Chemical Principles II
Chemical Principles II CHEM 1212K
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Date Created: 11/02/15
Chapter 12 Chemical Kinetics Thermodynamics does a reaction take place Kinetics how fast does a reaction proceed Reaction rate is the change in the concentration of a reactant or a product with time Ms A gt B AA AA change in concentration ofA over rate At time period At Al3 AB change in concentration of B over rate At time period At Because A decreases with time AA is negative 131 I v v 391 I if i r39 2 I x V t new 00 6 we a o 5 d We ac ggeoea 909909 00930 9 Coo 0 on i o a a gogo 3 300699 g O O 700 fang 0 g 90 69 00 one oooeogno o 0 6 o 0 39o gt e on e o 9 a o 9 o 0 0 n G 050 o 99 e a e a 90 a Gas 99 3 go one 0 99990 3990 9 so 0 9 06 9 0 0 o i 9 o so vo a 990 0 00 90909 o 110099 0 6900 0 9 0 o 939 09 a 0 o o o e 9 0 09 o 90 50 009090 a d o 0 one time 40 A molecules U A 3 t 3 3 ra e 2 At 395 B molecules 397 77 c 0 L 1 MB a rate 5 m 20 30 quot3 131 Br2 aq HCOOH aq gt2Br aq 2H aq CO2 g V quot7 E x 91 393 nm Detector light 1 300 400 500 600 ABr2 0c AAbsorption Wavelength nm Br2 aq HCOOH aq gt2Br aq 2H aq C02 9 Time s 00 500 1000 1500 2000 2500 3000 3500 4000 instantaneous rate rate for specific instance in time Br2 M 00120 00101 000846 000710 000596 000500 000420 000353 000296 average rate BiJmn 39 0011601 1111120 00101 0111181quot slope of mum tangent slope of 000300 tangent U 100 200 3 1 400 Its ABr2 Br2fina Br2initia At tfinal 39 1initial 131 TABLE 131 Hales of he Heac ion Between Molecular Brumine and Farmi Acid at 25quot rate 71 Tlrne s Bra MD Hate Mls quot 152 5 01 011120 4211 x 11gtquot 350 x 10quot 500 nm 152 x10 34 x10quot 10110 000846 295 10 1 50x 10 1500 000710 249 x 10quot J 51x 10quot 211110 000595 2 119 x In 351x 10 25110 0011501 1 75 x 10 350 gtlt 10 31100 100421 1411 x 10 5 352 x 10quot 35110 100353 121 X Inquot 3411 x 10 4000 1100296 1 04 x 10 5 151 x 10quot 5011 x In 7 400 x lo 5 rate 01 Brz A rate k Brz XIX IO39 7 12 rate 5 7 MIX 10 7 rate constant quot Brz 100 x IO 350 x103 5391 1 000201 000600 BrI M 1 1 00100 measure AP over time 2quot1202610I 2H20 I 02 9 PVnRT P RT 02RT 021 P E L AP rate At RT At 2Hzoz 3 7 2HzO I 02 9 IZU Inn snarl 012 mmHgmin nu 1 39 n m1 400 mm 800 mun l Imim IZUII Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed 1 AW AB rate 39 E Tt rate At aAbB gtcCdD ratel l lAClAD a At b At 0 At d At 131 i rate Write the rate expression for the following reaction CH4 9 2 g 39 002 9 20 g ACH4 l A02 A302 1 At 2 At At 2 At AH20 1 Example 141 T Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products a 17114 0Cltl gt CIaq OIaq b 302g gt 2038 0 4NHJg 5020 4 4N0g 6H20g 1 Example 141 T Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products a 17114 0Clth gt Claq OI th b 302g gt 2038 C 4NHg 502g gt 4N0g 6H20g Solution 21 Because each of the stoichiometric coef cients equals 1 AI AOC1 ACl AOI Ar Ar 7 Ar 7 Ar rate e b Here the coef cients are 3 and 2 so te71A0211A031 m a 3 At 2 Ar c In this reaction rate 1 ANH3 1 AOz lANO lAU IZO 4 A 5 AI 4 Ar 6 Ar Practice Exercise Write the rate expression for the following reaction CH4g 2028 gt C02g 2H20g 1 Example 142 T Consider the reaction 4N028 028 H 2N205g Suppose that at a particular moment during the reaction molecular oxygen is reacting at the rate of 0024 Ms a At what rate is N205 being formed b At what rate is N02 reacting 1 Example 142 T Consider the reaction 4N028 028 H 2N205g Suppose that at a particular moment during the reaction molecular oxygen is reacting at the rate of 0024 Ms a At what rate is N205 being formed b At what rate is N02 reacting Strategy To calculate the rate of formation of N205 and disappearance of N02 we need to express the rate of the reaction in terms of the stoichiometrie coef cients as in Example 131 t 7 71AtN02 7 A02 7 1 AMOS m e 4 Ar Ar 2 At We are given 24M I 00 s Continued where the minus sign shows that the concentration of 03 is decreasing with time Solution 3 From the preceding rate expression we have og 1 Athosj Ar 2 Ar Therefore NTioSI 2 0r024 Ms 0048 Ms b Here we have 1 AlNozl Am 4 Ar A1 so Al 4 0024 Ms 0t096 Ms Practice Exercise Consider the reaction 4PH3lt gt gt P4g 6H2g Suppose that at a particular moment during the reaction molecular hydrogen is being formed at the rate of 0078 Msr a At what rate is P4 being formed b At what rate is PH reacting The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers aAbB gtcCdD Rate k AW reaction is xth order in A reaction is yth order in B reaction is x yth order overall 132 F2 9 203902 9 gt2FC02 g Table 132 Rate Data for the Reaction between F2 and CIOZ F2M Cl02M Initial Rate Ms an my MD W Double F2 with CIOZ constant Rate doubles x1 Quadruple CIOZ with F2 constant rate k FZHC39Oz Rate quadruples y1 132 Rate Laws Rate laws are always determined experimentally Reaction order is always defined in terms of reactant not product concentrations The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation F2 9 23mg 9 gt2FCI02 g 1 rate k F2CI02 132 g Determine the rate law and calculate the rate constant for the following reaction from the following data 3208239 aq 3 aq 2304239 aq I3 aq Experiment 820821 I39 In REEgate 1 008 0034 22x 10394 2 008 0017 11 X 10394 3 016 0017 22x 10394 Double l39 rate doubles experiment 1 amp 2 rate k 8208239Xl39y y 1 X 1 rate k 8208239l39 Double 8208239 rate doubles experiment 2 amp 3 rate 22 x10394 Ms Szosz39lll39 008 M0034 M 008M S 132 A Example 143 T The reaction of nitric oxide with hydrogen at 1280 C is 2N0ltggt 2H2g gt Nzg 2H20g From the following data collected at this temperature determine a the rate law b the rate constant and c the rate of the reaction when N0 120 X 10 3 M and 3 H2 6 0 X 10quot M Earnerimam N0 MI H2 iM Initial Rate mm I 5 x m 3 20 x m 3 I x H 2 Mix IO 20 x mquot3 50x 1039 5 3 10 x m 40 x m 100 gtlt m 5 Salutiun 3 in perinmw I and 2 show than when we Enuhue we cnlncemmlmn anquot N 1 constanl comcmruliun U I Hg tlu ml quadruplcs Takimg the ll ul39 Ihu mlcs mm theme 1W4 experimenls mm a 50 x m ws 4 g mm x l jM l x m lm mm 13 x m mm M51334 m 3L11v39qj21u 4 m m Thcrt39l m t 3 I l39ltm x m m 2 4 41513 in ID 39i M39JIL UT 1 2 that u martian is second murder in ND Experiments 2 and 3 indicate thnl nimblqu H z at unnatural ND dnuhlea uhe rule Here we write the min as mm Hm x lib 5 Mfs a kc mn x mf Wq mo 2 uni Mm mtg2 511x mum quot mm rs IU UJHLU x Inf Mr Therefore 4126 m 3Mquot 11 ltrl r g quot N E quot 120 x m My my a 1 than has I111 mamim is rst UTILMT in 2 Hence 1 mu law given by me MNDHHIM FirstOrder Reactions A gt product rate rate k A AIA rate Ms 1 kA k A M 15 or 5 At A is the concentration ofA at any time t A0 is the concentration ofA at time 10 A A0expkt nA nA0 kt A lniA Decomposition of N205 400 5 2034 243039 5 2150 6200 i 39 o 500 1000 1500 3000 25m 3000 3500 5 133 g The reaction 2A B is first order in Awith a rate constant of 28 x 10392 3391 at 8000 How long will it take forA to decrease from 088 Mto 014 M lnA lnA0 kt A0 088 M A 014 M kt lnA0 lnA n amp In 088M lnAo InA A 014M t 6 s k k 28x 102 3397 133 FirstOrder Reactions The halflife tyz is the time required for the concentration of a reactant to decrease to half of its initial concentration tZ t when A A02 n We t 2 k k k g What is the halflife of N205 if it decomposes with a rate constant of 57 x 104 s4 0693 k 57 x 10394 91 How do you know decomposition is first order units of k s1 133 tyz 1200 s 20 minutes COHCCI l39llIiOll T l 2 7 Numhcr nl hallllivcs clapsu 4 l Firstorder reaction A gt product of halflives A A0n 1 2 2 4 3 8 4 16 AID AL 1912 Alum 7 A1018 133 SecondOrder Reactions MA A gt product rate Tt rate k A2 rate Ms AiA 2 k A12 M2 1MS Tt kA i L kt A is the concentration oant any timez A A0 AO is the concentration ofA at time i0 t t when A A02 t 1 No 133 ZeroOrder Reactions Mm A gt product rate Tt rate k A0 k rate MA k APMS Tt k A is the concentration ofA at any time 2 A No kt AO is the concentration ofA at time i0 t t when A A02 t 2k 133 Summary of the Kinetics of ZeroOrder FirstOrder and SecondOrder Reactions ConcentrationTime Order Rate Law Equation Half Life A 0 rate k A A0 kt tZ amp 2k t m2 1 rate MA MA InA10 kt 12 T 2 ratekA2 iLkt t 1 A A10 kAo 133 1 Example 144 T The conversion of eyelopropane to propane in the gas phase is a rstorder reaction with a rate constant of 67 X 10 4 s at 5 0 C CH CHZiCH gt CH iCHCHg cyulupmpunc propum a If the initial conccnlralion of cyclopropanc was 025 M what is me conecnnmion after 88 min I How long in minutes will 39II take for the concenlmlion of cyclopl opzmc I0 decrease from 025 M to 015 M C New long in minules will il take to convert 74 pen cent of the starting material Solution a w nolc um became k i givun m m w l w must hut camen R R mm In secnndx R R mm X GD 1 mm We write In A flu In AirV r 467 x mquot 528 n In 025 4174 Hence IAIV quot 7 UBM we we cunnm um he ogarllhm nan lngijz 7457 x mquot 1 A 76 x 1on x 62 IS mm m l 67 x noquot 2gtlt1 x 33mm 1 0 Us 61 Example 145 The rate If decomposition of aznrnethane is studied by monitoring the partial pressure nl the reaclam as a l unclion of lime Cl IJ NzN Cll g gt 1 42ng CalMg The data obtained at 300T are shown in the following table Pa rtlal Press re of Time 5 Annmelllane mmHgl 391 234 I 220 150 193 200 l 70 250 150 330 132 Are these values consistenl with 39 rst nrder kinelica If so delermine lhe rate cumstam CEHSHS C2H4 1 HIg Time min 331151 M 0 036 15 030 30 025 48 039 19 75 013 Determine the order of reaction and the rate constant J Example 146 T The decomposition of ethane CZHE to methyl radicals is a rstorder reaction with a rate constant of 536 X 10 4 squot at 700 C CZH68 H 2CH38 Calculate the halflife of the reaction in minutes J Example 146 The decomposition of ethane C2H5 to methyl radicals is a rstorder reaction with a rate constant of 536 X 10 4 squot at 700 C C2H5g gt 20138 Calculate the halflife of the reaction in minutes Solution For a rstorder reaction we only need the rate constant to calculate the half life of the reaction 0693 536 X 10 4squot 1min 129 X 1035 X 60 215 min 1 Example T Iodine atoms combine to form molecular iodine in the gas phase Kg 1g gt 12g This reaction follows secondorder kinetics and has the high rate constant 70 X lOgM s at 23 C a If the initial concentration MI was 0086 M calculate the concentration after 20 min b Calculate the halflife of the reaction if the initial concentration of I is 060 M and if it is 042 M a I la Alu W 603 l gtlt 7 v t 39gtlt 70 lOM s20mtn l ingt039086M TL where A is the concentration at t 20 min Solving the equation we get A 12 x 10 2 M This is such a low concentration that it is virtually undetectable The very large rate constant for the reaction means that practically all the 1 atoms combine after only 20 min of reaction time b For I0 060 M L k A10 70 x IOquotM gt s060 M 24 gtlt10quotquots For I0 042 M 21 2 70 X 109M s042 M 34 x10quot s AB CD Exothermic Reaction Endothermic Reaction Activated Aciivulcd complex complex Polcnlial energy Reaclinn progrch Rcuclion progress The activation energy Ea is the minimum amount of energy required to initiate a chemical reaction Rate constant Temperature Dependence of the Rate Constant Temperature k A exp EaRT Arrhenius equation Ea is the activation energy Jmol R is the gas constant 8314 JKmol T is the absolute temperature A is the frequency factor E 1 ka n R 7nA Ink ulml n InA 000 39 l 00 200 300 i 400 7 500 I 20gtlt10 3 130x I03 l40gtlt10 3 l I39Kquot gt K 399 No products formed 134 l Example 148 T The rate constants for the decomposition of acetuldehyde CHJCHOQg gt CH4g com were measured at ve different temperatures The data are shown in the table Piot in k versus UT and determine the activation energy in kJmol fqr the reaction Note that the reaction is g order in CHJCHO so k has the units of 1M3 i k lM39 s T K 0011 00 o 035 730 0 105 750 0 343 790 a E 4 slope 7F 209 X 10 K Ea g 8314JK mol209 gtlt 10 K 174 x 105Jmol 174 X 102 kJmol Check It is important to note that although the rate constant itself has the units lMf s the quantity ln k has no units we cannot take the logarithm of a unit Practice Exercise The secondorder rate constant for the decomposition of nitrous oxide N30 into nitrogen molecule and oxygen atom has been measured at different temperatures k1M s t C 187 x 10 3 600 00113 650 00569 700 0244 750 Determine graphically the activation energy for the reaction 1 Example 149 T The rate constant of a rstorder reaction is 346 X 10 39 1 at 298 K What is the rate constant at 350 K if the activation energy for the reaction is 502 kJmol Strategy A modi ed form of the Arrhenius equation relates two rate constants at two different temperatures see Equation 1312 Make sure the units of R and En are consistent Solution The data are k 246 gtlt10 k2 T298K T2350K Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions The sequence of elementary steps that leads to product formation is the reaction mechanism 2N0 g 02 g gt2N02 g N202 is detected during the reaction Elementary step NO NO gtl Elementary step M 02 2N02 Overall reaction 2N0 02 2N02 135 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation An intermediate is always formed in an early elementary step and consumed in a later elementary step Elementary step NO NO Elementary step 02 gt 2N02 Overall reaction 2N0 02 gt 2N02 The molecularity of a reaction is the number of molecules reacting in an elementary step Unimolecular reaction elementary step with 1 molecule Bimolecular reaction elementary step with 2 molecules Termolecular reaction elementary step with 3 molecules 135 Rate Laws and Elementary Steps Unimolecular reaction A gtproducts rate k A Bimolecular reaction A B products rate k AB Bimolecular reaction A A gt products rate k A2 Writing plausible reaction mechanisms The sum of the elementary steps must give the overall balanced equation for the reaction The ratedetermining step should predict the same rate law that is determined experimentally The ratedetermining step is the slowest step in the sequence of steps leading to product formation r a 1 L 135 The experimental rate law for the reaction between NO2 and CO to produce NO and 002 is rate kN022 The reaction is believed to occur via two steps Step 1 yo No2 gtNO Step 2 CO gt10 2 CO2 What is the equation for the overall reaction N02 00 NO 002 What is the intermediate NO3 What can you say about the relative rates of steps 1 and 2 rate kN022 is the rate law for step 1 so step 1 must be slower than step 2 135 iExample 1410 T The gasphase decomposition of nitrous oxide N20 is believed to occur via two ele mentary steps Step 1 NZOLgtNZ 0 Step 2 N20 04an2 02 Experimentally the rate law is found to be rate kNzO a Write the equation for the overall reaction b Identify the intermediates 0 What can you say about the relative rates of steps 1 and 2 Continued Strategy 21 Because the overall reaction can be broken down into elementary steps knowin the elementary steps would enable us to write the overall reaction b What are the characteristics of an intermediate Does it appear in tlte overall reaction c What determines which elementary step is rate determining How does a knowledge of the rate determining step help us write the rate law of a i39eaction397 Solution a Adding the equations for steps I and 2 gives the overall reaction 2N20 a 2N2 02 b Because the O atom is produced in the rst elementary step and it does not appear in the overall balanced equation it is an intermediate c If we assume that step 1 is the rateedetermining step that is if k2 gt k1 then the rate of the overall reaction is given by rate k N30 andk k Check Step I must be the rateidetermining step because the rate law written from this step matches the experimentally determined rate law that is rate kN3 Practice Exercise The reaction between N02 and CO to produce NO and C02 is believed to occur via two steps Step 1 N02 N02 gtNO N03 Step 2 NO CO gt N02 CO The experimental rate law is rate kN02Z 21 Write the equation for the overall reaction b Identify the intermediate c What can you say about the relative rates of steps 1 and 2 C C Potential varioerF Chemistry In Action Femtochemistry H2 CH2 CH2 H2 6H2 H2 ZCHFCHz H2 CH2 CH2 CH2 CH2 CH2 ZCHFCH2 3 U1 I a I m E 4i C CD 4 o oi Reaction progress Reaction progress 135 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed kAexpEaRT E31 k I Polcmiul energy CD C D catalyzed Rr nclinll pmgrrs uncatalyzed Renriinn prngri s s ratecatalyzed gt rateuncatalyzed I Ea lt Ea 136 In heterogeneous catalysis the reactants and the catalysts are in different phases Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters In homogeneous catalysis the reactants and the catalysts are dispersed in a single phase usually liquid Acid catalysis Base catalysis 136 Haber Process 9 b a awe FeAI203K20 N2 9 3H2 9 catalyst 2NH3 9 Ostwald Process 4NH3 9 502 9 Pt cata39ySt 4N0 g 6H20 9 2N0 g 02 g 2N02 g 2N02 g H20 I gtHN02 aq HNO3 aq 39 Hot Pt wire PtRh catalysts used over NH3 solution in Ostwald process 136 Catalytic Converters Exhaustpipc Tailpipc I Air ccmprcssm source of secondary air Camlylic converters COU b dH d b o tam co HO n urne y rocar OHS 2 converter 2 2 catalytic 2NO 2N02 converter 2N2302 Enzyme Catalysis Enzyme
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