K THUT PHN NGA. Bi tp:T s ha lu c tnh theo cng thc n = Qv/Qo Cc em tham kho thm mt s dng bi tp sau:1. Khi kho st qu trnh ln men Lacticngi ta thu c bng s liu thc nghim sau: tgian 50 100 150 200 250 300Nng lactose45 43 25 13 8 5Vtoc R 0,5 0,55 0,51 0,49 0,47 0,44a) Hy trnh by phng php xc nh cc thng s ng hc, khi qu trnh c m t bng ng hc Michaelis Menten: S mSmC K CR R+ ?b) chuyn ha 90 Lactose cn tin hnh phn ng trong thi gian bao lu?c) Khitinhnhphnng sau 280phtthi nng ngcnlilbao nhiu?Gii:a) xc nh cc tham s ca phng trnh ng hc Michaelis Menten m S mmR1C1RKR1+ b axR1C1RKyR1m S mm+ + Trong o '2 22 22. .. . .x xy x y xbx xy x x x ya

ml mgbaaR Kph VSV g mgbRm mm/. /1 Da vo bng s liu tnh cc h s a, b t suy ra Km v Rm.b)Khi thc hin phn ng sau 280 pht hm lng ng cn st c xc nh theo: 2MAX X m SVm SVMAX X m SVS2R C K CK C2R C K CC

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+ + =C) lm bi ton ngc li2. Khi tin hnh phn ng Isomer Ete etyl Axit Axetyl Axetic theo phn ng:Thu c s bin thin nng (phn mol) Etanol theo thi gian nh sau: , h 0 71,8145,5215,8264,3333,3383,5478,3x 0,366 0,2770,2150,174 0,1520,1300,1210,106Khi cn bng x = 0,078. Hy trnh by: a) Phng php cc nh thng s ng hc bng thc nghim? b) ng dng xc nh hng s tc ?c) Xy dng th m phng? Gii:b) Gi thit phn ng thun nghch bc 1Khi tc phn ng c th vit dng: -RA = k1CA = - k-1CB Tch phn thu c: Khi Da vo s liu t bng ta tnh cc gi tr k ti cc thi gian khc nhau. T tnh gi tr k trung bnh. c) V th m phngCu 3Phn ng: Tin hnh trong ru khan. Khi phn tch ly 20 cm30,1 N dung dch axit ban u.Thu c s liu thc nghim sau.103, s 0 0.6 1.2 1.8 2.4 3.64.86.6V Ba(OH)2 0,1N, cm3 21.00 17 14 11.000 9 64 3V Na(OH) 0,1N, cm3 0.00 2 3.5 5.000 6 78 9Hy trnh by: a) Phng php cc nh thng s ng hc bng thc nghim? b) ng dng xc nh hng s tc ?c) Xy dng th m phng? Gii:b) Gi thit c hai phn ng u c bc 1, khi tch phn phng trnh ng hc ta c: ( )2 10ln k kCCAA+ Xy dng quan h ( ) fCCAA 0ln khi ly CA l lng dung dch Ba(OH)2 0,1 N dng phn ng vi CO2 sinh ra, cn CA0 l lng Ba(OH)2 0,1 N ban u (= 0). Theo th (hay dng phng php bnh phng cc tiu) tm c h s gc: k = k1 + k2 = tg= 3,4.10-4 1/s Xy dng th CB = g(CR) vi CB v CR l lng sn phm trong phn ng 1 v 2Nng CR t l vilng dung dch NaOH 0,1 N dng trong thy phn ete to thnh, cn nng CB tnh nh sau: V Ba(OH)2 0,1 N ( = 0) V Ba(OH)2 () V NaOH 0,1 N () Kt qu thu c:99 , 0 tgkk221 T gii h tm k1v k2.4. Trong mt thit b thc xy ra phn ng bc 1 c khi lng ring thay i (tng 1,18 ln khi chuyn ha hon ton). Khi tin hnh thc nghim vi cht ch th, thu c kt qu sau: , s 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100C, vn 0 3 15 26 35 37 35 29 30 16 12 9 7 5 3 0Xc nh chuyn ha khik = 1,3.10-21/s nu thi gian lu trong thit b bng thi gian lu ca cht ch th?Gii:Nng trung bnh trong thit b c xc nh theo: ( ) d E C C0A ATrong thit b khuy i vi phn ng bc 1 c khi lng ring thay i th: ( )( ) [ ] + 110 AAk exp 1 1k expCCKt hp ta c: ( )( ) [ ]( ) ( ) + AE Ek exp 1 1k expCC110 AA y:1525 , 018 , 118 , 1 11 U1 U 0 U , sCi Ci.101 E().102AAE().102102AE() 30 2 6 0.1674 0.7314 0.1224 0.122435 14 49 1.1715 0.6933 0.8122 0.934740 25 100 2.0921 0.6570 1.3744 2.309145 34 153 2.8452 0.6223 1.7707 4.079850 36 180 3.0126 0.5894 1.7755 5.855355 34 187 2.8452 0.5580 1.5875 7.442860 28 168 2.3431 0.5281 1.2374 8.680165 20 130 1.6736 0.4997 0.8363 9.516470 15 105 1.2552 0.4727 0.5933 10.109775 11 82.5 0.9205 0.4470 0.4115 10.521280 8 64 0.6695 0.4227 0.2830 10.804285 6 51 0.5021 0.3996 0.2006 11.004890 4 36 0.3347 0.3776 0.1264 11.131295 2 19 0.1674 0.3568 0.0597 11.1909Nh vy: ( )( ) [ ]( ) ( ) + AE EkkCCAA110exp 1 1expThay vo ta c: 00 00111 11AAAAAA AAACCCCCC CUUUCC ++ +Bng trn ch c gi tr tham kho s liu khng ng vi bi.6. nhit u ra ca thit b phn ng c tnh theo cng thc: ( ) ( )PB BCC C Ht 0Vy nhit u ra cua hn hp l : tR = tV + tB. L THUYT:K thut phn ng d th :Cu 1.5Phn ng d th (h rn kh) c phn loi nh th no? Yu t no quyt nh tc phn ng trong mi trng hp? Lm th no iu chnh tc phn ng trong nhng trng hp ny?Trong thc t khi nghin cu ng hc xc tc d th rn kh, ch o c ng hc hiu dng nn dng thc nghim nh gi cc m hnh trn l rt kh. xc nh chnh xc cn loi b nh hng ca qu trnh khuch tn v truyn nhit bn trong v ngoi mao qun ca xc tc. Thng cn c vo 3 tiu chun v nh lng v tnh cht vt l nh gi:1) Phn mol ca cht tham gia phn ng l khng i, chuyn ha bng 0: xA= const; UA= 0. Tc u ph thuc vo p sut R0 = g(p);2) XA = const; UA = 0 t s ( ) P fRP0;3) p sut lm vic khng i, chuyn ha bng 0 tc u ph thuc vo nng : P = const; UA = 0 R0 = g(xA);4) Khi c P = const, x0 cho trc th c quan h: R = g(xA);Xt ht xc tc hnh cu, cn bng vt cht: ( ) ' ' * * kCFFC k C Ctn Vi: C nng nhn dng chy;C nng b mt xc tc; k* - hng s tc mt ngoi; k hng s tc ca h;Fn, Ft din tch mt ngoi v trong ca xc tc;T : Tc phn ng: R rng tc phn ng ph thuc vo nhit v s khuch tn, c 3 vng:- Khi nhit cha cao tc phn ng mt ngoi quyt nh th ka = k*;- Khi nhit tng cao th ka = do khuch tn chm quyt nh;Trng hp trung gian: thay i (iu chnh) tc phn ng: - Khi xy ra trong min ng hc: dng nhit , cht xc tc;- Khi xc ra trong min khuch tn: Ngoi dng khuy trn;Trong nhit v p sut;1.6 Trnh by cc gi thuyt c bn ca thuyt Langmuir khi thit lp ng hc phn ng d th rn kh?Thuyt hp ph ng nhit ca Langmuir c thit lp trn c s cc gi thit sau: 1) Qu trnh hp ph v nh hp ph c m t bng c ch phn ng n gin (bc phn ng bng h s t lng), tc l ton b tm hot ha c cng tnh cht: cng lc ht phn t, cng ln, cc phn t c hp ph khng tng tc ln nhau;2) Lng tm hot ha l khng i: C = C(A) + C= const;3) Ngoi cu t hp ph, khng hp ph cu t khc, khi c i cu tcngthamgiaphnngthlngtmhothacng khng i: C = C() + C(i) = const thit lp phng trnh ng hc Langmuir cn tha mn 3 gi thit: - Tha mn cc iu kin t 1 3;- C mt bc vi tc chm s quyt nh tc qu trnh, cc giai on khc trng thi cn bng;- Tc giai on chm nht l tc chung ca qu trnh;Trn c s thit lp cPhng trnh ng hc tng qut c dng c 3 thnh phn c trng cho: - ng hc;- ng lc;- Hp ph; V d phn ng dng A + B R + S khi giai on 3 chm l: C: kIIIKAKBC2c trng ng hc; CACB KC CS R c trng ng lc hc;( )ni iC K+ 1c trng hp ph;n = 2 s m;1.7 Trnh by cc gi thuyt c bn ca thuyt Temkin khi thit lp ng hc phn ng d th rn kh?C mt s im khc bit so vi nghc Langmuir:- Giai onchmquyt nhca phnngc thaybng nguyn l Bodenstein (dng c ch phn ng n gin);- Dng cht trung gian l nhng cht k bn ch tn ti trng thi cn bng hp ph; - Ch c cht trung gian c hp ph vo tm hot ha;- B mt cht xc tc khng ng nht, nhit hp ph gim dn t l nghch vi lng tm hot ha;Xt phn ng dng: A + B R + S Theo Temkin: A + () R + (J) B + (J) S + ()Ch c cht trung gian c hp ph vo tm hot ha.Nu b mt cht xc tc ng nht (ch c thay i nh ca cc tm hp ph) th: Theo nguyn l Bodenstein hp cht trung gian km bn nn: Tng s tm hot ha: C = C() + C(J) = const C() = C C(J) Do ( )( )C C C CkkKAJ RI 11 IKkk11 Tng t: IIKkk22 Thay vo ta c: R = k1CA[C C(J)] k-1(CRC(J) + k-2CS[C C(J)] k2CBC(J) = 0T :

Khi b mt ng nht ta c: Khi nghin cu phn b thi gian lu bng thc nghim, ngi ta cho cht ch th vo thit b phn ng v coi n nh hp en, ri o hm p ng u ra theo s chung hnh v Ty theo c im v cch thc hot ng ca thit b m dng bin u vo dng khc nhau, nn kt qu o c hm p ng u ra cng khc nhau:- Cho tnh hiu vo dng bc th xc nh hm tch phn phn b;- Cho tn hiu vo dng xung th thu c hm vi phn phn b;Tn hiu c th c cho vo nhiu dng: hnh sin, rng ca, tamgic, vunggc, khi, tns, t bindng(bc), Dirac (xung), t bin m C hai dng tn hiu hay c s dng, l:- Tn hiu dng bc: xem hnh v '