Chapter 14 Chemical Kinetics

Transcription

1 # of paricles 5/9/4 Chemical Kineics Raes of Reacions Chemical Kineics is he sudy of he rae of reacion. How fas does i ae place? Very Fas Reacions Very Slow Reacions Chaper 4 Chemical Kineics Acid/Base Combusion Rusing Eplosive Decomposiion Decomposiion of H O Radioacive Decay Collision Theory Reacion raes are no consan. They are acually dependan on he condiions under which he reacion aes place. How can we mae a reacion go faser? How do we increase he rae of a reacion? Collision Theory Rule #: When molecules collide, hey mus be in an opimal orienaion. Rule #: For a reacion o ae place, reacans mus collide!! Bu even if hey do collide, i does no guaranee a reacion!! Unfruiful Collisions (no reacions) RXN! (maybe) Correc orienaion for a reacion Rule #3: When molecules collide, hey mus have a cerain imum amoun of energy in order for he reacion o occur. These wo reacans canno reac because hey are no in conac wih one anoher. They may evenually run ino each oher, bu hey need ime o diffuse Number of paricles ha acually reac No enough energy (no reacions) Alhough his collision seems o be in he correc orienaion, he reacans do no have enough energy RXN! Enough energy for a reacion Collision Theory and Acivaion ( ) Eohermic vs. Endohermic Reacions Reacans *Acivaed Comple/Transiion Sae The acivaion energy ( ) for a reacion is he energy barrier ha mus be overcome for he reacion o ae place. Similar o a roller coaser Reacion Progress Producs Muli-sep reacion * * Reacion Progress Producs have lower energy han reacans released during reacion Reacion Progress Producs have higher energy han reacans absorbed during reacion A muli-sep reacion has wo and wo ransiion saes - EXOTHERMIC + ENDOTHERMIC

2 # of paricles # of paricles 5/9/4 Concenraion and Rae Surface Area and Rae Lower concenraion of reacans Higher concenraion of reacans Higher concenraion of reacans allow more opporuniies for collisions. Can only collide wih paricles on he surface of he solid If he paricles are made finer, here are many more opporuniies for collisions Higher surface area of reacans increase he rae of a reacion Temperaure and Rae Why does emperaure affec rae? More collisions: Since paricles move faser a higher emperaures here is more opporuniy o inerac/collide. More energy: More paricles will have enough energy o overcome Disribuion of energies of paricles a differen emperaures T Area = number of paricles ha reac a lower emp Reacans Caalyss and rae *Acivaed Comple/Transiion Sae Since he reacion aes a differen roue, a differen ransiion sae and differen mechanism will be involved T Area = number of paricles ha reac a higher emp Noice ha increasing he emperaure does no change he Reacion Progress Producs Noice ha he of he caalyzed and unanalyzed reacions are he same Lower Higher Caalyss allow a differen pah for he reacion o ale place. A pah wih a lower. More reacan paricles reach = faser reacion Caalyss are regeneraed in a reacion. Tha is, hey are no consumed Caalyss Caalyss allow more paricles o reac per uni of ime Biological Caalyss / Caalysis Many reacions in he human body are caalyzed by proeins called enzymes. Enzymes caalyze reacions by binding o reacan paricles called subsraes. Caalyzed (new rn roue) Disribuion of paricles ha reach in a caalyzed vs. non-caalyzed reacion Caalysis lowers allowing more paricles o reac (less energy is needed) Allows unfavorable reacions o ae place under biological condiions. Kineics of enzymes are much more complicaed!! More reacan paricles reach = faser reacion Area = number of paricles ha reac a lower emp Area = number of paricles ha reac a higher emp

3 Amoun of reacan [Phenyl Aceae] Concenraion 5/9/4 Chemical Kineics Wha does a graphical represenaion of differen reacion rae loo lie? Le s say we have wo differen decomposiion reacions; one faser han he oher. Reacion A Reacion B ime Slower Reacion Faser Reacion While fas and slow are good qualiaive words when comparing raes, how do we quaniaively deere his? [reacan( s)] [produc(s)] Raeof reacion Mos ofen, especially for aqueous soluions, we define amoun as concenraion (M or mol/l) denoed by braces, [ ]. Time can be in any unis. Very ofen, we use seconds (s). How do we deere rae of reacion? We can measure he concenraions of reacans or producs various ways. In some soluions, we can use visible specroscopy o deere concenraion of reacans or producs, usually epressed in M or mol/l. Gases can be epressed in parial pressures (am). [reacan( s)] [produc(s)] Raeof reacion.m = s A B.5M = 5s Δ = final - iniial.5m = 5s (.5M.M).5M M (.5M.5M).5M M Rae. Rae. (5s- s) 5s s (5s-5s) 5s s Chemical Kineics Raes of Reacions Chemical Kineics Raes of Reacions A B [reacan( s)] [produc(s)] Raeof reacion Δ = final - iniial M/ Δ Amoun of reacan 5. Δ Time Time () [Reacan] (.5M 5.M).5M Avg Rae 5. (.5-.).5 (.6M.63M).47M Avg Rae.47 (.5-.5). Avg Rae (.8 M.6 M).8 M. (3. -.5).5 M M M M/ Time ().73 M/ We can calculae an average rae of he reacion beween any wo daa poins during he reacion Overall Avg Rae (5. M.8 M) 4.9 M.6 ( -3.) 3. M Wha abou an insananeous rae? Le s say a.5 ues? Chemical Kineics Raes of Reacions [Phenyl Aceae] Time (s) Wha is he overall average reacion rae M/s Wha is he average rae from =.s o =5.s M/s Time (s) Should he average from =5.s o 3.s be higher or lower han =.s o =5.s??? The rae from =5.s o 3.s should be slower Rae of Disappearance of Reacans and Appearance of Producs aa + bb c C Generic rae of reacion Noice ha he reacans have a - and he producs do no. Δ Δ[B] Δ[C] Raeof reacion a Δ b Δ c Δ Δ Δ[B] Δ[C] Raeof reacion a Δ b Δ c Δ Rae of change in A Rae of change in B Rae of change in C If he coefficiens are differen, he raes will be differen 3

4 5/9/4 Rae of disappearance of reacans and appearance of producs based on coefficiens A + B C Δ Δ[B] Δ[C] Raeof reacion Δ Δ Δ If C is appearing a M/s a wha rae is A disappearing? Δ Δ[C] Δ Δ Δ 5. 5 M/s Δ If A is disappearing a -. M/ a wha rae is B disappearing? Δ 5. 5 M/s Δ Δ Δ[B] Δ[B] Δ[B] Δ[B]. M/s.M/s. M/s Δ Δ Δ Δ Δ If B is disappearing a -.44 M/ a wha is he (generic) reacion rae? Rae of Disappearance of Reacans and Appearance of Producs In he reacion N O 5 (g) 4NO (g) + O (g) he appearance of NO is 3. - M/s. Wha is he rae of disappearance of N O 5?.56 - M/s In he synhesis of Ammonia, N (g) + 3H (g) NH 3 (g), if he rae of disappearance of H is M/s, wha is he rae of appearance of ammonia, NH 3? M/s Δ[B] Raeof reacion Raeof reacion -.44M/ Δ Raeof reacion. M/ Chemical Kineics and Concenraion Rae Law Each reacion can be epressed by wha is called a Rae Law The rae law describes how each reacan s concenraion has an effec on he reacion rae Deering Rae Law Deering he Order of each reacan While i seems lie i should wor, we should never deere he Rae Law based on he balanced equaion!! This is because reacions can have muliple inermediae seps no accouned for in he balanced equaion. We can deere he Rae Law using eperimenaion daa of iniial raes. General Rae Law Rae = [B] y = concenraion of reacan A [B] = concenraion of reacan B = rae consan (depends on oher unis) Rae = usually in M/s or mol/l s and y = order of each reacan Only reacans ha affec he rae of he reacion are included in he rae law. In oher words, no every reacan ha appears in he balanced equaion will be in he rae law! Trial Trial Trial Iniial Iniial [B] Iniial Rae.M.3M. M/s.M.6M. M/s 3.3M.3M 9. M/s Rae = [B] y y. M/s.M.6M. M/s.M.3M A + B Z Firs, find wo rials where one reacan has he same concenraion. This why we can cancel he effec of ha reacan has on he rae (For eample, rials &). Plug everyhing you now ino he rae equaion. y Ignore he reacan wih he same concenraion and ignore, for now. y y = Solve for he eponen Deering Rae Law Deering he Order of Each Reacan Deering Rae Law Shor Cu So Far Now, we go hrough he same process for A: Trial Iniial Iniial [B] Iniial Rae.M.3M. M/s Trial Iniial Iniial [B] Iniial Rae.M.3M. M/s Rae = [B] Trial Iniial Iniial [B] Iniial Rae.M.3M. M/s.M.6M. M/s 3.3M.3M 9. M/s.M.6M. M/s 3.3M.3M 9. M/s We normally do no show he as an eponen.m.6m. M/s 3.3M.3M 9. M/s.6M.3M M. s M. s.3m.m M 9. s M. s Trial 3 y 9. M/s.3M.3M. M/s.M.3M Trial 9 3 y 9 3 = concenraion doubles y rae doubles concenraion riples 3 9 rae nonuples (9) Therefore, he rae law can be wrien as: Rae = [B] The reacion is second-order wih respec o A and firs-order wih respec o B + = 3 This reacion is 3 rd order overall s order Therefore, he rae law can be wrien as: Rae = [B] nd order The reacion is second-order wih respec o A and firs-order wih respec o B + = 3 This reacion is 3 rd order overall 4

5 5/9/4 Deering Rae Law Deering he Rae Law Consan () Since does no change during he reacion, we can use any rial o calculae he value of Trial Iniial.M.M Iniial [B].3M.6M Iniial Rae. M/s. M/s 3.3M.3M 9. M/s Cancel unis Rae = [B]. M/s = [.] [.3] (.M/s) (.M) (.3M) Trial orm M s s - Trial (.M/s) Rae = [B]. M/s = [.] [.6] (.M) (.6M) orm s M s Rae Law Rae 333 M s [B] A = nd -order B = s -order Overall reacion = 3 rd order = 333 /M s Deering Rae Law Pracice Use he following daa o deere he rae law and he value of for he reacion beween NO and O. Iniial [NO] Iniial [O ] Iniial Rae Trial (M/s) Trial Iniial Iniial [B] Iniial [C] Rae (M/s) Trial Deering Rae Law Trics Up My Sleeve Iniial Iniial [B] Rae (M/s) M.M concenraion changes. s order M. s M.67 s. y. rae changes. Trial Iniial Iniial [B] Rae (M/s) M.M concenraion changes.3 M 3.67 s M.67 s rae changes. Deering Rae Law Trics Up My Sleeve #.3. Tae naural log of boh sides.3 ln. ln(a ) = ln(a).3. Rearrange using solve for WTF?! ln(a ) = ln(a) =.96 3 ln.3 ln. ln.3 ln. ln Deering Rae Law Group Pracice Iniial P A Iniial P B Rae Trial (am) (am) (am/) Inegraed Rae Laws - Rae and Time Wih he help of calculus, we can use anoher form of he rae law ha allows us o calculae how concenraion of a reacan changes wih ime. I is called he inegraed rae law. s -order rae law Rae = Remember ln -or- ln ln Iniial Iniial [B] Rae Trial (M/s) *This one may iniially loo impossible, bu i s no. Hin: Thin abou wha you already now. nd -order rae law Rae = h -order rae law Rae = Wha abou a reacion where he concenraion of he reacans ha do no affec he rae. Does i eis? Yup. 5

6 (mol/l) (mol/l) (mol/l) 5/9/4 Inegraed Rae Laws s order A cerain decomposiion reacion is s order overall. Wha is he rae consan if he concenraion of A goes from an iniial concenraion of 5.55M o.m in 3. ues?.m ln (3.) 5.55M -.69 (3.) ln =.54 - Inegraed Rae Laws s order More naural log rics Wha would be he concenraion afer 3 ues? - ln (.536 )(3) 5.55M How do we ge rid of naural log (ln)? ln How long would i ae for he concenraion o reach 5% of is original concenraion? 5% ln (.536 ) % (.536 ) Using % direcly in he equaion will only wor wih s order since doesn include a concenraion uni. For oher orders ( nd and h ), use acual concenraions in he equaions. =.3 ln e ln() = ln ln -lny y ln M -(.536 )(3) e e - (.536 )(3) 5.55M CAUTION! ln ln DOES NOT y lny M =.5M - ln-ln5.55m (.536 )(3) ln 5.54 =.5M Inegraed Rae Laws nd order A second-order decomposiion has a rae consan of.5 M - -. Assug you sar wih 3.M, how long will i ae for only 5.% of he reacan o remain. ln Half-lives The ime i aes for a reacan o reach half of is iniial amoun Unlie s order, you canno simply use % s. You mus use specific concenraions or amouns for your reacan. 5.% 3.M.6M.% 6M 8M 4M M M Original concenraion So.. - (.5 M ).6M 3.M (.5 M ) % 5 % 5 %.5 % 6.5 %.M If he same reacion oo 65s o reach.m, wha was he original concenraion? Wach for ime unis. MUST be he same in and for - (.5 M )(.4) 65s.4 6s M / / / / / half-life half-lives 3 half-lives 4 half-lives General Equaion for Half-life Where = number of half-lives Half lives 4 s half-life 3 s half-life nd half-life nd half-life 3 rd half-life 4 h 3 half-life rd half-life 4 h half-life 7 6 s Order nd Order 5 Inegraed Rae Laws Half lives s half-life 3 nd half-life 3 rd half-life 4 h half-life Time (s) h Order Time (s) Half lives are equal Time (s) Half-lives increase over ime Half lives decrease ln() / / / 6

7 ln / 5/9/4 s Order nd Order Half-life equaions General Equaion for Half-life Half-lives are relaed o he rae consan,, by he following equaions ln h Order Where = number of half-lives / / ln() / Half-Life Eamples A cerain reacion has a rae consan of.5 hr -. Wha is he half life of his reacion in hours? Minues?.555hr How many half lives will i ae for he same reacion above o go from 6.M o.m? 8. half-lives If a nd order reacion has a half life of 3. days, how many hours will i ae for.% of he reacan o disappear if he concenraion sars a.m? / / ln() / Equaion of a sraigh line s -order rae law nd -order rae law Graphing Inegraed Rae Laws ln ln y = m + b slope y-inercep ln vs. / vs. ln / ime ime Deering order by graphing To deere he order of he reacan, you mus graph he daa hree differen ways: Time vs. Conc.; Time vs. ln(conc.); and Time vs. /Conc. Time (s) Time (s) ln Time (s) / /M Time vs. Conc. Time vs. ln(conc.) Time vs. /(Conc.) h -order rae law vs. ime Mulisep Reacions Reacion Mechanisms NO + CO NO + CO NO + CO NO + CO? Mulisep Reacions Reacion Mechanisms Reacions ha have muliple seps are made of individual seps called elemenary seps. Elemenary seps are single (concered) seps. Sep : NO NO + NO 3 Sep : NO 3 + CO NO + CO NO + CO NO + CO Inermediaes are chemical species ha are creaed during he reacion, bu no a par of he overall reacion equaion. Inermediaes are creaed in an elemenary sep and consumed in anoher Elemenary seps mus add up o he overall reacion equaion. This reacion happens in muliple seps. Bu wha are hey? And how can we figure i ou? Firs we need o learn abou hese seps. Sep : NO N O 4 Sep : N O 4 + CO NO + CO These are no plausible seps because he seps add up o a differen overall reacion CO CO 7

8 5/9/4 Moleculariy and Rae Laws of Elemenary seps Rae laws for elemenary seps can be deered direcly from he balanced equaion The moleculariy of an elemenary sep is based on he rae law for ha sep Overall reacion raes of muli-sep reacions Two-sep reacion Moleculariy Equaion Rae Law Unimolecular Bimolecular Termolecular A Produc(s) A + A Produc(s) A + B Produc(s) A + A +A Produc(s) A + B + B Produc(s) A + B + C Produc(s) Rae = Rae = Rae = [B] Rae = 3 Rae = [B] Rae = [B][C] 6, isses per hour 5, isses per hour Slow Sep Fas Sep Elemenary Seps Since elemenary seps are single seps, rae laws for each sep can be deered from he balanced chemical equaion Sep : NO NO + NO 3 Elemenary Sep Rae = [NO ] Boh of hese elemenary seps are bimolecular Sep : NO 3 + CO NO + CO Elemenary Sep Rae = [NO 3 ][CO] How many isses can be processed per hour overall??? The overall reacion can go no faser han he slow sep. This sep is called he rae-deering or rae-limiing sep. Jus remember ha reacion raes are no saic; reacion raes are variable based on concenraion. Muliple-sep Reacions Elemenary Seps Rae-limiing Sep s A X Sep Rae Law Rae = Slow-sep s Rae < Rae Rae laws for elemenary seps can be deered direcly from he balanced equaion A X + C Sep Rae Law Rae = [B][X] Overall Chemical Formula A + A X A + B C The elemenary-sep formulas mus add up o he overall chemical formula If he slow, rae-deering, sep is firs, he overall rae law is he same as he slow sep Sep Rae Law Rae = Overall Rae Law = Rae = Noice ha he concenraion of B does no affec he rae of reacion Overall reacion raes of muli-sep reacions 5, isses per hour Fas RXN 5, isses per hour Two-sep reacion, isses per hour Slow RXN Elemenary Seps How many isses can be processed per hour overall??? Alhough i seems lie he overall rae should = he slow sep, in realiy, reacion raes are variable based on concenraions of reacans. Muliple-sep Reacions Elemenary Seps Rae-limiing Sep nd Muliple-sep Reacions Elemenary Seps Rae-limiing Sep nd Slow-sep nd 3 rd Overall Chemical Formula If he slow sep is second, he bolenec creaes an equilibrium in he firs sep. A X Rae > Rae A + A X A + B C The elemenary-sep formulas will add up o he overall chemical formula A X Slow-sep nd 3 rd Rae > Rae A X Sep Rae Law Rae = = - [X] Sep Rae Law Rae = [B][X] Equilibrium Sep Rae Law Rae = Sep Rae Law Rae = [B][X] A X + C However, unlie he rae-limiing sep being firs, If he slow, raedeering, sep is second i s much more complicaed Some producs in Sep rever bac ino reacans before coninuing o sep Sep Rae Law Rae = = - [X] Solve for he inermediae [X] Subsiue Inermediaes canno be in he overall rae law. The overall rae law canno be based solely on he Sep rae law. We mus combine boh rae laws. Rae [B] Combine s Overall Rae Law Rae [B] 8

9 # of paricles 5/9/4 Muliple-sep Reacions Elemenary Seps Rae-limiing Sep Noice ha he same reacion wih he same elemenary seps can give us differen overall rae laws because of he locaion of he slow sep. Slow-sep s A + B C Slow sep nd 3 rd Muliple-sep Reacions Elemenary Seps NO + CO NO + CO has an eperimenally deered rae law of Rae = [NO ] NO + CO NO + CO? A X A X Which of he following is mos liely o represen he mechanism for he reacion? Rae < Rae Rae > Rae Overall Rae Law Rae = Overall Rae Law Rae [B] How do we now which one is correc for his reacion? Eperimenally Deered Rae Law NO + CO NO + CO NO NO + NO 3 (slow) NO 3 + CO NO + CO (fas) NO NO + NO 3 (fas) NO 3 + CO NO + CO (slow) Caalyss # Mechanisms Arrhenius Equaion Caalyzed (new rn roue) H O H O + O H O + I - IO - +H O Svane Arrhenius deered ha acivaion energy ( ) of a reacion is relaed o he reacion consan. Tha is a any emperaure he lower he acivaion energy, he faser a reacion will proceed. Ae -E RT a H O + IO - I - + H O + O E a ln lna- R T Area = number of paricles ha reac a lower emp Area = number of paricles ha reac a higher emp More reacan paricles reach = faser reacion Noice ha he caalys, I -, is regeneraed in he end. There is no ne change in caalys in he reacion. The inermediae, IO -, is compleely used in he reacion (as i should be). E a ln - lna R T y = m + b Arrhenius Equaion We can also do calculaions wihou graphing, bu combining wo equaions ino one. If we now boh and, along wih T and T we calculae he energy of acivaion Ea ln lna- RT Ea ln lna- RT ln E a R T R 8.34 J molk T In a cerain equaion, he rae consan a 7K was measured as.57m - s - and ha a 895 K was measured as 556M - s -. Find he acivaion energy of he reacion. 556 ln.57 Ms Ms Ea 8.34 J mol K 7K 895K E a J K J/mol mol K 9