EXAMPLES OF EVALUATION OF RATE FORMS FROM KINETIC DATA IN BATCH SYSTEMS

Transcription

1 HE 47 LETURE 5 EXMPLES OF EVLUTION OF RTE FORMS FROM INETI DT IN BTH SYSTEMS EXMPLE : Deermine he reacion order and he rae consan or a single reacion o he ye roducs based on he ollowing exerimenal inormaion obained a isohermal condiions a V cons. ( min) mol li The las daa oin simly indicaes ha aer a very long ime (several hours as comared o minues) racically no is ound. Thus, a he exerimenal condiions used he reacion is racically irreversible.. Dierenial nalysis: i We can orm i and i i i + rom he able, and i i i We can lo now smooh curve so ha i vs. i as a sewise curve shown below. Now we have o ass a 0

2 HE 47 LETURE mol li min (min) he area under he sewise curve and he smooh curve are aroximaely equal. From d he smooh curve we now read o he corresonding values o a desired measured concenraions i. i d Now we can lo d log log + α log i.e. we lo d vs. on a log-log lo. From he sloe o he sraigh line ha we managed o ass hrough he daa oins we ind ha α.05. From any oin o he line now we could deermine. For examle: log log +.05 log 0. log log log li mol

3 HE 47 LETURE 5 However, we should quicly realize ha he esimaed order is only.5% rom nd d order and, hus, mos liely he reacion is o order wo. α.. We can now evaluae or every daa oin and hen average hem ou. d d log sloe log

4 HE 47 LETURE 5 The mean value o urns ou o be li Thus a he emeraure mol o he exerimen we have deermined mol r 0.50 li Noe: lhough he values o he rae consans calculaed rom various daa oins vary considerably, he variaion is random and shows no rend wih concenraion level indicaing ha he seleced order is correc. Inegral Mehod: Suose ha we have aemed o solve he same roblem by he inegral mehod. Since0 he soichiomery is roducs, we will ry irs wheher a irs order rae orm can i he daa (hoing or an elemenary reacion). ssume α : d d o o o ln o log (.306 ) log log.306 o We should hen lo he exerimenal daa on a semi log lo ( on he log scale and on he linear scale). I he assumed order o one is correc we should be able o obain a sraigh line hrough he daa oins. I is clear rom he enclosed igure ha a sraigh line canno be obained since he daa show a deinie curve here - convex owards he boom. I we connec he irs and he las daa oin by a sraigh line all he oher daa are below he line indicaing ha he concenraion dro is aser han rediced by irs order behavior. 3

5 HE 47 LETURE 5 mol li (min) Try nd order α d + o Plo vs. on a linear lo. This ime a sraigh line is obained which roerly inersecs he ordinae a li. From he sloe o he line we ge 0.50 o mol. I is insrucive again o evaluae 's rom he individual daa oins. o

6 HE 47 LETURE sloe (min) li The mean value o urns o be 0.46 he dierence beween his mol mean value and he one obained by "eye iing" he line hrough daa oins is.7% and is negligible as ar as engineering alicaions are concerned. Noice ha he dierence beween he larges 0.54 and smalles 0.4 is only 0.0 or less han 8.5% based on he smalles -value. For he same daa he 's evaluaed by he dierenial mehod varied beween a low o 0.34 and a high o 0.68, he dierence being or 5% based on he smalles. Thus, or he same qualiy daa he inegral mehod ends o smooh ou he errors and give more consisen esimaes or he rae consan. Here by inegral mehod we have deermined also: mol r li EXMPLE : 0.50 Gas hase decomosiion o di--buyl eroxide is moniored in a bach reacor o consan volume a isohermal condiions o 70. The run is sared wih ure di-buyl eroxide and he change o oal ressure o he sysem was recorded in ime. From he daa below ind he rae exression and he rae consan. ime min P mm Hg

7 HE 47 LETURE 5 The reacion is: H O O H H H O H P + Q V cons, T cons The hyohesized rae is o he orm: mol r li α We have seen beore ha we can also exress he rae in erms o change o he arial ressure: where mm Hg r% P α RT α In a bach sysem o consan volume: r% dp P We have shown beore ha P α P o ξ RT s + ν ξ j PT PTo Po T To j Since in his case Po P To PT RT here ν j 3 P 3 P To P T dpt 3 PTo PT α dpt ( α ) [ 3 PTo PT] α 6

8 HE 47 LETURE 5 DIFFERENTIL NLYSIS: PT Evaluae underneah as he sewise curve. Evaluae, lo vs. a sewise curve, rom a smooh curve ha has he same area dpt and calculae he corresonding dpt 3 P To P T. Plo vs. 3 P To P T on a log-log scale. The sloe gives α ; rom he daa ind and and heir mean value. The augmened able is shown below as well as he wo igures (nex age) PT dpt PT 3 PTo PT min dpt I seems ha more han one sraigh line can be assed hrough he oins on he log vs. (3 P To P T ) lo. The maximum sloe seems o be. and he minimum This indicaes ha robably α.0. dpt 3 P P To T These values are given in he las column o he above able. 7

9 HE 47 LETURE 5 P T (min) 0 - dp T sloe sloe log 8 log 0 8 log 5.0 log P To P T From he above igures and able we ind mm Hg mol r% 0.08 P 0.08 P or r P li INTEGRL METHOD: Suose we assumed zero-h order dp T " P P + " T To Daa shows deinie curvaure and reacion is no zero-h order. 8

10 HE 47 LETURE 5 0 P T 0 0 ssume nd order: (min) dpt " 3 PTo PT + 3 PTo PT PTo 443 y " { x 0.3 3P To P T (min) gain daa show a deinie curvaure and he reacion is no nd order. ssume s order 9

11 HE 47 LETURE 5 dpt " 3 P P To P To ln " 3 PTo PT T To T To log ( 5 3.) log 3 P P log P.306 " sloe P To P T (mm Hg) sloe log ( 5 3. ) (min) Now we do ge a sraigh line. ".306 sloe min Direcly rom daa P ln To PTo PT min

12 HE 47 LETURE 5 mol r P li mm Hg r% 0.08 P Noice again ha he variaion in he rae consan based on inegral mehod is much less han i dierenial mehod is used. Examle 3: soluion onsider he reacion beween suluric acid and diehylsulae in aqueous H SO4 + H 5 SO 4 H5 SO4 H + B P o isohermal condiions ( T.9 ) saring wih equimolar mixure o he reacans and wih no roduc he daa resened below were obained. Iniial reacan concenraion was 5.5 ( mol li ) or each o hem. Find he rae exression. min mol li min several days mol li Since or each mole o reaced one ges wo moles o P i he reacion wen o comleion one would ind ( mol li) o P. Since only 5.80 ( mol li) o P are ound his indicaes ha he reacion is reversible. P eq P eq eq B eq o o eq P x e eq o i e P eq o o 0.57 Le us assume ha he reacion is nd order in boh direcions d R B b 3

13 HE 47 LETURE 5 Since we sar in soichiomeric raio hereore o Bo B o d o b ; 0 0 Le us use inegral analysis. From he above rae exression a equilibrium 4 x e B b eq ( x ) e 4.98 We can searae he variables in he above dierenial equaion: d o b o o The inegral on he le hand side can be romly evaluaed by using a se o inegraion ables. For an exercise we will inegrae i here: b o + o 4 Find he roos o he denominaor:, m o 4 The inegrand can now be wrien as: B + o o 4 4 o o Using arial racions and evaluaing or B 4 o 0 and B we ge: 3

14 HE 47 LETURE 5 Thus: d d d o 0 o o + o o o ln 4 o o er some reorganizaion we ge: ln ( ) + o o o Using he reviously esablished relaionshi 4 x e ( x ) We can rewrie his in he orm: ln e ( ) xe xe x xe o xe x xe Using he already evaluaed value o and nown o we should lo i.e ln y vs. on a semi-log lo y y y y

15 HE 47 LETURE 5 Suose ha we have assumed a irs order reversible reacion: d b b Peq eq xe x e uon inegraion.3 ln + o + o P ln P or in alernaive orm: ln x e x x x e e Now i we exec his rae orm o hold we should also ge a sraigh line on a semi-log lo o y 0.7 he revious age. vs.. The values o y are also calculaed in he able on Boh orms are loed on semi-log aer on he nex age and boh yield a reasonable sraigh line!? arenly rom he exerimenal daa given we are unable o disinguish beween a reversible nd order reacion in boh direcions and a reversible irs order reacion in boh direcions. I we consider again he wo inegraed orms wrien in erms o conversion we can readily see ha when x e 0.5 he wo orms become idenical and indisinguishable rom each oher. Since under he condiions o he exerimen equilibrium conversion was x e 0.57, which is close o 0.5, due o exerimenal scaer we canno disinguish beween he wo orms. I we erormed he exerimens a dieren T so ha x e is ar rom 0.5; or i we used nonsoichiomeric raio o reacans, we would ind ha he rae indeed is nd order in each direcion. I is imoran o deermine he roer order since when designing a larger reacor we may be oeraing a condiions when x e >> 0.5 and he redicions o he reacor size or a desired roducion rae will dier vasly based on he rae orm. 34

16 HE 47 LETURE 5 Plo or nd order reversible: y sloe log (min) 35

17 HE 47 LETURE 5 Plo or s order reversible: y sloe log From he sloe or he nd order rae orm we ge he value o he rae consan. (min) Then b sloe li 4.99 mol b li mol The rae in ( min) o mol li is given by he above exression a T.9. The emeraure deendence o he consans would have o be ound by erorming exerimens a dieren emeraures. 36

18 HE 47 LETURE 5 auion: Noe : Noe : Quesion: nswer: noe o cauion is here in order. The inegraed exression ha we used or a reversible nd order reacion o he ye + B P was: ln ( ) xe xe x xe o xe x xe This exression is only valid when he exerimen is erormed wih soichiomeric raio o and B. I is no valid when o Bo. The exression or he same reacion ye is reored by Levensiel (age 63, equaion 56) and i loos almos exacly he same as he one above exce ha i has an exra acor o on he righ hand side. Did I mae a misae or did he mae a misae? Noe ha my and his (which are rae consans or he reacion orward) will dier by a acor o wo!!? Neiher o us made a misae bu is based on roducion o P while r is based on disaearance o. Since due o soichiomery r Poenial Trouble: Noe 3: his imlies which indeed is he case. The choice o he subscri or ec. does no indicae on which comonen he rae consan is based. Thus, i one has only an inegraed orm o wor wih, one has no way o nowing wheher is based on reacan or roduc, ec. lariy ha whenever ossible. Since he above menioned ambiguiy abou rae consans always exiss ry o: a. use inegraed orms only when you now wha he 's are based on; b. develo your own inegraed orms by he hel o inegral ables. This las choice aer all is no ha diicul. In he roblem ha we jus solved we had o inegrae d 0 0 o d o + o 4 quic loo in he R Mahemaical Tables shows ha we have a roblem o he ye 37

19 HE 47 LETURE 5 dx y where x y a + bx + cx a o b c o 4 The answer is: dx cx + b q ln y q cx + b + q where q 4 ac b Thus in our case q 4 o. Subsiue roer erms or q, c, a, b and x in he above exression. auion: Do no orge o evaluae he above exression a he uer and lower limi o inegraion (exression a uer limi - exression o lower limi) since you sared wih a deinie inegral and R Tables gives you he answer or an indeinie one. You should ge he exression used in he roblem. 38