Chemical Kinetics. Reaction Rate. Reaction Rate. Reaction Rate. Reaction Rate. Chapter 13: Chemical Kinetics: Rates of Reactions

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1 Chemical Kinetics The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products Chapter 3: Chemical Kinetics: Rates of Reactions 2008 Brooks/Cole 2008 Brooks/Cole 2 Reaction Rate Reaction Rate Combustion of Fe(s) powder: 2008 Brooks/Cole Brooks/Cole 4 Reaction Rate Change in [reactant] or [product] per unit time. Cresol violet (Cv + ; a dye) decomposes in NaO(aq): Cv + (aq) + O - (aq) CvO(aq) change in concentration of Cv rate = + = elapsed time Δ [Cv + ] Δt Reaction Rate Average rate of the Cv + reaction can be calculated: Time, t [Cv + ] Average rate (s) (mol / L) (mol L - s - ) x x x x x x x x x x x x x x x x x Brooks/Cole Brooks/Cole 6

2 Reaction Rates and Stoichiometry Cv + (aq) + O - (aq) CvO(aq) Stoichiometry: Loss of Cv + Gain of CvO Rate of Cv + loss = Rate of CvO gain Another example: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Reaction Rates and Stoichiometry For any general reaction: a A + b B c C + d D The overall rate of reaction is: Rate = Δ[A] = Δ[B] = + Δ[C] = a Δt b Δt c Δt + Δ[D] d Δt Reactants decrease with time. Negative sign. Products increase with time. Positive sign Negative rate Positive rate Rate of loss of N 2 O 5 divided by -2, equals rate of gain of O Brooks/Cole Brooks/Cole 8 Reaction Rates and Stoichiometry For: 2 (g) + I 2 (g) 2 I (g) the rate of loss of I 2 is mol L - s -. What is the rate of formation of I? Rate = Δ[ 2] = Δ[I 2] = Δt Δt Δ[ Rate = 2 ] = ( ) = Δt + Δ[I] 2 Δt + Δ[I] 2 Δt Δ[I] So = mol L Δt - s - Average Rate and Instantaneous Rate Graphical view of Cv + reaction: [Cv + ] (mol/l) 5.0E-5 4.0E-5 3.0E-5 2.0E-5.0E t (s) 2008 Brooks/Cole Brooks/Cole 0 Average Rate and Instantaneous Rate Effect of Concentration on Reaction Rate Rate may change when [reactant] changes. Cv + example shows this. For Cv + the rate is proportional to concentration. t [Cv + ] Rate of Cv + Rate/[Cv + ] (s) (M) loss (M / s) (s - ) x x x x Brooks/Cole 2008 Brooks/Cole 2 2

3 Rate Law and Order of Reaction Determining Rate Laws from Initial Rates A general reaction will usually have a rate law: rate = k [A] m [B] n... The orders are usually integers (-2, -, 0,, 2 ), but may also be fractions (½, ⅓ ) 2008 Brooks/Cole Brooks/Cole 4 Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base: C 3 COOC 3 + O - C 3 COO - + C 3 O Dividing the first two data sets: 4.5 x 0-4 M/s = k (0.040 M) m (0.080 M) n 2.2 x 0-4 M/s = k (0.040 M) m (0.040 M) n Rate law: rate = k [C 3 COOC 3 ] m [O - ] n raised to any power = 2008 Brooks/Cole Brooks/Cole 6 Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates Use experiments 2 & 3 to find m: 9.0 x 0-4 M/s = k (0.080 M) m (0.080 M) n 4.5 x 0-4 M/s = k (0.040 M) m (0.080 M) n 2008 Brooks/Cole Brooks/Cole 8 3

4 Determining Rate Laws from Initial Rates If a rate law is known, k can be determined: k = Using run : rate [C 3 COOC 3 ][O - ] k = 2.2 x 0-4 M/s (0.040 M)(0.040 M) k = M - s - = L mol - s - Could repeat for each run, take an average But a graphical method is better. The Integrated Rate Law Calculus is used to integrate a rate law. Consider a st -order reaction: A products Δ[A] rate = = k [A] Δt d [A] = = k [A] dt (as a differential equation) Integrates to: ln [A] t = k t + ln [A] 0 y = m x + b (straight line) 2008 Brooks/Cole Brooks/Cole 20 The Integrated Rate Law Zeroth-order reaction The Integrated Rate Law The reaction: A products doesn t have to be st order. Some common integrated rate laws: Order Rate law Integrated rate law Slope [A] ln[a] slope = -k time t First-order reaction slope = -k time t Rate data for the decomposition of cyclopentene C 5 8 (g) C 5 6 (g) + 2 (g) were measured at 850 C. Determine the order of the reaction from the following plots of those data: 0 rate = k [A] t = -kt + [A] 0 -k rate = k[a] ln[a] t = -kt + ln[a] 0 -k 2 rate = k[a] 2 [A] t [A] 0 +k /[A] Second-order reaction slope = k y The most accurate k is obtained from the slope of a plot. x time t The reaction is first order (the only linear plot) k = - x (slope) of this plot Brooks/Cole Brooks/Cole 22 alf-life alf-lives are only useful for st -order reactions. Why? alf-life For a st -order reaction: ln[a] t = -kt + ln[a] 0 When t = t /2 [A] t = ½[A] 0 Then: ln(½[a] 0 ) = -kt /2 + ln[a] 0 ln(½[a] 0 /[A] 0 ) = -kt /2 {note: ln x ln y = ln(x/y)} ln(½) = -ln(2) = -kt /2 {note: ln(/y) = ln y } 2008 Brooks/Cole Brooks/Cole 24 4

5 alf Life t /2 of a st -order reaction can be used to find k. For cisplatin (a chemotherapy agent): ½ the cisplatin lost after 475 min. (0.000 M M) 0.00 [cisplatin] halves every 475 min [cisplatin] (mol/l) k = ln = 475 min t /2 =.46 x 0-3 min - Calculating [ ] or t from a Rate Law Use an integrated rate equation. a) [reactant],600.s after initiation. b) t for [reactant] to drop to /6 th of its initial value. c) t for [reactant] to drop to mol/l t (min) 2008 Brooks/Cole Brooks/Cole 26 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = mol/l and t /2 = 400.s (a) Calculate [reactant],600.s after initiation. Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = mol/l and t /2 = 400.s (b) Calculate t for [reactant] to drop to /6 th of its initial value. st order: k = ln 2/ t ½ = 0.693/(400. s) =.733x0-3 s - and ln [A] t = -kt + ln [A] 0 so ln[a] t = -( s - )(600 s) +ln(0.500) ln[a] t = = [A] t = e = mol/l [reactant] 0 [reactant] 0 2 t /2 [reactant] 0 [reactant] t /2 4 [reactant] 0 [reactant] 8 0 t /2 [reactant] 0 [reactant] 0 t 8 6 /2 4 t /2 = 4 (400 s) = 600 s Note: part (a) could be solved in a similar way. 600 s = 4 t /2 so M Brooks/Cole Brooks/Cole 28 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = mol/l and t /2 = 400.s (c) Calculate t for [reactant] to drop to mol/l? Nanoscale View: Elementary Reactions From part (a): k =.733 x 0-3 s - then ln [A] t = -kt + ln [A] 0 ln (0.0500) = -( s - ) t + ln(0.500) = -( s - ) t t = s - t =.33 x 0 3 s 2008 Brooks/Cole Brooks/Cole 30 5

6 Elementary reactions Unimolecular Reactions 2-butene isomerization is unimolecular: unimolecular bimolecular 3 C C=C C 3 (g) 3 C C=C C 3 (g) unimolecular unimolecular 2008 Brooks/Cole Brooks/Cole 32 Unimolecular Reactions Potential energy (0-2 J ) Initial state E a = 435 x 0-2 J transition state or activated complex E a is the activation energy, the minimum E to go over the barrier. ΔE = -7 x 0-2 J Final state Reaction Progress (angle of twist) Exothermic overall cis-trans conversion twists the C=C bond. This requires a lot of energy (E a = 4.35x0-9 J/molecule = 262 kj/mol) Even more (4.42x0-9 J/molecule) to convert back. Transition State 2008 Brooks/Cole Brooks/Cole 34 Bimolecular Reactions e.g. Iodide ions reacting with methyl bromide: Bimolecular Reactions I - (aq) + C 3 Br(aq) IC 3 (aq) + Br - (aq) transition state I - must collide in the right location to cause the inversion. I - must collide with enough E and in the right location to cause the inversion Brooks/Cole Brooks/Cole 36 6

7 Bimolecular Reactions Potential energy (0-2 J ) transition state Reactants (initial state) Reaction Progress (changing bond lengths and angles) Also has an activation barrier (E a ). Forward and back E a are different. ere the forward reaction is endothermic. E a = 26 x 0-2 J Products (final state) ΔE = 63 x 0-2 J Temperature and Reaction Rate Increasing T will speed up most reactions. igher T = higher average E k for the reactants. = larger fraction of the molecules can overcome the activation barrier. number of molecules 25 C 75 C kinetic energy E a Many more molecules have enough E to react at 75 C, so the reaction goes much faster Brooks/Cole Brooks/Cole 38 Temperature and Reaction Rate Reaction rates are strongly T-dependent. Data for the I - + C 3 Br reaction: Temperature and Reaction Rate k (L mol - s - ) T (K) T (K) k (L mol - s - ) x x x x x x 0 - -E a / RT k = A e Quantity Name Interpretation and/or comments A Frequency factor ow often a collision occurs with the correct orientation. E a Activation energy Barrier height. e -Ea/RT Fraction of the molecules with enough E to cross the barrier. T Temperature Must be in kelvins. R Gas law constant 8.34 J K - mol Brooks/Cole Brooks/Cole 40 Determining Activation Energy Take the natural logarithms of both sides: ln ab = ln a + ln b -E a / RT ln k = ln A e -E a / RT ln k = ln A + ln e ln k = ln A + ln k = E a R E a RT T ln e + ln A ln e = A plot of ln k vs. /T is linear (slope = E a /R). ln k Determining Activation Energy The iodide-methyl bromide reaction data: intercept = slope = x 0 3 K /T (K - ) E a = -(slope) x R = -(-9.29 x0 3 K) 8.34 J K mol = 77.2 x 0 3 J/mol = 77.2 kj/mol A = e intercept = e A = 2.28 x 0 0 L mol - s - (A has the same units as k) 2008 Brooks/Cole Brooks/Cole 42 7

8 Rate Laws for Elementary Reactions Reaction Mechanisms When [ 3 O + ] is between 0-3 M and 0-5 M, rate = k [I - ][ 2 O 2 ] 2008 Brooks/Cole Brooks/Cole 44 Reaction Mechanisms Reaction Mechanisms 2 I - (aq) + 2 O 2 (aq) O + (aq) I 2 (aq) O(l) slow Shows the bonding in 2 O 2 fast fast overall 2 I O O + I O 2008 Brooks/Cole Brooks/Cole 46 Reaction Mechanisms Reaction Mechanisms A good analogy is supermarket shopping: You run in for item (~ min = fast step), but The checkout line is long (~0 min = slow step). Time spent is dominated by the checkout-line wait. In a reaction, a slow step may be thousands or even millions of times slower than a fast step. The overall rate is expected to be rate = k [ 2 O 2 ][ I - ] as observed! 2008 Brooks/Cole Brooks/Cole 48 8

9 Mechanisms with a Fast Initial Step Consider: 2 NO (g) + Br 2 (g) 2 NOBr (g) Mechanisms with a Fast Initial Step The generally accepted reaction mechanism is: 2 NO + Br 2 2 NOBr NO + Br 2 NOBr 2 fast, reversible 2008 Brooks/Cole Brooks/Cole 50 Mechanisms with a Fast Initial Step k NO + Br 2 NOBr 2 reversible, equilibrium k - rate forward = rate back k [NO][Br 2 ] = k - [NOBr 2 ] [NOBr 2 ] = k [NO][Br 2 ] k Brooks/Cole 5 Mechanisms with a Fast Initial Step The earlier rate law: becomes: rate = k 2 [NOBr 2 ] [NO] k [NO][Br 2 ] rate = k 2 [NO] 2008 Brooks/Cole 52 k - k k rate = 2 [Br k 2 ][NO] 2 - Now only contains starting materials - can be checked against experiment. Summary Elementary reactions: the rate law can be written down from the stoichiometry. unimolecular rate = k[a] bimolecular rate = k[a] 2 or rate = k[a][b] Catalysts and Reaction Rate 2008 Brooks/Cole Brooks/Cole 54 9

10 Catalysts and Reaction Rate 2-butene isomerization is catalyzed by a trace of I 2. 3 C C=C C 3 (g) 3 C No catalyst: rate = k [cis-2-butene] C=C A trace of I 2 (g) speeds up the reaction, and: rate = k [ I 2 ] ½ [cis-2-butene] k uncatalyzed k C 3 (g) Catalysts and Reaction Rate I 2 is not in the overall equation, and is not used up. The mechanism changes! 3 C C=C C 3 I 2 splits into 2 atoms. Each has an unpaired e -. (shown by the dot) I attaches and breaks one C-C bond 3 C C 3 I C C 2008 Brooks/Cole Brooks/Cole 56 Catalysts and Reaction Rate Rotation around C-C 3 C C 3 I C C 3 C I C C C 3 3 C 3 C I C C C 3 C=C C 3 Loss of I and formation of C=C I 2 is regenerated + I Catalysts and Reaction Rate I 2 dissociates to I + I Reactants (initial state) Rotation around C-C I adds to cis-2-butene, (double single bond) E a = 262 kj/mol Transition state for the uncatalyzed reaction E a = 5 kj/mol ΔE = -4 kj/mol Reaction Progress I leaves; double bond reforms I + I regenerates I 2 Products (final state) 2008 Brooks/Cole Brooks/Cole 58 Catalysts and Reaction Rate Enzymes: Biological Catalysts 2008 Brooks/Cole Brooks/Cole 60 0

11 Enzyme Activity and Specificity Enzyme Activity and Specificity 2008 Brooks/Cole Brooks/Cole 62 Enzyme Activity and Specificity Enzymes are effective catalysts because they: Bring and hold substrates together while a reaction occurs. old substrates in the shape that is most effective for reaction. Can donate or accept + from the substrate (act as acid or base) Stretch and bend substrate bonds in the induced fit so the reaction starts partway up the activation-energy hill. Enzyme kinetics Potential energy, E Formation of the enzyme-substrate complex Reactants (initial state) E a E' a ΔE Reaction Progress Transition state for the uncatalyzed reaction Transformation of the substrate to products Products (final state) Activation energy E' a is much smaller than E a and so the enzyme makes the reaction much faster 2008 Brooks/Cole Brooks/Cole 64 Enzyme Activity and Specificity Enzyme catalyzed reactions: Catalysis in Industry C 3 O(l) + CO(g) RhI 3 C 3 COO(l) auto exhausts are cleaned by catalytic converters: 2 CO(g) + O 2 (g) Pt-NiO 2 CO 2 2 C 8 8 (g) + 25 O 2 Pt-NiO 6 CO 2 (g) O(g) 2 NO(g) catalyst N 2 (g) + O 2 (g) 2008 Brooks/Cole Brooks/Cole 66

12 Controlling Automobile Emissions forms a bond with the Pt surface Converting Methane to Liquid Fuel Methane is hard to transport. It can be converted to methanol: NO approaches the Pt surface dissociates into N and O atoms (each bonded to Pt) they form N 2 and O 2 and leave the surface. N and O migrate on the surface until they get close to like atoms C 4 (g) + ½ O 2 (g) CO(g) (g) CO(g) (g) C 3 O(l) A Pt-coated ceramic catalyst allows the st reaction to occur at low T Brooks/Cole Brooks/Cole 68 2