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Chng 4
CN BNG PHA
4.1. Mt s khi nim c bn
Pha: l tp hp nhng phn ng th ca mt h,
c cng thnh phn ha hc v tnh cht l ha
mi im. S pha k hiu l f
S cu t: l s ti thiu hp phn to ra h.
K hiu l k
t doca mt h l thng s nhit ng c lp
xc nh h cn bng. K hiu l c.
4.2. Qui tcpha Gibbs
Bc t do ca h:
c = k - f + nTrong :
k: s cu t
f: s pha
n: s thng s bn ngoi tc ng ln h
4.3. Gin pha v cc qui tc cn bng pha
4.3.1. Biu din thnh phn ca h 2 cu t
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Thnh phn ca cc cu t trn gin pha
thng dng l phn mol xihay phn trm khi lng yi.
Trong h hai cu t, dng mt on thng c chia
thnh 100% nh sau:
A BxB
0 1
Hnh 4.1. Gin pha h hai cu t
Trn trc to ch cn biu din cho mt cu t v
thnh phn ca cu t cn li c xc nh theo cngthc: xA+ xB= 1 hay y1+ y2= 100%
Khi im biu din ca h cng gn cu t no th
hm lng ca cu t cng ln.
4.3.2. Biu din thnh phn ca h 3 cu t
Thnh phn ca h 3 cu t thng c biudin bng mt tam gic u nh sau:
80
80
80
60
60
60
40
20
20
20 40
40
100
100100
A
BC
P
(%C)
(%B)
(%A)
I
Hnh 4.2. Gin pha h ba cu t
Ba nh ca tam gic l ba im h ca cc cu t
nguyn cht A, B v C.
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Ba cnh ca tam gic biu din ba h hai cu t
tng ng l AB, AC v BC.
Mi imtrong tam gic biu din h 3 cu t.
Cch biu din im P(40%A, 40%B, 20%C) trn
gin tam gic u ABC.
Trn cnh AC, ta v ng thng i qua im 40%
v song song vi cnh BC.
Trn cnh AB, ta v ng thng i qua im 40%
v song song vi cnh AC.
Trn cnh BC, ta v ng thng i qua im 20%v song song vi cnh AB.
Ta thy 3 ng thng trn ct nhau ti P. Vy P l
im biu din ca h c thnh phn (40%A, 40%B,
20%C).
4.4. Cc qui tc ca gin pha
4.4.1. Qui tc lin tc
t (pht)
T0C
Chuyn pha
Hnh 4.3. Gin nhit - thi gian ca cht nguyncht.
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Cc ng hoc cc mt trn gin pha biu
din s phthuc giacc thng s nhit ng ca h
s lin tc nu trong h khng xy ra s bin i cht,
s thay i s pha hoc dng cc pha.
Nh vy ta c th suy ra, nu trong h c s thay i v
pha hay s thay i v dng pha th trn cc ng hay
cc mts xut hin cc im gy, lm cho th
khng cn lin tc.
4.4.2. Qui tc ng thng lin hp
Trong iu kin ng nhit v ng p nu h
phn chia thnh hai h con (hay c sinh ra t hai h
con) th im biu din ca ba h ny phi nm trn
cng mt ng thng, ng thng ny gi l ng
thng lin hp.
A B
HM N
Hnh 4.4. Minh ha quy tc ng thng lin hip
V d: h H = h M + h N. Th im biu din cc
h H, M v N nm thng hng.
4.4.3. Qui tc n by
Nu c ba im h lin hp M, H v N th lng
tng i ca chng c tnh theo qui tc n by
nh sau:
A B
HM N
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Hnh 4.5. Minh ha quy tc n by
p dng quy tc n by, ta c:
HM
HN
g
g
N
M
Trong : gM: Khi lng ca h M
gN: Khi lng ca h N
4.4.4. Qui tc khi tm
Nu mt h gm n h con th im biu din ca
n phi nm khi tm vt l ca a gic c nh l cc
im biu din ca n h con.
V d: H H gm ba h con l H1, H2v H3. vi khi
lng tng ng l:
g = g1 + g2 + g3
A
B C
H1
H2
H3
K
H
Hnh 4.6. Minh ha quy tc khi tmNh vy, H phi nm khi tm vt l ca tam
gic H1H2H3.utin ta xc nh imbiu din ca h
K, tha mn iu kin:
H K = h H1 + h H2
vKHKH
gg
1
2
2
1 .
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Tip theo ta xc nh im H tha mn iu kin
sau:
H H = h K + h H3
vKH
HHg
gggg 3
3
21
3
K
4.5. nh hng ca p sut n nhit chuyn
pha
Phng trnh Clausius Claypeyron I:
VT.dPdT
Trong :
T: nhit chuyn pha (K)
: nhit chuyn pha (cal/mol hoc J/mol)
V = V2V1: bin thin th tch (ml)
Nu V c tnh bng ml, c tnh bng cal
v 1cal = 41,3 ml.atm, nn phng trnh Clausius
Claypeyron tr thnh:
41,3.
VT.
dP
dT
4.6. nh hng ca nhit n p sut hi boha
Phng trnh Clausius Claypeyron II
2RT
dT
dlnP
Ly tch phn 2 v, ta c:
121
2
T
1
T
1
R
P
Pln
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Trong :
T: nhit (K)
P: p sut (atm)
: nhit ha hi (cal/mol hoc J/mol)
R: l hng s kh
4.7. Bi tp mu
V d 1. 00C nhit nng chy ca nc l 1434,6
cal/mol. Th tch ring ca nc v nc lng ln
lt l 1,098 v 1,001 ml/g. Xc nh h s nh hng
ca p sut n nhit nngchy ca nc v tnh
nhit nng chy ca nc 4 atm.
Gii
p dng phng trnh:
VT.
dP
dT
Vi: V = VlngVrn= 1,0011,098 = - 0,097 (ml/g)
Hoc: V = 18.(- 0,097) = -1,746 (ml/mol)
0,008141,31434,6
1,746273
dP
dT
(K/atm)
Nh vy, c tng p sut ln 1 atm th nhit nng chy ca nc gim 0,0081K. Mt cch gnng, 4atm, nhit nng chy ca nc l:
T = 273 + (-0,0081) x (4 - 1)= 272,9757K = - 0,02430C
V d 2. Tnh nhit si ca nc 2 atm, bit nhitha hi ca n l 538,1 cal/g (coi nhit ha hi khngthay i trong khang t 1 atm n 2 atm).
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Gii
Nhit ha hi: = 538,1x18 = 9685,8 (cal/mol)
p dng cng thc:
121
2
T
1
T
1
R
P
Pln
273100
1
T
1
1,987
9685,8
1
2ln
T tnh c: T = 394K = 1210C
V d 3.Tnh nhit nng chy ca 1 mol diphenylaminnu 1kg diphenylamin nng chy lm tng th tch ln
9,58.10-5m3cho bit dT/dP = 2,67.10-7K.m2/N. Nhit nng chy ca diphenylamin l 540C, khi lng molca cht ny l 169.
Gii
p dng cng thc:
VT.
dP
dT
3
7
5
19,83.102,67.10
9,58.101000
16954273
dP
dT
VT.
(J/mol)
V d 4. p sut hi bo ha ca axit xyanhydric HCN
ph thuc vo nhit theo phng trnh:
T
12377,04lgP(mmHg)
Xc nh nhit si v nhit ha hi ca n
iu kin thng.
GiiNhit si ca axit HCN p sut 760 mmHg:
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Ta c:T
12377,04lg(760)
T = 297,4K
Vy nhit si ca axit HCN l 24,40C.
Ly o hm hai v phng trnhtheo T, ta c:
2T
1237
dT
dlgP
M:24,575.T
dT
dlgP
Suy ra: 22 T1237
4,575.T
= 5659 (cal/mol)
V d 5. Trn 200g hn hp gm 3 cht A, B, C cha
20% A, khi cn bng hn hp chia lm hai lp.
Lp th nht c khi lng 60g v bao gm 50%A
v 20% B. Lp th hai cha 80%B.
Hy xc nh im biu din ca ba cu t A, B, C
trn gin tam gic u trong hai lp trn.
Gii
c gi t v hnhKhi lng ca cht A trong hn hpban u:
mAo= 20%200 = 40 (g)
Phn trm ca cht C trong lp th 1:
%C = 100 - 50 - 20 = 30 (%)
Vy imbiu din calp 1: I1(50%A, 20%B, 30%C)Khi lng lp th 2:
m = 20060 = 140 (g)
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Khi lng ca cht A trong lp th 1:
mA1= 50%60 = 30 (g)
Khi lng ca cht A trong lp th 2:
mA2= 40 - 30 = 10 (g)
Khi lng ca cht B trong lp th 2:
mB2= 80%140 = 112 (g)
Khi lng ca cht C trong lp th 2
mC2= 140 - 122 = 17 (g)
im biu in lp 2: I2(7,15%A, 80%B,12,85%C)
V d 6. Khi lng ring ca phenol dng rn v dng
lng ln lt l 1,072 v 1,056 g/ml, nhit nng chy
ca phenol l 24,93 cal/g, nhit kt tinh ca n 1
atm l 410C. Tnh nhitnng chy ca phenol 500
atm.
Gii
p dng phng trnh:
VT.
dP
dT
Vi: 0,0141,072
1
1,056
1V (mol/g)
Ta c: 34,26.1041,324,93
0,014314
dP
dT
(K/atm)
Nh vy, c tng p sut ln 1 atm th nhit
nng chy ca phenol tng 4,26.10-3K. Mt cch gn
ng, 500atm, nhit nng chy ca phenol l:
T = 314 + 4,26.10-3(500 - 1)
= 316,13K = 43,130C
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V d 7. p sut thng, nhit si ca nc v
cloroform ln lt l 1000C v 600C, nhit ha hi
tng ng l 12,0 v 7,0 kcal/mol. Tnh nhit m
2 cht lng trn c cng p sut?
Gii
Gi T l nhit m ti 2 cht lng c cng p
sut:
Ta c:
T
1
T
1
T
1
T
1
2
2
1
1
Vi: 1= 12 (Kcal/mol), T1= 273 + 100 = 3730K
2= 7 (Kcal/mol), T2= 273 + 60 = 3330K
Th cc gi tr vo phng trnh trn, ta c:
T
1
333
17
T
1
373
112
Suy ra: T = 448,40K
4.8. Bi tp t gii
1. Xc nh nhit ha hi ca H2O 4atm nu
1000C nhit ha hi ca nc bng 2254,757 J/g.
2. Xc nh nhit si ca benzoatetyl (C9H10O2) p sut 200 mmHg bit rng nhit si chun
ca benzoatetyl l 2130C v nhit ha hi bng
44157,52 (J/mol).
S: T = 433,10K
3. Nhit nng chy chun ca Bi l 2710C. iu
kin khi lng ring ca Bi rn v lng l 9,673
v 10 g/cm3. Mt khc khi p sut tng ln 1 atm
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th nhit nng chy gim i 0,00354K. Tnh
nhit nng chy ca Bi.
S: 11 kJ/mol.
4. Ti 1270C HgI2b chuyn dng th hnh t dng sang dng vng. Nhit chuyn ha l 1250 J/mol,
V = 5,4 cm3.mol-1, dng c t trng ln hn
dng vng. Xc nh dT/dP ti1270C.
S: -1,73.10-6K/Pa
5. Khi un nng lu hunh rombic chuyn thnh lu
hunh n t km theo bin thin th tch V =
0,0000138 m3/kg. Nhit chuyn ha chun bng
96,70C v dT/dP = 3,25.10-7 K/Pa. Xc nh nhit
chuyn pha.
S: = 501,24 kJ/kg
6. Xc nh th tch ring ca thic lng ti nhit
nng chy chun 2320C nu nhit nng chy ring
l 59,413 J/g, khi lng ringca thic rn l 7,18
g/cm3v dT/dP = 3,2567.10-8K/Pa.
S: 0,147 cm3/g
7. 200 mmHg metanol si 34,70C cn khi tng p
sut ln gp i th nhit si l 49,90C. Tnh
nhit si chun ca metanol.
S: 65,40C
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8. Tnh p sut cn thit nhit si ca nc t
c 1200C. Cho bit nhit ha hi ca nc l
539 cal/g.
S: P = 2 atm
9. Cho gin pha ca h 3 cu t (hnh di). Xc
nh thnh phn ca A, B, C khi im h chung l
im P v hy kt lun v thnh phn ca A, B khi
im h dch chuyn theo ng thng ni t nh
C vi im I.
80
80
80
60
60
60
40
20
20
20 40
40
100
100100
A
BC
P
(%C)
(%B)
(%A)
I
S: %A = 40%, %B = 40%, %C = 20%
10. Nc nguyn cht c th tn ti 9 dng pha khc
nhau (kh, lng v 7 dng rn). Tnh s pha ti a
ca nc c th ng thi nm cn bng vi nhau.
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Chng5
DUNG DCH V CN BNG
DUNG DCH - HI
5.1. Cch biu din thnh phn ca dung dch
- Nng phn trm khi lng (%):
100(%)g...gg
gC
n21
i%,i
- Nng mol/lit:V
nCM
- Nng ng lng gam (lg/l):V
n'CN
- Nng molan (Cm): 1000m
nC
dm
ctm
- Nng phn mol:
i
i
in
nx
- CN= z.CM (z: sin tch trao i trong phn
ng)
5.2. S ha tan ca kh trong cht lng
5.2.1. nh hng ca p sut n tan ca cc
kh trong cht lng
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nh lut Henry: nhit khng i, ha tan
ca mt kh trong mt cht lng t l thun vi p sut
ring phn ca kh trn pha lng
xi= kH.Pi
Trong :
kHl hng s Henry
Pil p sut hi ca pha kh trn pha lng
5.2.2. nh hng ca nhit n ha tan ca
kh trong cht lng, phng trnh Sreder
Xt cn bng:
i (kh) = i (dung dch c nng xi) + Hha tan
Hng s cn bng:(kh)x
(dd)xK
i
i
Do ta c:2RT
i
T
plnK
Ly tch phn phng trnh, ta c:
0T
1
T
1
Ri
ilnx
Vi: T0l nhit ngng t (nhit si)
5.3. S ha tan ca cht lng trong cht lng v cnbng dung dch - hi
5.3.1. H dung dch l tng tan ln v hn
5.3.1.1. p sut hi - nh lut Raoul
p sut hi bo ha ca mi cu t bt k t lthun vi phn phn t ca n trong dung dch.
liRi .xkP
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Khi dung dch ch c cu t i (dung dch i nguyn
cht): xi = 1 v kR= Pi0li
0ii .xPP
ivi dung dch thc, nh lut Raoult ch c th
p dng cho dung mi ca dung dch v cng long:l1
011 .xPP
5.3.1.2. Gin p sut - thnh phn (P - x)
p dng nh lut Raoult cho dung dch l tng ca hai
cu t (A- B): lB
0A
lA
0AA x1.P.xPP (1)
lB
0BB .xPP (2)
p sut tng ca h l:
P = PA + PBlB
0B
lB
0A .xPx1.P
lB0A0B0A .xPPP (3)Nu ta biu din cc phng trnh (1), (2) v (3) ln
th p sut - thnh phn (P- x) ta c hnh 5.1.
A B
(3)
(1)
(2)
xB
P P0B
P
0
A
Hnh 5.1. Gin p sut hi (P - x) ca dung dch 2
cu t l tng
5.3.1.3. Thnh phn pha hi nh lut Konovalop I
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Xt h dung dch l tng ca hai cu t A v B
nm cn bng vi pha hi ca chng.
Theo nh lut Raoult ta c:
lAx
l
B
x
.lAx
l
B
x
0AP
0
B
P
hAx
h
B
x
(nh lut Konovalop I)
Trong 0
A
0B
P
P v c gi l h s tch hay h s
chng ct
5.3.1.4. Gin thnh phn hi thnh phn lng
T nh lut Konovalop I,ta bin i rt ra biu thc:
lB
.x)lBx(1
lB
.x
lB
.xlAx
lB
.x
hBxh
Ax
hBx
lB
.x11
lB
.xh
Bx (4)
Biu din phng trnh (4) ln th (x- x) ta c cc
ng trn hnh 5.2.Bh
Bl
C
A xl
xh
Hnh 5.2. Gin (x-x) ca h hai cut A-B5.3.2. H dung dch thc tan ln v hn
5.3.2.1. p sut hi
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- Dung dch sai lch dng c p sut hi trn dung
dch ln hn p sut hi tnh theo nh lut Raoult.
- Dung dch sai lch m c p sut hi trn dung
dch nh hn p sut hi tnh theo nh lut Raoult.
5.3.2.2. Thnh phn pha hi, nh lut Konovalop II
i vinhng h c thnh phn ng vi im cc
tr trn ng p sut hi tng cng (P - x) th pha lng
v pha hi cn bng c cng thnh phn.hB
lgB xx
5.3.3. H hai cht lng hon ton khng tan ln
5.3.3.1. Tnh cht
- Thnh phn ca pha hi cng ch ph thuc vo
nhit m khng ph thuc vo thnh phn ca
hn hp lng.
f(T)P
P
P
P
x
x0A
0B
A
B
hA
hB
- Nhit si ca hn hp cng khng ph thuc
vo thnh phn, n nh hn nhit si ca mi
cu t v ch ph thuc vo p sut bn ngoi.
- Trong qu trnh si, nhit si ca hn hp s
gi nguyn cho n khi mt trong hai cu t
chuyn ht thnh hi, th nhit si ca h s
tng vt n nhit si ca cu t cn li.
5.3.3.2. Chng ct li cun theo hi nc
A0A
0OH
OHM
18
P
Pg 2
2
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Trong : 0 OH2P v0
AP ln lt l p sut hi ca
nc v ca cht A.
5.3.3.3.nh lut phn b
nhit v p sut khng i, t s nng camt cht tan trong hai dung mi khng tan ln l mt
hng s khng ph thuc vo lng tng i ca cht
tan v dung mi.
KC
C
Y/B
Y/A
CY/A, CY/B: l nng ca cht tan Y trong dung
mi A v trong dung mi B.
K: h s phn b
5.4. Bi tp mu
V d 1: Tnh p sut hi ca dung dch ng
(C12H22O11) 5% 1000C v nng % ca dung dch
glycerin trong nc c p sut hi bng p sut hi
ca dung dch ng 5%.
Gii
p sut hi ca dung dch ng: OH0OH 22.xPP
758
342
5
18
9518
95
760P
(mmHg)
Dung dch glycerin:
OH0OH 22.xPP 0,997
760
758
P
Px
0OH
OH
2
2
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M: 0,997
92
m
18
m18
m
xglyOH
OH
OH
2
2
2
Suy ra: OHgly 20,014mm Nng phn trm ca dung dch glycerin
100mm
mC%
OHgly
gly
2
1,38100m0,014m
0,014m
OHOH
OH
22
2
(%)
V d 2. 123,30C bromobenzen (1) v clorobenzen (2)
c p sut hi bo ha tng ng bng 400 v 762
mmHg. Hai cu t ny to vi nhau mt dung dchxem
nhl tng. Xc nh:
a. Thnh phn dung dch 123,30C di p sut khquyn 760mmHg.
b. T s mol ca clorobenzen v bromobenzen trong
pha hi trn dung dch c thnh phn 10% mol
clorobenzen.
GiiHai cu t ny to vi nhau mt dung dch l tng
nn: l1020102l202l10121 xPPP.xP.xPPPP
a. Thnh phn hn hp 123,30C di p sut kh
quyn 760mmHg0,00552
762400
762760
PP
PPx
02
01
02l
1
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0,9948x1x l1l2
Vy thnh phn ca Bromobenzen l: 0,00552
thnh phn ca Clorobenzen l: 0,9948
b. T s mol ca clorobenzen v bromobenzen
0,210,9
0,1
400
760
x
x
P
P
x
xl1
l2
10
20
h1
h2
V d 3.Benzen v toluen to vi nhau mt dung dch
xem nhl tng. 300C p sut hi ca benzen bng
120,2 mmHg, ca toluen bng 36,7 mmHg.Xc nh:a. p sut hi ring phn ca tng cu t.
b. p sut hi ca dung dch.
Nu dung dch c hnh thnh t s trn 100g benzen
v 100g toluen.
Giia. p sut hi ring phn ca tng cu t
Phn mol ca benzen:
54,0
92
100
78
10078
100
nn
nx
TB
BB
Phn mol ca toluen:
46,0
92
100
78
10092
100
nn
nx
TB
TT
p sut hi ca Benzen: 64,9080,54120,2.xPP B0BB (mmHg)
p sut hi ca Toluen:
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16,8820,4636,7.xPP T0TT (mmHg)
b. Xc nh p sut hi ca dung dch81,7916,88264,908PPP TB (mmHg)
V d 4.Etanol v metanol to thnh dung dch xem nhl tng. 20oC p sut hi bo ha ca etanol v
metanol ln lt l 44,5 v 88,7 mmHg.
a. Tnh thnh phn mol cc cht trong dung dch
cha 100getanol v 100g metanol.
b. Xc nh cc p sut ring phn v p sut tngca dung dch.
c. Tnh phn mol ca metanol trong pha hi nm cn
bng vi dung dch trn.
Gii
a. Phn mol mi chtS mol etanol: )2,1739(mol
46
100nE
S mol metanol: 3,125(mol)32
100nM
Phn ca etanol: 0,413,1252,1739
2,1739xE
Phn ca metanol: 0,593,1252,1739
3,125xM
b. g)18,245(mmH0,4144,5xPP lE0
EE
(mmHg)333,520,597,88xPP lM0
MM
P = 18,245 + 52,333 = 70,578 (mmHg)
c. Phn mol ca metanol trong pha hi:
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0,7415
0,59144,5
88,71
0,5944,5
88,7
1)x(1
.xx
lM
lMh
M
V d 5.Hn hp SnCl4(1) v CCl4(2) tun theo qui lut
ca dung dch l tng. 90oC p sut hi bo ha
ca SnCl4 v CCl4 ln lt l 362 mmHg v 1112
mmHg. Di p sut chun 760mmHg, SnCl4 si
1140C v CCl4 si 77oC:
a. Xy dng gin thnh phn - p sut ca cc
cu t v xc nh trn gin p sut P1, P2v P
ca hn hp c phn mol ca CCl4l 0,7.
b. Xc nh thnh phn hn hp SnCl4 - CCl4 si
900C di p sut 760mmHg.
c. Xc nh thnh phn hi ti 900
C.Gii
a. Xy dng gin thnh phn - p sut
SnCl4
CCl4
362
P
(mmHg)1112
1
2
3
x
108,6
779,1
0,7
Hnh 5.3. Gin p sut - thnh phn P - x
p sut ca SnCl4:
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p dng cng thc:
121
2
T
1
T
1
RP
Pln
i vi cht A:
Ta c:
12
A
0298A
0373A
T
1
T
1
RP
P
ln (1)
Tng t i vi cht B:
12
B
0298B
0373B
T
1
T
1
RP
Pln (2)
Ly phng trnh (1) (2), ta c:
12
AB0298A
0 298B
0373B
0 373A
T
1
T
1
R
1
P
P
P
Pln
373298
37329850007000
987,1
1
P
P
P
Pln
0298A
0298B
0373B
0373A
507,0
P
P
3
1
0 373B
0373A
52,1P
P
0373B
0373A
V d 7. 800C p sut hi bo ha ca A nguyn cht
v B nguyn cht ln lt l 100 v 600 mmHg.
a. Hy v th p sut - thnh phn (P - x) ca
dung dch l tng A - B.
b. Tm thnh phn ca A v B sao cho ti p sut
ca A v B bng nhau.
Giia. Gin P - x (hnh 5.4)
lA
lA
0AA x100x.PP (1)
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lB
lB
0BA x600x.PP (2)
lA
lA
0B
0A
0Bt x500600xPPPP (3)
Hnh 5.4. Gin p sut - thnh phn P - x
b. Thnh phn cu t A v B.
Ta c: PA= PB
lAlA x1600x.100
857,.0x lA v 143,0xlB
5.5. Bi tp t gii
1. 250C p sut hi bo ha ca nc nguyn cht
l 23,7 mmHg. Tnh p sut hi trn dung dchcha 10% glyxerin trong nc nhit .
S: 23,2 mmHg
2. 500C, dung dch l tng bao gm 1 mol cht A
v 2 mol cht B c p sut tng cng l 250
mmHg. Thm 1 mol cht A vo dung dch trn thp sut tng cng l 300 mmHg. Hy xc nh p
sut hi bo ha ca A v B nguyn cht 500C.
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S: 450 v 150mmHg
3. Xem dung dch ca CCl4 v SnCl4 l dung dch l
tng. Tnh thnh phn ca dung dch si 1000C
di p sut 760 mmHg v tnh thnh phn cabong bng hi u tin, bit rng 1000C p sut
hi bo ha ca CCl4v SnCl4ln lt l 1450 v
500 mmHg.
S: 0,274 v 0,522
4. Xt dung dch toluen benzen cha 70% khi
lng benzen 300C. Hy xc nh:
a. Cc p sut phn v p sut tng cng ca
dung dch
b. Thnh phn ca pha hi nm cn bng vi
dung dch trn.Bit rng 300C p sut hi bo ha ca benzen
v toluen ln lt l 120,2 v 36,7 mmHg.
S: a. 88,2 ; 9,8 ; 98,0 mmHg; b. 0,9 ; 0,1
5. Mt dung dch cha 0,5 mol propanol v 0,5 mol
etanol c chng cho n khi nhit si cadung dch l 900C. p sut hi ca phn ngng t
thu c l 1066 mmHg (cng o nhit 900C).
Xem dung dch l l tng v bit rng 900C p
sut hi bo ha ca propanol v etanol ln lt l
574 v 1190 mmHg. Hy tnh:
a. Thnh phn mol ca dung dch cn li trong
bnh chng
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b. Thnh phn mol ca phn ngng t.
c. S mol etanol ha hi.
S: a. x = 0,3 ; b. x = 0,8 ; c. 0,32 mol etanol
6. 500C p sut hi ca n - hecxan v n - pentan
ln lt l 400 v 1200 mmHg.
a. Tnh p sut hi ca dung dch cha 50%
(khi lng) ca n-pentan.
b. Xc nh phn mol ca n - hecxan trong pha
hi.c. Xc nh thnh phn ca hai cu t trn trong
pha lng p sut hi ca chng bng nhau.
7. Tnh p sut hi bo ha ca dung dch 5g ng
glucose (C6H12O6) trong 180g nc 200C. Bit
rng nhit ny p sut hi bo ha ca nc17,5 mmHg.
8. 200C p sut hi bo ha ca dung dch cha
52,8g A v 180g H2O l 16,5 mmHg. Xc nh khi
lng phn t ca A, bit rng nhit ny p
sut hi bo ha ca nc l 17,5 mmHg.
9. Xc nh phn mol ca dung dch cha 20% A (M =
46), 30% B (M =18) v 50% C (M = 60) v khi
lng.
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Chng6
CN BNG GIA DUNG DCH
LNG V PHA RN
6.1. Tnh cht ca dung dch long cc cht tan
khng bay hi
6.1.1. gim p sut hi ca dung dch
gim tng i p sut hi ca dung dch bng
tng phn phn t ca cc cht tan khng bay hi trong
dung dch.
xP
P
P
PP00
0
Trong :
P0
: p sut hi ca dung mi nguyn chtP: p sut hi ca dung dch
x: phn mol ca cht tan
6.1.2. tng im si v h im kt tinh
ng OC m t p sut hi trn dung mi rn
nguyn cht. ng OA m t p sut hi trn dung mi lng
nguyn cht.
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ng OB m t nh hng ca p sut bn
ngoi n nhit nng chy ca dung mi
nguyn cht.
tng im si v h im kt tinh ca cc
dung dch cht tan khng bay hi t l thun vi nng
ca dung dch.
T = K. Cm
Vi:
Cm: Nng molan ca dung dch.
K: Hng s nghim si Ks hay hng s nghimng K
T TS
Pngoai
P
B1
B
O
O1C
A A
T
Hnh 6.1. Gii thch tng im si v h im
kt tinh
1000
.MR.TK
20
Trong :
R: hng s kh
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T0: nhit chuyn pha
M: phn t lng
: nhit chuyn pha
6.1.3. p sut thm thup sut thm thu ca dung dch c nng xc
nh l p sut ph phi tc ng ln mt mng bn
thm nm phn cch gia dung dch v dung mi
nguyn cht dung dch ny c th nm cn bng
thytnh vi dung mi (qua mng bn thm).= CRT
Trong :
: p sut thm thu
C: Nng dung dch (mol/l)
R: hng s kh
T: nhit tuyt i
6.2. S kt tinh ca dung dch hai cu t. H khng
to dung dch rn, khng to hp cht ha hc
6.2.1. Gin nhit - thnh phn (T- x)
Cc im a, b tng ng vi nhit kt tinh cacc cu t A v B nguyn cht.
ng aeb c gi l ng lng.
ng arArBb c gi l ng rn.
Vng nm trn ng lng h ch c mt pha lng
LA-B Vng nm pha di ng rn, h bao gm hai
Pha rn: rn A v rn B (RA, RB).
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Vng nm gia ng lng v ng rn h tn
ti cn bng ca hai pha: RAL hoc L - RB.
Hnh 6.2. Gin (T-x) ca h hai cu t, cn bng lng
rn.
6.2.2. Kho st qu trnh a nhit
Ti nhit T2: H Q2 = lng l2 + rn r2
Lng rn B
Lng lng l2=
gB
gl=
Q2l2
Q2r2
T
A
e
B
b
a
RB
r2
TeRCHRA
r1
E
T2
T1
l2
l1
Q
Q2
t (thi gian)
T
q
r
s t
u
Hnh 6.3. Qu trnh a nhit ca h Q
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H H = pha lng e + h rn chung RC
Lng lng e
Lng rn chung=
HRC
EH
H rn chung RC= pha rnA + pha rn B
Lng rn A
Lng rn B=
RCRB
RCRA
6.2.3. Hn hp eutecti
p sut khng i, hn hp eutecti s kt tinh nhit khng i theo ng thnh phn ca n. Hn
hp eutecti c tnh cht ging nh mt hp cht ha
hc, song n khng phi l mt hp cht ha hc m n
ch l mt hn hp gmnhng tinh th rt nh, rt mn
ca hai pha rn A v rn B nguyn cht kt tinh xen k
vo nhau.
6.2.4. Php phn tch nhit
T
A B
b
a
t
T
1
1 23 4
5 6
2 3 4 5 6
x y z t x y z t
Hnh 6.4. Minh ha php phn tch nhit
Lp 6 h c cng khi lng vi thnh phn ca
cu t B thay i t 0% n 100%. Lm nng chy tng
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h ri h dn nhit , quan st s thay i nhit
theo thi gian v v cc ng ngui lnh (T - t).
6.3. H hai cu t khng to dung dch rn, khi kt
tinh to thnh hp cht ha hc bn
A B
d
e1
e2
b
DE1 E2
RA - LL - RD
RA va RD
RD - L L - RB
RD va RB
T Q1
h
a
Q l1 l2 l3
rDrB
Hnh 6.5. Gin (T-x) h 2 cu t to hp cht hoa
hc bn
D l hp cht ha hc ca A v B.
ng ae1l ng kt tinh ca cht rn A.
ng e1de2l ng kt tinh ca cht rn D.
ng e2b l ng kt tinh ca cht rn B.
Hai im e1 v e2 tng ng l cc im eutecti
ca h A - D v h D - B.
6.4. Bi tp mu
V d 1.Bng im ca dung dch nc cha mt cht
tan khng bay hi l -1,50C. Xc nh:
a. Nhit si ca dung dch.b. p sut hi ca dung dch 250C.
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Cho bit hng s nghim lnh ca nc l 1,86 vhng
s nghim si ca nc l 0,513. p sut hi ca nc
nguyn cht 250C l 23,76 mmHg.
Giia. h im ng c ca dung dch: T= 1,50C
Ta c:
m .CKT
0,8061,86
1,5
K
TC
m (mol/1000g)
tng im si:0,4140,8060,513.CKT mSS (
0C)
Nhit si ca dung dch:
Tdd= 100 + 0,414 = 100,414 (0C)
b. p sut hi ca dung dch
23,42
0,80618
1000181000
23,76.xPP 0
(mmHg)
V d 2. 200C p sut hi nc l 17,54 mmHg vp
sut hi ca dung dch cha cht tan khng bay hi l
17,22 mmHg. Xc nh p sut thm thu ca dung dch
400C nu t trng ca dung dch ti nhit ny l
1,01 g/cm3 v khi lng mol phn t ca cht tan l
60.
Gii
Ta c: 0,01817,54
17,2217,54
P
Px
0
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M 0,018
18
m
60
m
60
m
xOHct
ct
2
(1)
Gi s 100mm OHct2
(g) (2) ctOH m100m 2
T (1) v (2), ta c: mct= 5,75 (g)
Th tch ca dung dch: 991,01
100
d
mV dd (ml)
Nng ca dung dch:
0,968100099
605,75
V
nCM (mol/l)
p sut thm thu:
= CRT = 0,968x0,082x(273 + 40) = 24,84 (atm)
V d 3. Gin kt tinh (T-x) ca h hai cu t A - Bc cho trong hnh sau.
T0C
e
A B
0,650,4
xB
Rc
0 1
H
Q
0,2
I
II
a
b
a. Tnh s pha v bc t do ca h ti cc vng I, II,
III v ti im eutecti.b. Lm lnh 90g h Q, khi im h nm ti H, c A v
B kt tinh mt phn v im rn chung (gm c
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rn A v rn B) nm ti RC. Tnh lng rn A v
rn B kt tinh v lng lng eutecti cn li.
Gii
a. S pha v bc t do ca hVng I: f = 1, c = kf + 1 = 2
Vng II: f = 2, c = kf + 1 = 1
Ti im eutecti: f = 3, c = k f + 1 = 0.
b. Khi lng ca pha rn v pha lng
5
4
0,25
0,2
He
HR
m
m c
r
l
Ta c : ml+ mr= 90 g
ml = 40 g; mr= 50 g
Khi lng ca rn A v rn B
4
0,2
0,8
RR
RR
m
m
AC
BC
R
R
B
A
Ta c RA+ RB= 50 g
RA= 40 g
RB= 10 g
V d 5.Gin kt tinh (T - x) ca h hai cu t A - B
c cho trong hnh sau. Lm lnh 110 gam h Q.
a. Xc nhs pha v bc t do ca h ti cc vng I,
II v nhit kt tinh ca cu t A, B nguyn cht.
b. Xc nh nhit bt u kt tinh ca h Q. Khi
im h nm ti H c A v B kt tinh mt phn
v im rn chung (gm rn A v rn B) nm tiRC. Tnh lng rn A v rn B kt tinh v lng
lng eutecti cn li.
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c. Tnh lng lng eutecti ti a thu c t h trn.T
a
bQ
200e H Rc
BA
0 0.2 10.80.60.4
400
600
800
I
II
III100
RBRA
Gii
a. S pha v bc t do ca cc vng
Vng I: f = 1, c = 2
Vng II: f = 2, c = 1
Nhit kt tinh A: 4500C
Nhit kt tinh B: 7000C
b. Nhit bt u kt tinh: 6000C
Ta c h pt:
38
0,150,4
HRHe
m
m
cle
Rc
mRc+ mle= 110
Gii hta c: mRc= 80 (g), mle= 30 (g)
Khi lng rn A v rn B.
Ta c h pt:
17
3
0,85
0,15
RR
RR
m
m
AC
BC
R
R
B
A
mRa+ mRb= 80
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Gii h ta c: mRb = 68 (g); mRa= 12 (g).
c. Khi lng eutecti ti a khi RCtrng vi RB
4
3
0,4
0,3
He
HR
m
mB
Rc
l(e)
ml(e)+ mRc= 110
Gii h ta c: ml(e) = 62,86 (g)
6.5. Bi tp t gii
1. Gin kt tinh ca Sb v Pb c dng nh hnh
v. Lm lnh 200g h Q.a. M t gin pha ca h hai cu t trn.
b. Xc nh bc t do ca vng (I), (II) v ti
im e.
Sb Pb
40 70 850 100
3270C
6360C
Q
LHR
N2460C H1
I
II
e
T0C
c. Khi im h Q trng vi im H. Hy xc nh
khi lngca pha lng v pha rn.
d. Khi h Q kt tinh hon ton, hy xc nh
lng eutectic thu c.
S: c. mr= 85,7g; ml= 114,3g ; d. me= 94,11g
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2. V gin pha ca h Sb - Pb da vo cc d
kin thc nghim sau:
Thnh phn hn hp lng
(% khilng)
Nhit bt u
Kt tinh (0C)Sb Pb
100 0 632
80 20 580
60 40 520
40 60 433
20 80 300
10 90 273
0 100 326
a. Xc nh thnh phn eutecti.b. Xc nh khi lng Sb tch ra nu 10kg hn
hp lng cha 40% Pb c lm ngui ti
4330C.
S: a. 87%Pb v 13%Sb; b. mSb= 5kg
3. Gin kt tinh ca h A - B c dng nh hnhdi. Lm lnh 100g h Q. Khi im h nm H.
C A v B kt tinh mt phn. im rn chung
nm ti R.
a. Xc nh lng A v B kt tinh v lng
lng eutecticn li.b. Tnh lng eutectiti a thu c.
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c. Phi trnA v B vi thnh phn nh th no
thu c hn hp A v B ng thi kt
tinh.
A B
40 60 800 100
Q
H
T0
C
a
b
e
20
R
S: a. 12g; 48g v 40g; b. 57,14g
4. Xc nh nng mol v nng molan ca dung
dch cha 20g CH3COOH trong 100g nc 250C.Bit nhit ny khi lng ring ca dung dch
1,01 g/cm3.
S: Cm= 3,33 molan, CM= 2,8M
5. Tnh nhit kt tinh, nhit si, p sut thm
thu ca dung dch cha 9g ng glucose
(C6H12O6) trong 100g nc 250C. Cho bit
nhit ny p sut hi ca nc l 23,76mmHg,
khi lng ring ca dung dch l 1g/cm3, hng s
nghim lnh v hng s nghim si ca nc
tng ng 1,86 v 0,513.
S: Tkt= -0,930C; Ts= 100,260C; = 11,2 atm
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6. Benzen ng c 5,420C v si 81,10C. Nhit
ha hi ti im si bng 399J/g. Dung dch cha
12,8g naphtalen trong 1kg benzen ng c
4,910C.
a. Xc nh nhit si ca dung dch ny.
b. Tnh p sut hi ca benzen trn dung dch
81,10C.
c. Tnh nhit nng chy ring ca benzen.
S: a. 81,360
C; b. 754.1mmHg; c. 128,24 J/g
7. Acid acetic k thut ng c 16,40C. Bng im
ca acid acetic nguyn cht l 16,70C. Hng s
nghim lnh ca acid nguyn cht l 3,9. Xc nh
nng molan ca tp cht trong acid k thut.
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Chng 7
IN HA HC
7.1. Khi nim v dung dch in ly
7.1.1. Dung dch cc cht in ly
Mt cht tan khi ha tan vo dung mi, to thnh
dung dch m dung dch c kh nng dn ingi l
dung dch in ly.
7.1.2. S in ly tng im si v gim im ng c ca
dung dch in ly cao hn so vi dung dch l tng hay
dung dch khng in ly.
T = i.K.Cm
Trong :i: l h s Vant Hoff
K: hng s
Cm: nng molan
p sut thm thu ca dung dch in ly cng cao
hn p sut thm thu ca dung dch l tng hay dung
dch khng in ly.
in ly= i.C.R.T
Trong :
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: p sut thm thu
C: nng mol/l
R hng s kh
T: nhit tuyt i
H s b chnh i l t s gia tng s tiu phn
thc sc trong dung dch v s tiu phn ban u: 11i
Trong :
: phn ly
= m + nVi m, n l h s ca phng trnh:
AmBn = mAn+ + nBm-
7.2. Bi tp mu
V d 1. Tnh nhit kt tinh ca dung dch cha7,308g NaCl trong 250g nc cho bit 291K p sut
thm thu ca dungdch l 2,1079.106N/m2, khi lng
ring ca dung dch l 1g/cm3, nhit nng chy ca
nc nguyn cht l 333,48.103J/kg.
Gii
Ta c:T= i.k.Cm (1)
Vi:2 2
od
RT M 8,314.273 .18k 1,86
1000 1000.333,48.18
Trong : 0,50,2558,5
7,308Cm
(mol/kg)
Ta li c: i.CRT Nng ca dung dch:
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i vi dung dch in ly yu, ta c: 11
5K 1,82.10
0,019C 0,05
Khi dung dch c nng 0,05M th phn ly ca
dung dch l 1,9%.
V d 3. Dung dch cha 4,355 mol ng ma trong 5
lt dung dch 291K c cng p sut thm thu vi
dung dch cha 2 mol NaCl trong 4 lt dung dch. Xc
nh phn ly ca dung dch NaCl v h s Vant Hoff.Gii
Dung dch ng l dung dch khng in ly:
20,785
2910,0824,355CRT
(atm)
i vi dung dch NaCl ta c: iCRT
1,7422910,0820,5
20,78
CRT
i
in ly:i 1 1,742 1
0,7421 2 1
Vy in ly ca dung dch NaCl l 74,2%.
V d 4. Tnh p sut thm thu ca dung dch NaCl0,15M 370C bit phn ly ca dung dch l 95%.
Gii
Ta c: iCRT
M:i 1
1
i 1) 1 0,95(2 1) 1 1,95
p sut thm thu ca dung dch l:
iCRT 1,95.0,15.0,082.310 7, 43atm
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9.2. Phng trnh hp ph ng nhit Freunlich
ng hp ph ng nhit gn vi dng parapol,
do Freunlich ngh phng trnh thc nghim:
x = b.p1/n
Trong :
x: hp ph.
p: p sut kh cn bng trn cht hp ph.
b v n: l cc hng s.
Hoc: lgpn
1lgblgx
9.3. Phng trnh hp ph ng nhit Langmuir
Gi p l p sut kh, l phn b mt ti thi im
no b phn t kh chim, phn b mt cn trng s
l 1 - .
Ta s c: 1
2 1
k .pk k .p
Nu t:1
2
k
kA
mx
x
Vi: x l hp ph mt thi im no .xml hp ph cc i.
m m
p A 1p
x x x
9.4. Phng trnh hp ph BET (Brunauer- Emmett -
Teller)
0 m m 0
P 1 C 1 P
V(P P) V .C V .C P
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Vi:
P0: p sut hi bo ha
V: th tch kh hp ph p sut P
Vm: th tch kh b hp ph lp th nht (lpn
phn t)
C: tha s nng lng.
th0
P
V(P P)theo P/P0 l mt ng thng, t
c th xc nh Vmv C. Bit Vm, ta c th tnh c
b mt cht hp ph.m m
0
0
V .N.WS
V
Trong :
N : s Avogadro(= 6,023.1023)
Wm : b mt chim bi cht b hp ph lp n
phn t.
V0: th tch ca 1 mol kh iu kin chun
(22.400 cm3/mol).
9.5. S hp ph trn ranh gii b mt pha lng - rn
Lng cht b hp ph x (mmol/g) b hp phn t
trn b mt cht rn trong dung dch c tnh bng
cng thc:
0 1(C C )V
x 100m
Trong :
C0v C1l nng ban u v cn bng ca cht
b hp ph (mol/l).
V l th tchtrong xy ra s hp ph (l).
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m l lng cht hp ph (g).
9.6. Bi tp mu
V d 1. Tnh lng ru etylic b hp ph 150C trn
b mt dung dch c nng 0,12M cho bit 150C
sc cng b mt ca nc l 73,49.10-3 N/m v ca
dung dch trn l 63,3.10-3 N/m.
Gii
p dng cng thc:dm dd
.
G RT Th cc s liu vo cng thc ta c:
32dm dd
. 10,19.10G mol.m
RT 8,314.(15 273)
V d2. Xc nh ngng keo t ca dung dch in ly
K2Cr2O7nng 0,01M i vi keo nhm. Bit rng keo t 1 lt keo phi thm vo mt lng cht in ly
l 0,0631 lt.
Gii
Ta c cng thc tnh ngng keo t nh sau:
= C.V
1000
Trong :
C: Nng ca dung dch in ly (mol/l)
V: th tch ca dung dch cht in ly (ml)
: th tch ca dung dch keo (ml)
Th cc gi tr c c vo cng thc trn ta c:3C.V 0,01.0,0631.1000 .1000 0,631.10 mol / lit
1000
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V d 8.Ngng keo t ca Al2(SO4)3i vi keo As2S3
l = 96.10-6kmol/m3. Hi cn bao nhiu ml dung dch
Al2(SO4)3nng 0,01 kmol/m3 keo t 0,1 m3dung
dch keo As2S3ni trn.
Gii
Ta c cng thc tnh ngng keo t nh sau.
=C.V
1000
Trong :
C: Nng ca dung dch in ly (mol/l)
V: th tch ca dung dch cht in ly (ml)
: th tch ca dung dch keo (ml)Th cc gi tr c c vo cng thc trn ta c:
6 20,01.V96.10 V 96.10 ml100
NGN HNG CU HI MN HC
HA L
78. t do ca h c ngha:
a. cho bit s thng s nhit ng c lp ti
thiu dng xc lp h trng thi cn
bng.
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b. cho bit s thng s nhit ng ph thuc ti
thiu dng xc lp h trng thi cn
bng.
c. cho bit s thng s nhit ng c lp ti
thiu dng xc lp h trng thi khng
cn bng.
d. cho bit s thng s nhit ng ph thuc ti
thiu dng xc lp h trng thi khng
cn bng.
79. Pha l khi nim dng m t:a. mt tp hp nhng phn ng th c trong
h.
b. mt tp hp nhng phn ng th tn ti
trong h.
c. mt tp hp nhng phn ng th c trong h
m c cng tnh cht l ha mi im.
d. mt tp hp nhng phn ng th c trong h
m tnh cht vt l v hahc l ng nht.
80. Hn hp FeO v CuO c s pha bng:
a. 2.
b. 1.c. 0.
d. 3.
81. Cu t:
a. l s hp phn ti thiu to ra h v khng th
tch ra khi h.
b. l s hp phn ti thiu to ra h v c th
tch ra khi h.
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c. l s hp phn c mt trong h v khngth
tch ra khi h.
d. l s hp phn c mt trong h v c th tch
ra khi h.
82. t do tnh theo qui tc pha Gibbs:
a. c = k - f + n.
b. c = k - n + f.
c. c = f - n + k.
d. c = k - f - n.
112. Cho h: NaIO3(r) = NaI(r) + 3/2O2(k). Bit NaIO3v NaI to dung dch rn.Vy s pha f ca h:
a. 3.
b. 2.
c. 1.
d. 0
113. S thng s bn ngoi n tc ng ln h:
a. 0.
b. 2.
c. 2 (P= hng s).
d. 2 (T= hng s).
114. Cho h: NaIO3(r) = NaI(r) + 3/2O2(k). S phngtrnh lin h v nng q l:
a. 0.
b. 1.
c. 2.
d. 3.
115. Cho h: NaIO3(r) = NaI(r) + 3/2O2(k). Bit NaIO3v
NaI to dung dch rn. t do ca h l:
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a. 0.
b. 1.
c. 2.
d. 3.
116. Thng qua gin pha ta s:
a. nh tnh c cc qu trnh chuyn pha.
b. nh lng cc qu trnh chuyn pha.
c. nh tnh v nh lng cc qu trnh.
d. nh tnh v nh lng cc qu trnh chuyn
pha.117. Cho gin pha:
A BM
0,8
Qua gin pha ta thy:
a. hm lng ca cu t A ln hn cu t B.
b. hm lng ca cu t B ln hn cu t A.c. hm lng ca cu t B bngcu t A.
d. a, b, c u sai.
118. Cho gin :
A BM
0,8
H M c thnh phn:a. xA= 0,2
b. xB = 0,2
c. xA = 0,8
d. a, b, c u sai
119. Cho gin pha h ba cu t nh sau:
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A
C B
M
Qua gin ta c:
a. Cc im h thuc cnh song song vi BC s
biu din cho h c cng thnh phn cu t
A.
b. Cc im h thuc cnh song song vi BC sbiu din cho h c cng thnh phn cu t
B.
c. Cc im h thuc cnh song song vi BC s
biu din cho h c cng thnh phn cu t
C.
d. Cc im h thuc cnh song song vi BC sbiu din cho h c cng thnh phn cu t B
v C.
120. Cho gin pha h ba cu t nh sau:A
C B
M
Khi tng nng cu t A th im h M s:
a. di chuyn v nh A.
b. di chuyn v nh B.c. di chuyn v nh C.
d. ng yn khng i.
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121. Khi mt h ban u tch thnh 2 h con th:
a. cc im h phi thng hng.
b. cc im h tm thnh mt tam gic.
c. cc im h trn mt ng trn.
d. a, b, c u sai.
122. Cho qu trnh sau: NH4Cl(r) = NH3(h) + HCl(k).
t do ca h:
a. 2.
b. 1
c. 0.d. 3.
123. H c t do c = 1, trong bit h chu s tc
ng bi 2 yu t nhit (T) v p sut (P). Vy
ta c th ni:
a. s tm c mt hm s biu din quan h
hai thng s T, P ca h.
b. ng vi mi gi tr ca T ta s c mt gi tr
ca P v ngc li.
c. m t ton hc ca h l mt hm ch c mt
bin vi min xc nh l R.
d. a, b v c ng.124. Thp l mt hp kim gia Fe v C, vy s pha ca
thanh thp trn bng:
a. f = 1.
b. f = 2.
c. f = 3.
d. f = 0.
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d. 0.
129. Tam gic ABC trn c bao nhiu h 3 cu t:
a. c 3 h.
b. c 33h.
c. c 3! h.
d. c v s h.
130. Theo qui tc ng thng lin hp th t mt h M
khi tch pha thnh hai h con, th cc im h phi:
a. nm trn cng mt mt phng.
b. nm trn cng mt ng thng.c. nm trn cng mt ng cong.
d. khng nm trn cng mt ng thng.
131. Trn gin pha, khi im h chy v pha cu t
no th:
a. hm lng ca cu t tng ln.
b. hm lng ca cu t gim xung.
c. hm lng ca cu t khng thay i.
d. hm lng c th tng, c th gim.
132. Hin tng thm thu l qu trnh vt l:
a. chuyn cht qua mng bn thm.
b. chuyn dung mi qua mng bn thm.c. chuyn cht tan qua mng bn thm
d. chuyn dung mi v cht tan qua mng bn
thm
133. Qu trnh thm thu khi cn bng s to ra mt p
sut p, p sut c nghal:
a. p sut cn tr qu trnh thm thu xy ra.
b. p sut ca mi trng cng vi h.
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c. p sut thytnh ca ct dung mi
d. p sut ca kh quyn
134. Mng bn thm c tnh cht:
a. chuyn dung mi theo 2 chiu.
b. chuyn dung mi theo 1 chiu.
c. thm t mt bn.
d. thm theo mt hng.
135. Qu trnh rt mt cht no ra khi hn hp bng
mt dung mi thch hp gi l:
a. qu trnh chit.b. qu trnh li cun bng dung mi.
c. qu trnh trch li.
d. a,b,c u ng.
136. Qu trnh chit da trn nh lut no sau y:
a. nh lut Raoult.
b. nh lut Hess.
c. nh lut bo ton khi lng.
d. nh lut phn b Nernst.
137. Dung dch l mt h c tnh cht:
a. ng nht c t hai cu t tr ln.
b. ng th c t hai cu t tr ln.c. ng nht gia hai pha: pha phn tn v pha
lin tc.
d. ng th gia hai pha: pha phn tn v pha
lin tc.
138. Dung dch nc muicha bo ha:
a. l h d th.
b. l h ng th.
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c. l h vi d th.
d. l h 2 pha.
139. Cho p sut hi bo ho ca HCN theo nhit
nh sau: lgP (mmHg) = 7,04 - 1237/T. Nhit si
ca HCN iu kin thng l:
a. 14,20C.
b. 24,40C.
c. 34,20C.
d. 44,20C.
140. Cho p sut hi bo ho ca HCN theo nhit nh sau: lgP (mmHg) = 7,04 - 1237/T. Nhitchuyn pha ( hhcp ) ca HCN c gi tr:
a. 5659 cal/mol.
b. 5569 cal/mol.
c. 5695 cal/mol.
d. 5965 cal/mol
141. Tnh cht ca dung dch l tng l:a. iVV .b. BBAAABBA ffff .
c. Bin thin cc i lng nhit ng bng
khng.d. c a, b, c u ng.
142. Dung dch v cng long c tnh cht:
a. nh dung dch l tng.
b. nh dung dch thc.
c. nh dung dch keo.
d. nh dung dch rn.
143. Tnh cht cadung dch thc l:
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a. BBAAABB-A ffff .
b. iVV .c. 0H .
d. 0U .
144. Cho kh: G(kh) = G(dung dch) v dng kh G lnguyn cht n nguyn t.Vy htH ?
a. solvatephanlint HH .
b. solvatephaloangnt HH .
c. solvatent H .
d. a v b ng
145. Cho kh: G(kh) = G(dung dch). Hng s cn bng
ca phn ng c biu din cho pha lng nh
sau:
a. Kx= xlG.
b. Kx= xhG.
c.hG
lG
xx
xK
d.lG
hG
xx
xK
146.nh lut Raoult p dng cho:
a. dung dch l tng.b. dung dch v cng long.
c. dung dch thc.
d. a v b ng.
147. Ni dung ca nh lut Raoult th hin qua m t
ton hc nh sau:a. Pi= P0i.xli.
b. Pi= P0i.xhi.
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c. Pi= Ki.xli.
d. Pi= Ki.x hi.
148.nh lut Konovalop I m t ton hc nh sau:
a. lB
lBh
B1)x(1
x
x .
b.lB
lBh
B1)x(1
xx
.
c.hB
hBl
B1)x(1
xx
.
d. hB
h
BlB
1)x(1xx
149. ngha vt l ca l:
a. h s tch.
b. kh nng tch ri tng cu t.
c. kh nng bay hi ca tng cu t.
d. kh nng phn li
150. Gin nhit thnh phn ca h Al - Si khng
ng hnh biu din nh sau:T0C
e
Al Si
103
1,5.103
I
II
III
0,1 0,45 0,85
M
250
Vng III c tnh cht:
a. bo ho Al.b. cn bng gia Al (r) v Al (l).
c. bt u kt tinh Si.
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153. Gin nhit thnh phn ca h Al - Si khng
ng hnh biu din nh sau:T
0C
e
Al Si
1031,5.10
3I
II
III
0,1 0,45 0,85
M
250
H c thnh phn XSi = 0,45 th khi tin hnh anhit s:
a. bo haAl trc.
b. bo ha Si trc .
c. bo ha c hai.
d. khng th bo ha Al.
154. Gin nhit thnh phn ca h Al - Si khng
ng hnh biu din nh sau:T
0C
e
Al Si
103
1,5.103
I
II
III
0,1 0,45 0,85
M
250
Khi h c thnh phn XSi= 0,45 th tinh th u tin
xut hin nhit :
a. 14000C.
b. 15000C.
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c. 16000C.
d. 15500C.
155. Gin nhit thnh phn ca h Al - Si khng
ng hnh biu din nh sau:T
0C
e
Al Si
103
1,5.103
I
II
III
0,1 0,45 0,85
M
250
nhit 15000C v c thnh phn XSi= 0,85 th
h c tnh cht nh sau:
a. Si kt tinh mt phn .
b. Al cha kt tinh .
c. dung dch bo ho Si.
d. a, b, c ng.
156. Gin nhit thnh phn ca h Al - Si khng
ng hnh biu din nh sau:T
0C
e
Al Si
103
1,5.103 I
II
III
0,1 0,45 0,85
M
250
Qu trnh kt tinh s kt thc ti nhit :
a. nhit eutecti.
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b. ti 5000C.
c. ti 10000C.
d. ti 12500C.
157. p sut hi bo haca niken cacbonyl 00C v
130C ln lt bng 129 mmHg v 224 mmHg.
Nhit hahi ca niken cacbonyl l:
a. 6585 cal/mol.
b. - 6585 cal/mol.
c. 6585 kcal/mol.
d. a, b v c u sai158. p sut hi bo haca niken cacbonyl 00C v
130C ln lt bng 129 mmHg v 224 mmHg.
Nhit si ca nikencacbonyl iu kin thng:
a. 2830K.
b. 2380K.
c. 3280K.
d. 3820K.
159. Cho gin pha h hai cu t A- B nh sau:T0C
e
A B
0,65 0,80,4
xB
Rc
0 1
H1
H
Q
t do ti im Q l:
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a. 0.
b. 1.
c. 2.
d. 3.
160. Nhit s thay i th no khi thc hin qu trnh
chuyn pha h mt cu t nguyn cht :
a. khng thay i.
b. thay i theo thi gian.
c. ch thay i khi c tp cht.
d. a v c ng.161. Dung dch l tng c to thnh t:
a. cc phn t cht ging nhau v tnh cht vt
l.
b. cc phn t cht ging nhau v tnh cht ha
hc.
c. cc phn t cht ging nhau c v tnh cht
vt l v tnh cht ho hc.
d. a, b, c u sai.
162. Dung dch thc khc vi dung dch l tng c
im:
a. tng lc tng tc gia cc phn t bngkhng.
b. lc tng tc gia cc phn t khc khng.
c. lc tng tc gia cc phn t bng nhau v
bng khng.
d. lc tng tc gia cc phn t khng ging
nhau v khc khng.
163. Dung dch l tng l dung dch c tnh cht:
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a. tng lc tng tc gia cc phn t bng
khng.
b. lc tng tc gia cc phn t khc khng.
c. lc tng tc gia cc phn t bng nhau v
bng khng.
d. lc tng tc gia cc phn t khng ging
nhau v khc khng.
164. H ng th c s pha (f) bng:
a. f = 1.
b. f = 2.c. f = 3.
d. f = 0.
165. Qui tc u tin khi chn dung mi ha tan phi
da vo:
a. phn cc ging nhau.
b. phn cc khc nhau.
c. m in ging nhau.
d. m in khc nhau.
166. Hin nay vt cht c bao nhiu trng thi tn ti:
a. 1.
b. 2.c. 3.
d. 4.
167. Khi tin hnh chng ct mt h c im si ng
vo thnh phn ca im ng ph, th nhit ca
h s:
a. tng.
b. gim.
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c. khng thay i.
d. a, b, c u sai
168. Ti im eutecti ca h 2 cu t, t do C ca h
bng:
a. 0.
b. 1.
c. 2.
d. 3
169.nh lut Konovalop I ch p dng cho dung dch:
a. thc.b. l tng.
c. dung dch keo.
d. dung dch rn.
170. Hng s trong cng thc ca nh lut Konovalop
I, gi l:
a. h s chng ct.
b. h s tch.
c. h s lng - hi.
d. a, b u ng
171. Hng s trong cng thc ca nh lut Konovalop
I cng ln th:a. nhit si hai cht cng gn nhau.
b. nhit si ca hai cht cng xa nhau.
c. nhit si ca hai cht bngnhau.
d. a, b v c u sai.
172. Khi ha tan cht rn vo chtlng to thnh dung
dch, tnh cht ca dung dch s thay i nh th
no:
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a. nhit si ca dung dch tng so vi nhit
si ca dung mi nguyn cht.
b. nhit si ca dung dch gim so vi nhit
si ca dung mi nguyn cht.
c. p sut hi ca dung dch gim so vi p sut
hi ca dung mi nguyn cht.
d. c a v c ng.
173.p sut thm thu ca dung dch ph thuc vo
yu t no:
a. nng ca dung dch.b. trng thi ca dung dch.
c. p sut hi ca dung dch.
d. c b v c ng.
174.p sut thm thu ca dung dch s gim khi:
a. nhit gim.
b. nhit tng.
c. nng dung dch tng.
d. in ly gim.
175.p sut thm thu ca dung dch tng khi:
a. nhit dung dch tng.
b. nhit dung dch gim.c. p sut hi ca dung dch gim.
d. c a v c u ng.
176. Nhit si ca dung dch cha cht tan khng
bay his thay i nh th no nu nng ca
dung dch tng.
a. tng.
b. gim.
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c. khng nh hng.
d. cha xc nh c.
177. Nhit kt tinh ca dung dch cha cht tan
khng bay hi s thay i nh th no nu nng
ca dung dch tng.
a. tng.
b. gim.
c. khng nh hng.
d. cha xc nh c.
178. Trong mt h gm hai h con. xc nh thnhphn ca cc h con phi s dng qui tc no:
a. qui tc lin tc.
b. qui tc ng thng lin hp.
c. qui tc n by.
d. qui tc khi tm.
179.p sut hi ca dung dch ph thuc vo yu t
no:
a. nhit , bn cht ca dung mi v cht tan.
b. thnh phn ca cc cu t trong pha lng.
c. p sut tng.
d. c a, b, c u ng.180. Nhit chuyn pha ca mt cu t ph thuc vo
yu t no:
a. nhit .
b. p sut.
c. th tch ring.
d. c a, b, c u ng.
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181. S ha tan ca cht kh vo trong lng ph thuc
vo yu t no:
a. nhit p sut v bn cht ca cht kh v
lng.
b. nhit dung ring ca cht kh v lng.
c. nhit ho hi ca cht lng.
d. nhit ngng t ca cht lng.
182. Xc nh nhit si ca nc 2 atm. Bit nhit
ha hi ca nc 9702 (cal/mol).
a. 120,90Cb. 2000C
c. 206,20C
d. 80,50C
183. Xc nh p sut hi ca dung dch cha 2 mol A
v 1 mol B. Cho bit p sut hi ca A v B
nguyn cht ln lt l 120,2 v 36,7 mmHg.
a. 277,1 mmHg
b. 193,6 mmHg
c. 92,37 mmHg
d. 64,53 mmHg
184. Xc nh nhit kt tinh ca dung dch cha 5gur (M = 60 g/mol) trong 100g nc. Cho bit hng
s nghim lnh ca nc l 1,86.
a. -1,550C
b. 1,550C
c. 1,480C
d. - 1,480C
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185. Xc nh p sut thm thu ca dung dch cha
10g ng glucose (M=180g/mol) trong mt 100ml
dung dch 300C.
a. 0,0138 atm
b. 13,8 atm
c. 0,0137 atm
d. 33,44 atm
186. Mt dung dch c xem l dung dch l tng
phi c c im g:
a. lc tng tc gia cc phn t cng loi vcc phn t khc loi l nh nhau.
b. khi to thnh dung dch khng c hiu ng
no (V = 0, U=0, H = 0).
c. thnh phn ca cht tan rt b so vi thnh
phn ca dung mi.
d. C a v b u ng.
187. Mt dung dch c xem l dung dch v cng
long phi c c im g:
a. lc tng tc gia cc phn t cng loi v
cc phn t khc loi l nh nhau.
b. khi to thnh dung dch khng c hiu ngno (V = 0, U=0, H = 0).
c. thnh phn ca cht tan rt b so vi thnh
phn ca dung mi.
d. C a v b u ng.
188. S dng phng php no tch hai cu t
nc v etanol tan ln vo nhau.
a. chng ct.
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b. trch ly.
c. chit tch.
d. kt tinh.
189. Phng trnh hp ph Langmuir ch p dng cho:
a. hp ph n lp.
b. hp ph a lp.
c. hp th a lp.
d. hp th n lp.
190. Hin nay xc nh din tch b mt ring cho
cht rn ngi ta dng phng php hp ph vgii hp ph Nitlng. Vy thuyt hp ph no cho
kt qu ng tin cy nht:
a. Langmuir.
b. B.E.T.
c. Brunauer.
d. Freundlich
191. Qu trnh hp ph vt l khc vi hp ph ha hc:
a. nhit hp ph nh.
b. l thun nghch.
c. khng lm bin i cht hp ph.
d. a, b v c ng.192. Trong h d th cc phn t trong lng mt pha c
tnh cht khc vi cc phn t trn ranh gii cc
pha l:
a. cn bng v ngoi lc.
b. khng cn bng v ngoi lc.
c. lun hng v b mt phn chia pha.
d. l cht hot ng b mt.
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193. Cht hot ng b mtl cht ch c tc dng:
a. trong lng pha.
b. trn ranh gii ca b mt phn chia pha.
c. bt c ni no cah.
d. b, c u ng.
194. Sc cng b mt:
a. l nng lng t do b mt tnh cho mt n
v din tch b mt phn chia pha.
b. l nng lng b mt tnh cho mt n v din
tch b mt.c. l nng lng t do b mt tnh cho mt n
v din tch b mt ring.
d. l nng lng b mt tnh cho mt n v din
tch b mt ring.
195. Qu trnh hp ph s:
a. lm gim G ca pha kh.
b. lm gim G ca h.
c. l qu trnh ta nhit.
d. a v c u ng.
196. Vai tr ca cht hot ng b mt:
a. lm gim sc cng b mt.b. lm gim nng lng t do.
c. to nh ha.
d. to mi - xen.
197. Sc cng b mt chi phi:
a. kh nng thm t.
b. kh nng ha tan.
c. kh nng thm thu.
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d. kh nng to bt.
198. Trong hp ph kh v hi trn b mt cht rn th:
a. hp ph l s tng nng ca kh (hi) trn
b mt phn chia pha.
b. hp ph l stng nng ca kh (hi) trn
b mt pha rn.
c. cht b hp ph l cht thc hin qu trnh hp
ph.
d. c a,b, c u ng.
199. Chn pht biu ng nht:a. Cht b hp ph l cht thc hin s hp ph.
b. Cht b hp ph l cht b thu ht ln trn b
mt cht hp ph.
c. Cht hp ph l cht c b mt thc hin s
hp ph.
d. C b v c.
200. Trong hp ph da vo lc hp ph ta chia hp
ph thnh:
a. hp ph ion v hp ph trao i.
b. hp ph vt l v hp ph ha hc
c. hp ph ha hc v hp ph trao i.d. hp ph vt l v hp ph ion
201. Chn pht biu ng nht:
a. trong hp ph, khi nhit tng th hp
ph gim do qu trnh hp ph thng thu
nhit.
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b. trong hp ph, khi nhit tng th hp
ph tng do qu trnh hp ph thng thu
nhit.
c. trong hp ph, khi p sut v nng cht b
hp ph tng th hp ph tng nhng c
gi tr gii hn.
d. trong hp ph, khi p sut v nng cht b
hp ph tng th hp ph tng nhng
khng c gi tr gii hn.
202. Chn pht biu ng:a. H phn tn l h bao gm cc ht phn b
trong mt mi trng no , cc ht lun
lun l mt cu t.
b. H phn tn l h bao gm cc ht phn b
trong mt mi trng no , cc ht lun
lun l nhiu cu t.
c. H phn tn l h bao gm pha phn tn v
mi trng phn tn, pha phn tn lun lun
l nhiu cu t.
d. H phn tn l h bao gm pha phn tn v
mi trng phn tn vi pha phn tn c thl mt hoc nhiu cu t.
203. Cu to ca mixen keo bao gm:
a. Nhn, lp hp ph v ion to th.
b. Nhn, lp hp ph v lp khuych tn.
c. Nhn, ion to th v lp khuych tn.
d. Nhn ion i v ion to th.
204.in tch ca ht mixen keo c quyt nh bi:
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a. nhn keo.
b. lp khuych tn.
c. ion to th.
d. ion i.
205. Thc hin phn ng trao i iu ch keo AgI
khi cho d AgNO3:
AgNO3+ KI = AgI + KNO3. K hiu keo s l:
a. [ mAgI nNO3-(n-x)Ag+].xAg+.
b. [ mAgI nAg+(n-x)NO3-].xNO3-.
c. [ mAgI nAg+(n+x)NO3-].xNO3-.d. [ mAgI nNO3-(n+x)Ag+].xAg+.
206. Keo hydronol st (III) c iu ch bng cch cho
t tFeCl3vo nc si. K hiu ca keo l:
a. [ mFe(OH)3. nFe3+( 3nx) Cl-].xCl-
b. [ mFe(OH)3. Fe3+( 3nx) Cl-].xCl-
c. [ mFe(OH)3. nFe3+( 3n + x) Cl-].xCl-
d. [ mFe(OH)3. nFe3+( n - x) Cl-].xCl-
207. Keo hydronol st (III) c iu ch bng cch cho
t tFeCl3vo nc si. Ion to th l:
a. Cl-
b. Fe3+c. OH-
d. H+
208. Keo hydronol st (III) c iu ch bng cch cho
t t FeCl3vo nc si. Ht keo mang in tch
l:
a. m
b. dng
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213. Nu dung dch keo c kch thc trung bnh ca
ht phn tn l 20A0, phn tn ca dung dch
keo ny l:
a. 0,05
b. 0,005
c. 200
d. 0,2
214. iu ch dung dch keo n phn tn bng
phng php ngng t t dung dch thc, th mi
quan h gia tc to mm (V1) v tc phttrin mm (V2) phitha mn iu kin sau:
a. V1> V2
c. V1= V2
d. V1V2
215. H keo ch c kh nng phn tn nh sng khi mi
quan h gia bc sng nh sng () v ng
knh ht phn tn (d) tha mn iu kin sau:
a. d
b. = d
c. < dd. > d
216.nh sng b phn tn mnh qua h keo khi n c
bc sng nh sng :
a. ln
b. trung bnh
c. nh
d. a, b, c u ng
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217. Ngng keo t l:
a. Nng ti a ca cht in ly cn thit
gy ra s keo t vi mt tc n nh.
b. Nng ti thiu ca cht in ly cn thit
gy ra s keo t vi mt tc n nh.
c. Nng ti thiu ca cht phn tn cn thit
gy ra s keo t vi mt tc n nh.
d. Nng ti a ca cht phn tn cn thit
gy ra s keo t vi mt tc n nh.
218. Cc tnh cht in hc ca h keo bao gm:a. tnh cht in di v inthm
b. tnh chy v sa lng
c. tnh cht in di v sa lng
d. a, b u ng.
219. Trong cc mi tng quan gia cc p sut thm
thu ca cc dung dch sau y, mi tng quan
no l ng?
a. dd l tng > dd in ly > dd keo
b. dd l tng< dd keo < dd in ly
c. dd keo < dd l tng< dd in ly
d. dd l tng< dd in ly < dd keo220.nh lut tc dng khi lng ch c p dng
cho:
a. tc cht tham gia phn ng.
b. phn ng n gin, mt giai on.
c. phn ng nhiu giai on ni tip nhau.
d. a, b u ng.
221. Hng s tc phn ng ph thuc ch yu vo:
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a. nhit .
b. p sut.
c. nng .
d. th tch.
222. Chn pht biu ng:
a. ng ha hc l mt phn ca ha l nghin
cu v tc , c ch ca cc qu trnh ha
hc v cc yu t nh hng n tc qu
trnh ha hc.
b. ng ha hc nghin cu v chiu hng vgii hn ca cc qu trnh ha hc.
c. ng ha hc v nhit ng hc u c
phng php nghin cu ging nhau l u
da vo trng thi u v cui ca qu trnh.
d. ng ha hc nghin cu v chiu hng v
cc yu t nh hng n chiu hng v
gii hn ca qu trnh.
223. Chn pht biu ng:
a. Phn ng ng th l phn ng c cc cht
tham gia phn ng khng cng pha vi
nhau cn phn ng d th l phn ng nhiupha.
b. Phn ng ng th l phn ng c cc cht
tham gia phn ng cng pha vi nhau cn
phn ng d th l phn ng c cc cht
khc pha vi nhau.
c. Khi phn ng xy ra trong iu kin ng tch
v ng nhit th bin thin nng mt cht
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bt k tham gia phn ng trong 1 n v thi
gian c gi l tc phn ng.
d. b v c u ng.
224. Chn pht biu ng:
a. Cht xc tc l cht lm thay i vn tc phn
ng v bin i v cht khi phn ng xy ra.
b. Cht xc tc l cht lm thay i vn tc phn
ng v khng bin i v cht khi phn ng
xy ra.
c. Cht xc tc l cht lm thay i vn tc phnng v khng bin i v cht v lng khi
phn ng xy ra.
d. Cht xc tc l cht lm thay i vn tc phn
ng v khng bin i v lng khi phn ng
xy ra.
225. Xc tc lm tng vn tc phn ng v:
a. lm tng nng lng hotha ca phn ng.
b. lm gim nng lng hotha ca phn ng.
c. lm tng s phn t hotng.
d. lm gim s phn t hotng.
226. Nhit lm tng tc phn ng v:a. lm tng nng lng hotha ca phn ng.
b. lm gim nng lng hotha ca phn ng.