tieu luan hoa li nang cao-truong thi phuong
TRANSCRIPT
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
PHN I: M U
Ha l v cu to cht ng vai tr ht sc quan trng trong vic nghin cu c
s l lun cc qu trnh ha hc. S hiu bit v cu trc, nng lng v c ch phn ng
l gii cc qui lut din bin ca mt qu trnh ha hc l nhim v hng u ca mn
hc Ha l. Ni cch khc, nm chc cc kin thc Ha l s gip cc nh khoa hc hiu
su sc hn bn cht ca qu trnh ha hc.
Ha l hin i l s kt hp phc tp ca cc chuyn ngnh hp nh: ng ha
hc, in ha hc, nhit ng ha hc, l thuyt v dung dch, l thuyt v xc tc, ha
hc bc x, ha keo, .
Nhit ng lc hc l mt ngnh ca vt l hc, nghin cu mi quan h nhit,
cng v cc dng nng lng. C s ca nhit ng lc hc l cc nguyn l I, II, III vnguyn l 0. p dng cc nguyn l c bn ca nhit ng lc hc vo qu trnh ha hc
hnh thnh nn nhit ng ha hc. Nhit ng ha hc ch yu nghin cu v:
- Nhit ha hc: Nghin cu hiu ng nhit ca cc phn ng ha hc v cc qu
trnh ha l nh ha tan, hp ph, chuyn pha.
- Cn bng ha hc: Nghin cu kh nng, chiu hng v gii hn ca phn ng,
ngha l trong nhng iu kin no mt phn ng ha hc c th xy ra, t xy ra theo
chiu no, khi phn ng t cn bng, hiu xut phn ng l bao nhiu, nh hng ca
cc yu t bn ngoi n s chuyn dch cn bng phn ng nh th no...
- Cn bng pha: Nghin cu cc iu kin cn bng pha trong cc h d th,
nhng quy lut ca cc qu trnh chuyn pha.
- Ngoi ra nhit ng ha hc cn nghin cu cc tnh cht v qui lut c bn ca
cc loi dung dch v ng dng thc t ca chng.
Nghin cu nhit ng hc trong cn bng pha l mt nhim v quan trng ca
Ha l v trong thc t chng ta hay gp cc h c nhiu pha, c bit l h hai pha.
Nghin cu gin pha ca h bc hai gip ta hiu r c tnh, tnh cht ca cc qutrnh bay hi, kt tinh, nng chy, ca cc dung dch, cc h kim loi, m ta thng
gp trong cng nghip ho hc, luyn kim v mt s ngnh khoa hc khc.
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
PHN II: C S L THUYT1. Mt s khi nim:1.1 Pha
L tp hp nhng thnh phn ng th c cng tnh cht l ha.Cc pha c phn cch nhau bng b mt phn chia
R L H1.2 Cu t
S ha cht c kh nng to nn h.Cu t k hiu: nCh xt nhng cu t c kh nng tn ti c lp.
VD: H2O - NaClH2 + I2 2HI c th xem h 2 cu t2HI H2 + I2 H 1 cu t
S cu t = S ha cht s phng trnh lin h1.3 Bc t do f
Tc l s thng s trng thi ( T, P, C) c th thay i mt cch ty nhng khng lm thay i s pha trong h.2. Qui tc pha (Qui tc Gibb)
Xt h n cu t u c mt trong k pha nm cn bng vi nhau. Khi h nm trng thi cn bng n tha mn cc iu kin cn bng nhit ng hc.
+ Cn bng nhit T1 = T2 = ... =Tk
+ Cn bng c hc P1 = P2 = ... = Pk
+ Cn bng ha hc k12
1
1
1 ... ===k
2
2
2
1
2 ... ===.....................
k
nnn === ...21
S phng trnh lin h (k-1).nS thng s trng thi (n-1).k + 2S bc t do f = (n-1)k + 2 (k-1)n
f = n k + 2Qui tc pha S bc t do bng s cu t trong h tr i s pha v cng 2+Nu T = const f = n k + 1+Nu T, P = const f = n k
+Nu thm 1 thng s cng f = n k + 3Da vo qui tc pha ca Gibb c cch phn loi h
a)Phn loi theo cu tn = 1 h 1 cu tn = 2 h 2 cu tn = 3 h 3 cu t
b)Phn loi theo s pha
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k = 1 h 1 phak = 2 h 2 phak = 3 h 3 pha
c)Phn loi theo bc t dof = 1 h 1 bin
f = 2 h 2 binf = 3 h bt bin
3. H hai cu t3.1- Nhn xt chung
Khc vi h mt cu t, i vi h hai cu t c hon ton xc nh bi3 thng s: p sut, nhit v nng mt cu t (v bit nng ca 1 cu tth s suy ra nng ca cu t kia)
Nh vy c 3 iu kin thay i l T, P, nng (x). Ta nhn thy biudin thnh phn ca h th nn chn cch biu din % khi lng hoc nng
phn s mol l tin li nht, v nh vy khi bit nng ca cu t ny ta c th
suy ra nng ca cu t kia ngay.Mt khc ta nhn thy rng biu din gin pha ca h ta cn 3 trc: T,P, nng (x) th ta s c gin pha khng gian. l iu gy bt tin, n gin v ph hp vi iu kin thc t ngi ta thng nghin cu h hai cut vi 2 iu kin thay i c th l: P x; T x; nh i vi h gm cc cht rnhoc lng, cn pha kh, th thng ch nghin cu s thay i pha theo T x(ngha l P = const).
VD: Gin h hai cu t khi P = constP = const
T
Hi
Dth
DthDth
LngoSAT
A B
nc
AT0
x3.2- H hai cu t tn ti th lng3.2.1 Dung dch lng l tng: cha 2 cu t
- C bn cht ging nhau hoc gn ging nhau.- C lc tng tc fA-A = fB-B = fA-B (f lc tng tc).- Qu trnh trn ln V = 0, H = 0.- Tun theo nh lut Raoult+ Xt hn hp A, B.
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-Ta thy hai cu t A, B ca dung dch lng l tng tun theo nh lutRaoult iii xpp
0= .
Pi: p sut hi bo ha trn b mt cht lng i.0
iP : p sut hi bo ha ca i nguyn cht.
Ta c: AAAxpp 0
=BBB xpp
0=
Pchung = PA + PB = BABA xPpp )(000 + (1) , vi xA = 1 - xB, xA, xB
ln lt l nng phn mol ca A, B th lng.Biu thc (1) biu din s ph thuc ca p sut chung vo thnh phn
th lng.Gi s 0Ap 1; (3) >>1Ta c Pchung = PA + PB = BABA xPpp )(
000 + = f(x)
T (2) x =y
y
).1(
P = ).(000
ABA PPP + yy
).1( = f(y)Gin :
P T = const
L
H
P = f(x)
P = f(y)
-ng lng
-ng hi
A B
HL
0
AP
3.2.2 Dung dch thc- Sai lch dng (+)+ Tng tc gia cc cu tfA-B P l tng+ V > 0, H > 0
- Sai lch m (-)+ Tng tc gia cc cu tfA-B > fA-A, fB-B.+ P < P l tng+ V < 0, H < 0
0
BP
0
AP
P
A BSai lch (+)
T = const
0
AP
A B
Sai lch (-)
P T = const
3.3 Quy tc n by:
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Gi s un nng 1 mol dung dch c nng x0 a ln vng d th imtng trng l k, lc ny c m mol dung dch chuyn thnh hi, (1-m) mol nm lng.
k
y
x
ab
xB
PP0B
P0A
AB
T = const P = f(x)
P = f(y)
S mol ban u ca B l: 1.x0S mol ca B L l: (1-m)xS mol ca B H l: myTheo nh lut tc dng khi lng: x0 = (1- m)x + my
xy
xxm
= 0 v 1 m =
xy
xy
0 0
0
1 xy
xx
m
m
=
=kb
ka
lng hi/ lng lng =kb
ka
3.4 Cc nh lut Konovalov
)1(. yx
xy
dx
dP
dx
dP B
=
Ta c du cadx
dPph thuc vo y x
3.4.1 nh lut Konovalov 1:
y > x dx
dP> 0 dP, dx ng bin.
y < x dx
dP< 0 dP, dx nghch bin.
Cu t no c thnh phn th hi ln hn th lng th khi thm vo trongdung dch th p sut chung ca h s tng ln nhit khng i hoc nhit
si ca h s gim xung p sut khng i.3.4.2 nh lut Konovalov 2:
y = x dx
dP= 0 hoc
dx
dT= 0
Ti im cc tr ca p sut hi bo ha ( hay nhit si) phn lng vphn hi nh nhau. Ta c th m t bng gin sau:
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T = constP
A B
0
AP
0
BP
n P = constT
A B
0
AT
3.5. Phng php chng ct3.5.1- Hn hp khng to ng ph: Hnh 1
Nu un nng dung dch c thnh phn N1 th dung dch s si khi t nhit t1, hi cn bng vi pha lng c thnh phn N3 v giu cu t B hn, v vy saukhi bay hi mt phn th phn dung dch cn li s c thnh phn N2 giu cut A hn so vi dung dch ban u. Dung dch s si nhit t 2 cao hn t1.Hi cn bng vi pha lng lc ny c thnh phn N4, n cng giu cu t B hn sovi pha lng. V vy, phn dung dch s giu cu t A hn v si nhit caohn. Nu tip tc chng ct th phn cn cui cng ch gm cu t A tinh khit vnhit si t n tA.
tA
BA N2N1 N N4 N3 N5
DK
C
tB
Hi
lng
Thnh phnA BN1 N2 CN4 N3
Hi
lng
Thnh phn
tB
tA
Hnh 1: Biu nhit si- thnh phn cahhai cu t
Hnh 2: Biu nhit si - thnh phnca hhai cu t c im ng ph
Nu xt n phn hi, gi s phn hi c thnh phn N3 c ngng t liv em chng ct th n s si nhit t3 v hi thu c c thnh phn N5 s
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cng giu cu t B. Nu tip tc qu trnh chng ct v ngng t nh vy th cuicng trong bnh ngng ta s thu c cu t B tinh khit.3.5.2- Hn hp to ng ph: Hnh 2
i vi hn hp to ng ph th ta khng th tch ring bng phng phptrn c. Gi s ta c dung dch c thnh phn N1 nm gia A v C th dung dch
s si nhit t1 v hi nhn c c thnh phn N2 giu cu t B hn so vipha lng. phn dung dch cn li giu cu t A hn s si nhit cao hn. Nutip tc chng ct phn cn cui cng s l cu t A tinh khit. Nhng t dungdch ban u c thnh phn N1 ta khng th tch c cu t B tinh khit m tatch c dung dch ng ph C.
VD: H nc ru etylic. H ny c nhit si cc tiu 78,130C ngvi thnh phn 95,57% theo khi lng ru ( khong 960) .- nng cao hm lng cht cn tch t hn hp ng ph ngi ta c th
ph ng ph bng mt s phng php sau:+ Thay i hnh dng gin :
VD: Xt vi h etanol nc ta h p sut ngoi t 760mmHg xung100mmHg th t l ru trong dung dch ng ph s tng ln n 99,6% mol(99,60) v nhit si ng ph h n 34,20C.
34,20C
960 99,60H2O C2H5OH
1000C
P = 1atm
T0C
T0C
+ Thm cu t th ba vo: Cu t ny c tc dng ngn cn s hnh thnhdung dch ng ph hoc n to ng ph vi mt trong hai cu t v b tch ratrong qu trnh chng ct.
VD: Xt vi h etanol nc cho thm CaCl2 vo lc ny nc s solvatha ion Ca2+ mnh hn cn C2H5OH d dng bay ra v ta thu c C2H5OH trn960.4. Phng php phn tch nhit
Mt phng php thng dng nghin cu s thay i pha ca h 2 cut l phng php phn tch nhit.
Ni dung ca phng php ny theo di s thay i nhit chuyn pha(s xut hin pha mi) ca tng h 1 cu t vi nhng thnh phn khc nhau.
Mt phng php thng dng l o nhit lin tc theo thi gian, ricn c vo s thay i nhit xt ti s xut hin pha mi. Khi c pha mixut hin s km theo s thay i nhit khng u hoc t ngt. y l
phng php phn tch nhit hay dng trong phng th nghim. Cch lm nh sau:Chng hn xt h gm 2 cu t A v B vi thnh phn:
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Cu t A 0% 10% . 100%Cu t B 100% 90% . 0%
em un nng cho ho lng hon ton ri lm lnh t t theo di s
thay i nhit bng nhit k. C qua thi gian nht nh (khong 30 giy) thghi nhit , sau v ln th 2 trc (trc tung ghi T, trc honh ghi thi giant). Lc ta s c ng cong nhit thi gian khc nhau gi l ngngui ca cu t nguyn cht hay hn hp. Trn ng cong s c s thay ikhc thng tng ng vi s xut hin pha mi.
Ta s xt c th cc dng c bn ca ng ngui .4.1- ng ngui ca cu t nguyn cht
ng ngui l th biu din mi quan h gia nhit theo thi giantrong qu trnh lm ngui mt cht hay hn hp.
ng ngui ca cu t A v B c dng nh (hnh 4.5)
Hnh 4.5- ng ngui ca cu t nguyn cht
Lc u ta ch c 1 pha lng, nhit gim dn, ng vi on ab, lc :f = 1 1 + 1 = 1
n b nhit dng li, cu t bt u kt tinh nn chnh l nhit kt tinh ca cu t.
on bc, nhit khng thay i theo thi gian v trong qua trnh lmlnh c to ra lng nhit, lng nhit ny n b p li lng nhit ca h mt i trong qua trnh kt tinh, nn nhit khng thay i theo thi gian. p
dng quy tc pha cho on bc ta c:f = 1 2 + 1 = 0.Ngha l khi kt tinh nhit khng th thay i c.n c, kt tinh ht, lc ny h l 1 pha rn.Do , f = 1 1 + 1 = 1.
Ngha l nhit li tip tc gim, ng vi on cd v h kt tinh honton.
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To
K
a
TA
TB
b c
d
a/
b/
c/
d/
Nguyn cht A
Nguyn cht B
t (thi gian)
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Ch : Cc cht khc nhau u c ng ngui tng t nhau nhng khcnhau nhit kt tinh.4.2- ng ngui ca hn hp
ng ngui ca hn hp c dng nh (hnh 4.6)
Hnh 4.6- ng ngui ca hn hpTrn th v 2 ng ngui ca 2 hn hp c thnh phn khc nhau.
Sut qu trnh ng vi on ab (hay a/b/), h ch c 1 pha lng v nhit gimdn, tip theo ta thy 2 ln thay i tc gim ca nhit .
T T1T/1 l nhit kt tinh ca 1 trong 2 cu t. Cu t no kt tinhtrc l tu thuc vo thnh phn ca n so vi thnh phn c hn hp c bit(gi l hn hp tecti)
n b (hay b/
) h bt u c 2 pha: hn hp lng v cu t kt tinh. V cs kt tinh 1 cu t km theo ta nhit nn nhit gim chm i, theo quy tcpha: C = 2 2 + 1 = 1. Ngha l vn c th thay i 1 thng s (nhit hocnng ). Sut qu trnh kt tinh bc (hay b /c/) h c 2 pha, v mt cu t kt tinhnn thnh phn pha lng thay i lin tc.
T2 : gi l nhit kt tinh ca hn hp tecti.ng ngui ca hn hp tecti
tecti l hn hp ng nht gm cc tinh th ca 2 cu t, khi kt tinh chai cu t kt tinh ng thi theo mt t l nht nh.
Nh vy hn hp tecti kt tinh vi thnh phn pha lng v pha rn nh
nhau, trn th ng vi on cd (hay c
/
d
/
). Sut qu trnh hn hp tecti kt tinhht c 3 pha: 1 pha lng v 2 pha rn (tinh th ca 2 cu t) nn :f = 2 3 + 1 = 0, nn T2 = Te = const v c gi l nhit kt tinh ca hn hptecti.
Ta nhn thy rng, nhit kt tinh ca hn hp tecti l thp nht vchung cho mi trng hp.
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To
K
T1
a/
b/
c/ d/
e/
t (thi gian)
T/1
T2
a
b
cd
e(1) (2) (3)
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iu ny c ngha l mt h hai cu t ch c 1 hn hp tecti ng vithnh phn c nh, nn d thnh phn ban u khc nhau nhng cc hn hp us t n thnh phn ca hn hp tecti.
n nhit T1 cu t no c thnh phn nhiu hn so vi thnh phn can trong hn hp tecti th cu t kt tinh trc.
Th d: hn hp tecti c thnh phn 70%A + 30%B. V nu hn hp lngc 60% A v 40%B th B s kt tinh trc.
Khi pha lng tecti kt tinh (n d) h ch cn 2 pha rn v s bc t doC = 2 2 + 1 = 1, do nhit li tip tc gim.
Nh vy nu hn hp ban u c thnh phn hn hp tecti th ngngui cng ch c 1 im dng ng vi nhit Te (ng (3) trn hnh v), cngha l ng ngui tecti cng ging nh ng ngui nguyn cht, ch khc lnhit thp nht.4.3 Gin pha ca h hai cu t
Sau khi c c cc ng ngui cn c vo nhit kt tinh, thnh
phn ca hn hp ta s c gin pha ca h.C th thy cch lm qua (hnh 4.7)a cc ng ngui ca h ln trc to :Vi trc honh l trc thnh phn,Trc tung l nhit ,Ta s c gin pha ca h 2 cu t.
Hnh 4.7- Gin pha ca h 2 cu t
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(1)
(2)
(3)
(4)(5)
ToK T0K T0K(1
)(2) (3) (4) (5)
Lng (A+B)
Rn A vLng (A+B)
Rn B vLng (A+B)
Rn A + Rn B
TA
A(%) %Bt
TB
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PHN III: BI TP VN DNGBi 1: Gin bc hai ng p etanol benzen th hin mt cc tiu ng ph ng p di p sut 1 bar.
65
70
75
80
67,5
0 0,2 0,4 0,6 0,8 1
t(0C)
AZ
xB
a. Xc nh ta ca AZ di p sut p = 1 bar.b. Xc nh nhit bt u ha lng v kt thc khi hn hp hi c thnh phn0,2; 0,9.
Bi gii:
a. T gin ta thy xAz = 0,54 v tAz = 67,50
Cb.Vi xB = 0,2, s ha lng bt u nhit khong 740C v kt thc
nhit khong 720CVi xB = 0,9, s ha lng bt u nhit khong 760C v kt thc
nhit khong 730C
Bi 2: Gin bc hai ng p ca h liti clorua natri clorua:a. Xc inh bn cht cc pha c mt trong mi min A, B, C v D.
b. Xc nh cc ta ca cc tiu M.
Bi gii:a. H c hp thnh: A: t mt hn hp ng th NaCl v LiCl nng chy. B v C: t mt dung dch rn nm cn bng vi mt dung dch lng v c
hai u to thnh bi KCl v NaCl. D: t mt dung dch rn NaCl v KCl.
b. Da vo hnh v ta c c xM = 0,39 v tM = 5200C.
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A
B
C
D
toC
LiCl NaCl
P = 10bar
0
500
600
700
800
0,51
Gin bc hai liti clorua - natri clorua
M
Bi 3: Cho gin trng thi ca h KCl LiCl nh sau:
361
400
450
500
600
650
700
800
600
b
p M N
E
0 20 40 60 80 100
%LiCl
HLiCl - NaCl
B
c0
0a
o
oe
od
to
c
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a. S pha, thnh phn pha v nhit ti cc im a, b, c, d, e v E.b. Thnh phn ca h c biu din bng im M trn gin bng qui tc nby.
Bi gii:a.
im a: Mt pha lng cha 40% LiCl, 7000C.im b: Hai pha cha 80% LiCl 5100C nm cn bng vi tinh th LiCl.im c: Mt pha nng chy ca LiCl 6500C.im d: Mt pha tinh th KCl 6250C.im e: Hai pha rn gm tinh th LiCl 80% v KCl 20%.im E: tecti, h 3 pha ( 2 pha tinh th + 1 pha lng) pha lng cha
58,5% LiCl. Nhit tecti 3610C.b.
im M ng vi h 2 pha 4500C, pha lng cha 20% LiCl v 80% KCl.
Lng pha rn KCl/ Lng pha lng =20
30=
MP
MN
Bi 4: Khi lng ring ca phnol rn l 1,072.103 kg.m-3, ca phnol lng l
1,056.103 kg.m-3. Nhit nng chy ca n l 1,044.105 kgJ . Nhit kt tinh l
314,20K. TnhdT
dP v nhit nng chy ca phnol P = 5,065.107 N.m-2.
Bi gii:
p dng :( )
( )28
3
65
10525410056107212314
1005610721100441 =
=
= m.N.,
.,,.,
.,.,.,
V.T
H
dT
dP nc .-
1
tnh T0nc p sut ngoi cho nh trn, ta phi coi gi tr
m.N.,dP
dT 18105254 = .. trong khong p sut 1,044.105 5,065.107 (N.m-2) l
khng i v bng m.N., 18105254 ., nn:
=2
1
2
1
8105254P
P
T
T
dP..,dT
K,.,..,,T078
2 43161006551052542314 =+=
Bi 5: V gin pha ca h Sb Pb da vo cc d kin cho bng sau:
Thnh phn hn hp lng,% khi lng
Nhit bt uthot tinh th, oC.
Sb Pb
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
100 0 632
80 20 580
60 40 520
40 60 43320 80 300
10 90 273
0 100 326
a. Xc nh thnh phn tecti.
b. C bao nhiu Sb tch ra nu 10kg hn hp lng cha 40% Pb, c lm nguiti 4330C?
Bi gii:
Gin :
toc
A
K
N
E
B
DC
F
M400
600
200
520
433
326
10080 876040200Sb
%Pb
Gin pha ca h Sb Pba. Hai ng lng AE v BE ct nhau ti E ng vi thnh phn owtecti 87%Pb v13%Sb.
b. Nu hn hp lng cha 40% Pb c lm ngui ti 4330C th ng vi im Mtrn gin . Ti thnh phn 40% Pb, tinh th u tin tch khi hn hp lng lSb ti 5200C (im K trn gin ). Dn theo s tch Sb, hn hp lng cng giuPb hn do im K chuyn dch ti N chng hn (60%Pb). Gi L l lng pha
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
lng ti im K v R l lng Sb tinh th tch khi hn hp lng ti K. Theo quytc n by FMR. = MNL. hay
R =FM
MNL = 10.
40
20= 5 kg
Bi 6:
Cho biu hai thnh phn ng p ca h gecmani silic theo phn molca silic xSi = x2.
1000
1100
1200
1300
1400
1500
1600
1000
1100
1200
1300
1400
1500
1600
0,32 0,5 0,6 0,67 1 x2 (Si)
toC
Lng
dung dch rn
1414
0
Ta xt h thu c khi trn ln 8,43g silic vi 14,52g gecmani.
a. Tnh thnh phn chung ca h.b. Hn hp ny c a ti 12000C, xc nh thnh phn ca hai pha cn bng.c. Xc nh lng v khi lng ca tng pha cn bng.
Bi gii:Theo nh ngha phn mol
x2 = xSi =GeSi
Si
nn
n
+ =60,0
6,72
52,14
1,28
43,81,28
43,8
=
+
xSi = 0,60 v xGe = 0,40.
Theo gin t = 12000
C ct ng lng x2(l) = 0,32 v ng rn x2(r) = 0,67.
n = n(r) + n(l) = 50,06,72
52,14
1,28
43,8=+
Ta li c:
1,032,067,0
60,067,0.5,0
)()((
))((.)(
22
22 =
=
=lxrx
xrxnln
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
n(r) = n n(l) = 0,40Thnh phn mol ca hai pha:Trong pha lng:
nSi(l) = xSi(l).n(l) = 0,032 molnGe(l) = n(l) - nSi(l) = 0,068 mol
Khi lng ca pha lng khi l:m(l) = nSi(l).MSi + nGe(l).MGe = 5,84g.
Trong pha rn:nSi(r) = xSi(r).n(r) = 0,268 molnGe(r) = n(r) - nSi(r) = 0,132 mol
Khi lng ca pha rn khi l:m(r) = nSi(r).MSi + nGe(r).MGe = 17,11g.
Bi 7:
Hn hp bc hai toluen benzen c th c coi nh l l tng. Gin ng p ca n di p = 1,0bar cho hnh sau:
90
100
110
120
70
80
0 0,2 0,4 0,6 0,8 xB
NM
Q
P = 1,0bar
t0c
Ta xt hn hp thu c khi trn 4,0mol benzen vi 6,0 mol toluen.a. Xc nh nhit bt u si ca hn hp ny.
b. Tnh cc lng cht lng v hi cn bng 1000C.c. Tnh lng toluen lng c mt trong h 1000C.
Bi gii:a. H c phn mol benzen xB = 0,4, trn gin nhit bt u si c c
trn ng cong si t = 940
C.b. Ta c
n(h) = nNQ
MN. v n(l) = n
NQ
MQ.
Theo gin NM = 0,2, MQ = 0,087 v NQ = 0,287
n(h) = =10.287,0
2,07,0 mol
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
V n(l) = 10.287,0
087,0= 3,0 mol
c. ng ng nhit 1000C ct ng cong si xB(l) = 0,22 nn xT(l) = 0,78, nhvy lng toluen trong pha long bng nT(l) = xT(l). n(l) = 0,78.3,0 = 2,3 mol.
Bi 8: Xc nh nhit ho hi ca thu ngn, nu nh p sut hi bo ha ca n 3300C l 0,65.105N.m-2. Cho bit nhit si ca thu ngn p sut P = 1,013.105
N.m-2 l 3570C.
Bi gii:
Theo phng trnh Clausius Clapeyron cho qu trnh bay hi:
V.T
H
dT
dP cp
= hay :
h
h
V.T
H
dT
dP =
PT.R.T
H
dT
dP h=
hay : 2T.RdT
HP
dPh
=
Trong khong nhit they i T1T2 v coi Hh = const trong khongnhit trn, nn ta ly tch phn tng ng : T1T2 v P1P2
=
=
211
2
2
112
1
2
1
TTR
H
P
Pln
T
dT.
R
H
P
dP hT
T
hP
P
Suy ra:
( ) ( )
60363010650
100131273357273330103148
5
53
12
1
221
++
=
= .,
.,ln...,
TTP
Pln.T.T.R
Hh
Bi 9: 3700K noc s si di p sut l bao nhiu? Bit nhit ho hi ring
ca nc l 2254,75 gJ
Bi gii:
p dng :
=
=
211
2
2
112
1
2
1
TTR
H
P
Pln
T
dT.
R
H
P
dP hT
T
hP
P
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
y:
( )
( )
=
=
=
==
K.m o lJ.,R
KT
KT
m oJ.,..,Hh
3
02
0
1
33
1 03 1 48
3 7 3
3 7 0
1 05 8 54 01 01 87 52 2 5 4
Suy ra: ( )25121
1221 10910
=
= m.N.,P
T.T
TT
R
HPlnPln h
Bi 10: Tnh bin thin p sut cn c thay i im ng bng ca nc
khong 10C. Cho bit nhit nng chy 00C l 334 gJ . Khi lng ring ca
nc lng v nc ln lt l : 0,9998ml
g ; 0,9158ml
g .
Bi gii:
Ta c : kgcm,
gcm,
,VL
332100000021
99980
1===
kgcm,
gcm,
,VR
338109009081
91680
1===
kgm.,,,VVV RL3
5100692100081090
===
Theo phng trnh Clausius Clapeyron cho qu trnh nng chy:
( )atmm.N.,
kgm.,.
kgJ.
VV.T
H
dT
dP
V.T
H
dT
dP
RL
ncnc13410341
10069373
1033427
35
3
==
=
=
=
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TIU LUN HA L NNG CAO GVHD: PGS.TS TRN THI HA
TI LIU THAM KHO1. Trn Hip Hi, Nguyn Vn Du; Bi tp ha l, Nh xut bn Gio dc, 1985.2. Trn Vn Nhn v cng s; Ha l tp 1, tp 2, tp 3; Nh xut bn Gio dc H Ni,1999.3. o Qu Chiu; Ha hc nm th nht, Ha hc nm th hai; Nh xut bn Gio dc,2006.4. Lm Ngc Thim, Trn Hip Hi, Nguyn Th Thu; Bi tp ha l c s, Nh xut bnkhoa hc v k thut H Ni, 2003.
5. Nguyn nh Hu; ng ha hc v xc tc; Nh xut bn Gio dc, 1990.6. Ren DIDIER; Ngi dch: Nguyn nh Bng V ng ; Ha i cng tp 1,tp 2, tp 3; Nh xut bn Gio dc, 1997.7. Hong Nhm; Ha hc v c tp 1; Nh xut bn Gio dc, 2003.8. TS. Trn Thi Ho; Bi ging Ho L nng cao; HKH Hu - 2010.9. V ng ; C s l thuyt cc qu trnh ho hc; NXBGD 2006.10. V ng ( Ch bin ) - Trnh Ngc Chu - Nguyn Vn Hi; Bi tp C sl thuyt cc qu trnh Ha hc; NXBGD 2006.11. F. Cotton G. Wilkinson, Ngi dch: L Mu Quyn L Ch Kin; C sHo hc V C - Phn I; NXB i hc v Trung hc chuyn nghip H Ni 1984.
HVTH T Th Ph H K18 20