CH16_kin_notes.W7_8

CH16_kin_notes.W7_8 - 03 concentration (moi/L x 105)...

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Unformatted text preview: 03 concentration (moi/L x 105) CHEMICAL KINETICS ‘ ' Chapter 16 - a knowledge of the rate, or time dependence, of chemical change is of critical importance for the successful syn- thesis of new materials and for the utilization of the energy generated by a reaction. During the past century it has become clear that all macroscopic chemical processes consist of many elementary chemical reactions that are themselves simply a series of encounters between atomic or molecular species to understand the time-dependence chemical kineticists have focussed on sorting out all of the elementary chemical reactions involved in a macro— scopic chemical process and determining their respective rates. Yuan T. Lee, Nobel Prize in Chemistry, 1986 ATOMS, MOLECULES, AND IONS MUST COLLIDE IN ORDER TO REACT Rates of Chemical Reactions factors that affect'rate rate expressed in terms of reaction stoichiometry EX 1. a) Consider the combustion of hydrogen gas: 2 H2(g) + 02(g) —> 2 H20(g) If hydrogen is burning at the rate of 0.85 mol 3—], what is the rate of consumption of oxygen? What is the rate of formation of water vapor? b) The reaction 2NO(g) + C12(g) —> 2NOCl(g) is carried out in a closedvessel. If the partial pressure of NO is decreasing at the rate of'23 mm Hg min‘l, what is the rate of change of the total pressure of the vessel? how rate measured: average, instantaneous, initial rate FIG 11 - Appearance of Product N02(g) + C0(g) --9 N0(g) + C02(g) FIG I - Disappearance of Reactant C2H4(g) + 03(8) —‘> C2H4O(g) + 02(8) 4.00} Rate (mol/L-s) a 10.0x10-7 3m \ X , 350,104 0-0350 '"§;g$;:°gg;;fa“°°s Airfl- _ = - - - = —5 -1 —1 c 7.80x10 7 150 ~ 50" 9'8 “0 m°l L 5 At d 2.50x1o-7 0-9300 " e 1.30x10-7 a‘\ \ 0.0250 f‘ 4 00200 ‘ 2.00 6 AER}. .1; | g "0.0174: N : “.8qu Ms‘ % 0.0150 INVI’IAI. Rave ' I t 00100 I“ . Average rate _ g “I w AENO] . \ 100 : e - 0.0050 = 0'0155 ' 0 =1.28 x 10—4 mol L-1s-1 . a 50 — o M? Pr—l—l—L—r—T—l \ 0.0000 .’ 0 50 150 250 -2- Rate Laws - Effect of Concentration on Rate (a way to categorize reactions) examples rate law (from experiment!) H203) + 12(3) —9 2H1(g) r = k[H2][12] 2No(g> + 020;) —> 2N02(g) r = k1N02121021 2NzOsQE') -> 4N02(g) + 02(8) r = klNzosl H2P03(aq) + OH_(aq) —> HPO§(aq) + H2(g) r = k[HZPO§][OH—]2 3/2 H20) + Br2<g> —> 2HBr(g> r = leZHBrZ] [Brz] + k'[HBr] order rate constant determination (see Rate Laws on website for mathematical desCription) initial rate method: EX 2. Consider the gas-phase reaction between nitric oxide and bromine at 273°C. 2 NO(g) + Br2(g) —> 2 NOBr(g) The following data for the initial rate of appearance of NOBr were obtained. Experiment [NO] [Brz] Initial Rate 1 0.10 M 0.20 M 24 M 5'1 2 0.25 0.20 150 3 0.10 0.50 60 4 0.35 0.50 735 a) Determine the rate law b) Calculate the average value of the rate constant from the four data sets. c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2? d) What is the rate of disappearance of Br2 when [N0] = 0.075 M and [Brz] = 0.25 M?‘ FIG III - Rate Law from Linear Regression on Initial Rate vs Concentration Data NHZWI) + NOEWI) —> N2(g) + 2 H200) Experiment Initial NH4+ MCnuufloinitiai No; MLWa‘LObserved Initial Jar. Number Concentration (M) Concentration (M) Rate (M/s) 1 0.0100 4.605 0.200 — 1.6m 5.4 x 10-7 ‘ NAB 2 0.0200 £53196 0.200 “ 10.8 x 10“7 ‘ I33" 3 0.0400 «3.2161 0.200 “ 21.5 x 10'7 . " I105 4 0.0600 1.313 0.200 " 32.3 x 10—7 ‘ In.“ 5 0.200 - (.000 0.0202 $309; 10.8 x 10—7 ‘ '33“ 6 0.200 " 0.0404 -3.a,o<1 21.6 x 10—7 -‘ ‘3-05 7 0.200 N 0.0606 - 2.303 32.4 x 10‘7 ‘ I'Ma‘l 8 0.200 “ 0.0808 ‘1516 43.3 x 10‘7 ‘ R35 r = a [manna-11 51m. = not; + xLWLNHLp‘] + 712m, mag] ‘ -3- Integrated Rate Laws (Dependence of Concentrationon Time) 1. first-order reaction FIG IV - Plot of a First-Order Reaction N205(g) —> 2 NQz(g) + 1/2 02(g) Time (s) , 1x105 2x105 3x105 1.0- “M..-” r .V“ 77.. .,,. .,, V r .. _. . V . .. h .-..r......[ integrated rate [Nzog] (mol L") t'la} ll": [*1 l x'“‘er€¢9‘?!9.ITa9§1n 7 half—life i In [N205]- o'a 0- 1x105 2x105 3x105 Tlme(s)~ EX 3. The first—order rate constant for the decomposition of N205 N205(g) -> 2 N02(g) + 1/2 02(g) at 70°C is 6. 82 x 10'3 5—1. Suppose we start with 0.0250 mol of N205 (g) in 2.0 L. a) How many moles of N205 will remain after 2.5 min? b) How many minutes will it take for the quantity of N205 to drop to 0.010 mol? 0) What is the half-life of N205 at 70°C? 2. second-order reaction FIG V - Plots of a Second-Order Reaction ' N02<g) —> NO(g) + 1/2 02<g> integrated rate 1n[NOz] r ‘5'80 100 200 300 Time (s) half—life 1/ [N02] O 100 200 300 Time (s) EX 4. At 300°C the rate constant for the second order reaction V N02(g) —) NO(g) + 1/2 02(g) is 0.543 M‘1 5—1. If the initial concentration of N02 in a closed vessel is 0.0500 M, what is the concentration remaining after 0.500 hr? 3. zero-order reaction 1 integrated rate -5- FIG VI - A Reaction Plotted as Various Orders 2 N02(g) -> 2 N0(g) + 02(g) 0 W — t 0.010 tn]: $31.9 *‘ [no o'ooe ‘ MR]; 0.002 [MUM om 0.00“ 0.0055 own 0.00“ h I s 0 so I09 3.00 300 FIG VII ,- A Comparison of Different Rate Laws For a general chemical reaction: aA -—> products where a denotes the stoichiometric coefficient. The rate law, integrated rate law, half—life, and form of concentration data giving a linear plot are summarized for various orders the reaction could follow. In the table ka = ak where k is the rate constant and a the stoichiometric coefficient. . Hanna, ln 2 . or [A] = [Abe—"at i — ~1~ M r [A] — [A10 “ ka [A10 [A] —1—— - ~1— + 2k t 3 ; vs t [A]2 ‘ [A13 “ 2k. [A13 [A12 1 2'1—1 — 1 1 + (n — 1)kat m——_1 (n — 1)ka [A10 VSt [A]"“ =1A10-1 1A1"-l 4. fractional lives _6_ How Do Chemical Reactions Occur? Explain Effect of Temperature on Rate collision model - kinetic theory interpretation: to react molecules need 1. to collide FIG VIII - Collision of C] with NOCl 1) NOCl(g) —> no reaction 2) ClON(g) ——> C12(g)+NO(g) Before collision Collision A After collision flame-new Cl(g) + { 2. to be properly oriented (a) Effective collision 3. to have a minimum translational kinetic energy or activation energy (minimum collision energy reactants need in order to form products) “cfmewmsim Collision » mAflercollisinn 0 m. moo (b) lncflcclivc collision FIG IX - Pressure vs Time for FIG X - Reaction Coordinate for Methyl Isocyanate Rearrangement Methyl Isocyanate Rearrangement CH3NEC(g)A —-> CH3CEN(g) CH3NEC(g) —> CH3CEN(g) 160 ' 3140 r ~- V 120 %100 5 30 — A—— c a 60 ——-- ~-~~ H3041} a 40 T (Activated -V g 20 00 10,000 20,000 30,000 5-2 Time (s) U 5-0 E) 4.8 r 4.6 r 4.4 A ~ a 4,2 ——— —— g 4.0 e 3.8 — —— —— — .9 3.6 3.4 0 10,000 20,000 30000 Time (s) where energy comesvfrom: FIG XI - Maxwell-Boltzmann Distribution FIG XII - Maxwell-Boltzmann Distribution of Molecular Speeds of Molecular Kinetic Energies Lower temperature éeu moving at about 400 m/s than at any other speed ~ Higher temperature 0 0 Minimum kinetic -‘ energy required Many more molecules ar moving at 1600 mls whe /02 at 25 °C the sample is at 1000 °c than when it is at 25 °C 02 at 1000 Dc ‘3», .EF '5 . L- o .n E :i Z 0 200 400 600 800 1000 1200 1400 1600 1800 Molecular speed (mls) Kinetic energy ————> -7- kinetic theory rate model for elementary reaction: B + C ——> products — to react molecules need 1. to collide- 2. to be properly oriented 3. to have a minimum kinetic energy, EEl then the rate, r becomes r = (total # of collisions) >< (fraction oriented) >< (fraction with Ea) [B] [C] (size) (speed) (fraction oriented) {Ea/RT k [B] [C] = A 6’53"” [B] [C] Arrhenius equation EX 5. The rate constant for the first—order isomerization of cyclopropane, C3H6, to propene is 6. 7 x 10—4 5—1 at 500°C. What is its value at 300°C? A = 1. 6 x 1015 “, Ea = 272 kJ/mol EX 6. Solve for k(300) for the preceding example only using k(500) = 6. 7 X 10—4 s‘1 and Ea = 272 kJ/mol (do not use the value for the pre-exponential factor, A) -8- How Do Chemical Reactions Occur? Explain Effect of Concentration on Rate (Reaction Mechanisms) elementary steps molecularity principle of detailed balance rate-determining step reaction coordinate example Gas phase reaction of chlorine with‘chloroform is described by the stoichiometric equation C12(g) + CHC13(g) —> HCl(g) + CC14(g) ' V A proposed mechanism for the reaction is C12 <=> 2 fast k1/k_1 c1 + CHC13 —> HCl + ccr3 slow k2 CCI3 + C1 —> CC14 fast k3 working with reaction mechanisms: sequence of elementary steps given steps sum to overall reaction (stoichiometric equation) kf(Ea’f), kr(Ea’r) for each step know rate law for each step from molecularity overall rate law found if needed, rate-determining step approximation'used prior rapid equilibria used to determine concentration of species not in overall stoichiometry intermediates, catalysts, inhibitors identified rate law compared with experiment, kgxp identified with rate constants and equilibrium constants of individual steps -9- Reaction Mechanisms, Reaction Coordinates, and Activation Energy, Ea I. CH3NC(g) ——> CH3CN(g) EX 7. In this isomerization of methyl isonitrile to acetonitrile the rate was experimentally observed to obey the rate law r = kexp[CH3NC]. ’ a) Show that the following mechanism is consistent with the rate law and identify kexp. proposed mechanism: CH3NC(g) —> CH3CN(g) b) From the data in the table deduce the activation energy E, and the frequency factor A in the Arrhenius equation. Be sure to check the correlation coefficient on your calculator. 2.52 x 10‘5 5.25 x 10—5 6.30 x 10*4 3.16 x 10‘3 [cc = —o.9997, 158 kJ/mol, 1.82 x 1013 S4] 11. 03(g) + NO(g) L> 02(5)) + N02(g) (Fig 16.17, prob. 16.68) EX 8. For this reaction of the nitric oxide depletion of ozone in the upper atmosphere the rate. was experimentally observed to obey the rate law r = kexp[NO][O3]. a) Show that the following mechanism is consistent with the rate law and identify kexp. proposed mechanism: 03(8) + NO(g) —> 02(8) + N02(g) k (M'1 s“) T(K) b) From the data in the table deduce the activation energy Ed and the frequency factor A in the Arrhenius equation. [cc = -O.9970, 11.9 kJ/mol, 1.55 x 1012 M‘1 s—1] 0) If AH0 = —198.8 kI/mol for the reaction what is Ea for the reverse reaction? [210.8 kJ/mol] III. N02(g) + CO(g) —> NO(g) + C02(g) (sample prob. 16.3, section 16.7, prob. 16.71) EX 9. The experimental rate law below 500 K is r = kexp[N02]2. a) Show that the following mechanism is consistent with the rate law and identify kexp. N02 + N02 —) N03 + SI OW k 1 N03 + —> N02 + C02 fast k2 b) What is the function of N03? (3) Determine AH ° for the reaction from the following data. CO AH (kJ/mol) -10- IV. 2 NO(g) + 02(g) —> 2 N02(g) (Table and sample prob. 16.2, section 16.7, prob.16.64) EX 10. For this important smog reaction the experimental rate law is r = kexp[NO]2[02]. a) Show that the following mechanism is consistent with the rate law and identify kexp. proposed mechanism: NO + NO <=> N202 fast kl/k_1 N202 + 02 ——) slow [(2 b) What is the function of N202? 0) Activation energies are: 82 (r1), 205 (r_1) , and 82 (r2) kJ/mol. If the Arrhenius frequency factors are all approximately the same what happens to the overall rate as the temperature is increased? [r decreases with T ] V. 2Ce4+(aq) + T1+(aq) a 2 Ce3+(aq) + Tl3+(aq) EX 11. The experimental rate law is r = kexp[Ce4+][Ag+][Tl+]/[Ce3+]. a) Show that the following mechanism is consistent with the rate law and identify kexp. proposed mechanism: Ce4+ + Ag+ <=> Ce3+ + Ag2+ fast k1/k_1 T1+ + Ag2+ —> T12+ + Ag+ slow k2 T12+ + C6” —> T1“ + Ce3+ fast k3 b) What is the function of Ag“? -11- VI. I'(aq) + CID—(aq) —)' IO_(aq) + Cl“(aq) (prob.16.62) EX 12. For this reaction in alkaline solution the experimental rate law is r = kexp[F][ClO‘]/[OH—]. a) Show that the following mechanism is consistent with the rate law and identify kexp. proposed mechanism: ClO‘ + H20 <=> HOCl + OH‘ fast Iq/k_1 r + HOCI"—> H01 + Cl‘ slow k2 OH‘ + HOI <=> 10‘ + H20 fast k3/k_3 b) What is the function of OH‘? VII. H2(g.) + 12(g) —> 2 HI(g) (sample prob. 16.3 follow—up prob., prob. 16.81) EX 13. The following two mechanisms have been proposed for this reaction. Show that they give the proper stoichiometry, determine the rate law they predict, give the function of all species, state the molecularity of each step, and comment on each mechanism. mechanism I: mechanism H: 12 <=> 21 fast kI/k_1 12 <=> 21 H2 '1‘ '—> slow [(2 H2, + I <=> H21 H21+I —> 2H1 -12- VIII. 03(g) + O(g) —> 202(g) (sample prob. 16.9, section 16.8, prob. 16.68) EX 14. Depletion of the ozone layer is believed to be caused by chlorine—catalyzed O3 decomposition. For the uncatalyzed reaction Eaif = 19 kJ/mol. For the following mechanism, Ealif = 2. 1 and Eazyf = 0.4 kJ/mol. proposed mechanism: Cl+O3 ——-> ClO+02 C10 + O —-> C1 + 02 a) What is AH0 for the reaction? {-390 kJ/mol] b) What is Em, for the uncatalyzed reaction? [409 kJ/mol] c) Which step in the above mechanism is faster? d) If AH ° for the first reaction is ~165 kJ/mol, sketch the energy profile for the reaction 03 (:1 AH}(kJ/mol) 142.3 121.7 IX. Reaction Coordinate for aReaction A ——> B (prob. 16.67) EX 15. Based on the following reaction profile Energy Reaction pathway a) What are the intermediates formed in the reaction? b) How many transition states are there? Identify them. 0) How many activated complexes are there? Identify them. d) Which step is the fastest? 6) Which step is the slowest? 1) Is the reaction A ——> B exothermic or endothermic? ...
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This note was uploaded on 04/13/2010 for the course CHEM 114 taught by Professor Jursich during the Spring '08 term at Ill. Chicago.

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CH16_kin_notes.W7_8 - 03 concentration (moi/L x 105)...

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