Chemical Kinetics and Equilibrium

Transcription

1 Chemical Kinetics and Equilibrium Part 1: Kinetics David A. Katz Department of Chemistry Pima Community College Tucson, AZ USA

2 Chemical Kinetics The study of the rates of chemical reactions and how they occur. The conditions that affect the rate or speed at which reactions occur. The reaction mechanism (how the reaction occurs at the atomicmolecular level).

3 Reaction Rates Reaction rate = change in concentration of a reactant or product with time. concentration Rate= time Rate can be expressed as initial rate average g rate instantaneous rate

4 Determining a Reaction Rate

5 Factors Affecting Reaction Rates Physical State t of the Reactants t In order to react, molecules must come in contact t with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. Gaseous reactions are preferred Solutions are most convenient

6 Factors Affecting Reaction Rates Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Burning an iron nail Burning iron in air Burning iron in oxygen

7 Lycopodium powder burning: a) In an evaporating dish b) In air

8 Factors Affecting Reaction Rates Temperature At higher temperatures, reactant At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

9 Factors Affecting Reaction Rates Catalysis Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts t are not consumed during the course of the reaction. Negative catalysts (inhibitors) can be used to slow down the rate of a reaction.

10 Collision Theory of Reaction Rates In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

11 Collision Theory of Reaction Rates The possible collisions for three sets of molecules that would result in a possible reaction

12 Collision Theory of freaction Rates For a collision to be effective (i.e., result in a reaction): Molecules must collide with the correct orientation (or alignment) There must be enough energy to overcome repulsions so molecules do not rebound Bonds must break and new bonds formed.

13 Collision Theory of Reaction Rates NO + NO 3 2 NO 2 Ineffective means that this has a lower probability of reacting.

14 Activation Energy 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 les possess sufficient energy to get over the activation energy barrier.

15 Reaction Energy Diagrams This diagram relates the energy for the rearrangement of methyl isonitrile. To visualize the energy changes throughout a process we use a reaction energy or a reaction coordinate diagram

16 Reaction Energy Diagrams The height of the hill is called the activation energy, E a The species present at the transition state is called the activated complex. This is the energy of the reactants, E react This is the energy of the products, E prod This is the net energy This is the net energy for the reaction, E

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18 Reaction energy diagram for CH3Br + OH- CH3OH + Br -

19 Reaction energy diagram for CH 3 Br + OH - CH 3 OH + Br - The transition state or activated complex has definite bond lengths and bond angles

20 Entropy on the Molecular Scale Ludwig Boltzmann ( ) described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample. An animation of the Maxwell-Boltzman distribution for molecular speeds in a gas can be found at

21 Entropy on the Molecular Scale Molecules exhibit several types of motion: Vibrational: Periodic motion of atoms within a molecule. (This is the lowest energy state) Vibrations occur at absolute zero. Called zero-point energy. Rotational: Rotation of the molecule on about an axis or rotation about bonds. Translational: Movement of the entire molecule from one place to another. (This is the highest energy state)

22 Entropy on the Molecular Scale Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules. He referred to this sampling as a microstate of the thermodynamic system.

23 Maxwell Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide At any temperature there is a wide distribution of kinetic energies.

24 Maxwell Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules l has higher energy.

25 Maxwell Boltzmann Distributions 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 res lt the As a result, the reaction rate increases.

26 Maxwell Boltzmann Distributions This fraction of molecules can be found through the expression f = e E a/rt where R = the ideal gas constant (8.31x kj/k mol) T = the absolute temperature in K

27 Arrhenius Equation 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. A must be determined experimentally

28 Arrhenius Equation Starting with k = Ae EE a/rt Taking the natural logarithm of both sides, the equation becomes ln k = -E ( a 1 RT ) + ln A A graph of ln k vs. 1/T is a straight line of the form y = mx + b

29 Arrhenius Equation 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 slope =- Ea R R = gas constant = 8.31 J/mol K

30 Arrhenius Equation If k is determined experimentally at two or more temperatures, T 1 and T 2, you can select two points from your data or graph, and E a can be calculated from the equation: a[ 2 [ ] [k ] E 1 1 ln = - - [k 1] R T T 2 1 R = gas constant = 8.31 J/mol K T = temperature in K

31 Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

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

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

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

35 Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) 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.

36 Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) All reactions slow down over time. Therefore, the rate of a reaction is generally stated as the instantaneous rate near the beginning.

37 Reaction Rates and Stoichiometry C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) In this reaction, the mole 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 = - [C 4H 9 Cl] t = [C 4H 9 OH] t

38 Reaction Rates and Stoichiometry What if the ratio is not 1:1? Then: 2 HI(g) H 2 (g) + I 2 (g) Rate = 1 2 [HI] t = [I 2] t The ½ represents the stoichiometery of the reaction

39 Reaction Rates and Stoichiometry To generalize, for the reaction aa + bb cc + dd Rate = 1 [A] = 1 [B] = 1 [C] 1 [D] = a t b t c t d t O i i f ti b t th t f ti One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration

40 Concentration and Rate For the reaction NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Data is collected for several experiments: Note that the concentrations are always changed, one at a time, in a systematic manner.

41 Concentration and Rate NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Comparing Experiments 1 and 2, when [NH 4+ ] doubles, the initial rate doubles. Note: [NO 2 ] is constant. Also, comparing Experiments 2 and 3, when [NH 4+ ] doubles, the initial rate doubles.

42 Concentration and Rate NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Compare Experiments 1 and 2: (Note: put the larger value in the numerator) rate 2 knh rate1 k [ NH ] [ NO ] Substitute data from the table: m n [ 4 ] 2 [ NO2 ] 2 4 m n m k [0.0200] 0200] [0.200] 7 m k [0.0100] [0.200] Cancel terms to get: 2=22 m m =1 n 2 2 n 1 1 This tells us that the reaction is first order with respect to [NH 4+ ] (i.e., only one molecule of NH + 4 is involved in the reaction mechanism)

43 Concentration and Rate NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Comparing Experiments 5 and 6, when [NO 2 ]d doubles, the initial rate doubles. Note: [NH 4+ ] is constant. Also, comparing Experiments 6 and 8, when [NO 2 ] doubles, the initial rate doubles.

44 Concentration and Rate NH 4 + (aq) + NO 2 (aq) N2 (g) + 2 H 2 O(l) Compare Experiments 5 and 6: rate k NH NO rate knh NO m 6 [ 4 ] [ 2 ] m n 5 [ 4 ] [ 2 ] Substitute data from the table: 7 m n k [0.200] [0.0404] 0404] m n k [0.200] [0.0202] Cancel terms to get: n 5 5 2=2 n n = 1 This tells us that the reaction is first order with respect to [NO 2- ] (i.e., only one molecule of NO 2- is involved in the reaction mechanism)

45 Concentration and Rate 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, the constant of proportionality, is called the rate constant.

46 Rate Laws A rate law equation shows the relationship between the reaction rate and the concentrations of reactants. Rate = k [NH 4+ ] [NO 2 ] The exponents of the concentration terms tell the order of the reaction with respect to each reactant. For a zero order reaction, the rate is independent of concentration of reactants For a first order reaction, the rate depends on the concentration of one reactant For a second order reaction, the rate depends on the concentration of two molecules of reactant(s)

47 Rate Laws For the reaction NH (aq) NO 2 (aq) N2 N 2 (g) + 2H 2 O(l) Rate = k [NH 4+ ] [NO 2 ] This reaction is First-order in [NH 4+ ] (recall: m =1) First-order in [NO 2 ] (recall: n = 1) The overall reaction order can be found by adding the exponents on the reactants in the rate law. m + n = = 2 This reaction is second-order overall.

48 Rate Laws For the reaction NH (aq) NO 2 (aq) N2 N 2 (g) + 2H 2 O(l) Rate = k [NH 4+ ] [NO 2 ] To calculate the value of k, substitute data from any experiment Using Exp. 1: 5.4 x 10-7 M/s = k [0.0100] [0.200] Solve for k k = 2.7 x 10-4 M -1 s -1 All the k values should be approximately the same at the same temperature.

49 Integrated Rate Laws Using calculus to integrate the rate law for a first-order process 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.

50 Integrated Rate Laws Manipulating this equation produces ln [A] t [A] 0 = kt Separate this equation to get ln [A] t ln [A] 0 = kt Solve for ln [A] t to get which is in the form y ln [A] t = kt + ln [A] 0 = mx + b

51 First-Order Processes ln [A] t = -kt +ln[a] 0 Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.

52 First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NC CH 3 CN

53 First-Order Processes CH 3 NC CH 3 CN This data was collected for this reaction at C.

54 First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative slope: 5.1 x 10-5 s 1.

55 Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, (Note: the process is not shown here) we get 1 = kt + 1 [A] t [A] 0 also in the form y = mx + b

56 Second-Order Processes 1 [A] t = kt + 1 [A]0 So if a process is secondorder in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

57 Second-Order Processes 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

58 Second-Order Processes 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 ]

59 Second-Order Processes Graphing ln 1/[NO 2 ]vs vs. t, however, gives this plot. Time (s) [NO 2 ], M 1/[NO 2 ] Because this is a straight line, the process is second- order in [A].

60 Summary of the Integrated Rate Law Graphs

61 Showing graphs for a 1 st order reaction Note: only one graph will result in a straight line

62 Half-Life 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.

63 Half-lifelife

64 Half-Life: First Order Reaction For a first-order process, the integrated rate law is 0.5 [A] ln 0 [A] = kt 1/2 0 Which reduces to Calculate the value for ln 0.5 ln 0.5 = kt 1/2 Solve for t 1/ = kt 1/ k = t 1/2 NOTE: For a first-order process, the half-life does not depend on [A] 0.

65 Half-Life: First Order Reaction To calculate the concentration at time t, or the length of time needed for the concentration of reactant to reach a specific concentration, two equations may be needed: The equation for half-life (to determine the value of k) = t k 1/2 and the integrated rate equation ln [A] t [A] 0 = kt

66 Half-Life: Second Order Reaction For a second-order process, the integrated rate law is 1 1 = kt 0.5 [A] 1/2 + 0 [A] 0 Divide the first term: 2 [A] 0 = kt 1/2 + Subtract 1/[A] 0 from both sides 1 [A] = = kt 1/2 [A] [A]0 0 1 Solve for t 1/2 = t 1/2 1 k[a] 0

67 Half-Life: Second Order Reaction To calculate the concentration at time t, or the length of time needed for the concentration of reactant to reach a specific concentration, two equations may be needed: The equation for half-life (to determine the value of k) 1 = t k[a] 1/2 0 and dthe integrated t rate equation 1 [A] t = kt + 1 [A]0

68 Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

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

70 Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

71 Reaction Mechanisms The molecularity of a process tells how many The molecularity of a process tells how many molecules are involved in the process.

72 Multistep Mechanisms 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.

73 Slow Initial Step NO 2 (g) +CO(g) + NO (g) +CO 2 (g) 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.

74 Slow Initial Step 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. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

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76 Fast Initial Step 2NO(g) 2 +Br 2 (g) 2 NOBr (g) 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.

77 Fast Initial Step A proposed mechanism is Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (slow) Note: Step 1 includes the forward and reverse reactions.

78 Fast Initial Step 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 2] ]?

79 Fast Initial Step NOBr 2 can react two ways: With NO to form NOBr 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

80 Fast Initial Step 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 ]

81 Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives Rate = k 2 k 1 k 1 [NO] [Br 2 ] [NO] Collecting terms and consolidating the Collecting terms and consolidating the constants into a single constant, the equation becomes Rate = k [NO] 2 [Br 2 ]

82 Reaction energy diagram for the two-step NO 2 -F 2 reaction

83 The Rate-Determining Step of a Reaction Mechanism The overall rate of a reaction is related to the rate of the slowest, or rate-determining step. Correlating the Mechanism with the Rate Law The elementary steps must add up to the overall equation. The elementary steps must be physically reasonable. The mechanism must correlated with the rate law.

84 A summary of rate law information for zero order, first order, and second order reactions

85 Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

86 A Homogeneous Catalyst Bromide ion catalyzes the reduction of hydrogen peroxide in acid solution 2 H 2 O Br H + H 2 O H 2 O + Br 2 2 H 2 O + O Br H

87 A Heterogeneous Catalyst The hydrogenation of ethene (ethylene): H 2 C=CH 2 (g) + H 2 (g) CH 3 -CH 3 (g)

88 CATALYSIS Catalytic converters in auto exhaust systems The catalytic converter contains a ceramic surface that is coated with a thin later of platinum, palladium, and/or rhodium. Older catalytic converters used ceramic beads. Newer catalytic converters use a honeycomb structure. Reduction of exhaust gases, mixed with air, takes place on the metal surface: 2CO + O 2 2CO 2 2 NO N 2 + O 2

89 Enzymes Enzymes are catalysts in biological systems. In one type of reaction, the substrates fit into the active site of the enzyme much like a key fits into a lock. In a second type of reaction the substrates fit into a more generalize active site of the enzyme