BI 1- PHN 2. IU CH DUNG DCH KEOV XC NH NGNG KEO T.I. MC CH :iu ch dung dch keo Fe(OH)3. Xc nh ngng keo t ca dung dch ny vi cht in li Na2SO4.II. KIN THC L THUYT :-Ht keo l ht c kch thc nm gia ht v m v ht vi m.V vy, ta c th iu ch dung dch keo bng cch ngng t t dung dch thc hoc phn tn t h phn tn th.-Phng php ngng t: tp hp cc ht c kch thc nh thnh ht c kch thc ln nh ht keo.C th lm bng phn ng ha hc hoc thay th dung mi-Phng php phn tn: dng nng lng c, in, cht in li phn tn cc ht ln v kch thc ht keo.V d:iu ch keo AgI bng phng php ngng t:+Keo m: dng pipet cho vo bnh nn 20ml dung dch KI 0,05N. T buret cho tng git AgNO3 0,05N va thm va lc cho n ht khong 18- 20ml AgNO3 0,05N, ta c ht keo m c cu to:[(AgI)m.nI (n-x)K+]x- .xK++Keo dng: cho vo bnh nn 20ml dung dch AgNO3 0,05N. T buret cho tng git KI 0,05N va thm va lc cho n ht khong 15- 18ml KI, ta c ht keo dng c cu to:[(AgI)m.nAg (n-x)NO3-].xNO3-Ch: thdu ca keo bngcch ly1mnh giy lc, mt u nhngvaof dung dch keo. V giy lc lm t xenlulo khi thm nc s tch in m. Nu ht keo tch in m th n s theo mao qun ca giy lc m lan i xa. Nu ht keo tch in dng th n b gi li u ch nhng ca giy lc.1TR LI CU HI:1)Dung dch keo l g? C my cch iu ch dung dch keo?-Dung dch keo l h phn tn trong cc ht ca cht phn tn c ng knh c 10-7 n 10-9.-C 2 phng php iu ch dung dch keo:+Phn tn t h phn tn th.+Ngng t t dung dch thc.2)Vit cu to ht keo dng AgI v ht keo m H2SiO3?-Keo dng AgI: [(AgI)m.nAg (n-x)NO3-].xNO3-.-Keo m H2SiO3: [(H2SiO3)m.nSiO32-.(2n-x)K+].xK+.3)Ngng keo t l g? N ph thuc vo nhng yu t no?-Ngng keo t: l nng ti thiu ca dung dch keo c th xy ra qu trnh keo t.-Cc yu t nh hng n qu trnh keo t:+Nng .+Nhit .+Khuy trn.+Thi gian.+Cht in li quan trng v c ngha nht.+Yu t bn ngoi.III. PHN THC NGHIM V KT QU:1.Dng c v ha cht:-n cn :1 ci-Cc chu nhit 250ml :1 ci-Pipet1, 2, 10ml :2 ci mi loi-ng nghim :12 ci-King st + li Amiang :1 b-Dung dch FeCl3 10%-Dung dch Na2SO4 0,002M2-Nc ct2.Cch tin hnh: iu ch dung dch keo Fe(OH)3:-Cho 95ml nc ct vo cc chu nhit 250ml un si trn n cn. -Dng pipet ht chnh xc 5ml dung dch FeCl310% ri nh tng git vo cc nc ang si cho n ht.-un si tip 5 pht ri ly ra ngui bng nhit phng ta c dung dch keo Fe(OH)3.Hnh nh minh haXc nh ngng keo t:ng 1 2 3 4 5 6 7 8 9 10H2O 9.0 8.9 8.8 8.7 8.6 8.5 8.4 8.3 8.2 8.1Na2SO40.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Fe(OH)31 1 1 1 1 1 1 1 1 1Hin tng - - - - + + + + + +Tng cng mi ng l 10mlQuan st ng no c(keo t) nh du (+), ng no trong nh du (-).Ly ng c u tin tnh kt qu theo cng thc[ ]1000 ..24 2 1VSO Na V Trong : V1 l th tch ca dung dch Na2SO4 V2 l th tch dung dch keo l ngng keo t (mmol/l)IV.X L S LIU V TNH TON KT QU THC NGHIM:[ ]1000 ..24 2 1VSO Na V =8 , 0 1000 .1002 , 0 . 4 , 0(mmol/l)Vy ngng keo t l: 0,8(mmol/l)3Nhn xt: Nng cht in li ng vai tr quan trng nht n qu trnh keo t.S khc bit gia dung dch thc v dung dch keo l:-Ht keo c kch thc ln hn phn t nn tc khuch tn chm hn phn t.-Ccht keockchthcnhnnthngkhngbnvmt nhit ng c xu hng lin kt li vi nhau thnh cc ht c kch thc ln hn dintchbmt gimnnglngbmt gim(cli vmt nng lng)V. NHN XT CA GIO VIN HNG DN:BI 2: XC NH HNG S CN BNG CA PHN NGKI +

I2 = KI3 TRONG DUNG DCH NCI.MC CH:4pdngnhlut phnbxcnhhngscnbngtrong nc ca phn ng: KI2 + I2 = KI3 II. KIN THC L THUYT:Ta bit rng, khi hai cht lng khng tan vo nhau th hn hp s c tch lm hai (lp nng hn di v nh hn trn, chnh xc hn l: cht c t trng ln hn nm di). Nu ta thm 1 cht th 3 no c kh nng ho tan trong hai cht kia th cht s phn b gia hai cht lng to thnh hai dung dch cn bng c nng khc nhau.y chng ta gi thit rng gia cht tan v mi trng khng c phn ng ho hc xy ra.Theo nhit ng hc, iu kin cn bng, th ha hc ca cht tan trong hai pha phi khc nhau.1 = 01 + RTlna1 2 = 02 + RTlna2 01 v 02 l th ho tiu chun ca cht tan trong dung mi tng ng: a1, a2 l hot ca cht tan trong hai dung mi. Theo iu kin cn bng ta vit: 01 + RTlna1 = 02 + RTlna2 02-01 = RT ln 21aa nhit khng i ta c th vit:ln 21aa = RT2 1 = Const hay : 21aa = exp1]1

RT2 1 = KK c gi l h s phn b, K ch ph thuc vo bn cht cc cht v nhit .i vi dung dch long c th thay hot ca cht tan bng nng Cch tnh trn ch ng khi cc cht khng lm thay i s tab ln ca hai cht lng v cc cht tan khng c s phn ly hoc kt hpNu c s phn ly ca cht tan trong 1 pha:gi s cht tan KA phn b gia dung mi 1 v 2, trong dung mi 1 cht K khng phn ly (Hnh 3)5 KAK+ +A-KANu gi nng ca KA trong dung mi 1 l C1 v phn ly ca ca KA l th nng ca tng ion trong dung mi l C1. Nng KA khng phn ly l (1 ) C1 .Khi hng s phn ly ca KA l:KPL=- ) 1 (1212CC

) 1 (12C (1-)= PLKC12 KPL l hng s phn lyMt khcphntKAkhngphnlyli nmcnbngvi KA trong dung mi 2 h s phn b:KPB =21) 1 (CC = MK CC2212 hay KPB KPL = 2212CC = KKPB l hng s phn bNu c s kt hp ca cht tan trong l pha th K=PLPBKKTR LI CU HI:Cu 1: Pht biu v vit biu thc ca nh lut phn b.i nh lu t phn b: nu ta thm m t ch t thba nao ocokhanng hoa tan trong 2 ch t kia thich t osephn bgi a hai ch t long tao tha nh 2 dung dich cn bng con ng kha c nhauNu C1 vaC2 lanng cu a cu tch t tan trong hai dung mi 1 va2, khi cn bng nhi t khng i, thitysC1/C2 lam t hng svac go i lahsphn bKK=21CC6Cu2:Chngminhrngvicngmtlngdungminutin hnh chit nhiu ln th c li hn chit mt ln.Cng mt lng dung mi nu tin hnh chit nhiu ln th c li hn chit mt ln v: Chit mt ln lng chit c l: m=a.(1-cKV VV+00)- Ln chit th nht:VcNng cht tan trong dung mi chit:Vcx a1V0,aNng cht tan trong dung dch:0VxK=011VxVcx a K.cVx aVx101x1=a.cV K VV.00-Ln chit th 2:Nng cht tan trong dung mi chit:cVx x2 1 Nng cht tan trong dd: 02VxK=022 1VxVx xcx2=x1.cV K VV.00+Thay x1 tnh c trn vo cng thc tnh x2:x2=a.(200).cV K VV+7Sau ln chit th n:xn=a.(ncV K VV).00+Lng cht chit c sau n ln chit l: m=a.11]1

,_

+ncKV VV001 Chit n ln c li hn chit 1 ln.Cu 3: Vit biu thc tnh K phn b v K cn bng, cho bit chng ph thuc vo nhng yu t no.Cng thc tnh hng s phn b:K=exp1]1

RT01 02 =21aaHng s phn b ch ph thuc vo bn cht cc cht v nhit .Cng thc tnh hng s cn bng i vi p trong bi ny:KCB= [ ][ ][ ]23I KIKI Hng s cn bng ph thuc nng , nhit , p sut v bn cht ca cc cht.Cu 4: Ti sao khi chun b I2 trong dung dch H20 th khng nn cho h tinh bt vo ngay t u v khi chun I2trong dung dch CCl4 phi thm dung dch KI vo.-Khi chun I2 trong dd H2O th khng nn cho HTBvo ngay t u v: HTB c cu to dng xon hp th I2 gy sai s chun .-Phi thm dd KI vo ha tan I2 v:KI + I2 KI3 Khi thm KI vo [ ] KItng cn bng dch chuyn theo chiu thun l chiu to KI3.III. PHN THC NGHIM:p dng nh lut phn b xc nh hng s cn bng ca phn ng8KI + I2 KI3 trong ncTrong mi trng nc c hng s cn bng KCB= [ ][ ][ ]23I KIKI (1)Cn bng d th ca I2 gia hai lp tun theo nh lut phn b:K=[ ][ ]4 222CCLO HIINh vy nu ta pha 1 trong hai hn hp I2 trong nc v CCL4 ri chun lng l2 trong 2 lp ta setinh c KPB.Ta sexa c inh c l ng I2 trong l p n c.Chu n I2 trong l p n c ta sethu c lng (I2) t ng c ng bao g m I2 tdo vaI2 a trong KI3 (t c laKI3).KI + I2 KI3Phn ng KI + I2 KI3 trong ncDo ota tinh c:(KI3)=(I2)TC-(I2)H2O I2Bi t n ng [KI] ban u la[KI0] ta se tinh c nng [KI] khi cn b ng : [KI]cb =[KI0]-[KI3]Bi t (I2)H2O , [KI3]cb , [KI]cb ta setinh K theo (1)1.Dung cuva hoa cht:Ph u chi t 250ml : 2 ciBi nh no n 500ml (nut ma i) : 2 ciBi nh no n 100ml (nut ma i) : 2 ciC c 100ml : 2 ciBi nh no n 250ml : 2 ciPipet 1,2,5,10,25 ml : 2 ciBuret 25 ml hoc 50 ml : 6 ciKI 0,1N va1N :1 ci9I2 bao hoa trong CCl4:2 ciNa2S203 0,01 va0,001NHtinh bt2.Ca ch tin hanh:Hnh nh minh haCho va o 2 bi nh no n nut ma i m i binh 10ml I2 trong CCl4 .Bi nh thnh t thm va o 10ml n c c t ,bi nh thhai thm va o 50mlKI 0,1N. y n p ,lc kytrong vo ng 30 phu t ,thi nh thoa ng mnu t cho hi CCl4bay ra. Sau khi l c xong cho va o hai ph u chi t yn trn giacho hta ch l p.Chi t ly l p CCl4(ph n di) va o hai bi nh no n nut ma i 100ml. Ph n con lai v n nguyn trong ph u chi tTi n ha nh chu n :-Chu nI2trongCCl4bng Na2S2030,01N: M i l n lychu n 1ml + 10ml KI 0,1N chu n cho ma u dung dich chuy n tnu sang va ng rm m i cho htinh bt va o.Chu n n khi dung dich m t ma u xanh thid ng la i (chuy : khi chu n c n lc ma nh I2 trong CCl4 chuy n ln h t l p KI). Chu n 3 l n l y k t quatrung bi nh- Chu n I2 trong lp nc bng Na2S203 0,001N :M i l n chu n ly 10ml10III/X L S LIU V TNH TON KT QUPhng trnh phn ng:Na2S3O2 + I2=Na2S4O6 + 2NaI Theo nh lut ng lng:Ta c: CN(Na2S2O3). V(Na2S2O3) = CN(I2). V((I2)CN(I2) =( )( )( )23 2 23 2 2IO S NaO S Na NVV CVy theo s liu ta c:Phu 1:[ ]2IH2O =2 . 1025 . 001 , 0 =1,25.10-3 M[ ]2ICCl4 =2 . 16 , 20 . 01 , 0 = 0,103 MPhu 2: [ ]2ITC =2 . 25 . 7 . 01 , 0 =0,01875 M[ ]2ICCl4 =2 . 183 . 5 . 01 , 0 =0,02915 MTheo nh lut phn b:KPB =[ ][ ]4222CClO HII =103 , 010 . 25 , 13 =0,01214Dung dch Na2SO4 Dung dch I2 (mu chun)Nng (N)V(ml) V(ml) Nng (N)Phu 1Lp H2O0,001N 25 10 1,25.10-3Lp CCl40,01N 20,6 1 0,103Phu 2Lp KI0,01N 7,5 2 0,01875Lp CCl40,01N 5,83 1 0,0291511Ta c:[ ]2IH2O = KPB . [ ]2ICCl4

(Phu 2) = 0,01214.0,02915=3,53881.10-4 M[ ]3KI=[ ]2ITC - [ ]2IH2O=0,01875 - 3,53881.10-4 =0,0184 M[ ]0KI

= 06 , 01 , 0 . 05 , 0=121M[ ] KICB = [ ]0KI - [ ]3KI =121- 0,0184 = 0,06493 MVy hng s cn bng ca phn ng: KI + I2 KI3KCB =[ ][ ] [ ]23I KIKICB= 410 . 53881 , 3 . 06493 , 00184 , 0=800,74Vy: KCB = 800,74IV.NHN XT CA GIO VIN HNG DN:...........................................................................................................................................................................................................................................................................................................................................................................BI 3:XC NH NNG LNG HOT HACA PHN NG HA HC.I. Mc ch:Xc nh nng lng hot ha E ca phn ng thy phn etyl axetat:CH3COOC2H5 +H2O +HCH3COOH + C2H5OH12II. Kin thc l thuyt: Xt phn ng: AB A +BPhn ng i qua trungthi trung gian bng cch to ra phc cht hot ng ABX* A +B hnh thnh phc trung gian phi cung cp cho h mt nng lng hot ha E1. Qu trnh ngc li phi cung cp cho h mt nng lng hot ha E2.Ta thy mc nng lng ca AB ln hn mc nng lng A+B l H = E2+E1. Nh vy c ngha l phn ng theo chiu thun l phn ng thu nhit. Vytrngthi X*ltrngthi chungchophnngthunvphnng nghch. Mc nng lng X gi l thm nng lng m mt phn ng mun tin hnh c phi vt qua mc nng lng . Nh vy nng lng hot ha l nng lng cn thit b sung vo h cho h c nng lng bng hoc ln hn mc X*.Bi ny xc nh nng lng hot ha ca phn ng:CH3COOC2H5+H2OCH3COOH+ C2H5OHTheo phng trnh Arrhenius: lg12TTkk=575 , 4E .1]1

1 21 1T TNh vy nu xc nh c hng s tc k nhit T1 v T2 ta s xc nh c E.Nu khi thc hin phn ng (1) ly lng nc l rt ln hn nhiu ln so vi etylaxetat th ta c th xem nng nc l khng thay i v phn ng tin hnh theo c ch bc 1.Phng trnh ng hc c dng : lgtkx CC303 . 2100Trong :C0l nng etylaxetat ban u. x l lng etylaxetat phn ng thi gian t13Vy mun xc nh k1 th ta phi xc nh nng cc thi im khc nhau ri t tnh ra.TR LI CU HI:Cu 1: Nng lng hot ha l g?- Nng lng hat ha lnng lng d ti thiu m h phn ng cn phi c so vi nng lng trung bnh ca h ban u phn ng c th xy ra. Cu 2: Nhit c nh hng nh th no n tc phn ng?- S nh hng ca nhit ti vn tc ca phn ng l: nu nhit phn ng cng cao th phn ng xy ra cng nhanh.Cu 3: Nng NaOH c nh hng g n chun ?- S nh hng ca nng NaOH ti chun l : khng nn dng nng qu long v kh pht hin im tng ng, khng nn dng nng cao d dn dn sai s ln. Nng NaOH hay dng l 0,1N.I II. Phn thc nghim: 1. Dng c ha cht: Bnh tam gic nt mi 250ml : 2 ciBnh tam gic nt thng 250ml : 6 ciBnh cu y trn 250 ml : 1 ciSinh hn thng: 1 cing ht 10ml : 2 ciBuret 25ml : 1 ciBp cch thy: 2 ciNhit k 1000C : 1 ciCH3 COOC2H5tinh khit NaOH 0,1NPhenolftalein2. Cch tin hnh:Tin hnh 2 th nghim song song ging nhau, ch khc nhau v nhit tin hnh. 141 th nghim tin hnh nhit phng, 1 th nghim tin hnh nhit phng + 100CCch lm: Cho vo hai bnh tam gic 250ml nt mi, mi bnh 100ml HCl 0.1N, mt bnh nhit phng, bnh cn li cho vo bp cch thy unnhit phng+100C. Sauchovomi bnh5ml etylaxetat nguyn cht. Lc u v tnh thi gian t .Xc nh V0 : Ly 10ml dung dch mt trong hai bnh tam gic chun bng NaOH 0,1N vi ch th phenolftalein(dung dch chuyn sang mu hng th dng ).Xc nh V: Ly 10ml dung dch bnh tam gic cn li cho vo bnh cu c sinh hn, t vo bnh cch thy un nhit 80-900C trong vng 120 pht (thnh thong lc v cho t nc ct qua sinh hn). Ly ra ngui bng nhit phng, ri em nh phn ton b bng NaOH 0,1N.Xcnh Vt: Trong hai bnh trn, c 20 pht ly ra mi bnh 10ml chun (nh cho thm 10ml nc ct v 3 git phenolftalein) chun bng NaOH 0,1N cho n khi mu hng xut hin. Lm 5-6 ln.3. Kt qu th nghim:V0 = 9,6ml, V= 36mlT01T(pht)Vt20 40 60 80 100 12010,3 11 11,6 12,6 13,3 13,7T02(T1+10)T(pht)Vt20 40 60 80 100 12011 12,2 13,6 15 16 17,24. X l kt qu:Gi V0 l th tch NaOH cn chun lng HCl ban u c trong 10ml ly mi ln chun. Nh vy gi tr (Vt - V0 ) l th tch cn thit chun lngCH3COOH sinh ra do phn ng thy phn sau thi gian t.T ta suy ra lng etylaxetat phn ng x sau thi gian t t l vi (Vt -V0 ), do vy ta vit x=m((Vt - V0 ), trong m l h s t l.15Mt cch tng t nh vy,lng CH3COOC2H5 phn ng thi imv cng s t l vi (V- V 0 ). Lng ny chnh bng lng CH3COOC2H5ban u C05 2 3 H COOC CH= m (V- V0 ).Do vy ta c lg (V- Vt)=lg(V- V 0) - 303 , 21t k

Tnh E theo cng thc:lg12TTkk=575 , 4E .1]1

1 21 1T TSuy ra:E=.. . 575 , 41 22 1T TT T lg12TTkkt(pht) 20 40 60 80 100 120lg(V-Vt)t =250 C1.410 1.398 1.387 1.369 1.356 1.348lg(V-Vt)t =350 C1.398 1.377 1.350 1.322 1.301 1.274V thi xc nh k1 T v k2 Tk1 T=-2,303tg,k2 T=-2,303tgTnh E theo cng thc:lg12TTkk=575 , 4E .1]1

1 21 1T T E=.. . 575 , 41 22 1T TT T lg12TTkk(cal/mol)16Biu xc nh K1 v K2y = -0.0013x + 1.4246y = -0.0006x + 1.42341.261.281.31.321.341.361.381.41.420 20 40 60 80 100 120 140Thi gian(t)Lg(Vc-Vt) Nhit phngNhit phng+10Linear (Nhit phng+10)Linear (Nhit phng)- Da vo th ta c: tg= -0,0006 1Tk= 1,3818.10-3 tg= -0.0013 2Tk=2,9939.10-3-Thay vo cng thc:E=.. . 575 , 41 22 1T TT T lg12TTkk suy ra: E=14100(cal/mol)Vy: Nng lng hot ha ca p ny trong mi trng axit l:E=14100(cal/mol) IV. NHN XT CA GIO VIN HNG DN: ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................BI 4: HP PH NG NHIT V PHNG TRNH FRENDLICH17I/MC CH:Xcnhshp phaxit axetic bi than hottnh trong dung dch ncII/ KIN THC L THUYT:S hp ph cht khng in ly hoc cht in ly yu trn ranh gii phn chiapharnlngthucvoloi ctnhhpphphntvtuntheo phng trnh Gibbs. Nu cht tan b hp ph mt lng ln hn dung mi th s hp ph gi l hp ph dng, cn ngc li l hp ph m. S hp ph ca cc cht hu c cng mt dy ng ng tun theo quy tc Traube: S hp ph gim vi s tng chiu di mch bi v vi s tng kch thc phn t tit din s dng ca cht hp ph hp ph b gim, cc phan t ln khng b ri vo l hp ca cht hp ph.S hp ph trn ranh gii rn lng tun theo quy tc thng bng phn cc ca Rebindis: Cc cht b hp ph trn ranh gii phn chia pha ch xy ra trong trn hp nu do s c mt ca n lp b mt lm cho hiu s phn cc ca pha b gim . o s hp ph, ngi ta cho mt lng cn chnh xc cht hp ph vo cc th tch nh nhau ca dung dch ch b hp ph. Lng cht b hp ph c tnh nh l hiu s nng u v nng cn bng trong cc th tch nh nhau ca dung dch. Trong cc dung dch long, thng thng ngi ta khng tnh s hp ph ca dung mi. Theo cc d kin thc nghim xy dng th s ph thuc khi lng cht b hp ph vo nng cn bng ca n trong dung dch.S hp ph trn vt xp t dung dch long c m t bng phng trnh thc nghim Frenlich:X/m = k CnTrong : X l s mol cht b hp ph (mol/l)m l khi lng cht hp ph (g)C l nng cn bng (mol/l hay mmol/l)18 k l hng s tng ng cho khi lng ca cht b hp ph nng cn bng.n l hng s (n = 0,1 0,5) tm hng s k v n th chuyn phng trnh trn thnh dng khc:lgmx = lgk + nlgC v tg = n = CmxlglgKhi th ph thuc lg(x/m) vo lgC l ng thng nh hnh 6 TR LI CU HI: Cu 1: Hp ph l g? C my loi hp ph ?-Hp ph l qu trnh cht cha cc phn t cht kh,lng hay cht ha tan ln b mt phn cch pha. B mt phn cch pha c th l kh- lng, kh- rn hay lng-lng, lng- rn*Phn loi hp ph:c 2 loi hp ph : l hp ph vt l v hp ph ha hc.Cu 2: Nhng thuyt no hc v hp ph?*C 3 thuyt hc v s hp ph- Thuyt hp ph ng nhit frendlich- thuyt ca Langmuir- Thuyt ca BetCu 3 : Gii thch cch tnh cc hng s k, n?-Gii thch cch tnh cc hng s n, K+Da vo th ph thuc gia Lg(x/m) v LgC ta bit c tg.Li c : tg= n.-Thay n vo phng trnh lgmxlgK+n.lgC. Ta rt ra c K.Cu 4 : Nhng sai s no c th gp trong qu trnh th nghim? -Nhng sai s no c th gp trong qu trnh th nghim.19+ Sai s ch quan : sai s trong qu trnh cn, chun , ong, tnh ton kt qu.+ Sai s khch quan : do dng c th nghim, ha chtV d : Chun l qu trnh i xc nh im tng ng, t im tng ng ta suy ra nng cht cn xc nh. Th nhng trong thc t ta khng th tm c ch th thay i mu ti im tng ng m ch c ch th thay i mu trc v sau im tng ng thi.Nh vy trong qu trnh chun tt yu phi xy ra sai s ch th.III/ PHN THC NGHIM: 1/Dng cBnh tam gic 250ml: :12 ciPhu lc 100 :6 cing ht 2,5,10,25ml :mi loi 1 ciBuret 25ml :1 ciGiy lc :1 hp2/Ha chtCH3COOH 0,4N :1 ltNaOH 0,1N :1 ltCh th Phenolftalein :1 l3/Cch lmCho vo 6 bnh tam gic theo t l:Bnh 1 2 3 4 5 6CH3COOH 0,4N 50 40 30 20 10 5Nc ct(ml) 0 10 20 30 40 45Than hot tnh (g) 1 1 1 1 1 1Lc mnh khong 30 pht, sau yn 15 pht cho cn bng ri lc dung dch vo 6 bnh khc. Chun mi bnh dung dch sau khi hp ph 2-3 ln ri ly kt qu trung bnh.Ch :Mi ln chun ly 10ml v cho thm 3 git phenolftalein, chun bng dung dch NaOH 0,1N n khi xut hin mu hng20Kt qu th nghim c ghi vo bng sau:Bnh 1 2 3 4 5 6VNaOH 1 35.5 27,7 20 12,9 5,7 2,32 35,6 28 20,1 12,8 5,8 2,33 35,4 27,9 20 12,7 5,7 2,3V NaOH35,5 27,87 20,03 12,8 5,73 2,3IV. X L KT QU TH NGHIMKhi lng axit b hp ph c tnh theo cng thc sau: A=VmC C.10002 1 Trong :C1 ,C2 l nng axt trc v sau khi hp ph.V:th tch dung dch ly ra hp ph(50ml)m:Khi lng vt b hp ph(than 1g)Tnh ton ta c bng s liu sau:C10,4 0,32 0,24 0,16 0,08 0,04C20,3550 0,2787 0,2003 0,128 0,0573 0,023C 0,045 0,0413 0,0397 0,032 0,0227 0,017310 .mx2,25 2,065 1,985 1,6 1,135 0,85Lg(x/m) -2.77 -2.69 -2.83 -2.8 -2.95 -3.07LgC -0.44 -0.56 -0.68 -0.9 -1.24 -1.6421 th ph thuc lg(x/m) vo lgCy = 0.3604x - 2.4858-3.1-3.05-3-2.95-2.9-2.85-2.8-2.75-2.7-2.65-2.6-1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0lgCLg(x/m)(lgC,lgx/m)Linear ((lgC,lgx/m))- Da vo th ta c: n=tg=0,3604 Lgk=-2,4858 k=3,27.10-3V. NHN XT CA GIO VIN HNG DN:............................................................................................................................................................................................................................................Hnh nh minh haBI 5: NHIT HA HI CA CHT LNG.22I. MC CH:-Xc nh nhit ha hi ca nc.II. KIN THC L THUYT:-Nghin cu hin tng chuyn pha ca cht tinh khit khng c phn ng ha hc km theo.-Mt trong nhng c trng quan trng nht ca cht lng l p sut hi bo ha, n c trng cho kh nng bay hi ca mt cht.-Phng trnh Clauperon-Claudiut nu ln mi quan h gia p sut hi bo ha v nhit nh sau: V THV V THdTdP) (1 2Trong :: H nhit ha hiV2: th tch ca pha hiV1: th tch ca pha lng-Nu cho rng nhit si T v p sut P, th tch mol ca cht lng v ca hi cn bng l V1 v V2 tng ng, th cng do h sinh ra thay i th tch s bng: A= P.(V2-V1) = P. Vo hm ca cng thc theo nhit ta c:dTdAdTdPdPVdPdTdAV11 (1)o hm ly theo iu kin V khng i v p sut ch ph thuc vo nhit , khnh ph thuc vo th tch. nhit si chng c cn bng L=H.V vy theo nguyn l th 2 ca nhit ng hc th nhit ha hi bng nhit ca qu trnh thun nghch:dTHdTdA Do biu thc (1) c th vit:) (1 2V V THV THdTdP(2)23 d dng trong vic nghin cu cn bng lng -hi hay tinh th- hi, phng trnh Clauperon-Claudiut c th n gin ha vi cc iu kin sau:Vhi>Vlng (hay V hi>V tinh th) do ta c th b qua v lng hay V tinh th trong (2). iu kin thng iu ny cng c m bo:Sau khi b qua V1 v thay V2=RT/P vo (2) ta c:2 2lnRTHdTP dhayRTHPdTdP Ly tch phn bt nh ca phng trnh ta c:lnP = -BRTP+hay lgP = -BTA+(1)Trong :A=575 , 4 0303 , 2 . 987 , 1H H R=1,987 Cal/mol , A c th nguyn l Phng trnh (1) ng vi s ph thuc tuyn tnh:LgP = f(1/T) (nh hnh 7)Ly tch phn hu hn phng trnh trn ta c :1]1

11 1. 023 , 2lg2 1222121T T RHPPRTHPdTdPTTPP(2) Bit p sut hi bo ho ca cht lng hai nhit si ta c th xc nh c nhit ha hi ca cht lng da theo phng trnh (2)TR LI CU HI: 1)Chuyn pha l g ? C my loi chuyn pha, cho v d ?-Chuyn pha l qu trnh chuyn t pha ny sang pha khc.-C 2 loi chuyn pha :+Loi 1 : l qu trnh chuyn pha km theo s bin thin nhy vt ca o hm bc 1 th nhit ng.+Loi 2 : l qu trnh chuyn pha km theo s bin thin nhy vt ca o hm bc 2 th nhit ng.2)Vi iu kin no thH c tnh theo cng thc :24lgP1/Ttg R A/ lnP = -BRTP+.p dng khi : cn bng lng hi hay tinh th - hi.3)Bnh n p c vai tr g trong phng th nghim ?Bnh n p c vai tr gi cho p sut n nh trong phng th nghim.III. PHN THC NGHIM: 1.Dng c :-Bnh cha cht lng-ng sinh hn-Nhit k-Bnh bo him-n p-Kha-n cn2. Cch lm : thit b p sut hi bo ha :Phn bay hi gm : Bnh 2 c cha cht lng nghin cu, mt c cm nhit k, mt c cm sinh hn hi lu.Phn thay i p sut v o p sut : Gm c my ht chn khng, bnh n p, cc kha c th ni h thng chn khng vi kh quyn hay phn bay hi.Tin hnh : ng kha 8, m kha 7 ri cho my ht chn khng chy p sut trng h gim dn. Trong khi h t c chn khng nht nh th ng kha 7 li, tt my. Khi p sut trong h bng p sut bn ngoi kh quyn th mc thy ngn 2 nhnh qua ng ch u ngang nhau. D chnh lch mc thy ngn trn p k chnh l chnh lch gia p sut kh quyn v p sut ca h.P = (Pk h) mmHg Pk : p sut kh quynGi trPk c trn p k c trong phng th nghim.25 nguyn cht lng trong h, dng n cn un nng cht lng trong bnh cu, theo di nhit si. Khi no thy nhit ngng thay i th ghi ly nhit vo bng ng thi ghi gi tr h trn p k. Sau tip tc un cht lng p sut cao hn bng cch iu chnh kha 7 tht t t cho khng kh vo h p sut ca h tng ln 40mmHg (tc l h gim xung 40mm) ri ng kha li. un cho cht lng si p sut mi. Ghi nhit v chiu cao h tng ng. C nh vy ghi cc nhit khc nhau cc p sut khc nhau cho n khi p sut trong h bng p sut ngoi kh quyn.Ch : Trc khitin hnh th nghim phi th kn ca h thng bng cch ht chn khng ca h thng ri kha cht van 7, van 8. yn trong 5-10 pht. Nu thy chiu cao mc thy ngn khng thay i xem nh h thng kn hon ton. Cho nc qua sinh hn lin tc lm lnh hi.h(mmHg) P = Pk - h lgP t0C T0K 1/T422 338 2.529 79 352 0.00284380 380 2.580 81 354 0.00283338 422 2.625 83.5 356.5 0.00281300 460 2.663 87.5 360.5 0.00277260 500 2.699 88 361 0.00277226 534 2.728 89.5 362.5 0.00276187 573 2.758 91.5 364.5 0.00274146 614 2.788 94 367 0.00273107 653 2.815 95.5 368.5 0.0027167 693 2.841 97.5 370.5 0.002700 760 2.881 100 373 0.00268IV.X L KT QU :Trn c s nhng d kin thu c xy dng th s ph thuc ca lgP vo 1/T26 th ph thuc ca lgP vo 1/Ty = -2100.7x + 8.5132.52.552.62.652.72.752.82.852.92.950.00265 0.0027 0.00275 0.0028 0.002851/TlgP(1/T,lgP)Linear ((1/T,lgP))-Da vo th ta c: A=2100.7 1H4,575.2100,7= 9610,7 (a)-Da vo cng thc ta cng tnh c :Nhit ha hi ca nc: 1 2122 12lg 575 , 4T TPPT TH =10122(cal/mol) (b)T (a) v (b) suy ra :Sai s gia tnh theo th thc nghim v theo cng thc l :tg= -2100.71/TlgP27S= % 05 , 5 100 .1012210122 7 , 9610100 .221 HH HVy: H 10122t511(cal/mol)V. NHN XT CA GIO VIN HNG DN:...........................................................................................................................................................................................................................................................................................................................................................................2829