Chapter 14. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Transcription

1 Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 14 John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc.

2 Studies the rate (speed) at which a chemical process occurs. kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Both rate and mechanism are characteristic of a particular reaction In order for a reaction to occur, reactant molecules must collide with enough energy (activation energy) AND with the proper orientation.

3 Outline by Ch 14 sections What you need to know about kinetics 1. 4 factors that affect all reaction rates & why 2. How different reaction rates are measured 3. How concentration affects different rates 4. How the rate of different reactions change over time 5. How temperature affects different rates 6. The sequence of events in different reactions 7. How catalysts speed up reactions

4 1. Factors That Affect Reaction Rates Physical State/Form of the Reactants Ø In order to react, molecules must come in contact with each other. Ø The more homogeneous the mixture of reactants, the faster the molecules can react. Concentration of Reactants Ø As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

5 1. Factors That Affect Reaction Temperature Rates Ø At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Presence of a Catalyst Ø Catalysts speed up reactions by changing the mechanism of the reaction. Ø Catalysts are not consumed during the course of the reaction.

6 2. Measuring Reaction Rates p. 577 Rates of reactions can be determined by monitoring the change in concentration (molarity) of either reactants or products as a function of time.

7 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times.

8 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate = Δ[C 4 H 9 Cl] Δt

9 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

10 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 A plot of concentration vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

11 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

12 2. Measuring Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9OH(aq) + HCl(aq) p. 579 In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. Rate = -Δ[C 4 H 9 Cl] Δt = Δ[C 4 H 9 OH] Δt

13 2. Measuring Reaction Rates p. 580 What if the ratio is not 1:1? Therefore, 2 HI(g) H 2 (g) + I 2 (g) Rate = 1 2 Δ[HI] Δt = Δ[I 2 ] Δt

14 2. Measuring Reaction Rates p. 581 To generalize, then, for the reaction aa + bb cc + dd Rate = 1 a Δ[A] Δt = 1 b Δ[B] Δt = 1 c Δ[C] Δt = 1 d Δ[D] Δt

15 3. Determining Rate Info for a Particular Reaction One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

16 3. Determining Rate Info for a Particular Reaction NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) p. 582 Comparing Experiments 1 and 2, when [NH 4+ ] doubles, the initial rate doubles.

17 3. Determining Rate Info for a Particular Reaction NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) p. 582 Likewise, comparing Experiments 5 and 6, when [NO 2 ] doubles, the initial rate doubles.

18 3. Determining Rate Info for a Particular Reaction p. 583 This means Rate [NH 4+ ] Rate [NO 2 ] Rate [NH + ] [NO 2 ] or Rate = k [NH 4+ ] [NO 2 ] This equation is called the rate law, and k is the rate constant.

19 3. Determining Rate Info for a Particular Reaction p. 583 A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4+ ] First-order in [NO 2 ]

20 3. Determining Rate Info for a Particular Reaction p. 583 The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

21 4a. First Order rate laws p.587 Using calculus to integrate the rate law for a first-order process (Rate = k [A]) gives us Where ln [A] t [A] 0 = kt [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t, during the course of the reaction.

22 4a. First Order rate laws p. 588 Manipulating this equation produces ln [A] t = kt [A] 0 ln [A] t ln [A] 0 = kt which is in the form ln [A] t = kt + ln [A] 0 y = mx + b

23 4a. First Order rate laws ln [A] t = -kt + ln [A] 0 p. 589 Therefore, if a reaction is firstorder, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.

24 4a. First Order rate laws p. 589 Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NC CH 3 CN

25 4a. First Order rate laws CH 3 NC CH 3 CN p. 589 This data was collected for this reaction at C.

26 4a. First Order rate laws p. 589 When ln P is plotted as a function of time, a straight line results. Therefore, Ø The process is first-order. Ø k is the negative slope: s 1.

27 4a. First Order rate laws p. 591 Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

28 4a. First Order rate laws p. 591 For a first-order process, this becomes 0.5 [A] ln 0 [A] = kt 1/2 0 ln 0.5 = kt 1/2 NOTE: For a first-order process, the half-life does not depend on [A] = kt 1/ k = t 1/2

29 4b. Second Order rate laws p. 590 Integrating the rate law for a process that is second-order in reactant A (Rate = k [A] 2 ) we get 1 [A] = kt + 1 t [A] 0 y = mx + b also in the form

30 4b. Second Order rate laws 1 [A] t = kt + 1 [A] 0 p. 590 So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

31 4b. Second Order rate laws p. 590 The decomposition of NO 2 at 300 C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields data comparable to this: Time (s) [NO 2 ], M

32 4b. Second Order rate laws p. 591 Graphing ln [NO 2 ] vs. t yields: The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO 2 ], M ln [NO 2 ]

33 4b. Second Order rate laws p. 591 Graphing ln 1/[NO 2 ] vs. t, however, gives this plot. Time (s) [NO 2 ], M 1/[NO 2 ] Because this is a straight line, the process is secondorder in [A].

34 4b. Second Order rate laws p. 591 For a second-order process, the half-life is: 1 1 = kt 0.5 [A] 1/2 + 0 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] [A] 0 = 1 [A] 0 1 k[a] 0 = kt 1/2 = t 1/2

35 5. Temperature and Rate p. 593 Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

36 5. Temperature and Rate p. 593 In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

37 5. Temperature and Rate p. 594 Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

38 5. Temperature and Rate p. 595 There is a minimum amount of energy required for reaction: the activation energy, E a. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

39 5. Temperature and Rate p. 595 It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

40 5. Temperature and Rate p. 595 It shows the energy of the reactants and products (and, therefore, ΔE). The high point on the diagram is the transition state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier.

41 5. Temperature and Rate p. 596 Maxwell Boltzmann Distributions like this one show the fraction of reactant molecules with a given kinetic energy Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. Because temp. is an average at any temperature there is a wide distribution of kinetic energies.

42 5. Temperature and Rate p. 596 As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

43 5. Temperature and Rate p. 596 If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

44 5. Temperature and Rate p. 596 This fraction of molecules can be found through the expression f = e E a/rt where R is the gas constant and T is the Kelvin temperature.

45 5. Temperature and Rate p. 596 Svante Arrhenius developed a mathematical relationship between k and E a : k = A e E a/rt where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

46 5. Temperature and Rate p. 598 Taking the natural logarithm of both sides, the equation becomes ln k = -E ( 1 ) a + ln A RT y = mx + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. 1/T.

47 6. Reaction Mechanisms p. 599 The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

48 6. Reaction Mechanisms p. 599 Reactions may occur all at once or through several sequential steps. Each of these steps within a reaction is known as an elementary reaction or elementary process.

49 6. Reaction Mechanisms p. 599 The molecularity (uni, bi, or tri) of a process tells how many molecules are involved in the process, which explains the concentration dependence of reaction rate.

50 6. Reaction Mechanisms p. 603 In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

51 6. Reaction Mechanisms NO 2 (g) + CO (g) NO (g) + CO 2 (g) p. 603 The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps.

52 6. Reaction Mechanisms A proposed mechanism for this reaction is Step 1: NO 2 + NO 2 NO 3 + NO (slow) Step 2: NO 3 + CO NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. Because CO is not involved in the slow, ratedetermining step, it does not appear in the rate law.

53 6. Reaction Mechanisms 2 NO (g) + Br 2 (g) 2 NOBr (g) P.605 The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism.

54 6. Reaction Mechanisms p. 605 A proposed mechanism is Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (slow) Step 1 includes the forward and reverse reactions.

55 6. Reaction Mechanisms p. 605 The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

56 6. Reaction Mechanisms p. 606NOBr 2 can react two ways: Ø With NO to form NOBr Ø By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

57 6. Reaction Mechanisms p. 606 Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k 1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k 1 k 1 [NO] [Br 2 ] = [NOBr 2 ]

58 6. Reaction Mechanisms p. 606 Substituting this expression for [NOBr 2 ] in the rate law for the ratedetermining step gives Rate = k 2 k 1 k 1 [NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

59 7. Catalysts P.609 Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. P.610 One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

60 7. Catalysts P.612 Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.