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) 0 5 0 5 0 5 30 35 40 mol li 0.58 0.4 0.3 0.5 0. 0.9 0.6 0.4 0 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 5 0 5 5 0 0 5 5 0 0 5 5 30 30 35 35 40 i i i 0.084 0.034 0.08 0.04 0.006 0.006 0.006 0.004 We can lo now smooh curve so ha i vs. i as a sewise curve shown below. Now we have o ass a 0
HE 47 LETURE 5-0.08 mol li min 0.04 0.0 0 0 0 30 40 (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 5 0 5 0 5 30 35 0.58 0.4 0.3 0.5 0. 0.9 0.6 d Now we can lo 0.055 0.04 0.05 0.005 0.0065 0.005 0.004 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 0.004 log +.05 log 0. log log 0.004.05 log 0. 0.8039 li 0 0.57 mol
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 0.58 0.4 0.3 0.5 0. 0.9 0.6 0.63 0.43 0.46 0.68 0.34 0.39 0.56 0. 9 8 7 6 3 4 5 6 7 8 9 5 4 3 d 0.0 9 8 0. log 0. 004 sloe log 0.8.05 0. 7 6 5 4 3 0.00 3 4 5 6 7 8 9.0
HE 47 LETURE 5 The mean value o urns ou o be li 0.50. 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
HE 47 LETURE 5 mol li - 9-8 - 7-6 - 5-4 - 3 - - - 0 0 30 40 (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 0.58 0.4 0.3 0.5 0. 0.9 0.6 0.4 0.45 0.44 0.4 0.50 0.4 0.4 0.50 0.54 4
HE 47 LETURE 5 8 6 4 sloe 7-0.50 40 0 0 0 0 30 40 (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 0.034 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 0.0.5 5.0 0.0 5.0 0.0 P mm Hg 7.5 0.5.5 5.8 7.9 9.4 5
HE 47 LETURE 5 The reacion is: H O O H H H O H P + Q V cons, T cons 3 3 6 + 3 3 3 3 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
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 0 7.5.48 5.0 0.0987 0.08 min.5 0.5.0 0.95.0 0.079 5.0.5 0.80 0.75 0.0 0.0750 0.0 5.8 0.66 0.5 6.7 0.0776 5.0 7.9 0.4 0.35 4.6 0.076 0.0 9.4 0.30 0.5 3. 0.0806 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 0.95. 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
HE 47 LETURE 5 P T. - 0.8-0.4-0 5 0 5 0 (min) 0 - dp T sloe sloe log 8 log 0 8 log 5.0 log 0.4. 0.95 0. -.0 0.0 0.0 3P To P T From he above igures and able we ind mm Hg mol r% 0.08 P 0.08 P or r 0.08 0.00 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
HE 47 LETURE 5 0 P T 0 0 ssume nd order: 5 0 5 0 (min) dpt " 3 PTo PT + 3 PTo PT PTo 443 y " { x 0.3 3P To P T 0. 0. 0 5 0 5 0 (min) gain daa show a deinie curvaure and he reacion is no nd order. ssume s order 9
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 0.034 0 3P To P T (mm Hg) 0.0-9 - 8-7 - 6-5 - sloe log ( 5 3. ) 0.034 0 4-3 -.0-0 5 0 5 0 (min) Now we do ge a sraigh line. ".306 sloe.306 0.034 0.0787 min Direcly rom daa 0.0893 0.08 P ln To 0.0806 3 PTo PT 0.0788 0.087 min 0.0788 30
HE 47 LETURE 5 mol r 0.08 0.00 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 0 4 48 55 75 96 7 46 6 mol li 0.8.38.63.4.75 3.3 3.76 3.8 min 80 94 67 38 368 379 40 several days mol li 4. 4.3 4.45 4.86 5.5 5.3 5.35 5.4 5.80 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 5.80.6 eq B eq o o eq P x e eq o i e P eq 5.8 5.5 o o 0.57 Le us assume ha he reacion is nd order in boh direcions d R B b 3
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 + 4 4 4 + Using arial racions and evaluaing or B 4 o 0 and B we ge: 3
HE 47 LETURE 5 Thus: d d d 0 4 0 o 0 o o + o o 4 4 4 + o 4 + 4 ln 4 o o + 4 4 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. 0.3 4.548 ln 4.99 4.3 4.548 y 0.3 4.548 vs. on a semi-log lo. 4.3 4.548 y y y y 0 4 48 55 75 96 7 46 6.40.95.369.595.853.56.74.809.55.3.390.67.897.3.83.90 80 94 67 38 368 379 40 3.98 3.733 4.4 5.884 8.48.46. 4.46 3.4 3.877 4.63 6.095 8.757.77.53 4.76 33
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 0.948 0.7 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
HE 47 LETURE 5 Plo or nd order reversible: 0.0 - y 0.0-9 - 8-7 - 6-5 - 4-3 - - sloe log 0.0 0.0089 450-00 00 300 400 500 (min) 35
HE 47 LETURE 5 Plo or s order reversible: 0.0 - y 0.0-9 - 8-7 - 6-5 - 4-3 - - sloe log 5.0 0.0084 400-00 00 300 400 500 From he sloe or he nd order rae orm we ge he value o he rae consan. (min) Then 4.99.306 b sloe 0.0089 0.0089.306 li 4.99 mol 3.35 0 b 3.35 0 3 li.7 0 4.98 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
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
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