated Rate Laws Finally a use for calculus! Text 692019 and Questions to 376071.

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Presentation on theme: "Integrated Rate Laws Finally a use for calculus! Text 692019 and Questions to 376071."— Presentation transcript:

1 Integrated Rate Laws Finally a use for calculus! Text and Questions to

2 What is a rate? Text and Questions to

3 A rate as a derivative Text and Questions to

4 Lets look at the rate law Text and Questions to

5 Solving the equation Text and Questions to

6 Solving the equation Text and Questions to

7 Solving the equation Text and Questions to

8 What it means… Text and Questions to

9 If you know k and the initial concentration, you could calculate the concentration at any time. For example, if I know k=0.015 s -1 and I start with M A, how much A is left after 1 minute? Beware the units. 1 minutes = 60 seconds. Since k is in s - 1, I need my time to be in seconds. Plug and chug, baby! Text and Questions to

10 Using the integrated rate law Text and Questions to

11 Compare the integrated rate law to the rate law Text and Questions to

12 Other uses of the integrated rate law Text and Questions to

13 For example, suppose I monitor [A] Time (seconds)[A] (M) Since this is a first order reaction, the data should obey my integrated rate law. So I plot the ln[A] vs time and I should get a straight line. Text and Questions to

14 For example, suppose I monitor [A] Time (seconds)[A] (M)ln[A] Now, I plot the last column against the first column and put the best fit straight line on it. Text and Questions to

15 LOOK! ITS A POINT! ITS A PLANE! NO!!! ITS A STRAIGHT LINE! Text and Questions to

16 So, whats the rate constant? y = -0.02x ln[A] final = - kt + ln[A] t=0 m= slope=-0.02 m=-k k=-(-0.02)=0.02 s -1 So, if I KNOW its a 1 st order reaction, I can make a graph to find the rate constant. I can also make a graph to find out IF it is 1 st order. Text and Questions to

17 Different reaction 2 H 2 + O 2 2 H 2 O Time (seconds)[H 2 ] (M)ln[H 2 ] Now, I plot the last column against the first column and put the best fit straight line on it to see IF IF IF it is actually a straight line. Text and Questions to

18 NOT a straight line – NOT a 1 st order reaction! Text and Questions to

19 Is it or isnt it a straight line? A.It is a straight line B.It is NOT a straight line C.I cant tell without error bars D.I really dont care its Monday E.Your mother! Text and Questions to

20 This works for other orders of reaction also. Text and Questions to

21 Hey! Its 2 nd order! Text and Questions to

22 Also, theres the rare zeroth order Text and Questions to

23 Those are the easy ones For more complicated mixed orders like: Rate = k[A][B] The math gets much more complicated, so well ignore them until you become a chemistry major. But you can do a similar thing. Text and Questions to

24 But a lot of reactions fall into those three categories. Text and Questions to

25 How do we use this? Time[N 2 (g)] (M) 0 min min min min min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text and Questions to

26 GRAPH IT! Text and Questions to

27 Graph It! Time[N 2 (g)] (M)Ln([N 2 ])1/[N 2 ] 0 min min min min min N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text and Questions to

28 Try all 3 and see which one fits! Text and Questions to

29 Not 0th order – unless it was a sloppy experiment Text and Questions to

30 Maybe 2 nd order – its not a horrible fit. Text and Questions to

31 Hey, Goldilocks! It Fits 1 st order! Text and Questions to

32 What if I dont want to or cant make a graph? A.Find someone who can make a graph. B.Copy the answer from the person next to me. C.Calculate the rate of the reaction and see if the rate is constant or if the ln(rate) is constant or 1/rate is constant. D.Calculate the slope between data points and see if they are constant. Text and Questions to

33 What if I dont want to make a graph? Time[N 2 (g)] (M) 0 min min min min min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text and Questions to

34 3 possibilities Text and Questions to

35 k is the rate CONSTANT and its the slope of the line Text and Questions to

36 Slope is all over the place except 1 st order Time (min) [N 2 (g)] (M) slope– 0 th orderslope - 1 st orderK – 2 nd order N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following date, determine the rate law. Text and Questions to

37 Problem recognition Whats the tell? How do I know how to handle the problem? Text and Questions to

38 Method of initial rates – Rates measured for different initial mixes The reaction: 2 I - (aq) + S 2 O 8 2- (aq) 6 I 2 (aq) + 2 SO 4 2- (aq) was studied at 25° C. The following results were obtained for the rate of disappearance of S 2 O 8 2- [I-] 0 (M) [S 2 O 8 2- ] 0 (M)Initial rate (M/s) x x x x x10-6 Text and Questions to

39 Integrated rate law – concentration at different times Time[N 2 (g)] (M) 0 min min min min min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following date, determine the rate law. Text and Questions to

40 Does that make sense? A.Yes B.No C.Maybe Text and Questions to

41 Once I know the order, hows it work…? Once I know the order of the reaction, I can use the integrated rate law to determine the concentration at any time. Text and Questions to

42 The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text and Questions to

43 Do I know the rate constant? A.Yes B.No C.Not directly but implicitly D.I have no clue E.You look beautiful today Text and Questions to

44 As soon as Im talking about TIME, its an integrated rate law problem. The order of the reaction was given. This actually tells me two things: The Rate Law The Integrated Rate Law Text and Questions to

45 Text and Questions to

46 Text and Questions to

47 The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text and Questions to

48 Text and Questions to

49 Text and Questions to

50 Text and Questions to

51 A.Yes B.No C.Maybe D.You look like crap E.You look beautiful Text and Questions to

52 The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text and Questions to

53 In a word… A.Exploitation B.Death C.Life D.Stoichiometry E.Integration Text and Questions to

54 Rate = k[Cl 2 ] Rate = 1.92x10 -3 s -1 (0.632 M) = x10 -3 M/s Text and Questions to

55 How much HCl? Text and Questions to

56 Cl 2 (g) + H 2 (g) = 2 HCl (g) Text and Questions to

57 Another fun little rate thing… Half-life! For a reaction, you start with a lot of reactants and you end up with less reactants and more product. The amount of reactants should always be decreasing. Lets look at our earlier example… Text and Questions to

58 Not 0th order – unless it was a sloppy experiment Text and Questions to

59 Hey, Goldilocks! It Fits 1 st order! Text and Questions to

60 So, it is 1 st order. It must obey the first order rate equation. The concentration of the reactants should be asymptotically approaching zero. So if I start with the maximum A, soon I have 90% left, then 80% left, then 70% left…eventually 50% left. The time it takes for ½ (50%) of the A to react is called the half-life. Text and Questions to

61 Text and Questions to

62 Text and Questions to

63 t 1/2 = 2 hours So, lets say I start with 1 mol of Cl 2. In 2 hours, how much Cl 2 is left? A.1 mol B.0.75 mol C.0.50 mol D.0.25 mol E.I dont know enough to calculate it. Text and Questions to

64 t 1/2 = 2 hours Suppose I come back the next morning and find that there is only mol Cl 2 left. In 2 hours, how much Cl 2 is left? A mol B mol C mol D.It depends on how much the rate has slowed down as the Cl 2 decreased. E.You look FAB-ulous! Text and Questions to

65 1 st order is special… Radioactive decays show 1 st order kinetics. Thats why you hear half-life when people are talking about reactivity. But half-life actually applies to any reaction: its the time it takes for ½ the reactants to react! Text and Questions to

66 It also doesnt have to be ½ life. Suppose Im arrogant, obstinate, and just a general pain in the patootie… I insist on using t 9/10 – the time it takes for 90% of the reactants to react. Again, if it is first order…. Text and Questions to

67 Text and Questions to


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