2/6/2012 Chuong 1- Khai niem mo dau 1K THUT PHN NGChemical Reaction EngineeringVuBaM inhBm M ay Thiet b H Bach K hoa TP.H CM2/6/2012 Chuong 1- Khai niem mo dau 2Ti liu thamkho1) V B Minh, Qu trnh & TB trong cngngh ha hc- tp 4 , Nxb i hc QG TP.HCM.2) Octave Levenpiels; Chemical Reaction Engineering, John Wiley&sons, 1999.3) H. Scot Foggler, Elements of Chemical Reaction Engineering,Internationalstudents edition, 1989.2/6/2012 Chuong 1- Khai niem mo dau 3Ti liu c thm1) E.B.Nauman, Chemical Reactor Design, John Wiley & sons, 1987.2) Stanley M. Walas, Reaction Kinetics for Chemical Engineers,Int. Student Edition, 1990.3) Coulson & Richardsons, Chemical Engineering Vol 6,Elsevier, 1979.4) Richard M. Felder, Elementary Principles of Chemical Processes, John Wiley & sons, 2000.2/6/2012 Chuong 1- Khai niem mo dau 4 K thut phn ng ng th(Homogeneous chemical reaction eng.) K thut phn ng d th(Heterogeneous chemical reaction eng.)2/6/2012 Chuong 1- Khai niem mo dau 5K thut phn ng ng th Chng 1: Khi nim m u(Introduction to Chem. Reaction Eng.) Chng 2: X l d kin ng hc (Interpretation of Chemical Kinetics Data) Chng 3: Phng trnh thit k (Design Equation) Chng 4: p dng phng trnh thit k (Application of Design Equation) Chng 5: Hiu ng nhit . (Temperature Effects) Chng 6: Dng chy thc (Real Flow)2/6/2012 Chuong 1- Khai niem mo dau 6Chng 1: Khi nimm uQUATR NHVAT LY HOA HOCTHUAN NGHCH K . THUAN NGHCHTH.NGHCHK . THUAN NGTRUYEN K HOI C, NHI ETCAN BANG PHACan bang P.2/6/2012 Chuong 1- Khai niem mo dau 7Chng 1: Khi nim m u_ Thit k thit b phn ng (T.B.P.) khng c khun mu, c th c nhiu bn thit k, bn thit k ti u v kinh t.Lin quan n nhiu lnh vc khc nhau: Nhit ng lc hc, ng ha hc, C lu cht, Truyn nhit; Truyn khi.2/6/2012 Chuong 1- Khai niem mo dau 8Cc qu trnh trong qui trnh sn xut cng nghip2/6/2012 Chuong 1- Khai niem mo dau 9Trc khi thit k cn tr li hai cu hi sau:1. Phn ng no l phn ng chnh ?2. Tc phn ng nh th no ?Cu hi 1 lin quan n nhit ng lc hc.Cu hi 2 lin quan n cc qu trnh vntc nh: ng ha hc, truyn nhit, truyn khi2/6/2012 Chuong 1- Khai niem mo dau 101.1. ng ha hc(Chemical Kinetics) ng hc l cch m thin nhin ngnnga mi qu trnh xy ra cng mtlc. S.E. LeBlanc2/6/2012 Chuong 1- Khai niem mo dau 111.1. ng ha hc(Chemical Kinetics) Nghin cu tc phn ng ha hc v c chphn ng.1. Phn ng n v phn ng a hp(single &multiple reaction)2.Phn ng s ng & khng s ng(elementary & non elementary reaction)3. Cn bng cho phn ng thun nghch s ng( Chemical Equilibrium)4. Bc phn ng (Reaction Order)5. S ph thuc nhit nh lut Arrhnius(Temperature dependency)2/6/2012 Chuong 1- Khai niem mo dau 121.2. Nhit ng lc hc(Chemical thermodynamics) Nhit hp thu hoc phng thch Mc phn ng1. Nhit phn ng(Heat of reaction)2. Cn bng ha hc(Chemical Equilibrium)2/6/2012 Chuong 1- Khai niem mo dau 131. Nhit phn ng) T (T ) C (N ) T (T ) C (N H H) C (N ) C (N CdT C H H0 tcpi 0pi T r, T r,p i p i pTTp0T r, T r,i i0i i0 + A = A = AA + A = A }sptc sp2/6/2012 Chuong 1- Khai niem mo dau 142. Cn bng ha hc||.|

\|A == AA= = A1 20T r,TT0T r,20T r,0T1 T1 RH

KKln constant HT RH

dTlnK dHoff tVan'K lnRT F012002/6/2012 Chuong 1- Khai niem mo dau 151.3. Phn lai phn ngTng hp ammoniacOxit ha ammoniacHNO3Phn ng cracking, reformingTng hp metanolPhn ng chy ca thanNung qungAxit + cht rnHp thu + phn ngPhn ng th keo Phn ng chy cu ngnlaD thPhn ng pha lng Phn ng pha khC xc tc Khng xc tc ng th2/6/2012 Chuong 1- Khai niem mo dau 161.4. nh ngha vn tc phn ng Da trn mt n v thtch hn hp phn ng Da trn mt n v thtch bnh phn ng Da trn mt n v dintch b mt tip xc pha Da trn mt n vkhi lng cht xc tch . m mol/,dtdN V1r3ii =h . mol/m ,dtdN V1r3ib'i=h . m mol/,dtdN S1r2i' 'i =h . kg mol/,dtdN W1ri' ' 'i=2/6/2012 Chuong 1- Khai niem mo dau 17Th d 1.1 Mt ng c ha tin t chy hn hpnhin liu gm H2v O2lng.Bung thnh tr c ng knh l 60cm,chiu di75cmv qu trnh t sinh ra sn phmchy 108kg/s.Nu qu trnh chy hantan, tm vn tc phn ng ca hydrogen voxygen.O H O21 H2 2 2 +2/6/2012 Chuong 1- Khai niem mo dau 18Th d 1.2 Mt ngi nng 75 kg tiu th khang 6.000 kJ thc phmmi ngy. Gi s tt c thcphml glucose v phn ng tng qut nhsau C6H12O6+6O2 6CO2+6H2O - Hr= 2.816 kJ Tnh tc bin dng theo s mol oxygen s dng trn m3c th trong mtgiy. Cho bit 2.816 KJ / mol glucose2/6/2012 Chuong 1- Khai niem mo dau 19Cc yu t nh hng n vn tc phn ngPhn ng ng thp sutNhit Nng Phn ng d th .Truyn khi gia hai phaTruyn nhitGiai an kim sat vn tc (rate controlling step)2/6/2012 Chuong 1- Khai niem mo dau 20Th d 1.3 Ha cht A 2R, thit b chmbng ht cukhng rng: a = 200m2/m3B = 2.908 kg/m3(bulk density) = 0,40 Xc nh n v cc i lng v cc dng phngtrnh vn tc khc ?h . kg mol/, C 0,1dtdN W1 - rAii' ' ' '= =2/6/2012 Chuong 1- Khai niem mo dau 211.5. Phn lai thit b phn ng1) Phng php vn hnh /hat ng (Mode of Operation) Gin an (batch/ unsteady state) Lin tc (continuous / steady state) Bn lin tc (semi continuous) 2) Hnh dng thit b phn ng Khuy trn l tng (Ideal Mixing-Stirred Tank) ng/ y l tng (Ideal Plug Flow Reaactor)3) S pha ca hn hp phn ng ng th (Homogeneous) D th (Heterogeneous)2/6/2012 Chuong 1- Khai niem mo dau 22Bnh phn ng khuy trn2/6/2012 Chuong 1- Khai niem mo dau 23Bnh ngThit bphan ngNaphtha va khhoan lu2/6/2012 Chuong 1- Khai niem mo dau 24Kt hp gia hnh dng v phngphp vn hnh s c bao nhiudng thit b phn ng?2/6/2012 Chuong 2-Xu ly du kien dong hoc 1Chng 2: X l d kin ng hc(Interpretation of Kinetic Data)2/6/2012 Chuong 2-Xu ly du kien dong hoc 2 Phng trnh vn tc: c trng cho phn ng c xc nh t: l thuyt, m hnh cho trc,thc nghim Hai giai an: ph thuc nng vs ph thuc nhit 2/6/2012 Chuong 2-Xu ly du kien dong hoc 3Thit b phn ng th nghim c th hat nggin an hoc lin tcTheo di mc phn ng:1. Nng ca mt cu t2. Tnh cht vt l ca hn hp3. p sut tng ca h ng tch4. Th tch ca h ng p.2/6/2012 Chuong 2-Xu ly du kien dong hoc 4Cc phng php x l s liu ng hc1. Phng php tch phn2. Phng php vi phn3. Phng php thi gian bn sinh(half-life time)4. Phng php tc phn ng ban u (Intial reaction rate)2/6/2012 Chuong 2-Xu ly du kien dong hoc 52.1. Thit b phn ng gin an c th tchkhng i (th tch hn hp phn ng) V = const(2.1) dtdC dtdC V dV C V1 dt) V d(C V1 dtdNV1ri i i i ii=+= = =(2.2) dtdp RT1rii =2/6/2012 Chuong 2-Xu ly du kien dong hoc 6Trong thc t, thng o p sut tng hn hp phnng trong pha kh theo di phn ng(2.4)) P (Pnrp RT Cp Rcho(2.3) ) P (Pnap RT Cphay VN N na VN V x a. N VN RTpC0 R0 R R0 A0 A A0 A0 A0 A AAA+ = =A = =A == = =2/6/2012 Chuong 2-Xu ly du kien dong hoc 72.1.1. Phng php tch phn1) Gi thit c ch vphng trnh vn tctng ng2) Sp xp li3) Xc nh gi trF(CA)theo thc nghim4) V F(CA) theo t5) Nu khng thng, githit lif(kC)dtdCrAA= = kf(C)dtdCrAA= = kdt) f(CdC AA= kt dt k) (C F) f(CdC t0ACCAAAA0= = = } }2/6/2012 Chuong 2-Xu ly du kien dong hoc 8Hnh 2.12/6/2012 Chuong 2-Xu ly du kien dong hoc 9(1) Phn ng khng thun nghch bc 1lai mt phn t A sn phmktCCln dt kCdC C k dtdC A0At0CCAAAAAA0= = = } }2/6/2012 Chuong 2-Xu ly du kien dong hoc 10 chuyn ha (conversion), XAl phn tc cht chuyn ha thnh sn phmkt ) X (1 ln dt k X 1dX) X k(1dtdX dX C dC) X 1 ( CV) X (1 N VNC) X (1 N NAt0 0AAAAA A0 AA A0A A0 AAA A0 A= = = = == = =} }AX2/6/2012 Chuong 2-Xu ly du kien dong hoc 11- dCA/dt = kCA0,6.. CB0,4 l bc mt nhng khngp dng cHnh 2.2. Phn ng bc mt2/6/2012 Chuong 2-Xu ly du kien dong hoc 12(2) Phn ng khng thun nghch bc 2 lai hai phn t A +B sn phmA0B0A A0 B0 A A0 A0AA0 AB B0 A A0B AB AACCM) X . C )(C X . C C ( kdtdXC rX C X C (2.12) C . C k dtdC dtdCr= = = == = = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 13} }= = = tXA0A00A AAA A2A0AA0 Adt kC) X )(M X (1dX) X . M )( X . 1 ( C kdtdXC r 2/6/2012 Chuong 2-Xu ly du kien dong hoc 14Nu CB0 >> CA0th CBgn nh khng i, phn ng xem nh gi bc mt1 M kt ) C (C kt 1) (M CM.CClnC . CC . Cln) X (1 MX MlnX 1X 1lnA0 B0 A0ABA B0A0 BAAAB= = == ==2/6/2012 Chuong 2-Xu ly du kien dong hoc 15Hnh 2.32/6/2012 Chuong 2-Xu ly du kien dong hoc 16Lu a) 2A sn phm ktX 1X.C1

C1

C1(2.14) ) X (1 kC C k dtdCrAAA0 A0 A2AA02 2AAA== = = = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 17(3) Phng trnh vn tc thc nghimc bcn( ) | | 1)kt (n1 X 1 C hay1 n 1)kt, n( CC . C k dtdCrn 1An 1A0n 1A0n 1AnAAA = = = = = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 18(4) Phng trnh vn tc thc nghimc bc 01 n kt,X C CCk dtdCrA A0 A A0AA= = = = = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 19(5) Phn ng khng thun nghch bc tng qut theothi gian bn sinh t1/2aA +bB sn phm C . C kdtdCrbBaAAA= = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 20Nu tc cht hin din theo t l lng ha hc, chng s gi t l trong sut qu trnh phn ng. Nh vy ti thi imbt k CB/ CA =b/a C1) (nk1 2 t C kdtdChay ....abk ... ) Cab.( C kdtdCrn 1A0'1 n 1/2nA' A.. b abbAaAAA+ +== |.|

\|= = = AC2/6/2012 Chuong 2-Xu ly du kien dong hoc 21(6) Phn ng song song khng thun nghch(parallel reaction)A R, k1A S, k2A 2SSA 1RRA 2 1 A 2 A 1AAC kdtdCrC kdtdCrC ) k (k C k C kdtdCr= + = += + = ++ = + = = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 22CA +CR+ CS= const21S SR R21SRSR2 1A0Akk

C CC Ckk dCdC

rr t ) k k (CCln00== =+ = 2/6/2012 Chuong 2-Xu ly du kien dong hoc 23Hnh phnng song song 2/6/2012 Chuong 2-Xu ly du kien dong hoc 24(7) Phn ng ni tip khng thun nghch(consecutive reaction)A R S , k1 v k2R 2SR 2 A 1RA 1AC kdtdCC k C kdtdCC kdtdC+ = + = =2/6/2012 Chuong 2-Xu ly du kien dong hoc 25Phng trnh tnh||.|

\|== += = 1 2t k t kA0 1 RkA0 1 R 2Rkt0A A 10AAk ke eC k Ce C k C kdtdCe C C hay tkCCln2 11t2/6/2012 Chuong 2-Xu ly du kien dong hoc 26Ti thi imbt k CA0 = CA + CR+ CS( )( )t kA0 S 2 1t kA0 S 1 2t k1 21t k1 22A0 S212 1e 1 C C k ke1 C Ck kek kk ek kk 1 C C = )) = ))||.|

\|+ =2/6/2012 Chuong 2-Xu ly du kien dong hoc 27Thi imnng R t cc i( )1 2 2k k /k21A0max R,1 212 tb logmaxkk

CCk kkkln

k1 t||.|

\|= ||.|

\|= =2/6/2012 Chuong 2-Xu ly du kien dong hoc 28(8) Phn ng thun nghch bc 1AR KC = K = hng s cn bng21A A0 A0A A0 R0AeReCCA0R0CAeAA A0 R0 2 A A0 A0 1R 2 A 1AA0A Rkk

X C CX C C CCK1 KCCK X0dtdC) X C (C k ) X C (C kC k C kdtdXCdtdC dtdC=+= =+= =+ = = ==ee2/6/2012 Chuong 2-Xu ly du kien dong hoc 29V ng biu din ln (1 XA/XAe) theo t tac ng thng c h s gc lk1(1 + 1/KC).Phn ng thun nghch c xeml khng thunnghch nu nng da trn CA0- CAe( )( )( )t k k t K11 kC CC ClnXX1 lnX X k kdtdX2 1C1Ae 0 Ae A AAeAA Ae 2 1A+ =||.|

\|+ =||.|

\| =||.|

\| + =2/6/2012 Chuong 2-Xu ly du kien dong hoc 302.1.2. Phng php vi phnGi thit c ch (-rA) = - dCA/dt = k.f(C)T ng cong C theo t xc nh (-dCA/dt) tinhng thi imkhc nhauLp bng gi tr CA, (-dCA/dt) theo t v tnh gi trhms f(C).V (-dCA/dt)theo f(C), nu l ng thng thphng trnh vn tc ban u ph hp vi thcnghim.Nu khng c ng thng qua gc ta , githit li c ch khc (hms f(C)).2/6/2012 Chuong 2-Xu ly du kien dong hoc 312.2. Thit b phn ng gin an c th tch(th tch hn hp phn ng) thay i( )0 X0 X 1 XAA A 0i ii i i iiAA AVV V X 1 V VdtdVVC dtdC

dtdV C VdC dtV) d(CV1 dtdNV1r== = =+ =+ =+= = =2/6/2012 Chuong 2-Xu ly du kien dong hoc 32Vit li cc biu thc( )( )( )( )( )dtdXX 1C

dtX 1 d NX 1 V1 dtdNV1rX 1X 1CX 1 VX 1 N VNCAA AA0A A0A A 0AAA AAA0A A 0A A0 AA +=+ = = +=+ = =2/6/2012 Chuong 2-Xu ly du kien dong hoc 332.3. Nhit v tc phn ng nh lut Arrhnius k = k0e- E / RT vi: k0: tha s tn s (frequency factor) E : nng lng hat ha (activation energy), J/mol R: hng s kh = 8,27 J/mol.K T: K2/6/2012 Chuong 2-Xu ly du kien dong hoc 34nh hng ca nhit ln tc phn ng2/6/2012 Chuong 2-Xu ly du kien dong hoc 35nh hng ca nhit ln tc phn nga) Bnh thngb) Phn ng d th do qu trnh truyn khi kimsat, (-r) tng chmtheo T.c) Phn ng n, (-r) tng nhanh ti nhit bc chy.d) Phn ng xc tc do tc hp ph kimsat (T tng lmgimhp ph) hay phn ng enzym.e) Phn ng phc tp c phn ng ph v tng ngk ti nhit tng.f) Phn ng thun nghch pht nhit2/6/2012 Chuong 2-Xu ly du kien dong hoc 362/6/2012 Chuong 2-Xu ly du kien dong hoc 37Qu trnh truyn nhit cho bnh phn ng2/6/2012 Chuong 2-Xu ly du kien dong hoc 382/6/2012 Chuong 2-Xu ly du kien dong hoc 39Th d 2.1.Xc nh tc phn ng cho phn ng sau:(CH3)3COO(CH3)3 C2H6 +2CH3COCH3Phn ng c thc hin trong bnh kn, gin an, ng nhit v ghi nhn p sut tng thay i theothi gian nh sau vi di-tert-butil-peroxid lnguyn chtThi gian, ph 0,02,5 5,010,0 15,0 20,0p sut tng, mmHg 7,510,5 12,5 15,8 17,919,4pA, mmHg2/6/2012 Chuong 2-Xu ly du kien dong hoc 40Cu hi n tp1. Cc phng php theo di phn ng?2. Phng php x l s liu ng hc?3. X l s liu ng hc cho pha kh2/6/2012 Chuong 3 - Phuong trinh thiet ke 1Chng 3: Phng trnh thit k2/6/2012 Chuong 3 - Phuong trinh thiet ke 23.1. Cn bng vt cht & nng lng tng qut1. Cn bng vt cht cho mt phn t thtch(Lng tc cht nhp vo) - (Lng tccht ra khi) - (Lng tc cht phn ng)= (Lng tc cht tch tu (bin i))2/6/2012 Chuong 3 - Phuong trinh thiet ke 3PHN TTH TCH (TCH T)VO RAPHN NG2/6/2012 Chuong 3 - Phuong trinh thiet ke 42. Cn bng nng lng(Nng lng cc dng nhp vo) -(Nng lng cc dng ra khi) + (Nng lng trao i vi mi trng ngai)= (Nng lng tch tu (bin i))2/6/2012 Chuong 3 - Phuong trinh thiet ke 5PHN TTH TCH (TCH T)VO RATRAO I2/6/2012 Chuong 3 - Phuong trinh thiet ke 6Cc dng bnh phn ng khuy trn2/6/2012 Chuong 3 - Phuong trinh thiet ke 73.2. Thit b phn ng khuy trn l tng1. Hot ng n nh Vi: XA0, XAfl chuyn ha ca dng nhp liuv dng sn phm v l lu lng ca dng nhp liu, m3/s( )) r (X X

C v.V

FVhay0 tV. r t) X 1 ( F t) X 1 ( FAfA0 AfA0 A0Af Af A A0 A0 0= == A X A A 2/6/2012 Chuong 3 - Phuong trinh thiet ke 8Xc nh nhit ca dng sn phm tnh vntc phn ng cn bng nhit lng mt: sut lng dng nhp liu, kg/s Cp: nhit dung ring dng nhp liu (sn phm), J/kg.0CHr: nhit phn ng, J/molK : h s truyn nhit, W/m2.0CS: din tch b mt truyn nhit, m20 ) T K.S.(T H V. ). r ( C ). T (T m hay0 ) T KS(T F H ) X (X C ) T (T mf n r A p f 0 tf n A0 r A0 Af p f 0 t= + A = + A 2/6/2012 Chuong 3 - Phuong trinh thiet ke 9Th d 3.1: Xc nh lu lng mi dng nhp liuPhn ng thun nghch pha lng s ngA+B=R+Svi k1= 7 l/mol.ph , k2 = 3 l/mol.phV = 120 ltHai dng nhp liu ring bit c lu lng bngnhau: Dng c nng 2,8 mol A/ lt Dng c nng 1,6 mol B/ lt chuyn ha ca tc cht gii hn t 75%2/6/2012 Chuong 3 - Phuong trinh thiet ke 10Th d 3.2: Xc nh phng trnh vn tc cho phnng phn hy pha kh A R + Sxy ra ng nhittrong bnh khuy trn hat ng n nh0,96 0,88 0,75 0,63 0,22 XA (vi CA0 =0,002mol/l192 44,0 13,5 5,10 0,423 =V/v ph5 4 3 2 1 S TN2/6/2012 Chuong 3 - Phuong trinh thiet ke 11Bng 3.10,021 0,96 192 50,064 0,88 44,0 40,143 0,75 13,5 30,227 0,63 5,10 20,639 0,22 0,423 1XAThnghimsV= ,phvAA1- X1+ XCoA AAX= (-r ), . ,..50 002 0 22104 100 423=, . ,, .,50 002 0 6324 7 105 10=, . ,, .,50 002 0 75111 1013 5=, . ,.50 002 0 884 1044=, . ,.50 002 0 961 10192=2/6/2012 Chuong 3 - Phuong trinh thiet ke 123.2. Thit b phn ng khuy trn l tng 2. Hot ng gin andtdXNV ) r (dtdXNdt)] X 1 ( [N d V ) r (dtdN V ) r (0 t ; N t V. ) r (AA0 AAA0A A0AAAA A= == = A A = A 2/6/2012 Chuong 3 - Phuong trinh thiet ke 13Sp xp li v ly tch phn( )( ) ( )( ) ( )} }} }}+ =+ =+ = == ==A AAA0AAX0A A AAA0X0A A 0 AAA0A A 0CCAAX0AAA0X0AAA0) X (1 rdXC) X (1 V rdXN t ) X (1 V V *rdC V rdXC t constV *V rdXNt 2/6/2012 Chuong 3 - Phuong trinh thiet ke 14Th d 3.3:a) Tnh thi gian phn ngb) Th tch bnh phn ng khuy gin anCH3COOH + C4H9OHCH3COOC4H9 + H2O 1000C, xc tc H2 SO4nng 0,032%k.l Nhp liu: 4,97 mol butanol/mol axit Phng trnh vn tc (-rA)= k.CA2vi k = 17,4 ml/ mol.ph = 0,75 g/ml = const XAf= 50%2/6/2012 Chuong 3 - Phuong trinh thiet ke 153.3. Thit b phn ng dng ng l tng( )( )}== A A= A A= A A A A A Af00 0X0AAA0A0A AA A AA A A A A A) r (dX

FV

F) r (

dVdX

0 VV, Chia0 V r X F hay 0 tV. r t) X X 1 ( F t) X 1 ( F2/6/2012 Chuong 3 - Phuong trinh thiet ke 16Th d 3.4: Tnh th tch bnh phn ng ng4PH3(k) P4(k)+6H2(k) (-rPH3) = (10 h-1) CPH3 Hat ng nhit 6500C, p sut 4,6 atm Xaf=80% Nhp liu c sut lng 2 kmol/h phosphinnguyn cht.2/6/2012 Chuong 3 - Phuong trinh thiet ke 17Th d 3.5:CH4+ 2S2 CS2+ 2H2S Bnh phn ng ng c V = 35,2ml Th nghim 6000C, 1atm thu c 0,10g CS2trong 10 ph. Sut lng n nh ca S2l 0,238 mol/h, mtan l 0,119 mol/ha) Xc nh rCS2b) Tmk vi (-rA)=kpCH4.pS2bng 2 phngphp2/6/2012 Chuong 3 - Phuong trinh thiet ke 18Bi tp 3.5.Cho phn ng to thnh etilen glycol nh sau:CH2OH-CH2Cl + NaHCO3(CH2OH)2+ NaCl + CO2 Phn ng s ng k = 5,2 l/ mol.h 820C C 2 dng nguyn liu c sn nh sau: Dd 15% k.l bicarbonat sodium Dd 30% k.l etilen clorhidrin trong nc2/6/2012 Chuong 3 - Phuong trinh thiet ke 19Cu hi cho bi tp 3.5a) Tnh th tch bnh phn ng khuy trnhot ng n nh sn xut 50kg/h etilen glycol vi dng hn hp nhp liung mol t chuyn ha 95%?b) Tnh th tch bnh phn ng ng phnng cng iu kin trn?c) Tnh th tch bnh phn ng khuy trnhot ng gin an sn xut 50kg/metilen glycol vi hn hp nhp liu ban u ng mol t chuyn ha 95%?2/6/2012 Chuong 3 - Phuong trinh thiet ke 20Bi tp 3.1. Phn ng pha kh ng th A 3R xy ra theobc 2 Vi lu lng nhp liu l 1.000 l/h A nguyn cht 5 atmv 3500C thit b phn ng th nghiml ng cng knh 25mm, di 1.800mm cho chuyn ha t60%. Thit b phn ng dng trong sn xut phn ng80m3/h hn hp nhp liu gm50%A v 50% kh tr 25 atmv 3500C t chuyn ha 80%a) Cn s dng bao nhiu ng ng knh 25mm, di1.800mmb) Cc ng ny c mc song song hay ni tip?Gi s l ng l tng, khng c gim p, kh l tng2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke1Chng 4: p dng phng trnh thit k2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke2 C th s dng mt trong nhiu dng bnhphn ng Thay i t l nng tc cht trong nhpliu ban u Yu t nh hng n la chn:Lai phn ngChi ph thit b v dng c oTnh n nh khi hat ngTnh linh ng ca thit b khi thay i iukin hat ng2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke3Hai thng s thit k nh hng n tnhkinh t ca qu trnh:1) Th tch thit b phn ng2) S phn phi sn phm chnhtrong phn ng a hp2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke44.1. So snh kch thc thit b phn ng n(1) Bnh phn ng khuy trn hat ng nnh & Bnh ngS dng trc tip phng trnhthit kS dng gin (hnh 4.1)2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke52/6/2012 Chuong 3 Ap dung phuong trinh thiet ke64.1. So snh kch thc thit b phn ng n(2) S bin i t l nng ban u ca 2 tc chttrong phn ng bc hai Bnh ng, hnh 4.21 M ,) X (1X C k 1 FV C 1 M ,) X M(1X Mln1) M ( C k 1 FV C 1CC MAAA01 MA0A01 MAAA01 MA0A01 MA0B0==||.|

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\|=) =====2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke72/6/2012 Chuong 3 Ap dung phuong trinh thiet ke8(2) S bin i t l nng ban u ca 2 tccht trong phn ng bc hai Bnh khuy lin tc, hnh 4.31 M ,) X (1XC k 1 FV C 1 M , ) X M ( ) X 1 ( C k X FV C 2AAA01 MA0A01 MA A A0A1 MA0A01 M==||.|

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\|=====2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke92/6/2012 Chuong 3 Ap dung phuong trinh thiet ke10Th d 4.1. Phn ng A + B sn phm(-rA) = (500 l/ mol.ph) CA. CB Bnh ng Vo= 0,1 lt; v = 0,05 l/ph CA0= CB0= 0,01 gmol/ lta) Xc nh XAf?b) Cng nng sut v XAf, tmVk?c) Cng nng sut, tmXAfnu c Vk= VoNu CB0= 0,015 gmol/ lt, CA0= 0,010 gmol/ ltd) Vi cng v, tmXAfcho bnh ng ?e) Vi cng XAfban u, tmt l gia tng nng sut?f) Tmv cho bnh phn ng c Vk= 100lt, XAf=99%2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke11Th d 4.2. Tm iu kin ti u A R 100 gmol R/h c sn xut t dd bo haA (CA0= 0,1 gmol/ l). rR= (0,2 h-1)CA$A= 500 /mol A$ b = 10 /h.l A khng phn ng c thi b Tmth tch, sut lng, chuyn ha, githnh ca R ti iu kin ti u ?2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke12Th d 4.3. Tm iu kin ti u A R Gi s A khng phn ng trong dng snphm c ti ch, han lu vi chi ph l$r = 125 /gmol A han lu. Tmth tch, sut lng, chuyn ha, githnh ca R ti iu kin ti u ?2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke13Th d 4.3: A cha phn ng c han lu2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke144.2. H nhiu bnh phn ng4.2.1. Bnh ng mc ni tip v/ hay song songj bnh phn ng ngmc ni tip c tngth tch V s cho chuyn ha bng chuyn ha trong mtbnh phn ng c thtch V( )} =AjX0AAA0rdX

FV2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke154.2. H nhiu bnh phn ng4.2.1. Bnh ng mc ni tip v/ hay song song Vi cc bnh phn ng ng mc song song, phn phi dng nhp liu sao chothnh phn dng ra trong mi nhnhl ging nhau, tc l V/F hay cho minhnh l ging nhau.2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke164.2. H nhiu bnh phn ng4.2.2. Bnh khuy lin tc bng nhau mc nitip(1) Phn ng bc mt (hnh 4.7)AA0/ 1AjA0iCCln k1

1CC kj j=(((

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\|= =ojjk 2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke174.2. H nhiu bnh phn ng4.2.2. Bnh khuy lin tc bng nhau mc nitip(2) Phn ng bc mt (hnh 4.8)k C 1CCj 1 i kC 4 1 2 1 2 1... 2 1k21 CA0AA0i A0jAj+ = =|.|

\|+ + + + =2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke18Th d 4.3. Bnh khuy mc ni tip Bnh phn ng khuy lin tc t chuyn ha 90% tc cht A R theo phn ng bc 2 D nh thay bnh ny bng 2 bnh c tng th tchbng bnh trca) Cng XAf= 90%, nng st tng bao nhiu?b) Nng sut nh c XAftng bao nhiu ? Mc ni tip 2 bnh, mi bnh c th tch bng bnhtrcc) Cng XAf= 90%, nng st tng bao nhiu?d) Nng sut nh c XAftng bao nhiu ?2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke194.3. Thit k cho phn ng a hp4.3.1. Phn ng song song(1) Kho st nh tnh s phn phi sn phma1 a2A12RSRSa2A 2SSka1A 1RRkCkk

dCdC

rrC kdtdC r(phu) S AC kdtdC r(chinh) R A21= == = = = 2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke20M hnh cho thy nng ca cc tc chtcao hay thp cho qu trnh n nh2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke214.3. Thit k cho phn ng a hp4.3.1. Phn ng ni tip(1) Kho st nh tnh s phn phi sn phm2/6/2012 Chuong 3 Ap dung phuong trinh thiet ke222/6/2012 Chuong 3 Ap dung phuong trinh thiet ke232/6/2012 Chuong 3 Ap dung phuong trinh thiet ke242/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 1Chng 5: Hiu ng nhit Temperature Effect5.1. Khi nim5.2. Bnh khuy l tng hot ng n nh5.3. Bnh khuy l tng hot ng gin on5.4. Bnh ng l tng hot ng n nh5.5. Khong nhit ti u2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 25.1. Khi nim v hiu ng nhit Phn ng khng thun nghch Thu nhit Pht nhitPhn ng thun nghch Thu nhit Pht nhit2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 3Vn tc phn ng theo chuyn ha trongbnh phn ng an nhit2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 4 chuyn ha nhit cho phn ng thun nghchbc 1 trong bnh khuy hat ng an nhit2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 55.2. Bnh khuy l tng hot ng n nhBnh khuy l tng hot ng n nh nn nhit khng i v do vn tc phn ng l hngs.Gii h 3 phng trnh:1) Tc phn ng;2) Cn bng vt cht (phng trnh thit k)3) Cn bng nng lng2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 6iu kin hot ng n nh cho bnh khuyl tng hot ng n nh-Phn ng bc 1 (5.3)T T) H ( FC m X(5.2)e k 1e k X0 fr A0p tAfE/RT0E/RT0Afff 2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 72/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 85.3. Bnh khuy l tng hot ng gin onPhng trnh cn bng nhit lng v qu trnh an nhit (5.6) X XC mN HT T(5.5) dX N ) H ( dT C m(5.4) ) T KS(TdtdX)N H (dtdTC mA0 AP tA0 r0A A0 r P ti 0AA0 r P t 2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 95.4. Bnh ng l tng hot ng n nhan nhit (5.14)) X X (C mF ) H (T T(5.13) dX H F dT C mA0 AP tA0 r0A r A0 P t 2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 105.5. Khong nhit ti u V, FA0, XAfTi a ho nng sutKhong nhit ti u c th:a) ng nhitb) thay i theo thi gianc) thay i theo chiu did) t bnh khuy ny sang bnh khuy khc2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 11Phn ng khng thun nghch / thu & phtnhit chuyn ho ti a khng ph thuc nhit Tc phn ng tng theo nhit Nng sut ti a t c ti nhit cao nht cth.Nhit b gii hn do vt liu ch to v phnng ph nu c.2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 12Phn ng thun nghch thu nhit Tng nhit lm tng chuyn ho cnbng v vn tc S dng nhit cao nht c th2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 13Phn ng thun nghch pht nhit Tng nhit s tng tc phn ngnhng lm gim chuyn ho cn bng, vngc li. Phn ng c thc hin vi nhit gimdn t lc bt u n khi chm dt. Xc nh khong nhit ti u (V, FA0, XAf)2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 14Vn tc phn ng l hm s theo XA v T cho phn ng thun nghch pht nhit2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 15Th d 5.4: Phn ng thun nghch bc mtA Ra) T 00C n 1000C xc nh chuyn ho cn bng XAetheo nhit cho phn ng trn. F0298K = - 1.744 cal/mol, Hr,298K= - 18.000 cal/mol.b) Phn ng thc hin trong bnh ng t = 10 pht. Xc nh chuyn ho nu thit b phn ng hot ng ng nhit lnlt 250C, 650C v nhit ti u.Th nghim ng hc trong bnh gin on:XA = 0,793 sau 19 pht 250CXA = 0,691 sau 8 pht 350Cc)Lp li cu b vi bnh khuy hot ng n nh2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 16Th d 5.4: Phn ng thun nghch bc mtA Rd) S dng khong nhit ti u cho bnh ng viCA0=1mol/lt, tnh t t XA = 0,76 v xc nh t l tngnng sut so vi hot ng ng nhit. Tmax = 650C.e) S dng khong nhit ti u cho h hai bnh khuymc ni tip, tnh t t XA = 0,60v xc nh t ltng nng sut so vi hot ng ng nhit v nhit , chuyn ho t c trong mi bnh.2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 172/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 18Vn tc phn ng theo XA ti nhng nhit khc nhau2/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 192/ 6/ 2012 Chuong 5- Hieu ung nhi etdo 20ChngChng 66 - - Dng ch Dng ch y th y th c c 1 1 2/6/2012 2/6/2012ChngChng 6: 6: DngDng chchyy ththcc2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 2 2 Nh Nh ng ng sai sai l l ch ch soso v v i i m m h h nh nh l l t t ng ng c c th th l l : : Dng Dng ch ch y y t t t t c c a a lu lu ch ch t t S S tu tu n n han han c c a a lu lu ch ch t t C C c c v v ng ng t t ng ng trong trong thi thi t t b b Trong Trong t t t t c c c c c c l l ai ai thi thi t t b b th th c c hi hi n n qu qu tr tr nh nhtruy truy n n nhi nhi t t,, truy truy n n kh kh i i,, k k thu thu t t ph ph n n ng ng n n u uc c c c c c hi hi n n t t ng ng trn trn s s l l m m gi gi m m kh kh nng nng h h at at ng ng ( (hi hi u u su su t t)) c c a a thi thi t t b b . . M M c c ch ch c c a a chng chng n n y y l l nh nh gi gi nh nh l l ng ng nh nh h h ng ng c c a a c c c c sai sai s s trn trn ln ln chuy chuy n n h h a a. .2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 3 3Nh Nh ng ng sai sai l l ch ch soso v v i i m m h h nh nh l l t t ng ng2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 4 41. KH 1. KH I NI I NI M V M V KHU KHU Y TR Y TR N V N V M HNH M HNH33 phng phng ph ph p p c c t t nh nh c c c c sai sai s s soso v v i i l l t t ng ng: :1) 1) X X c c nh nh s s phn phn ph ph i i th th i i gian gian lu lu th th c c t t ( (th th ch chh h p p cho cho thi thi t t b b ph ph n n ng ng d d ng ng ng ng,, ch ch y y dng dng v v ph ph n n ng ng b b c c m m t t). ).2) 2) M M h h nh nh phn phn t t n n theo theo phng phng tr tr c c ( (th th ch ch h h p p cho chothi thi t t b b ph ph n n ng ng c c ch ch ch ch y y r r i i) )3) 3) M M h h nh nh h h nhi nhi u u b b nh nh khu khu y y b b ng ng nhau nhau m m c c n n i iti ti p pM M t t m m h h nh nh kh kh c c l l thi thi t t b b ph ph n n ng ng d d ng ng ng ng c c dng dng han han lu lu. .2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 5 5Thi Thi t t b b ph ph n n ng ng d d ng ng ng ng c c dng dng han han lu lu2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 6 62. S 2. S PHN B PHN B TH TH I GIAN LU C I GIAN LU C A LUA LU CH CH T TRONG BNH T TRONG BNHa)a) h h m m phn phn b b th th i i gian gian lu lu c c a a lu lu ch ch t t trong trong b b nh nhd I 1 d I1 d I1100 2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 7 7H H m m phn phn b b th th i i gian gian lu lu c c a a lu lu ch ch t t trong trongb b nh nh2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 8 8b)b) h h m m phn phn b b th th i i gian gian lu lu trong trong b b nh nh c c a a lu lu ch ch t t trong trongdng dng ra rad E 1 d E1 d E2200 2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 9 93. C 3. C C PHNG PH C PHNG PH P TH P TH C NGHI C NGHI M MC C c c d d ng ng t t n n hi hi u u k k ch ch th th ch ch (stimulation)(stimulation) p p ng ng (response)(response) th th ng ng s s d d ng ng2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 10 104. S 4. S D D NG C NG C C THNG TIN PHN PH C THNG TIN PHN PH I TH I TH II GIAN LU GIAN LU1) 1) H H phiphi tuy tuy n n khng khng c c m m h h nh nh dng dng ch ch y y2) 2) H H phiphi tuy tuy n n v v i i m m h h nh nh dng dng ch ch y y3) 3) H H tuy tuy n n t t nh nh v v i i m m h h nh nh dng dng ch ch y y:: M M h h nh nh dng dngch ch y y c c s s d d ng ng tin tin an an chuy chuy n n h h a a cho cho h h tuy tuy n n t t nh nh2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 11 11C C hai hai m m h h nh nh cho cho t t n n hi hi u u p p ng ng gi gi ng ng nhau nhau v v t t c c ng ng tng tng t t cho cho ph ph n n ng ng b b c c 1,1, nhng nhng c c t t c c ng ngkh kh c c nhau nhau v v i i ph ph n n ng ng c c t t c c khng khng tuy tuy n n t t nh nhtheo theo n n ng ng 2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 12 12Th Th d d 6.1: 6.1:B B ng ng 6.1:6.1: K K t t qu qu p p ng ng c c a a t t n n hi hi u u xung xung0 0 35 351 1 30 302 2 25 254 4 20 205 5 15 155 5 10 103 3 5 50 0 0 0N N ng ng ch ch t t ch ch th th ,, g/l g/l t t Th Th i i gian gian t, ph t, ph2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 13 13 ng ng bi bi u u di di n n EE theo theo t ( t (tr tr i i)) v v theo theo ( (ph ph i i) )2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 14 14Th Th d d 6.2 6.22/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 15 15M M i i quan quan h h gi gi a a D/ D/uL uL v v ng ng cong Ccong C v v th th nguyn nguyn cho cho m m c c phn phn t t n n nh nh ,, ph.tr ph.tr (6.15) (6.15)2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 16 16 ng ng cong Ccong C trong trong b b nh nh k k n n cho cho c c c c m m c c khu khu y y tr tr n n kh kh c cnhau nhau theo theo m m h h nh nh phn phn t t n n2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 17 17K K t t qu qu th th c c nghi nghi m m s s phn phn t t n n c c a a lu lu ch ch t t c c a adng dng ch ch y y quaqua th th p p chm chm v v i i v v n n t t c c trung trung b b nh nhtheo theo phng phng tr tr c c u u2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 18 18 nh nh gi gi h h at at ng ng c c a a b b nh nh ph ph n n ng ng ng ng b b ng ng ng ng cong C:cong C: M M t t s s d d ng ng sai sai s s c c a a b b nh nh ng ng:: (a)(a) khng khng sai sai l l ch ch; (b); (b) c c dng dng ch ch y yt t t t v v v v ng ng t t ; (c); (c) c c hi hi n n t t ng ng tu tu n n han han; (d); (d) lu lu l l ng ng o o sai sai,, th th t t ch ch ch ch t t l l ng ng,, t t n n hi hi u u khng khng th th c c s s l l tr tr; (e); (e) hai hai dng dng ch ch y y songsong song song trong trong thi thi t t b b 2/6/2012 2/6/2012 ChngChng 66 - - Dng ch Dng ch y th y th c c 19 19 nh nh gi gi h h at at ng ng c c a a b b nh nh ph ph n n ng ng khu khu y y tr tr n nb b ng ng ng ng cong C:cong C: M M t t s s d d ng ng sai sai s s c c a a b b nh nh khu khu y yCN BNG HO HCCHNG 8CN BCN BNG HONG HOHHCCCHNG CHNG 88Copyright 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, HarcourtBrace & Company, 6277 Sea Harbor Drive, Orlando, Florida Phn ng thun nghch (phn ng khng hon ton): cng k, p xy ra ng thi theo hai chiu ngc nhauV d - H2(k)+ I2(k) 2HI (k)Phn ng mt chiu (phn ng hon ton): = hayV d - KClO3= KCl (r) + 3/2O2(k)Phn ng ng th - p trong th tch 1 phaHCl(dd)+NaOH(dd) = NaCl (dd)+ H2O(l)Phn ng d th - p din ra trn b mt phn chia phaZn (r) + 2HCl (dd)=ZnCl2(dd)+H2(k) Phn ng n gin - p din ra qua 1 giai on (1 tc dng c bn)V d: H2(k) + I2(k) = 2HI (k)Phn ng phc tp p din ra qua nhiu giai on( nhiu tc dng c bn)Cc giai on : ni tip , song song, thun nghchPhn t s - l s tiu phn ca cht phn ngtng tc gy nn bin i ho hc trong 1 tcdng c bn.(nguyn dng, s 3)nh lut tc dng khi lng (M.Guldberg v P. Waage ) nhit khng i, p ng th, n gin:aA + bB = cC + dDTc phn ng :v= k.CaA.CbBi vi p n gin PTS=1 p n phn t I2(k) =2I(k)PTS=2 p lng phn t H2(k) + I2(k) = 2HI (k)PTS=3 p tam phn t 2NO (k) + O2(k) =2NO2(k) Cn bng ha hcbBaA t tC C k v =dDcC n nC C k v =vt= vnt0vvtvnPhn ng ca h kh l tng (p n gin ):aA (k)+ bB(k) cC(k)+dD(k)t = 0 C0AC0B0 0(mol/l ) t | CA + Cb + Cc | CD |vt= vn(CA)cb=const (CB)cb=const(Cc)cb=const(CD)cb=const AG=0 (PA)cb=const (PB)cb=const(PC)cb=const (PD)cb=constcbNhn xt v trng thi cn bng ho hc Trng thi cbhh l trng thi cn bng ng. Trng thi cn bng ng vi AGp= 0 .(A=0) Du hiu ca trng thi cn bng ho hc: Tnh bt bin theo thi gian Tnh linh ng Tnh hai chiu.Examples of Examples of Chemical Chemical EquilibriaEquilibriaS S t t o o th th nh nh th th ch ch nh nh CaCO CaCO3 3(r) + H (r) + H2 2O(l) + CO O(l) + CO2 2(k) (k)Ca Ca2+ 2+(dd) + 2 HCO (dd) + 2 HCO3 3- -(dd) (dd)Hng s cn bng cho phn ng ng thaA(k) + bB(k) cC(k) + dD(k) (p n gin )Khi trng thi t cn bng: vt= vn K hng s nhit xc nh: hng s cn bng.( ) ( ) ( ) ( )cbdD cbcC n cbbB cbaA tC . C . k C . C . k =( )cb bBaAdDcCntCC CC CkkK = =( ) ( ( ) ( )( ) ( )) ( ) ( )( ) b a d ccbbBaAdDcCcb bBaAdDcCcbbBaAdDcCpRTC CC CRT C RT CRT C RT Cp pp pK += = =( )nC pRT K KA=Hng s cn bng cho phn ng ng thH kh l tngXc nh KX X c c nh nh K K2 2 NOCl(KNOCl(K) ) 2 2 NO(kNO(k) + Cl) + Cl2 2(k)(k)[[NOClNOCl]][NO][Cl[NO][Cl2 2]]BanBan u u 2.00 2.00 0 0 0 0Ph Ph n n ng ng +0.33 +0.33Cn Cn b b ng ngK[NO]2[Cl2][NOCl]2 K [NO]2[Cl2][NOCl]2=(0.66)2(0.33)(1.34)2=0.080- - 0.66 +0.66 0.66 +0.661.34 1.34 0.66 0.66 0.33 0.33Hng s cn bng cho phn ng ng th(Dung dch lng , long)aA(dd) + bB(dd) cC(dd) + dD(dd)( )cbbBaAdDcCCC CC CK =Phn ng d phaCaCO3(r)CaO(r)+ CO2(k)RT K Kc P =( )cbCaCOCO CaOp32pp pK='( )cb COCaOCaCOp p23pppK K ='=( )cbCO c2C K =Trong biu thc ca hng s cn bng K khng xut hin cc thnh phn sau: cht rn nguyn cht, cht lng nguyn cht, dung mi.Mg(OH)2(r) Mg2+(dd)+2OH-(dd)K = [Mg2+]cb.[OH-]2cb= T Mg(OH)2 - Tch s tanCH3COOH(dd)+H2O CH3COO-(dd) + H3O+| || || | COOH CHCOO CH O HK33 3a += Hng s in ly ca axitNH4OH (dd) = NH4+(dd)+OH-(dd)| || || | OH NHOH NHK44b +=Hng s in ly ca bazeCH3COO-(dd) + 2H2O CH3COOH (dd) + OH-(dd)| || || |=COO CHOH COOH CHK33t Hng s thu phnCH3COONa (dd) + 2H2O CH3COOH (dd) + NaOH(dd)NHN XT v Kpv KcL hng s nhit nht nh, ch ph thuc vobn cht p v nhit , ch khng ph thuc vonng hoc p sut ring phn ca cht pPh thuc vo cch thit lp cc h s trong ptp.Hng s cn bng Kp,Kc khng c th nguyn.Hng s cn bng khng ph thuc vo cht xc tcHng s cn bng c gi tr cng ln th hiu sut p cng cao.Vit biu thc hng s cn bngViVittbibiuuththcchhngngsscncnbbngngS(r S(r)+ O )+ O2 2(k) (k) SO SO2 2(k) (k)K [SO2][O2]NH NH3 3(dd) +H (dd) +H2 2O(l) NH O(l) NH4 4+ +(dd) +OH (dd) +OH- -( (dd dd) )K[NH4+][OH-][NH3]S(r S(r)+ O )+ O2 2(k) SO (k) SO2 2(k) (k) K K1 1= [SO = [SO2 2] / [O ] / [O2 2] ]SO SO2 2(k) +1/2 O (k) +1/2 O2 2(k)(k)SO SO3 3(k) (k)K K2 2= [SO = [SO3 3] / [SO ] / [SO2 2][O ][O2 2] ]1/2 1/2S(r S(r)+ 3/2 O )+ 3/2 O2 2(k)(k)SO SO3 3(k)(k) K =???? K =????Knet [SO3][O2]3/2=K1K2 Thay i h s t lngThayThayiihhssttllngngS(r S(r)+ 3/2 O )+ 3/2 O2 2(k) (k) SO SO3 3(k) (k)22 S(r S(r)+3 O )+3 O2 2(k)(k)2 SO 2 SO3 3(k) (k)Knew [SO3]2[O2]3

K [SO3][O2]3/2

Knew [SO3]2[O2]3=(Kold)2 i chiu phn ngiichichiuuphphnnngngS(r S(r)+ O )+ O2 2(k) (k) SO SO2 2(k) (k)SO SO2 2(k)(k)S(r S(r)+O )+O2 2(k) (k)K [SO2][O2]

Knew [O2][SO2]

Kthun = 1/KnghchQuan h gia hng s cn bng v GPHN NG NG THKh l tngaA + bB cC + dDccPPP0TbBaAdDcC 0T TKQln RTKQln RT Q ln RT Gp pp pln RT G G = = + A =||.|

\|+ A = AtKhi phn ng t trng thi cn bng: AGT= 0pcbbBaAdDcCTK RTp pp pRT G ln ln0 =||.|

\| = ADungdch lng,longccc0TbBaAdDcC0T TKQln RT Q ln RT GC CC Cln RT G G = + A =||.|

\|+ A = AtKhi phn ng t trng thi cn bng: AGT= 0CcbbBaAdDcCTK RTC CC CRT G ln ln0 =||.|

\| = A Kp= f(bc p, T) Kp = f(C)Quan h gia hng s cn bng v GPhn ng d pha : aA + bB cC +dDKQln RT Q ln RT G G0T T= + A = A| | |||| | |t((

=b ad cB AD CQ| | |||| | |cbb ad ccbB AD CQ K((

= =Cht kh [] P (atm)Dung dch long [] C (mol/l)Rn nc, lng nc, dung mi (H2O) 1KQln RT GT = A Nu Q < K AG < 0 phn ng xy ra theo chiu thun Nu Q > K AG > 0 phn ng xy ra theo chiu nghch Nu Q = K AG = 0 h t trng thi cn bngV d : Tnh hng s cn bng ca phn ng:2 NO2(k) N2O4(k) 298K khi bit Gii: K J v kJ Hpu/ 6 , 176 S 040 , 580298pu0298 = A = A-5412.3J 176,6 298 58040 - = + = A A = A029802980298S T H G185 , 2298 314 , 83 , 5412ln0==A =RTGKp9 , 8224 2= =NOO NpppKQuan h ca Kp vi nhit v nhit phn ngo o oS T H G A A = ApoK RT G ln = A||.|

\|A=2 10121 1lnT T RHKKA+A =0202lnRSRTHKA+A =0101lnRSRTHKV dNO(k)+ O2(k)NO2(k)Tnh Kp 3250C? Bit: AH0= -56,484kJ v Kp= 1,3.106 250C02 . 14325 =K64 . 2 ln325 =K437 , 1159812981314 , 85648410 . 3 , 1ln6598 =|.|

\| =K1 1ln598 2980298598||.|

\|A=T T RHKKNguyn l chuyn dch cn bng Le ChatelierPht biu: Mt h ang trng thi cn bng m tathay i mt trong cc thngs trng thi ca h (nng, nhit , p sut) th cnbng s dch chuyn theochiu c tc dng chng lis thay i .Henr i LeChtelier (1850-1936)Henr i LeChtelier (1850-1936) An =0 p sut chung khng nh hng n trng thi cn bng.N2(k) + 3H2(k) 2NH3(k) ; AH0 >0Lm lnhun nngma : lng khng kh kh s dng trong 1 gi (kg khng kh kh/gi)mp: khi lng cht kh nguyn liu c sy/giwa2: m tuyt i khng kh vo (kg H2O/kg khng kh kh)wa1: m tuyt i khng kh ra(kg H2O/kg khng kh kh)wp1: m nguyn liu vo theo% khi lng kh kg (H2O/kg cht kh)wp2: m nguyn liu ra theo% khi lng kh (kg H2O/kg cht kh)Cn bng vt cht i vi H2O h thng ng bin gii hn mpwp1+ mawa2= mpwp2+ mawa1ma(wa2 wa1) = mp(wp2-wp1)Cn bng vt cht qu trnh sySymaTa1wa1mpTp1wp1Vt liu Vt liu Khng khmaTa2wa2mpTp2wp2Khng khTh dMt thit b sy lin tc c s dng sy thc phm c m ban u l 80% (theo % khi lng t). Tc nhn sy l khng kh nng c mtuyti l 0,0094kgH2O/kgkhngkh kh.Saukhisy, sn phm ra khi my sy c m l 20%(theo % khi lng t), khngkh rakhimysyc mtuytil 0,0186kgH2O/kg khng kh kh Tnh khi lng sn phm c sy (tnh theokgcht kh/gi) v tnh lng nc thotra sovi1kg cht kh(kgH2O/kg cht kh.gi). Cho bit lng khng kh s dng l 1000 kg khng khkh/gi. S qa trnh sy c th trnh by theo hnh v nh sau : SymaTa1wa1mpTp1wp1Vt liu Vt liu Khng khmaTa2wa2mpTp2wp2Khng khPhng trnh cn bng vt cht i vi nc trong qu trnh symawa1+ mpwp1= mawa2+ mpwp2ma(wa2-wa1) = mp(wp1- wp2)mp= [ma(wa2-wa1)]/ (wp1- wp2)ma=1000kgkhngkh kh/gi llngkhngkh khs dng trong 1 gimp= khi lng cht kh/gi nguyn liu c sywa1= 0,0094kgH2O/kgkhng kh kh l mtuyti khng kh vowa2= 0,0186 kg H2O/kg khng kh kh l m tuyt i khng kh rawp1=0,8/(1-0,8)=4kgH2O/kgchtkhl mnguynliu vo (theo% khi lng kh)wp2= 0,2/(1-0,2) = 0,25 kg H2O/kg cht kh l m nguyn liu ra (theo% khi lng kh)Khi lng cht kh/gi nguyn liu c symp= 1000*(0.0186-0.0092)/(4-0,25) = 2,4533 kg cht kh/giKhi lng nc tch ra so vi 1kg cht kh4 - 0,25 = 3,75 kg H2O/kg cht kh.giCn bng nng lngSymaTa1wa1mpTp1wp1Vt liu Vt liu Khng khmaTa2wa2mpTp2wp2Khng khPhng trnh cn bng nng lng : mpHp1+ maHa2= mpHp2+ maHa1+ qHp1Hp2nhit do vt liu sy mang vo v mang ra h thng (kJ/kg cht kh)Ha1Ha2nhit do khng kh mang vo v mang ra h thng (kJ/kg kkh kh)q nhit tn tht trong h thngHa= Cs(Ta T0) +waHLCsnhit dung ring ca khng kh m (kJ/kgkkh kh0K) = 1,005 + 1,88waTanhit khng kh (0C)T0nhit qui c (00C)wa m tuyt i khng kh (kgH2O/kgkkh kh) HLn nhit bc hi ca nc (kJ/kgH2O)Hp= Cpp(Tp T0) + wpCpw(Tp T0)Cppnhit dung ring ca vt liu (kJ/kg0C)TPnhit ca vt liu (00C)T0nhit qui c (00C)Cpwnhit dung ring ca H2O (kJ/kg0C)Th d :Mt loi thc phm c m l 75% (theo % khi lng t), nhit l 240C c cho vo my sy sy cho n khi m l 16% (theo % khi lngt). Khng kh nngs dng syc nhit l 1200C c mtuytil 0,009kgH2O/kgkhngkh kh.Saukhisy khng kh sy c nhit l 480C trong khi nhit ca sn phm sau khisyl 390C.Tnhlngkhngkh cnthit symtlngvt liu l 40 kg cht kh/gi. Cho bit n nhit bc hi ca nc 480C l2387,56 kJ/kg H2O v 1200C l 2202,59 kJ/ kg H2O, nhit dung ring ca vt liu sy l 2 KJ/Kg0C v nhit dung ring ca nc l 4,180 KJ/ Kg0C. Mt mt nhit trong qu trnh xem nh khng ng k (q=0). S qa trnh sy c th trnh by theo hnh v nh sauSymaTa1wa1mpTp1wp1Vt liu Vt liu Khng khmaTa2wa2mpTp2wp2Khng khPhng trnh cn bng vt cht i vi nc : mawa1+mpwp1=mawa2+mpwp2 (1) 0,009ma+ 3*40 = mawa2+ 0.19*40120-7,6 = ma*(wa2 0,009) ma*wa2 0,009 ma= 112,4 (1)Phng trnh cn bng nng lng: maHa1+ mpHp1= maHa2+ mpHp2(2)ma l khi lng khng kh kh cn thit mp = 40kg cht kh/gi l khi lng vt liu c sy wa1= 0,009 kgH2O/kg khng kh kh l m tuyt i ca khng kh vowa2l m tuyt i ca khng kh rawp1=0,75/(1-0,75)=3kgH2O/kg cht kh l m tuyt i ca vt liu vowp2=0,16/(1-0,16)=0,19kgH2O/kg cht kh : m tuyt i ca vt liu ra Ha1l nhit hm ca khng kh vo Ha1=(1,005+1,88wa1)(Ta1-0) + wa1HL1Ha1 = (1,005+1,88*0.009)(120) + 0.009*2202,59=142,45 kj/kg kkh khHa2 l nhit hm ca khng kh ra Ha2=(1,005+1,88wa2)(Ta2-0) + wa2*HL2Ha2= (1,005+1,88wa2)(48) + wa22387,56 = 2477.8 wa2+ 48,24Hp1l nhit hm ca sn phm vo Hp1= Cps(Tp1-0) + wp1Cpw( Tp1-0)Hp1= 2*24 + 3*4.18*24 = 348.96 kj/kg cht khHp2l nhit hm ca sn phm ra Hp2= Cps(Tp2-0) + Wp2*Cpw( Tp2-0)Hp2= 2*39 + 0,19*4,18*39 = 108,9738 kj/kg cht kh Phng trnh cn bng nng lng :ma142,45 + 40*348.96 = ma(2477,8 wa2+ 48,24)+ 40*108,9738142,45ma+ 13958,4 = ma(2477,8wa2+ 48,24) +4358,95213958,4 - 4358,952 = 2477,8 mawa2+ 48,24 ma- 142,45ma2477,8 mawa2- 94,21 ma= 9599.448(2)Gii h thng phng trnh (1) v (2)mawa2 0,009 ma= 112.4 (1)2477,8 mawa2- 94,21 ma= 9599.448 (2) Khilngkhngkh khcnthitl ma=3739,48kgkhngkhkh/gi