Part One: Reaction Rates. 1. Rates of chemical reactions. (how fast products are formed and/or reactants are used up)

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1 A. Chemical Kinetics deals with: CHAPTER 13: RATES OF REACTION Part One: Reaction Rates 1. Rates of chemical reactions. (how fast products are formed and/or reactants are used up) 2. Mechanisms of chemical reactions. (steps by which they occur) B. Importance: 1. Even though a reaction is thermodynamically favorable it may not occur at all if it is kinetically very slow. C(diamond) + O 2 (g) CO 2 (g) ΔG = -397 kj/mol Diamonds combine with O 2 in air so slowly that to all purposes, not at all, even though such a reaction would proceed downhill in free energy. 2. CH 4 + 2O 2 CO 2 + 2H 2 O, burning of natural gas at high T, is kinetically very rapid as well as being thermodynamically favorable. C. Factors affecting the rate of reaction: 1. The Physical State of the Reactants. a. Rate depends on surface area of contact between the reactants. b. Gases typically react faster than solids or liquids because they more intimately mix and mix faster. c. Finely divided powders or aerosols suspended in air can react explosively because of large surface area exposed. (Exploding grain elevators.) 2. Concentrations of Reactants - the Rate Law. a. Collisions per second between reactant species is greater at higher concentration. b. Dependence of Rate on Concentrations is expressed in the Rate Law. Chapter 13 Page 1

2 c. A rate law expression takes the form: Rate = k[a] x [B] y... d. Expresses how rate varies with concentrations of species present. 3. Temperature: a. Reactions run faster at higher temperatures. 4. Presence of Catalysts. a. A catalyst is a substance that increases the rate of reaction without being consumed by the reaction. b. Example: MnO 2 catalyzes the decomposition of H 2 O 2. D. Definition of Rate. (Section 13.1) 1. Rate loosely is amount of product formed or reactant consumed per unit time. 2. Consider hypothetical reaction: aa + bb cc + dd 3. Rate defined in terms of rate of growth or loss of concentrations [ ] of species: Rate + 1 Δ[ C] = 1 c Δt c Δ[ C] = rate of formation of C Δt Alternatively: [A], [B], [C], [D] increase of C conc. short interval of time Rate = + 1 d Δ[ D] Δt or Rate = 1 a Δ[ A] Δt or Rate = 1 b Δ B [ ] Δt Chapter 13 Page 2

3 4. Analogy: 2 bread slices + 3 sardines + 1 pickle 1 sandwich Rate = Δ [ sandwiches ] = 1 Δt 2 Δ[ bread] Δt etc. 5. The Rate changes with time, in general: H 2 (g) + 2ICl(g) I 2 (g) + 2HCl(g) E. Experimental Determination of the Rate. (Section 13.2) 1. One must be able to measure the concentration of at least one of the reactants or products during the course of the reaction. 2. For slow reactions, one could withdraw samples and analyze them during reaction. 3. More conveniently, one could monitor a physical property of the sample which depends on the concentration of one of the reactants. Example: the absorption of light by one of the species measured by spectroscopy. Through Beers Law, we know that Absorbance concentration at a fixed wavelength. F. Dependence of Rate on Concentrations of Reactants - the Rate Law. (Section 13.3) 1. A rate law expression takes the form: Rate = k[a] x [B] y... Chapter 13 Page 3

4 2. Expresses how rate varies with concentrations of species present. 3. k = rate constant. 4. Powers x and y bear no necessary relationship with balancing coefficients but do depend on the mechanism of the reaction. Example: Rate of one-car accidents = k[cars] 1 Rate of two-car accidents = k [cars] 2 Rate of one-car accidents is said to be 1st-order with respect to concentrations of cars. Rate of two-car accidents is 2nd-order with respect to concentrations of cars. 5. Suppose you found that: Rate = k[h 2 O 2 ][I - ] Reaction is 1st-order with respect to H 2 O 2 1st-order with respect to I - and 2nd-order overall 6. Units of k depend on overall order. 7. k is independent of time, is independent of concentrations, but varies with T and presence of catalyst. G. Experimental Determination of Rate Law. (Section 13.3) 1. Method of initial rates: measure rate at beginning of reaction, when all concentrations are known repeat measurement, vary initial concentration on one species, holding the others fixed 2. Example: The iodine clock reaction at a given temperature. Rate depends on concentration of three species: Rate = k[io 3 - ] a [I - ] b [H + ] c Need to discover the exponents a, b, and c by experiment. Chapter 13 Page 4

5 Initial concentrations (M) Run Initial Rate [IO 3 - ]0 [I - ] 0 [H + ] x 10-5 M s x x x Find the rate law and the rate constant: First, compare Run 1 and Run 2. What changed? Only [IO 3 ] o It doubled and the rate doubled, therefore a = 1 Next, compare Run 1 and Run 3. What changed? Only [I - ] o It doubled and the rate quadrupled, therefore b=2 By comparing Run 1 and 4, we find that c=2 Therefore the Rate Law must be: - - Rate = k[io 3 ][I ] 2 [H + ] 2 Pick one run to solve for k: k = M s 1 ( 0.10M ) 0.10M ( ) 2 ( 0.10M ) = 2.00 M 4 s 1 2 H. Change of Concentrations with Time - Integrated Rate Laws. (Section 13.4) 1. Can solve for [reactant] as function of time if rate law known. 2. This generates an integrated rate equation = equation that relates concentration of species to its initial concentration and time elapsed t. 3. Can also determine t 1 2 = half-life of a reactant = time required for half of initial amount of reactant to be used up. Chapter 13 Page 5

6 4. First-Order Reactions: of the general form aa products Rate = 1 a Δ[ A] Δt = k[a] By integration of the above equation, we obtain the integrated first order rate law, which can be rearranged in a number of forms: [A] = [A] o e -akt [ A] A [ ] o = e akt ln [ A] = -akt A [ ] o ln [ A] o = +akt [ A] [ A] log o 10 = [ A] akt if [A] o, k, and t are known, can solve for [A]; usually a = Problem: The initial concentration of reactant species A in the 1st order reaction: A B is 2.0 M. What is the concentration after 10 seconds if the rate constant k = 0.30 s -1? [A] = [A] o e -akt a = 1, k = 0.30 s -1, t = 10 s, [A] o = 2.0 M [A] = 2.0 M e s-1 10 s = 2.0 M x 4.97 x 10-2 = 9.95 x 10-2 M =.0995 M 6. Derive relation of t 12 to k for 1st order reaction. Chapter 13 Page 6

7 Suppose t elapsed = t 12 Then [A] = 1 2 [A] o t 1 2 = ln 2 ak ln [ A] o [ ] o 1 A 2 = akt 12 ; ln 2 = akt Special Property of 1st Order Reactions: t 1 2 is a constant. Thus if concentration goes to 1 2 during first half-life, it goes to 1 4 in 2nd half-life, to 1 8 in 3rd half-life. Plot: Suppose [A] o = 1 M, t 1 2 = 10 seconds. 8. Radioactive isotope concentration decays by 1st order kinetics. 9. Since [A] = [A] o e -akt intercept b = ln[a] o ln [A] = ln [A] o - akt y = b + m x ln [A] slope = -ak straight line t 10. Second-Order Reactions having a rate law of form aa products 1 Rate - k[a] [ A] 1 2 = akt [ A] o Chapter 13 Page 7

8 Plot: t 1 2 = 1 ak A [ ] o does depend on [A] o I. Collision Theory of Reaction Rates. (Section 13.5) 1. Helps to expose other factors influencing rate: temperature catalysts 2. Collision theory = a. reactants must collide before they can react. b. only some of the collisions are effective in producing products: must have correct orientation (see Figure below) Figure Chapter 13 Page 8

9 must have sufficient energy (see Figure 13.13) Figure Rate = (collision frequency) x (energy factor) x (orientation factor) determined by conc., kinetic energies sizes of particles of molecules enter here 4. Energy factor = e E a RT where E a = activation energy Chapter 13 Page 9

10 5. Yields Rate Expression: Rate = A e E a RT (concentrations) where: A contains info on the orientation, size of particles e E a RT is the energy factor Rate constant is then identified as: (Section 13.6) k = A e E a RT (Arrhenius Equation) Says that rate constant, and hence the rate, depends exponentially on the T: k as T A = Arrhenius pre-exponential factor, called also the frequency factor ln k = ln A - E a RT 6. Another form of Arrhenius equation for comparing k at two T s: ln k 2 = E a k 1 R 1 1 T 1 T 2 J. Transition State Theory. (More generalized theory) 1. Reactants must pass through a short-lived, high energy intermediate state, called the transition state. Chapter 13 Page 10

11 2. Theory uses statistical arguments to calculate the concentration of species in this state: Rate = frequency factor x [transition state] 3. Result comes out virtually same as collision theory but can be applied to a much wider variety of reactions. Part Two: Reaction Mechanisms A. Relationship Between Reaction Mechanism and Rate Law. (Section 13.7) 1. Mechanism = step-by-step pathway by which reaction occurs. 2. Mechanism = series of elementary reaction steps. Example: decomposition of N2O5 2 N2O5 4 NO2 + O2 Mechanism: Figure Chapter 13 Page 11

12 3. For any single elementary step, the rate is proportional to the product of concentrations of each reactant molecule: A B + C Rate = k[a] A + B C Rate = k[a][b] unimolecular elementary rxn bimolecular elementary rxn 4. Rxn is never faster than its slowest step, called the rate-determining step. (i.e. the bottleneck ) 5. Example #1. Overall reaction: NO 2 (g) + CO(g) NO(g) + CO 2 (g) Experimentally found that: Rate = k[no 2 ] 2 What mechanism would be consistent with this observation? Consider reaction progressing by TWO elementary steps: (1) NO 2 + NO 2 N 2 O 4 (slow) (2) N 2 O 4 + CO NO + CO 2 + NO 2 (fast) NO 2 + CO NO + CO 2 (overall) Rate (step 1) = k[no 2 ][NO 2 ] = k[no 2 ] 2 Rate (step 2) = k[n 2 O 4 ][CO] But step 1 is slow, so it limits the rate. Rate = k[no 2 ] 2 Note that this mechanism proposed existence of reaction intermediate N 2 O 4. Other mechanisms can be proposed which also yield the correct rate law. Therefore, simply finding a mechanism which gives the correct rate law does not guarantee that this is the correct mechanism. Other mechanisms may be proposed that might also predict the same rate law. We can only eliminate inconsistent mechanisms, never prove the correct one. Chapter 13 Page 12

13 6. Example #2. Overall reaction: 2 NO(g) + Br 2 (g) 2 NOBr(s) Observed rate law: Rate = k[no] 2 [Br 2 ] Proposed mechanism: A single step involving simultaneous collision of 3 molecules. NO(g) + NO(g) + Br 2 (g) 2 NOBr(s) This immediately produces the observed rate law. However, likelihood of ternary collision is small. Another proposed mechanism: (1) NO + Br 2 NOBr 2 (fast, equilibrium) (2) NOBr 2 + NO 2 NOBr (slow) Since (2) is slow, it is rate-determining, so: Rate = k 2 [NOBr 2 ][NO] But NOBr 2 is a transient intermediate species, so [NOBr 2 ] needs to be written in terms of conc. of other species. Since (1) is in equilibrium: forward rate = reverse rate k 1 f [ NO] [ Br 2 ] = k 1r [ NOBr 2 ] and [ NOBr 2 ] = k 1 f k 1r Substituting back in: k 1 f Rate = k 2 k 1r [ NO] [ Br 2 ] NO [ ] = k[no] 2 [Br 2 ] [ NO] [ Br 2 ] This is consistent with the observed rate law, so this mechanism is plausible. Chapter 13 Page 13

14 7. A generalization can be made: The experimentally determined reaction orders indicate the number of molecules of those reactants involved in; (1) the slow step only, if it occurs first, or (2) the slow step and any fast equilibrium steps preceding the slow step. B. Catalysis. (Section 13.9) 1. Catalyst =substance that increase the rate of a given reaction without being used up by the reaction. 2. They operate by allowing reactions to occur by alternative pathways with lower activation energies. Figure Since rate constant depends on activation energy E a as follows: k = A e -E a RT if E a e -E a RT k rate 4. Two categories of catalysts: Chapter 13 Page 14

15 a. homogeneous - exists in the same phase as the reactants. b. heterogeneous - exists in different phase than reactants. 5. Example of homogeneous catalysis. Strong acids (i.e. the H + ion) catalyze hydrolysis of esters. Overall: Postulated mechanism: 6. Example of heterogeneous catalysis (contact catalyst). a. First step is usually adsorption of reactant on a solid surface (which acts as the catalyst): Figure Chapter 13 Page 15

16 b. Catalytic converters work this way, catalyzing oxidation of unburned fuel and CO. Figure Pt 2CO( g) + O 2 ( g) NiO 2CO 2 ( g) This reaction without catalysts would take thousands of years. 7. Enzymes are protein molecules which serve as catalysts for biochemical rxns. Figure Chapter 13 Page 16

17 8. CFC s (Chlorofluorocarbons) are thought to decompose in the upper atmosphere into ClO radicals that catalyze the decomposition of the ozone layer. 9. CFC s have been banned from many uses. (refrigerants, aerosol spray propellants) Chapter 13 Page 17

18 NOTES: Chapter 13 Page 18