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Chapter 14 Rates of Reactions Kinetics. I I. Introduction A) Demonstrations B) Chemical Kinetics is the study of the rates (speeds) of chemical reactions.

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Presentation on theme: "Chapter 14 Rates of Reactions Kinetics. I I. Introduction A) Demonstrations B) Chemical Kinetics is the study of the rates (speeds) of chemical reactions."— Presentation transcript:

1 Chapter 14 Rates of Reactions Kinetics

2 I I. Introduction A) Demonstrations B) Chemical Kinetics is the study of the rates (speeds) of chemical reactions and the mechanisms of chemical reactions.

3 C) The rate of a chemical reaction is a measure of how fast reactants are consumed and/or how fast products are made. D) The mechanism of a reaction is a detailed description of the way a reaction occurs. It is a sequence of elementary steps which lead from reactants to products.

4 Mechanisms can be proven wrong through _____________, but they can never be called _________________________ since they are, in general, educated guesses.

5 Practical reasons for studying kinetics: Some reactions we would like to speed up: drug delivery, paint drying, destruction of air pollutants in auto exhaust breakdown of materials in landfills. Some reactions we would like to slow down: food decay, rubber decay, human aging, destruction of the ozone layer, rusting, corrosion

6 E) Some reactions take place in a fraction of a second and other take many years. What variables affect the reaction rate? 1) The characteristics of the reactants and the products.

7 2) The concentration of the reactants – in some reactions the rate is unaffected by the concentration of one of the reactants as long as it is there in some amount.

8 3) The presence of a catalyst, a substance that … 4)The temperature at which the reaction occurs. Increasing the temperature usually increases the rate. A general rule is … 5) The surface area of a solid reactant or catalyst affects the rate.

9 G) Reaction rate is the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of a reactant per unit time.

10 For example, for the reaction: 2 N 2 O 5  4 NO 2 + O 2 The following table shows the concentration of N 2 O 5 as a function of time at 45 o C.

11 Time in min[N 2 O 5 ] mol/L

12 1) The rate of reaction can be written in the form: 2) Usually the rate is a rapidly changing quantity, as the reaction proceeds the reactants are used up and there remains less and less material to undergo reaction.

13 3)Generally we obtain for a reactant a curve which resembles the one below.

14 If we take the first 2 points from the table above, we can find as avg. rate of decomposition of N 2 O 5.

15 You should be able to see that the avg. rate is decreasing, hence the curve. Even though we obtain a negative value for the rate, since N 2 O 5 is decreasing, rates are reported as positive values.

16 Look at the graph again, And make the time interval smaller and smaller, we can obtain an instantaneous rate.

17 The instantaneous rate is equal to the slope of the line at that point. Calculus??? To what is the slope of the line at that point equal?

18 4) How is data obtained for a concentration curve? a) Monitor a color change. b) Measure pressure if a gas is produced. c) Monitor a change in pH if an acid or base reaction.

19 5) A look at the change in rate over time for another reaction: 2 NO 2(g)  2 NO (g) + O 2(g) at 300 o C

20 H) What does the balanced equation tell us about rates? The equation we will look at is: H 2 (g) + I 2 (g) ----> 2 HI (g) H 2 and I 2 must disappear at the same rate since 1 molecule of H 2 reacts with 1 molecule of I 2.

21 In the same amount of time, 2 molecules of HI must appear. The rate of appearance of HI must equal twice the rate of disappearance of H 2 (g) and I 2 (g).

22 The rate of disappearance of H 2(g) = the rate of disappearance of I 2(g) = ½ the rate of appearance of HI.

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24 In general, for the equation: aA + bB -----> cC + dD To obtain a rate equation of the rates of the substances in relation to each other we divide through by the coefficients.

25 But we usually want the first reagent in terms of all the others, so we multiply by the coefficient of the first reagent.

26 II. The Rate Law (Rate expression), Rate Constant Order of Reaction A) The following equation has been studied in the gaseous state and the data at 250 K may be summarized as follows:

27 F 2(g) + 2 ClO 2(g) ---> 2 ClO 2 F (g) B) From this data the answers to the following questions can be obtained:

28 1) What is the rate law of the reaction? 2) What is the order of the reaction? 3) What is the value of the rate constant k? 4) What is the rate of formation of ClO 2 F when [F 2 ]is mol/L and[ClO 2 ] is mol/L?

29 C) What is a rate law? 1) In 1864 it was discovered that the rate of a reaction is proportional to some power of the concentration of reactants at constant temperature.

30 2) In general, for the equation: aA + bB  cC + dD

31 rate law = rate equation = rate expression k is the specific rate constant which is independent of concentration.

32 k depends on the nature of the reactants; fast reactions have large k's and slow reactions have small k's ordinarily k ________________ with temperature.

33 THE EXPONENTS x AND y MUST BE EXPERIMENTALLY DETERMINED. They are not automatically obtained from the balanced equation.

34 Some experimentally determined rate laws for equations are as follows: a) 2 N 2 O 5 (soln)  2 N 2 O 4 (soln) + O 2 (g)

35 Notice that the exponent is NOT 2, the coefficient in the balanced equation, but has been experimentally determined to be 1. The reaction order with respect to a given species equals the exponent of the concentration of that species in the rate law as determined experimentally.

36 The order of the above reaction with respect to N 2 O 5 is 1. The order of a reaction is the sum of the exponents of the reactants in a rate law. The order of the above reaction, since 1 is the only exponent, is 1 as well. It is a first order reaction.

37 This means that when we double the concentration of N 2 O 5, we double the rate of reaction OR if we halve the concentration, we halve the rate. b) For the reaction: 2 NO + O 2  2 NO 2 What is the situation here?

38 D) The answers to the questions then are obtained in the following manner. The rate law for F 2(g) + 2 ClO 2(g) ---> 2 ClO 2 F (g) will look like the following.

39 Your job is to find the values of x and y from the experimental data.

40 You need to do two division problems to find x and y. What is the order of the reaction with respect to F 2 ? _______ What is the order of the reaction with respect to ClO 2 ? ______ What is the overall order of the reaction? __________

41 To find k, the rate constant, we take the experimentally determined rate law, put the data in from one of the experiments and solve for k. The units of k are important. k = _________________

42 To find the rate of formation of ClO 2 F when [F 2 ] is M and the [ClO 2 ] is M, we need to look at the relationship between the rate of formation of [ClO 2 ], and the rate of disappearance of F 2. Rate = __________________________

43 Transparency – in class problem – collect it

44 IV. A graphical method is often used to show the order of a reaction, or from a graph we can obtain the order of a reaction. A)For the general reaction of A ---> products B) If the reaction is first order, we can write

45 When we divide both sides by [A] and multiply both sides by dt we obtain the following: Those of you who have had calculus should recognize this as:

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47 Which can be changed to the following in log to the base 10.

48 The above equations can be rewritten in a more familiar form.

49 THIS MEANS THAT IF WE PLOT ln [A] vs. t AND OBTAIN A STRAIGHT LINE THE RATE IS FIRST ORDER. Or plot log [A] vs. t.

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51 Slope of the line is -k/ The rate constant k then is = to -slope of the line times If we plot ln[A] vs.time then the slope of the line is equal to -k. C) Half-life - t ½ is the time required for half of the original concentration of the limiting reactant to be consumed.

52 Half-life is _________________ to the rate constant k.

53 A fast reaction with a large k has a _________________ half-life, a slow reaction with a small k has a ___________________ half-life. D) For the reaction : A ----> B + C We obtain the following data.

54 Prove that the reaction is first order in A. Calculate the half-life of the reaction, t ½. What is our criterion for a first order reaction?

55 We make the following table. When we plot log [A] vs. t we obtain a _________________, therefore the reaction is first order.

56 The equation for the line is:

57 V. Second order reaction for a single reactant A) A ----> products B) The rate law is: rate = k[A] 2 C) The integrated form of the rate law gives us the following:

58 D) A straight line is obtained with slope = ____.

59 E) This means that we can tell the difference between a first order and second order reaction by ________________________________.

60 G) The half-life of a second order reaction depends on the initial concentration of the reactant as can be seen from the following derivation:

61 H) When we compare this to the formula for the half-life of a first order reaction, we notice that the t½ for the first order reaction is __________________of the initial concentration of the reactant. The t ½ for the second order reaction is ________________________to the initial concentration of the reactant.

62 I) The following equations will be given to you on an exam or quiz for you to apply to questions in an appropriate manner.

63 VI. Collision Theory A)Consider the elementary process A 2(g) + B 2(g) ---> 2AB B) Collision theory assumes that for gas molecules to react they must collide. C) Binary collisions per unit time at room temperature and 1 atm pressure are extremely large, approx. 1 x collisions per Liter-sec.

64 D) With this many collisions, why isn't there an immediate explosion when we put H 2(g) and O 2(g) together in a balloon? E) Why isn't every collision an effective collision; i.e. one that gives product? F) For a reaction to occur reactant molecules must collide with an energy greater than some minimum value and with the proper orientation.

65 G) The minimum energy of collision required for two molecules to react is called the activation energy, E a. The value of E a depends on a particular reaction. 1) The rates of almost all chemical reactions_________ when T is raised. Generally speaking for a _________ increase, the reaction rate ________. H) Effect of temperature

66 2) What is the relationship between temperature and collision theory? a) Approx. ___ of the reaction rate increase due to an increase in temperature is accounted for by the increase in the number of___________. b) Approx. ____of the reaction rate increase is accounted for by the increase in the number of high energy _______________.

67 3) Recall our graph of the number of molecules vs. speed in a gas sample.

68 J) Arrhenius in 1889 proposed the following equation which relates the rate constant k to the temperature: where k is the rate constant from the rate equation ( rate = k[A])

69 A is the frequency factor (the collision factor and the orientation factor). e is the base of the natural logarithms Ea is the activation energy in Joules per mole. R is the molar gas constant, J/K mol. T is the temperature in Kelvins.

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71 J) The most valuable use for this equation is in determining the activation energy of a reaction from rate experiments at different temperatures.

72 L) E a is best obtained graphically. This is of the form y = mx + b

73 If we plot the log of k vs 1/T we obtain the slope of the line which is equal to - Ea/ R. then Ea = R slope.

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75 L) The equation can be used to compare rates at various temperatures when Ea is known, OR the Ea can be obtained from a comparison of two sets of rate data. r 1 = k 1 [A] x at T 1 r 2 = k 2 [A] x at T 2

76 We divide equation 1 by equation 2, take the log of both sides, and substitute for log k –E a /2.303RT. This should give: We remove the negative sign and we get the equation which we will use and which will be given to you on the exam or quiz.

77 1) An example: The activation energy for the reaction involved in the souring of raw milk is 75 kJ. Milk will sour in about 8 hours at 21 o C (room temp-70 o F). How long will raw milk last in a refrigerator at 5 o C?

78 2) The gas phase reaction between methane and diatomic sulfur is given by the equation: CH 4(g) + 2 S 2(g) ---> CS 2(g) + 2 H 2 S (g) At 550 o C the rate constant, k 1, for the reaction is 1.1 L/mol sec. At 625 o C the rate constant, k 2, is 6.4 L/mol sec. What is the activation energy of this reaction?

79 N) Transition State Theory 1) The hypothetical elementary process of: A 2 + B 2 --> 2 AB can be represented as the following:

80 is an activated complex. It is a transition state species that __________________________. In the past, these could not be detected only postulated. Now with laser technology we can detect species which exist for only a femto second(1x ), and evidence for the existence of some intermediates is available.

81 2) We can graph the potential energy changes which occur during the course of a reaction.

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84 3) We have shown reactions in which  H is - (negative). The reactions are exothermic.  H rxn = E af - E ar 4) Potential energy diagrams are helpful in visualizing why it is that highly exothermic reactions can be very slow. If E a is large then __________________.

85 VII. The mechanism of a reaction A) The mechanism of a reaction is the detailed pathway followed when reactants are converted to products. B) It must agree with 1)____________________. 2)____________________. 3) ____________________.

86 C) What do we mean by the rate determining step? Let's look at a factory where toasted corn puffs are made.

87 What is the rate determining step? ___ How fast can the factory produce packages of corn puffs with a prize? _________________ What does the factory need to do to improve the rate of production? __________________

88 D) For the reaction: 2 NO + F 2 ---> 2 NOF nitric oxide plus fluorine gives nitrosyl fluoride 1) The experimentally determined rate law is: rate = k [NO][F 2 ] 2) A mechanism is proposed on the basis of the rate equation and other information gained from a study of similar reactions over a period of years.

89 3) Remember that the mechanism is ___. 4) A suggested mechanism for this reaction is: NO + F 2 --> NOF + F (slow step) NO + F --> NOF (fast step) The sum of these two reactions gives: 2NO + F 2 + F --> 2 NOF + F Subtracting F from both sides gives the

90 5) The rate determining step is the _____. 6) Each step is an elementary reaction and the rate equation can be written directly from a _________________ elementary reaction. 7) The rate law of our reaction is rate = k [NO][F 2 ]. The proposed mechanism satisfies the requirements ___________ __________________________________.

91 E) Let's look at another equation its experimentally determined rate law and the proposed mechanism to see if they are in agreement. 2 O 3 -->3 O 2 1) The experimentally determined rate law is:

92 2) The proposed mechanism is: The rate of the overall reaction is simply the rate of step ______.

93 Rate = k 2 [O 3 ][O] There is a problem. We do NOT report rate laws with the concentrations of species which are not in the _________. How can we convert Rate = k 2 [O 3 ][O] to

94 3) We need to use a combination of knowledge from chemistry and algebra to change one rate law to another. The chemical knowledge we bring to the situation is that at equilibrium, since both step 1 and step 2 are fast and 2 is the reverse of 1 we will make their rates equal. k 1 [O 3 ] = k -1 [O 2 ][O]

95 Then we can use algebra to solve for [O], since that is what we want to __________ We can then substitute this relationship of [O] into the rate law we obtained from ____________ to obtain

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97 F) Catalysts 1) The rates of many reactions are increased by the presence of a catalyst, a substance which increases the rate of reaction without being consumed by it. 2) How is this possible? 3) NO (nitric oxide) catalyzes the decomposition of O 3. The following mechanism is proposed for this reaction:

98 What should you notice about NO?

99 4) A catalyst acts by making available a new reaction mechanism with a lower activation energy, Ea.

100 Sometimes is orients the molecules for a successful collision, for example the lock and key model mechanism of enzymes.

101 Other times it provides an alternate pathway for the reaction.The depletion of ozone in the stratosphere by Cl atoms provides an example of the lowering of activation energy by a catalyst. The uncatalyzed reaction has such a large activation energy as can be seen from the following diagram that its rate is extremely slow.

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103 VIII. Review - Factors which affect the rate of reaction: A) Nature of the reactants - B) Concentration - C) Temperature - D) Catalysts -

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