Bai Tap Hoa Ly Full-libre

Bi tp ha l c s rt gn Su tm v trnh by: Don Trng C

1

BI TP HO L C S

MC LC

Chng 1: Nguyn l I nhit ng hc..2 Chng 2: Nguyn l II nhit ng hc..7 Chng 3: Cn bng ha hc..13 Chng 4: Cn bng pha..22 Chng 5: Dung dch v cn bng dung dch - hi..27 Chng 6: Cn bng gia dung dch lng v pha rn........34 Chng 7: in ha hc...40 Chng 8: ng ha hc118 Chng 9: Hp ph v ha keo.........................................58


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Chng 1

NGUYN L I NHIT NG HC

1.1. Nguyn l I nhit ng hc 1.1.1. Nhit v cng Nhit v cng l hai hnh thc truyn nng lng ca h. Cng k hiu l A v nhit k hiu l Q.

Quy c du Cng A Nhit Q H sinh > 0 < 0 H nhn < 0 > 0

1.1.2. Nguyn l I nhit ng hc Biu thc ca nguyn l I nhit ng hc: U = Q - A Khi p dng cho mt qu trnh v cng nh: dU = Q - A dng tch phn nguyn l I c th c vit: 2

1

V

V

PdVQU 1.1.3. p dng nguyn l I cho mt s qu trnh. 1.1.3.1. Qu trnh ng tch: V = const, dV = 0. 2

1

V

Vv 0PdVA

T ta c: QV = U 1.1.3.2. Qu trnh ng p: P = const, dP = 0. Ap = P.(V2 - V1) = P.V

Do : Qp = U + PV = (U + PV) = H 1.1.3.3. Qu trnh ng p ca kh l tng T phng trnh trng thi kh l tng: PV = nRT

Ta c: Ap = PV = nRT Up = Qp nRT 1.1.3.4. Qu trnh dn n ng nhit ca kh l tng

Bin thin ni nng khi dn n ng nhit (T = const) kh l tng l bng khng nn:

2

1

1

2TT P

PnRTln

VV

nRTlnAQ Trong : P1: p sut trng thi u. P2: p sut trng thi cui. 1.1.3.5. Nhit chuyn pha

TQ cp

Trong : cp: nhit chuyn pha (cal hoc J) nc = -, hh = -ngt

Ghi ch: R l hng s kh l tng v c cc gi tr sau:

R = 1,987 cal/mol.K = 8,314 J/mol.K R = 0,082 lit.atm/mol.K 1 cal = 4,18 J; 1 l.atm = 101,3 J = 24,2 cal

1.2. nh lut Hess 1.2.1. Ni dung nh lut Trong qu trnh ng p hoc ng tch, nhit phn ng ch ph thuc vo trng thi u v trng thi cui m khng ph thuc vo cc trng thi trung gian. Biu thc ca nh lut Hess:

QV = U v Qp = H Trong : U: nhit phn ng ng tch. H: nhit phn ng ng p.

Khi qu trnh xy ra iu kin tiu chun ta c nhit phn ng tiu chun: H0298, U0298. i vi cc qu trnh xy ra khi c mt cc cht kh (c xem l kh l tng), ta c:

H = U + RTn Vi n l bin thin s mol kh ca qu trnh.1.2.2. Cc h qu ca nh lut Hess Nhit phn ng nghch bng nhng tri du vi nhit phn ng thun.

Hnghch = - Hthun


3

Nhit phn ng bng tng nhit sinh ca cc cht to thnh tr i tng nhit sinh ca cc cht tham gia phn ng.

H phn ng = Hssp - Hstc Nhit phn ng bng tng nhit chy ca cc cht tham gia phn ng

tr i tng nhit chy ca cc cht to thnh. H

phn ng = Hchtc - Hchsp Ghi ch: Nhit to thnh tiu chun (H0298, tt), nhit t chy tiu

chun (H0298,c) c cho sn trong s tay ha l. 1.3. Nhit dung 1.3.1. nh ngha Nhit dung ng p:

PPp T

HdPQC

Nhit dung ng tch: VV

v TU

dTQC

Mi lin h: Cp - Cv = R Nhit lng Q c tnh: 21

T

T

CdTmQ hoc 21

T

T

CdTnQ

1.3.2. nh hng ca nhit n nhit dung S ph thuc vo nhit ca nhit dung c biu din bng cc

cng thc thc nghim di dng cc hm s: Cp = a0 + a1.T + a2.T2 Hoc Cp = a0 + a1.T + a-2.T-2

Trong : a0, a1, a2, a-2 l cc h s thc nghim c th tra gi tr ca chng trong s tay ha l. 1.2.2. nh lut Kirchhoff

Hiu ng nhit ca phn ng ph thuc vo nhit c biu din bi nh lut Kirchhoff:

pP

CT

H Hoc

v

VC

TU

Sau khi ly tch phn ta c:

T0

p0T dTCHH Nu ly tch phn t T1 n T2 ta c: 2

1

12

T

TpTT dTCHH

1.4. Bi tp mu V d 1: Tnh bin thin ni nng khi lm bay hi 10g nc 200C. Chp nhn hi nc nh kh l tng v b qua th tch nc lng. Nhit ha hi ca nc 200C bng 2451,824 J/g.

Gii Nhit lng cn cung cp lm ha hi 10g nc l:

Q = m. = 10. 2451,824 = 24518,24 (J) Cng sinh ra ca qu trnh ha hi l:

A = P.V = P(Vh - Vl) = PVh = 1353,332938,314

1810

nRT (J)

Bin thin ni nng l: U = Q A = 23165 (J) V d 2: Cho 450g hi nc ngng t 1000C di p sut khng i 1 atm. Nhit ha hi ca nc nhit ny bng 539 cal/g. Tnh A, Q v U ca qu trnh.

Gii Nhit lng ta ra khi ngng t l:

Q = m.ng. t = 450. (- 539) = - 242550 (cal) Cng ca qu trnh:

A = P.V = P. (Vl - Vh) = - P.Vh = - nRT = 18529(cal)3731,987

18450

Bin thin ni nng ca qu trnh l: U = Q A = - 224021 (cal) V d 3: Cho phn ng xy ra p sut khng i:

2H2 + CO = CH3OH(k)


4

nhit to thnh tiu chun 298K ca CO v CH3OH(k) bng -110,5 v -201,2 kJ/mol. Nhit dung mol ng p ca cc cht l mt hm ca nhit : Cp (H2) = 27,28 + 3,26.10-3T (J/mol.K) Cp (CO) = 28,41 + 4,1.10-3T (J/mol.K) Cp (CH3OH)k = 15,28 + 105,2.10-3T (J/mol.K)

Tnh H0 ca phn ng 298 v 500K? Gii

Nhit phn ng 298K l: H0298 = - 201,2 - (-110,5) = - 90,7 (KJ) Bin thin nhit dung: Cp = Cp(CH3OH) Cp(CO) 2Cp(H2) = - 67,69 + 94,58. 10-3T (J/K) Nhit phn ng 500K l : 500

298p

0298

0500 dTCHH

500298

33 dTT94,58.1067,6990,7.10

= - 96750,42 (J) V d 4: Cho 100g kh CO2 (c xem nh l kh l tng) 00C v 1,013.105 Pa. Xc nh Q, A, U v H trong cc qu trnh sau. Bit Cp = 37,1 J/mol.K. a. Dn n ng nhit ti th tch 0,2 m3. b. Dn ng p ti 0,2 m3. c. un nng ng tch ti khi p sut bng 2,026.105 Pa.

Gii a. Dn n ng nhit (T = const) ti th tch 0,2m3.

nRTPV

nRTlnVV

nRTlnAQ 21

2TT

70612730,082

44100

0,2.101273.ln8,31444

100 3 (J) U = 0

b. Dn n ng p (P = const) ti 0,2m3.

H = Qp = n.Cp. (T2 T1) nRPVnRPVn.C 12p

1

2730,08244

100

0,2.1010,08237,1 3

JA = PV = P(V2 V1) J15120

0,0828,314

1

2730,08244

100

0,2.101 3

U = Q A = 67469 - 15120 = 52349 (J) c. un nng ng tch (V = const) ti p sut bng 2,026.105Pa (2 atm)

A = 0 Cv = Cp - R = 37,1 - 8,314 = 28,786 (J/mol.K)U = Qv = n.Cv.(T2 T1) Ta c:

1

1

2

2

TP

TP

546K27312T

PPT 1

1

22

Suy ra: U = Qv = 1 28,786(546 - 273) = 7859 (J) H = U + PV = 7859 (J) V d 5: Mt kh l tng no c nhit dung mol ng tch mi nhit c Cv = 2,5R (R l hng s kh). Tnh Q, A, U v H khi mt mol kh ny thc hin cc qu trnh sau y: a. Dn n thun nghch ng p p sut 1atm t 20dm3 n 40dm3. b. Bin i thun nghch ng tch t trng thi (1atm; 40dm3) n

(0,5atm; 40dm3). c. Nn thun nghch ng nhit t 0,5 atm n 1 atm 250C.

Gii a. Dn n thun nghch ng p (P = const).


5

Tnh cng A: l.atm2020401.VVPPdVA 21

V

V12

20280,0828,31420 (J)

Tnh nhit lng Q: RVPRVPCTT.CdTCQ 12p12pTT pp 21 702040

R3,5R (l.atm)

70970,0828,31470 (J)

Bin thin ni nng: U = Q A = 5069 (J) Bin thin entapy H = Qp = 7097 (J)

b. Dn n thun nghch ng tch (V = const). A = 0 Nhit lng: RVPRVPCTT.CdTCQ 12v12vTT vv 21 5010,540

R2,5R (l.atm)

50690,0828,31450 (J)

U = Qv = - 5069 (J) c. Nn ng nhit (T = const) U = 0

171715,0ln298314,81

PP

nRTlnAQ2

1TT (J)

V d 6: Tnh nhit to thnh ca etan bit: Cgr + O2 = CO2 H0298 = -393,5 KJ H2 + 1/2O2 = H2O(l) H0298 = -285 KJ 2C2H6 + 7O2 = 4 CO2 + H2O(l) H0298 = -3119,6 KJ

Gii Cgr + O2 = CO2 (1) H2 + 1/2O2 = H2O(l) (2) 2C2H6 + 7O2 = 4CO2 + 6H2O(l) (3)

Nhit to thnh C2H6 l: 2C + 3H2 = C2H6 (4) H0298(4) = 4H0298(1) + 6H0298(2) - H0298(3) H0298(4) = 4(-393,5) + 6(-285) - (-3119,6) = 164,4 (KJ)

V d 7. Tnh Q, A, U ca qu trnh nn ng nhit, thun nghch 3 mol kh He t 1atm n 5 atm 4000K.

Gii Nhit v cng ca qu trnh:

16057(J)51400ln8,3143

PP

nRTlnAQ2

1TT

U = 0 V d 8. Cho phn ng: 1/2N2 + 1/2O2 = NO. 250C, 1atm c H0298 = 90,37 kJ. Xc nh nhit phn ng 558K, bit nhit dung mol ng p ca 1 mol N2, O2 v NO ln lt l 29,12; 29,36 v 29,86 J.mol-1.K-1.

Gii Hiu ng nhit ca phn ng 558K l: 558

298p

0298

0558 dTCHH

Trong : Cp = 29,86 1/2(29,12) 1/2(29,36) = 0,62 (J.K-1) H0558 = 90,37 + 0,62.(558 - 298).10-3 = 90,5312 (KJ) 1.5. Bi tp t gii 1. Xc nh bin thin ni nng khi lm ha hi 20g etanol ti nhit

si, bit nhit ha hi ring ca etanol bng 857,7 J/g v th tch hi ti nhit si bng 607 cm3/g (b qua th tch pha lng).

S: 2,54 kJ 2. Tnh H v U cho cc qu trnh sau y:

a. Mt mol nc ng c 00C v 1 atm;


6

b. Mt mol nc si 1000C v 1 atm. Bit rng nhit ng c v nhit ha hi ca 1 mol nc bng -6,01

kJ v 40,79 kJ, th tch mol ca nc v nc lng bng 0,0195 v 0,0180 lit. Chp nhn hi nc l kh l tng.

S: a. H = U = -6,01 kJ b. H = 37,7 kJ; U = 40,79 kJ

3. Nhit sinh ca H2O(l) v ca CO2 ln lt l -285,8 v -393,5 kJ/mol 250C, 1 atm. Cng iu kin ny nhit t chy ca CH4 bng -890,3 kJ/mol. Tnh nhit to thnh ca CH4 t cc nguyn t iu kin ng p v ng tch.

S: -74,8 kJ/mol; 72,41 kJ/mol 4. Tnh nhit to thnh chun ca CS2 lng da vo cc d liu sau: S(mon) + O2 = SO2 H1 = -296,9 kJ CS2(l) + 3O2 = CO2 + 2SO2 H2 = -1109 kJ C(gr) + O2 = CO2 H3 = -393,5 kJ S: 121,7 KJ 5. Trn c s cc d liu sau, hy tnh nhit to thnh ca Al2Cl6 (r)

khan: 2Al + 6HCl(l) = Al2Cl6(l) + 3H2 H0298 = -1003,2 kJ H2 + Cl2 = 2HCl(k) H0298 = -184,1 kJ HCl(k) = HCl(l) H0298 = -72,45 kJ Al2Cl6(r) = Al2Cl6(l) H0298 = -643,1 kJ S: 1347,1 kJ 6. Tnh nhit phn ng: H2(k) + S(r) + 2O2(k) + 5H2O(l) = H2SO4.5H2O(dd)

Bit nhit sinh ca H2SO4(l) l -193,75 Kcal/mol v nhit ha tan H2SO4(l) vi 5 mol nc l -13,6 Kcal.

S: -207,35 Kcal 7. Cho 100 gam kh nit iu kin chun (1atm, 250C), CP(N2) = 3,262

cal/mol.K. Tnh gi tr ca cc i lng Q, A v U trong cc qu trnh sau: a. Nn ng tch ti 1,5 atm. b. Dn n ng p ti th tch gp i th tch ban u. c. Dn n ng nhit ti th tch 200lt. d. Dn n on nhit ti th tch 200lt.

S: a. Qv = 2424 cal; b. QP = 8786 cal, AP = 1937 cal c. QT = AT = 1775 cal; d. U = A = 1480 cal

8. 250C phn ng tng hp NH3. N2(k) + 3H2(k) = 2NH3(k)

H0298,tt (kcal/mol) 0 0 -11,04 V nhit dung ca cc cht:

CP (N2) = 6,65 + 10-3T (cal.mol-1.K-1) CP (H2) = 6,85 + 0,28.10-3T (cal.mol-1.K-1) CP (NH3) = 5,92 + 9,96.10-3T (cal.mol-1.K-1)

Xc nh hm s H0T = f(T) v tnh H01000 ca phn ng? S: H0T = -18,22 15,36.10-3T + 8.10-6T2 (Kcal) H0 = -25,58 Kcal


7

Chng 2

NGUYN L II NHIT NG HC

2.1. Nguyn l II nhit ng hc 2.1.1. nh ngha entropy

Trong qu trnh thun nghch, bin thin entropy khi chuyn h t trng thi 1 sang trng thi 2 c xc nh bng phng trnh:

TQdS

hay TQS TN Entropy c o bng n v cal.mol-1.K-1 hay J.mol-1.K-1 2.1.2. Biu thc ton ca nguyn l II

TQdS

Du = khi qu trnh l thun nghch. Du > khi qu trnh l bt thun nghch. 2.1.3. Tiu chun xt chiu trong h c lp

Trong h c lp (on nhit) Nu dS > 0 : Qu trnh t xy ra Nu dS = 0 hay d2S < 0: Qu trnh t cn bng 2.1.4. Bin thin entropy ca mt s qu trnh thun nghch 2.1.4.1. Qu trnh ng p hoc ng tch 2

1

T

T TdTCS

Nu qu trnh ng p: 21

T

Tp T

dTCS

Nu qu trnh ng tch: 21

T

Tv T

dTCS

2.1.4.2. Qu trnh ng nhit Trong qu trnh thun nghch ng nhit, ta c th p dng:

TQS T

i vi qu trnh chuyn pha nh qu trnh nng chy, qu trnh ha hi

T

THS T

nc

ncnc T

S hay hh

hhhh T

S i vi kh l tng:

1

2T V

VnRTlnQ

Ta c: 2

1

1

2T

PP

nRlnVV

nRlnT

QS Bin thin entropy nhit bt k c th tnh bng phng trnh: nc

chph

2

chph

1

T

T nc

ncRp

chph

chphT

0

RpT T

T

dTCT

TdTCS

Thh

hh

nc T

kp

hh

hhT

T

lp T

dTCT

TdTC

hoc TTdTCS pT Trong :

1RpC : nhit dung trng thi rn 1


8

2RpC : nhit dung trng thi rn 2

Bin thin entropy tiu chun ca cc phn ng c xc nh bng phng trnh: 0298(tc)0298(sp)0298 SSS 2.2. Th nhit ng

Cc th nhit ng bao gm: ni nng, entapy, nng lng t do v th ng p.

Nng lng t do F v th ng p G c nh ngha bi cc phng trnh sau: F = U - TS G = H - TS Ti mt nhit xc nh, bin thin th ng p v ng tch c biu din bng phng trnh sau: F = U - TS G = H - TS V G = Gcui - Gu F = Fcui - Fu Th ng p to thnh tiu chun ca cc cht (G0298) c th tra trong s tay ha l.2.2.1. Xt chiu trong h ng nhit, ng p

Trong h ng nhit, ng p Nu dG < 0 : Qu trnh t xy ra Nu dG = 0 hay d2G > 0 : Qu trnh t cn bng 2.2.2. Xt chiu trong h ng nhit, ng tch

Trong h ng nhit, ng tch Nu dF < 0 : Qu trnh t xy ra Nu dF = 0 hay d2F > 0 : Qu trnh t cn bng 2.3. Bi tp mu V d 1. Tnh bin thin entropy khi un nng thun nghch 16 kg O2 t 273K n 373K trong cc iu kin sau: a. ng p b. ng tch

Xem O2 l kh l tng v nhit dung mol Cv = 3R/2. Gii

a. i vi qu trnh ng p Cp = Cv + R = 5R/2

cal/K7752733731,987.ln

25

3216.10

TdTCnS

3T

Tp

2

1

b. i vi qu trnh ng tch cal/K465

2733731,987.ln

23

3216.10

TdTCnS

3T

Tv

2

1

V d 2. Xc nh nhit lc cn bng nhit v bin thin entropy khi trn 1g nc 00C vi 10g nc 1000C. Cho bit nhit nng chy ca bng 334,4 J/g v nhit dung ring ca nc bng 4,18 J/g.K.

Gii Gi T (K) l nhit ca h sau khi trn. Gi s h l c lp.

Ta c phng trnh: Nhit lng ta ra = Nhit lng thu vo

- Qta = Qthu hay Q3 = Q1 + Q2 - 10.4,18.(T - 373) = 334,4 + 1.4,18.(T - 273) T = 356,64 (K) Bin thin entropy ca h: S = S1 + S2 + S3

Vi: 1,225(J/K)273

334,4TS

nc

nc1

1,117(J/K)TdT4,181.S

356,64

2732

1,875(J/K)TdT4,1810.S

356,64

3733 S = 0,467 (J/K)

V d 3. Tnh bin thin entropy ca qu trnh nn ng nhit, thun nghch. a. 1 mol oxy t P1 = 0,001atm n P2 = 0,01atm. b. 1 mol mtan t P1 = 0,1 atm n P2 = 1 atm. Trong hai trng hp trn kh c xem l l tng.

Gii


9

a. K)4,575(cal/11,987.ln0,PP

nRlnS2

1

b. K)4,575(cal/11,987.ln0,PP

nRlnS2

1

V d 4. Xc nh bin thin entropy ca qu trnh chuyn 2g nc lng 00C thnh hi 1200C di p sut 1 atm. Bit nhit ha hi ca nc 1000C l 2,255 (kJ/g), nhit dung mol ca hi nc Cp,h = 30,13 + 11,3.10-3T (J/mol.K) v nhit dung ca nc lng l Cp,l = 75, 30 J/mol K.

Gii

Bin thin etropy ca qu trnh S = S1 + S2 + S3 Vi 2,61(J/K)

TdT75,3

182S

373

2731

12,09(J/K)37322552S2 0,2(J/K)

TdTT11,3.1030,13

182S

393

373

3-1

S = 14,9 (J/K) V d 5. Mt bnh kn hai ngn, ngn th nht c th tch 0,1 m3 cha oxi, ngn th hai c th tch 0,4 m3 cha Nit. Hai ngn u cng mt iu kin nhit l 170C v p sut 1,013.105 N/m2. Tnh bin thin entropy khi cho hai kh khuch tn vo nhau.

Gii Khi hai kh khuch tn vo nhau, th tch ca hn hp V2 = 0,5 m3 Bin thin entropy ca h: S =S1 + S2

Vi S1: bin thin entropy ca kh Oxy khi khuch tn S2: bin thin entropy ca kh Nit khi khuch tn

K)13,32(cal/VV

nR.lnS1

21

)7,46(cal/KVV

nR.lnS'

1

22

Vy S = 20,78 (cal/K) V d 6. Tnh U, H v S ca qu trnh chuyn 1 mol H2O lng 250C v 1 atm thnh hi nc 1000C, 1 atm. Cho bit nhit dung mol ca nc lng l 75,24 J/mol.K v nhit ha hi ca nc l 40629,6 J/mol.

Gii

Nhit lng cn cung cp

hh

373

29821p 75,24dTQQQ

)46272,69(J40629,6298)75,24(373Qp Cng ca qu trnh

J3101,13738,3141nRTVP0AAA 221 Ni nng U = Q A = 43171,5 (J) H = Qp = 4627,6 (J) Bin thin entropy ca qu trnh

hh

hh373

298p21 T

T

dTCSSS J/K 125,8

37340629,6

29837375,24ln

V d 7. Cho phn ng c cc s liu sau: 3Fe(r) + 4H2O(h) = Fe3O4(r) + 4H2(k)

H0298 t.t (Kcal/mol)

0 -57,8 -267 0


10

S0298 (cal/mol.K)

6,49 45,1 3,5 32,21

Cp(Fe) = 4,13 + 6,38.10-3.T (cal/mol.K) Cp(H2Oh) = 2,7 + 1.10-3.T (cal/mol.K)

Cp(Fe3O4) = 39,92 + 18,86.10-3.T (cal/mol.K) Cp(H2) = 6,95 - 0,2.10-3.T (cal/mol.K) a. Tnh hiu ng nhit ng p v ng tch 250C v 1atm? b. Tnh hiu ng nhit ng p v ng tch 1000K? c. Xt chiu phn ng 250C v 1atm?

Gii Phn ng: 3Fe(r) + 4H2O(h) = Fe3O4(r) + 4H2(k) a. Tnh H0298 = -267 - 4.(-57,8) = - 35,8 Kcal.

Tnh U0298 = H0298 - nR.T vi n = 4 - 4 = 0 Do U0298 = H0298 = -35,8 Kcal b. Tnh H01000 = H0298 + 1000

298

Cp.dT Cp = [4.Cp(H2) + Cp(Fe3O4)] [4.Cp(H2O) + 3.Cp(Fe)] Cp = 44,53 - 5,08.10-3.T Ta c:

H01000 = -35800 + 1000298

3.T)dT5,08.10(44,53

= - 6854,37 (cal) U01000 = H01000 - nRT vi n = 4 - 4 = 0 U01000 = H01000 = - 6854,37 (cal) c. Xt chiu phn ng ktc t cng thc: G0298 = H0298 T.S0298. Trong : S0298 = (4x32,21 + 35) (4x45,1 + 3x6,49)

= - 36,03 (cal) G0298 = -35800 + 298x36,03 = - 25063,06 (cal) V: G0298 < 0 nn phn ng t din bin.

2.4. Bi tp t gii 1. Tnh bin thin entropy ca qu trnh un nng ng p 1 mol KBr t

298 n 500K, bit rng trong khong nhit : Cp(KBr) = 11,56 + 3,32.10-3T cal/mol.

S: 6,65 cal/mol.K

2. Tnh bin thin entropy ca qu trnh un nng 2 mol Nit (c xem l l tng) t 300K n 600K di p sut kh quyn trong 2 trng hp: a. ng p b. ng tch Bit rng nhit dung Cp ca Nit trong khong nhit 300 - 600K c cho bng phng trnh: Cp = 27 + 6.10-3T (J/mol.K). S: 41 J/K; 29,5 J/K

3. Tnh bin thin entopy ca qu trnh trn 10g nc 00C vi 50g nc lng 400C trong h c lp. Cho bit nhit nng chy ca nc bng 334,4 J/g, nhit dung ring ca nc lng bng 4,18 J/g.

4. Tnh bin thin entropy ca phn ng: 4 Fe + 3O2 = 2Fe2O3.

Cho bit S0298 ca Fe, O2 v Fe2O3 tng ng bng 27,3; 205 v 87,4 J/mol.K. 5. Hy d on du ca S trong cc phn ng sau:

a. CaCO3(r) = CaO(r) + CO2(r) b. NH3(k) + HCl(k) = NH4Cl(r) c. BaO(r) + CO2(k) = BaCO3(r)

S: a. S > 0; b. S < 0; c. S < 0 6. Tnh 0298G khi to thnh 1 mol nc lng bit cc gi tr entropy tiu

chun ca H2, O2 v H2O ln lt bng 130; 684; v 69,91 J/mol.K v nhit to thnh nc lng 250C l -285,83 KJ/mol.

S: 0298G = -237,154 kJ 7. Tnh 0298S , 0298H v 0298G ca phn ng phn hy nhit CaCO3

bit: CaCO3 = CaO + CO2

S0298 (J/mol.K) 92,9 38,1 213,7 KJ/mol)(H 0tt,298 -1206,90 -635,10 -393,50

S: o298S = 158,9 J/K; o298H = 178,30 kJ; o298G = 130,90 kJ


11

8. Cho phn ng: CO(k) + H2O(k) = CO2(k) + H2(k), c nhng gi tr bin thin entanpy v bin thin entropy tiu chun 300K v 1200K nh sau:

KJ/mol41,16H0300 KJ/mol32,93H01200

J/K42,40S0300 J/K29,60S01200 Phn ng xy ra theo chiu no 300K v 1200K?

S: J2590GKJ;28,44G 012000300 9. Cho phn ng: CH4(k) + H2O(k) = CO(k) + 3H2(k). Cho bit nhit to thnh chun ca CH4(k), H2O(h) v CO(k) ln lt l -74,8; -241,8; -110,5 KJ/mol. Entropy tiu chun ca CH4(k), H2O(h) v CO(k) ln lt l 186,2; 188,7 v 197,6 J/mol.K. (Trong tnh ton gi s H0 v S0 khng ph thuc nhit ).

a. Tnh G0 v xt chiu ca phn ng 373K. b. Ti nhit no th phn ng t xy ra.

S: a. G0= 1,26.105J/mol; b. T> 961K 10. Cho phn ng v cc s liu sau:

COCl2(k) = Cl2(k) + CO(k)

H0298 t.t (Kcal/mol) - 53,3 0 -26,42 S0298 (cal/mol.K) 69,13 53,28 47,3

Cp(CO) = 6,96 (cal /mol.K) Cp(COCl2) = 14,51 (cal /mol.K) Cp(Cl2) = 8,11 (cal /mol.K)

a. Tnh hiu ng nhit ng p v ng tch ca phn ng 250C?

b. Xt chiu phn ng 250C? c. Tnh hiu ng nhit ng p ca phn ng 1000K?

S: a. H0 = 26,88 Kcal, U0 = 26287,87 cal b. S0 = 31,45 cal/K, G0 = 17507,9 cal c. H0 = 26486,88 cal

11. Tnh nhit lng cn thit lm nng chy 90 gam nc 00C v sau nng nhit ln 250C. Cho bit nhit nng chy ca nc 00C l 1434,6 cal/mol, nhit dung ca nc lng ph thuc vo nhit theo hm s: Cp = 7,20 + 2,7.10-3T (cal.mol-1.K-1).

S: Q = 8169,4 cal

12. Tnh bin thin entropy ca qu trnh ng c benzen di p sut 1atm trong 2 trng hp: a. ng c thun nghch 50C bit nhit ng c ca benzen l

-2370 cal/mol. b. ng c bt thun nghch -50C. Bit nhit dung ca Benzen lng v rn ln lt l 30,3 v 29,3

cal/mol.K. S: a. S = 0 cal/K ; b. S = 0,31 cal/K

13. Cho phn ng v cc s liu sau: FeO(r) + CO(k) = CO2(k) + Fe(r)

H0298 t.t (Kcal/mol)

-63,7 -26,42 -94,052 0

S0298 (cal/mol.K)

1,36 47,3 51,06 6,49

Cp(Fe) = 4,13 + 6,38.10-3.T (cal/mol.K) Cp(CO) = 6,34 + 1,84. 10-3.T (cal/mol.K) Cp(FeO) = 12,62 + 1,50.10-3.T (cal/mol.K) Cp(CO2) = 10,55 + 2,16.10-3.T (cal/mol.K)

a. Tnh hiu ng nhit ng p v ng tch ca phn ng 2980K?

b. Tnh hiu ng nhit ng p v ng tch ca phn ng 10000K?

c. Xt chiu phn ng iu kin tiu chun. d. Xt chiu phn ng 1000K xem entropy khng thay i theo

nhit . S: a. H0298 = U0298 = -3932 cal

b. H01000 = U01000 = -4567 cal

14. Cho phn ng v cc s liu sau: C(r) + CO2(k) = 2CO(k)

S0298 (cal/mol.K) 1,36 51,06 47,3 0298 (Kcal/mol) 0 -94,052 -26,42 Cp(CO) = 6,96 (cal /mol.K) Cp(Cgr) = 2,07 (cal /mol.K) Cp(CO2) = 8,88 (cal /mol.K)


12

a. Tnh hiu ng nhit ng p v ng tch ca phn ng 250C v 1atm.

b. Xt chiu phn ng 250C v 1atm. c. Tnh hiu ng nhit ng p ca phn ng 1000K.

S: a. H0298 = 41212 cal; U0298 = 40619 cal c. H01000 = 43297 cal


13

Chng 3

CN BNG HA HC

3.1. Hng s cn bng 3.1.1. Cc loi hng s cn bng

Phn ng: aA(k) + bB(k) cC(k) + dD(k) Hng s cn bng tnh theo p sut :

cbbB

a

A

dD

c

CP

.PP

.PPK Hng s cn bng tnh theo nng mol/l:

cbbB

a

A

dD

c

CC

.CC.CCK

Hng s cn bng tnh theo phn mol:

cbbB

a

A

dD

c

Cx

.xx

.xxK Hng s cn bng tnh theo s mol:

cbbB

a

A

dD

c

Cn

.nn

.nnK Mi quan h ca cc hng s cn bng: n

cbin

nx

nCP n

P.K.PKRT.KK

n l bin thin s mol kh ca h. n = (c + d) (a + b)Nu n = 0 ta c Kp = KC = Kx = Kn

3.1.2. Phng trnh ng nhit Vant Hoff Xt phn ng: aA(k) + bB(k) cC(k) + dD(k)

Ti nhit khng i, ta c: P0TT RTlnGG Vi P0T RTlnKG b

BaA

dD

cC

p.PP.PP

Trong : PA, PB, PC, PD l p sut ring phn ti thi im bt k

P

PK

T RTlnG Nu P > KP: phn ng xy ra theo chiu nghch Nu P < KP: phn ng xy ra theo chiu thun Nu P = KP: phn ng t cn bng Ch :

n

in

nx

nCP

n

P.P(RT) 3.2. Cn bng trong h d th 3.2.1. Biu din hng s cn bng Nu cc phn ng xy ra trong cc h d th m cc cht trong pha rn hoc pha lng khng to thnh dung dch th biu thc nh ngha hng s cn bng khng c mt cc cht rn v cht lng. V d: Fe2O3(r) + 3CO(k) = 2Fe(r) + 3CO2(k) Hng s cn bng:

3CO

3CO

P PP

K 2 3.2.2. p sut phn ly

p sut hi do s phn ly ca mt cht to thnh l c trng cho cht mi nhit c gi l p sut phn ly.

V d: CaCO3(r) = CaO(r) + CO2(k) p sut phn ly: PCO KP 2

3.2.3. phn ly phn ly l lng cht phn ly so vi lng cht ban u:

on

n

n: lng cht phn ly


14

no: lng cht ban u 3.3. Cc yu t nh hng n hng s cn bng 3.3.1. nh hng ca nhit n hng s cn bng

T phng trnh ng p Vant Hoff

2P

RTH

dTdlnK

Trong khong nhit nh t T1 n T2, xem H khng i. Ly tch phn 2 v, ta c: 12Tp Tp T1T1RHKKln 12

Nu phn ng thu nhit, H > 0 0dT

dlnKP : nh vy khi nhit tng, gi tr Kp cng tng, phn ng dch chuyn theo chiu thun.

Nu phn ng ta nhit, H < 0, 0dT

dlnKP : nh vy khi nhit tng, gi tr Kp s gim, phn ng dch chuyn theo chiu nghch. 3.3.2. nh hng ca p sut

Ti nhit khng i ta c: const.PKK nxp Nu n > 0: Khi tng p sut P, gi tr Pn

cng tng, do Kx gim, cn bng s dch chuyn theo chiu nghch. Nu n < 0: Khi tng p sut P, gi tr Pn

gim, do Kx tng, cn bng dch chuyn theo chiu thun. Nu n = 0: th Kp = Kx = const. Khi p sut chung P khng nh hng g n cn bng phn ng. 3.4. Bi tp mu V d 1. Hng s cn bng ca phn ng: CO(k) + H2O(h) CO2(k) + H2(k) 800K l 4,12. un hn hp cha 20% CO v 80% H2O (% khi lng) n 800K. Xc nh lng hydro sinh ra nu dng 1 kg nc.

Gii Gi x l s mol ca H2O tham gia phn ng.

CO + H2O CO2 + H2

28250

181000

0 0

x x x x

( x28250 ) ( x

181000 ) x x

V n = 0, ta c hng s cn bng: 4,12

x18

1000.x

28250

x

.nn

.nnKK

2

OHCO

HCOnP

2

22

Gii phng trnh ta c: x = 8,55 (mol) Vy khi lng H2 sinh ra: m = 17,1 (g)

V d 2. 2000C hng s cn bng Kp ca phn ng dehydro ha ru Isopropylic trong pha kh:

CH3CHOHCH3(k) H3CCOCH3(k) + H2 bng 6,92.104 Pa. Tnh phn ly ca ru 2000C v di p sut 9,7.104Pa. (Khi tnh chp nhn hn hp kh tun theo nh lut kh l tng).

Gii Gi a l s mol ban u ca CH3CHOHCH3.

x l s mol CH3CHOHCH3 phn ly, ta c: CH3CHOHCH3(k) H3CCOCH3(k) + H2

a 0 0 x x x

(a x) x x Tng s mol cc cht lc cn bng: xani xa P.xax.xnP.KK

n

cbinP vi n = 1

692,0xa

0,97.x22

2 x = 0,764a


15

Vy phn ly: 0,764a

x

V d 3. un nng ti 4450C mt bnh kn cha 8 mol I2 v 5,3 mol H2 th to ra 9,5 mol HI lc cn bng. Xc nh lng HI thu c khi xut pht t 8 mol I2 v 3 mol H2.

Gii Gi x l s mol H2 tham gia phn ng:

H2 + I2 2HI Ban u 5,3 8 0 Phn ng x x 2x

Cn bng (5,3 x) (8 x) 2x Theo bi: 2x = 9,5 x = 4,75 (mol)

Hng s cn bng:

50,49x8x5,3 4x.nn nK 2IH 2HIn 22 Hn hp 8 mol I2 v 3 mol H2.

H2 + I2 2HI

Ban u 3 8 0 Phn ng y y 2y Cn bng (3 y) (8 y) 2y

V nhit khng i nn hng s cn bng cng khng i: 50,49y8y3 4yK 2n y = 2,87 S mol HI to thnh: nHI = 5,74 (mol) V d 4. Hng s cn bng ca phn ng:

PCl3(k) + Cl2(k) PCl5(k) 500K l KP = 3 atm-1. a. Tnh phn ly ca PCl5 1atm v 8 atm. b. p sut no, phn ly l 10%.

c. Phi thm bao nhiu mol Cl2 vo 1mol PCl5 phn ly ca PCl5 8 atm l 10%.

Gii a. Tnh phn ly ca PCl5

Gi a l s mol PCl5 ban u l phn ly ca PCl5, ta c:

PCl5(k) PCl3(k) + Cl2(k) Ban u a 0 0 Phn ng a a a Cn bng a(1-) a a Ta c 1a P1a anPKK 22ninP Vi n = 1, ni = a(1+)

31

1P.

2

2 3P2 = 1 - 2

3P11

Vi P = 1 atm 0,5 Vi P = 8 atm 0,2 b. p sut no phn ly l 10%

Ta c 31

1P.

2

2 31

0,11.P0,1

2

2 P = 33 atm c. Lng Cl2 cn thm vo

Gi b l s mol Cl2 cn thm vo:


16

PCl5(k) PCl3(k) + Cl2(k)

Ban u 1 0 b Phn ng 0,1 0,1 0,1

Cn bng 0,9 0,1 (b + 0,1)

Ta c: n

inP

n

PKK 31

1,1b8

0,90,1b0,1. b = 0,5 (mol)

V d 5. C th iu ch Cl2 bng phn ng 4HCl(k) + O2 = 2H2O(h) + 2Cl2 Xc nh HSCB KP ca phn ng 3860C, bit rng nhit v p sut 1 atm, khi cho mt mol HCl tc dng vi 0,48 mol O2 th khi cn bng s c 0,402 mol Cl2.

Gii Gi x l s mol O2 tham gia phn ng.

Tng s mol lc cn bng: x1,48ni ; n = -1 Theo bi ta c: 2x = 0,402 x = 0,201 (mol)

4HCl(k) + O2 2H2O(k) + 2Cl2(k) 1 0,48 0 0 4x x 2x 2x

(1 - 4x) (0,48 - x) 2x 2x

Hng s cn bng: n

cbinP

n

P.KK

ncbi

4

22

Pn

P4x1.x0,48

2x.2xK 81,21,27910,196.0,2790,402K 144P (atm-1) V d 6. Cho Fe d tc dng vi hi nc theo phn ng: 3Fe + 4H2O(h) = Fe3O4(r) + 4H2 2000C nu p sut ban u ca hi nc l 1,315 atm, th khi cn bng p sut ring phn ca hydro l 1,255 atm. Xc nh lng hydro to thnh khi cho hi nc 3atm vo bnh 2 lit cha st d nhit .

Gii Gi x l s mol H2O tham gia phn ng:

3Fe + 4H2O(h) Fe3O4(r) + 4H2

1,315 0 x x

(1,315 - x) x Theo bi ta c: x = 1,255 (atm) Hng s cn bng:

54

4OH

4H

P 1,91.101,2551,3151,255

PP

K2

2 Gi x l p sut ring phn ca H2 lc cn bng:

3Fe + 4H2O(h) Fe3O4(r) + 4H2

3 0 x x

(3 x) x V nhit khng i nn hng s cn bng cng khng i:

544OH

4H

P 1,91.10x3

x

PP

K2

2


17

x = 2,863 (atm) S mol kh H2 sinh ra:

0,1484730,08222,863

RTP.V

nnRTPV (mol) Khi lng kh H2 sinh ra: 0,2960,1482m 2H (g) V d 7. p sut tng cng do phn ng nhit phn 2FeSO4(r) = Fe2O3(r) + SO2(k) + SO3(k) nhit 929K l 0,9 atm. a. Tnh hng s cn bng KP 929K ca phn ng. b. Tnh p sut tng cng khi cn bng nu cho d FeSO4 vo bnh c

SO2 vi p sut u l 0,6 atm 929K. Gii

a. Hng s cn bng: 0,20250,450,45.PPK

32 SOSOp (atm2) b. p sut tng cng:

Gi x l s mol ca SO3 sinh ra: 2FeSO4 Fe2O3(r) + SO2 + SO3

0,6 0 x x

(0,6 + x) x V nhit khng i nn hng s cn bng cng khng i: 2025,0x0,6x..PPK

32 SOSOp x2 + 0,6x - 0,2025 = 0 x = 0,24 (atm) p sut ca hn hp:

08,108424,0PPP32 SOSO (atm)

V d 8. Tnh HSCB KP 250C ca phn ng CO + 2H2 = CH3OH(k) bit rng nng lng t do chun Go i vi phn ng CO + 2H2 = CH3OH(l) bng -29,1 KJ/mol v p sut hi ca metanol 250C bng 16200 Pa.

Gii

CO + 2H2 = CH3OH(k) (1) CO + 2H2 = CH3OH(l) (2) Ta c:

2HCO

P(2)2

.PP1K

P(2)OHCH2HCO

OHCHP(1) .KP

.PPP

K3

2

3 Mt khc: P(2)0(2) RTlnKG

RT

GexpK

0(2)

P(2)

1261682988,314

29,1.10exp

3 (atm-3) Suy ra: 20177126168

1,013.1016200K 5P(1) (atm-2)

V d 9. Hng s cn bng 1000K ca phn ng: 2H2O(h) = 2H2 + O2 l KP = 7,76.10-21 atm.

p sut phn ly ca FeO nhit l 3,1.10-18 atm. Hy xc nh HSCB KP 1000K ca phn ng

FeO(r) + H2 = Fe(r) + H2O(h) Gii

2H2O(h) = 2H2 + O2 (1) 2FeO(r) = 2Fe(r) + O2 (2) FeO(r) + H2 = Fe(r) + H2O(h) (3)

Ta c: 2p. (3) = p. (2) - p. (1) G0(3) = G0(2) - G0(1) P(1)P(2)P(3) RTlnKRTlnK2RTlnK P(1)P(2)P(3) lnKlnK2lnK

P(1)

P(2)2P(3) K

KK

M: Kp(1) = 7,76.10-21 (atm)


18

Kp(2) = 3,1.10-18 (atm)

Suy ra: 19,997,76.103,1.10

KK

K 2118

P(1)

P(2)P(3)

V d 10. Cho phn ng: CuSO4.3H2O(r) = CuSO4(r) + 3H2O(h)

bit hng s cn bng KP 250C l 10-6atm3. Tnh lng hi nc ti thiu phi thm vo bnh 2 lt 25oC chuyn hon ton 0,01 mol CuSO4 thnh CuSO4.3H2O.

Gii Gi x l mol H2O thm vo:

CuSO4.3H2O(r) CuSO4(r) + 3H2O(h) Ban u 0,01 x Phn ng 0,01 0,03 Cn bng 0,00 (x - 0,03)

Tng s mol ti thi im cn bng: 0,03xn i (mol)

Hng s cn bng:

633ninp 101VRT0,03x 1nPKK 21010,03x 2980,082 2 23,08.10x (mol) V d 11. Cho kh COF2 qua xc tc 1000oC s xy ra phn ng

2COF2(k) CO2 + CF4(k) Lm lnh nhanh hn hp cn bng ri cho qua dung dch Ba(OH)2 hp thu COF2 v CO2 th c 500 ml hn hp cn bng s cn li 200ml khng b hp thu. a. Tnh HSCB KP ca phn ng. b. Bit KP tng 1% khi tng 1oC ln cn 1000oC, tnh Ho, So v Go

ca phn ng 1000oC. Gii

a. Tnh HSCB KP ca phn ng

Gi x l s mol COF2 tham gia phn ng: 2COF2(k) CO2 + CF4(k) a x

2x

2x

(a x) 2x

2x

Tng s mol lc cn bng: an i Ta c: iCFiCF VVnn 44 500200a2

x 54ax

V n = 0, hng s cn bng:

425a254a

xa

2x

n

.nnKK 2

2

2

2

2COF

CFCOnP

2

42

b. Kp tng 10% khi tng 1oC ln cn 1000oC. Hng s cn bng KP 1001oC:

Kp = 4 + 0,04 = 4,04

Ta c: 1201000

)p(T

)p(T

T1

T1

RH

KK

ln1

2

32065

12731

12741

44,041,987ln

T1

T1

KK

RlnH

12

)p(T

)p(T

01000

1

2 (cal)

3507ln412731,987RTlnKG p(1000)01000 (cal) Ta li c: G01000 = H01000 - TS01000


19

27,941273

350732065T

GHS01000

010000

1000 (cal/K) V d 12. 1000K hng s cn bng ca phn ng:

C(gr) + CO2(k) 2CO(k) l Kp =1,85 atm v hiu ng trung bnh l 41130 cal. Xc nh thnh phn pha kh cn bng ti 1000K v 1200K bit p sut tng cng l 1atm.

Gii 1000K: gi xCO v 2COx l phn mol ca cc kh cn bng: Ta c, hng s cn bng: nx .KK vi n = 2 1 =1

Suy ra: 2CO

2CO

x

xK

M: 1xx2COCO COCO x1x 2

CO

2CO

x1

xK Vy 0K.xKx CO2CO (1) Vi Kp = 1,85 atm 01,851,85.xx CO2CO

Gii phng trnh ta c: xCO= 0,72 v 2COx = 0,28.

12,, 11RKKln 12 10001120011,987411301,85Kln ,1200 Ta tnh c:

KP,1200 = 58,28 atm Thay vo phng trnh (1) c: 058,2858,28.xx CO2CO

Gii phng trnh ta c: xCO = 0,98 2COx = 0,02 V d 13. Cho cc d kin sau:

CO CO2 Pb PbO H0298,tt (KJ/mol) -110,43 -393,13 0 -219,03 G0298 (KJ/mol) -137,14 -394,00 0 -189,14 Cp,298 (J/mol.K)

29,05 36,61 26,50 46,27

Chp nhn nhit dung khng thay i trong khong nhit 25 -1270C. a. Tnh G0, H0, Kp 250C ca phn ng:

PbO(r) + CO(k) = Pb(r) + CO2(k) b. Biu th 0 = f(T) di dng mt hm ca nhit . c. Tnh Kp 1270C.

Gii a. Tnh G0, H0, Kp 250C ca phn ng: H0298 = -393,13 + 0 + 110,43 + 219,03 = -63,67 (KJ) G0298 = -394 + 0 + 137,14 + 198,14 = -67,72 (KJ) Hng s cn bng: 27,33

2988,31467,72.10

RTGlnK

30298

P Kp = 7,4.1011 (atm) b. Biu th 0 di dng mt hm ca T. Cp,298 = 36,61 + 26,50 29,05 46,27

= - 12,21 (J/K) 298

0298

0 )21,12( d

)298(21,1263670)21,12(63670298 d

21,1242,60031 (J) c. Tnh Kp 1270C

12TP,TP, 11Rln 12


20

6,55298

1400

18,31463670

KK

ln1

2

Tp,

Tp, 9116,55

,400 1,055.107,4.10e (atm) V d 14. Cho phn ng v cc s liu tng ng sau:

Ckc(r) + 2H2(k) = CH4(k) H0298 (Kcal/mol)

0,453 0 -7,093

S0298 (cal/mol.K)

0,568 31,21 44,50

Cp (cal/mol. K)

2,18 6,52 4,170

a. Hy xc nh G0298 v Kp298 ca phn ng trn. b. 250C khi trn 0,55 mol kh CH4 vi 0,1 mol kh H2 trong bnh cha

Ckc rn (d), th phn ng xy ra theo chiu no nu p sut tng cng gi khng i 1 atm? Gii thch.

c. Kh H2 c nn vo bnh c cha Ckc rn d iu kin p sut 1 atm v nhit 298K. Hy xc nh p sut ring phn ca CH4 khi cn bng nhit p sut trn.

d. Thit lp phng trnh H0 = f(T) (phng trnh ch c s v T) v tnh H0 10000K. Gii

a. H0298(p) = -7,093 0,453 = -7,546 (Kcal) S0298(p) = 44,50 0,568 2x31,21 = -18,488 (cal) G0298 = -7546 + 298x18,488 = -2036,576 (cal) KP298 = 31,169 (atm-1)

b. )35,75(atm0,550,1

10,10,55

n

P 11

2

n

inp

p > Kp suy ra phn ng xy ra theo chiu nghch. c. 2HCH2

H

CHp 24

2

4 31,169PP31,169PP

K Ta c 01P31,169P1PP

2224 H2HHCH

Ta c 0,836(atm)P,0,164(atm)P42 CHH

d. 298)11,05(T7546dTCHHT

298p

0298

0T

Vy 4253,111,05TH0T (cal) 15303,14253,1100011,05H01000 (cal)

= - 15,3031 (Kcal)3.5. Bi tp t gii 1. Ti 500C v p sut 0,344 atm, phn ly ca N2O4 thnh NO2 l 63%. Xc nh KP v KC. S: Kp = 0,867 (atm); KC= 0,034 (mol/l) 2. 630C hng s cn bng KP ca phn ng:

N2O4 2NO2 l 1,27. Xc nh thnh phn hn hp cn bng khi:

a. p sut chung bng 1atm. b. p sut chung bng 10 atm.

S: a. 65,8% NO2; 34,2% N2O4 b. 29,8% NO2; 70,2% N2O4

3. un 746g I2 vi 16,2g H2 trong mt bnh kn c th tch 1000 lit n 4200C th cn bng thu c 721g HI. Nu thm vo hn hp u 1000g I2 v 5g H2 th lng HI to thnh l bao nhiu? S: 1582 g

4. Xc nh hng s cn bng Kp ca phn ng sau 700K SO2 + 1/2O2 = SO3

Bit rng 500K hng s cn bng Kp = 2,138.105 atm -1/2 v hiu ng nhit trung bnh trong khong nhit 500 700K l -23400 cal.

S: 2,6.10+2 atm-1/2 5. 1000K hng s cn bng ca phn ng: 2SO3(k) + O2(k) 2SO3(k)

C hng s cn bng KP = 3,5 atm-1. Tnh p sut ring phn lc cn bng ca SO2 v SO3 nu p sut chung ca h bng 1 atm v p sut cn bng ca O2 l 0,1 atm.

S: atm75,0P2SO , atm15,0P 3SO

6. Tnh G0 v hng s cn bng Kp 250C ca phn ng sau: NO + O3 NO2 + O2 .

Cho bit cc s liu sau: NO2 O2 NO O3


21

(KJ/mol)H 0 tt298, 33,81 0 90,25 142,12 (J/mol.K)S0298 240,35 240,82 210,25 237,42

S: Kp= 5.1034 7. 298K phn ng: NO + 1/2O2 = NO2, c G0 = -34,82 (KJ) v H0

= -56,34 (KJ). Xc nh hng s cn bng ca phn ng 298K v 598K.

S: Kp= 1,3.106 2980K v Kp= 12 5980K 8. nhit T v p sut P xc nh, mt hn hp kh cn bng gm 3

mol N2, 1 mol H2 v 1 mol NH3. a. Xc nh hng s cn bng Kx ca phn ng.

3H2(k) + N2(k) 2NH3(k) b. Cn bng s dch chuyn theo chiu no, khi thm 0,1 mol N2

vo hn hp cn bng T v P khng i. S: a. Kx= 8,33; b. Kx = 8,39

9. Hng s cn bng ca phn ng: PCl3(k) + Cl2(k) PCl5 (k) 500K l KP = 3 atm-1.

a. Tnh phn ly ca PCl5 2 atm v 20 atm. b. p sut no, phn ly l 15%.

S: a. 44,7%; 13%; b. 14,48 atm 10. Cho phn ng thy phn este axetat etyl. CH3COOC2H5 + H2O CH3COOH + C2H5OH

Nu ban u s mol ca este bng s mol nc th khi cn bng c 1/3 lng este b thy phn.

a. Xc nh hng s cn bng ca phn ng thy phn. b. Tnh s mol este b thy phn khi s mol nc ln gp 10 ln

s mol este. c. Tnh t l mol gia nc v este khi cn bng 99% este b

thy phn. S: a. Kn = 0,15; b. 75,9%; c. 393 ln

11. Cho phn ng: C2H4(k) + H2(k) C2H6(k)

Lp cng thc tnh s mol ca C2H6 trong hn hp cn bng theo s mol ban u ca C2H4 l a, ca H2 l b, hng s cn bng Kp v p sut cn bng ca h l P.

S. 1P.K

abPK4ba

2ba

P

P2

12. Cho phn ng: CO + Cl2 COCl2 C phng trnh m t s ph thuc ca Kp vo nhit T l:

lgKp(atm) = 5020/T 1,75lgT 1,158. a. Tm phng trnh m t s ph thuc nhit : G0T = f(T) v H0T = g(T).b. Tnh G0, H0, S0 v hng s cn bng KP, KC 700K. c. Hn hp phn ng sau s xy ra theo chiu no 1atm v

700K: 2 mol CO; 5 mol Cl2 v mol 3 COCl2. 0,4 mol CO; 1,6 mol Cl2 v 8 mol COCl2.

13. C th iu ch Cl2 bng phn ng: 4HCl(k) + O2 2H2O(h) + 2Cl2

Xc nh HSCB KP ca phn ng 3860C, bit rng nhit v p sut 1 atm, khi cho 1 mol HCl tc dng vi 0,5 mol O2 th khi cn bng s c 0,4 mol Cl2. S: Kp = 69,3 atm-1 14. 400C, hng s cn bng ca phn ng:

LiCl.3NH3(r) LiCl.NH3(r) + 2NH3(k) l Kp = 9 atm2, nhit ny phi thm bao nhiu mol NH3 vo mt bnh c th tch 5 lit cha 0,1mol LiCl.NH3(r) tt c LiCl.NH3(r) chuyn thnh LiCl.3NH3(r).


22

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 do ca mt h l thng s nhit ng c lp xc nh h

cn bng. K hiu l c. 4.2. Qui tc pha Gibbs

Bc t do ca h: c = k - f + n

Trong : 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 Thnh phn ca cc cu t trn gin pha thng dng l phn mol xi hay phn trm khi lng yi. Trong h hai cu t, dng mt on thng c chia thnh 100% nh sau:

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 cng thc: 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 biu din bng mt tam gic u nh sau:

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. Ba cnh ca tam gic biu din ba h hai cu t tng ng l AB, AC

v BC. Mi im trong 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


23

Hnh 4.3. Gin nhit - thi gian ca cht nguyn cht. Cc ng hoc cc mt trn gin pha biu din s ph thuc

gia cc 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 mt s 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.

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:

Hnh 4.5. Minh ha quy tc n by p dng quy tc n by, ta c:

HMHN

gg

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, H2 v H3. vi khi lng tng ng l:

g = g1 + g2 + g3

Hnh 4.6. Minh ha quy tc khi tm Nh vy, H phi nm khi tm vt l ca tam gic H1H2H3. u tin ta xc nh im biu din ca h K, tha mn iu kin:

H K = h H1 + h H2 v

KHKH

gg

1

2

2

1 . Tip theo ta xc nh im H tha mn iu kin sau:

H H = h K + h H3


24

v KH

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 = V2 V1: 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.

dPdT

4.6. nh hng ca nhit n p sut hi bo ha Phng trnh Clausius Claypeyron II

2RT

dTdlnP

Ly tch phn 2 v, ta c:

1212 T1T1RPPln 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 nng chy ca nc v tnh nhit nng chy ca nc 4 atm.

Gii

p dng phng trnh: VT.

dPdT

Vi: V = Vlng Vrn = 1,001 1,098 = - 0,097 (ml/g)

Hoc: V = 18.(- 0,097) = -1,746 (ml/mol)

0,008141,31434,6

1,746273dPdT (K/atm)

Nh vy, c tng p sut ln 1 atm th nhit nng chy ca nc gim 0,0081K. Mt cch gn ng, 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 nhit ha hi ca n l 538,1 cal/g (coi nhit ha hi khng thay i trong khang t 1 atm n 2 atm).

Gii Nhit ha hi: = 538,1x18 = 9685,8 (cal/mol) p dng cng thc: 1212 T1T1RPPln 273100 1T11,9879685,812ln T tnh c: T = 394K = 1210C V d 3. Tnh nhit nng chy ca 1 mol diphenylamin nu 1kg diphenylamin nng chy lm tng th tch ln 9,58.10-5m3 cho bit dT/dP = 2,67.10-7 K.m2/N. Nhit nng chy ca diphenylamin l 540C, khi lng mol ca cht ny l 169.

Gii

p dng cng thc: VT.

dPdT


25

3

7

5

19,83.102,67.10

9,58.10100016954273

dPdT

VT. (J/mol) V d 4. p sut hi bo ha ca axit xyanhydric HCN ph thuc vo nhit theo phng trnh:

T12377,04lgP(mmHg)

Xc nh nhit si v nhit ha hi ca n iu kin thng. Gii

Nhit si ca axit HCN p sut 760 mmHg:

Ta c: T

12377,04lg(760) T = 297,4K

Vy nhit si ca axit HCN l 24,40C. Ly o hm hai v phng trnh theo T, ta c:

2T1237

dTdlgP

M: 24,575.T

dTdlgP

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 hnh

Khi lng ca cht A trong hn hp ban u:

mAo = 20%200 = 40 (g) Phn trm ca cht C trong lp th 1:

%C = 100 - 50 - 20 = 30 (%) Vy im biu din ca lp 1: I1 (50%A, 20%B, 30%C)

Khi lng lp th 2: m = 200 60 = 140 (g)

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 nhit nng chy ca phenol 500 atm.

Gii

p dng phng trnh: VT.

dPdT

Vi: 0,0141,072

11,056

1V (mol/g) Ta c: 34,26.10

41,324,930,014314

dPdT (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 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:


26

Ta c: T1T1T1T1 2211 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: T133317T1373112 Suy ra: T = 448,40K 4.8. Bi tp t gii 1. Xc nh nhit ha hi ca H2O 4 atm 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 th nhit nng chy gim i 0,00354K. Tnh nhit nng chy ca Bi.

S: 11 kJ/mol. 4. Ti 1270C HgI2 b 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 ti 1270C.

S: -1,73.10-6 K/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 ring ca thic rn l 7,18 g/cm3 v dT/dP = 3,2567.10-8 K/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

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.

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.


27

Chng 5

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

gCn21

i%,i

- Nng mol/lit: VnCM

- Nng ng lng gam (lg/l): Vn'CN

- Nng molan (Cm): 1000m

nCdm

ctm

- Nng phn mol: iii nnx - CN = z.CM (z: s in 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

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 : kH l hng s Henry Pi l 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

iT

plnK

Ly tch phn phng trnh, ta c:

0T1T1Riilnx Vi: T0 l nhit ngng t (nhit si) 5.3. S ha tan ca cht lng trong cht lng v cn bng 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 l thun vi phn phn t ca n trong dung dch.

liRi .xkP

Khi dung dch ch c cu t i (dung dch i nguyn cht): xi = 1 v kR = Pi0

li

0ii .xPP

i vi dung dch thc, nh lut Raoult ch c th p dng cho dung mi ca dung dch v cng long: l1011 .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): lB0AlA0AA x1.P.xPP (1)

lB

0BB .xPP (2)

p sut tng ca h l: P = PA + PB

lB0BlB0A .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.


28

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 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

lBx.l

Ax

lBx

0AP

0BP

hAx

hBx (nh lut Konovalop I)

Trong 0A

0B

PP 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

hBx

hAx

hBx

lB.x11lB.xh

Bx (4) Biu din phng trnh (4) ln th (x - x) ta c cc ng trn hnh 5.2.

Hnh 5.2. Gin (x-x) ca h hai cu t A-B 5.3.2. H dung dch thc tan ln v hn 5.3.2.1. p sut hi 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 vi nhng 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)PP

PP

x

x0A

0B

A

BhA

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


29

A0A

0OH

OH M18

PP

g 22

Trong : 0 OH2P v

0AP 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 ca mt 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. K

CC

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: OH

0OH 22 .xPP

758

3425

1895

1895

760P (mmHg) Dung dch glycerin:

OH0

OH 22 .xPP 0,997760758PPx 0 OHOH 22

M: 0,997

92m

18m

18m

xglyOH

OH

OH2

2

2

Suy ra: OHgly 20,014mm Nng phn trm ca dung dch glycerin 100

mm

mC%

OHgly

gly

2

1,38100

m0,014m0,014m

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 dch xem nh l tng. Xc nh: a. Thnh phn dung dch 123,30C di p sut kh quyn 760mmHg. b. T s mol ca clorobenzen v bromobenzen trong pha hi trn dung

dch c thnh phn 10% mol clorobenzen.Gii

Hai 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 760mmHg

0,00552762400762760

PPPP

x 02

01

02l

1 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 0,21

0,90,1

400760

x

x

PP

x

xl1

l2

10

20

h1

h2


30

V d 3. Benzen v toluen to vi nhau mt dung dch xem nh l 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.

Gii a. p sut hi ring phn ca tng cu t

Phn mol ca benzen:

54,0

92100

78100

78100

nn

nx

TB

BB

Phn mol ca toluen:

46,0

92100

78100

92100

nn

nx

TB

TT

p sut hi ca Benzen: 64,9080,54120,2.xPP B0BB (mmHg)

p sut hi ca Toluen: 16,8820,4636,7.xPP T0TT (mmHg)

b. Xc nh p sut hi ca dung dch 81,7916,88264,908PPP TB (mmHg)

V d 4. Etanol v metanol to thnh dung dch xem nh l 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 100g etanol v

100g metanol. b. Xc nh cc p sut ring phn v p sut tng ca dung dch. c. Tnh phn mol ca metanol trong pha hi nm cn bng vi dung dch

trn. Gii

a. Phn mol mi cht

S mol etanol: )2,1739(mol46

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 lE0EE (mmHg)333,520,597,88xPP lM0MM P = 18,245 + 52,333 = 70,578 (mmHg) c. Phn mol ca metanol trong pha hi:

0,74150,591

44,588,71

0,5944,588,7

1)x(1.x

x 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, P2 v P ca hn hp c phn mol ca CCl4 l 0,7.

b. Xc nh thnh phn hn hp SnCl4 - CCl4 si 900C di p sut 760mmHg.

c. Xc nh thnh phn hi ti 900C. Gii

a. Xy dng gin thnh phn - p sut


31

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: 108,60,33620,71.PP 0SnClSnCl 44 (mmHg) p sut ca CCl4:

779,10,71112.xPP444 CCl

0CClCCl (mmHg)

p sut ca hn hp: P = 108,6 + 779,1 = 887,7 (mmHg)

b. Xc nh thnh phn hn hp SnCl4 - CCl4 Ta c: 1020102202101t xPPPxPxPP 0,47

11123621112760

PPPP

x 02

01

02t

1 Vy thnh phn ca SnCl4 l 0,47 Thnh phn ca CCl4 l 0,53

c. Xc nh thnh phn hi 3,464

0,470,53

3621112

x

x

PP

x

xl1

l2

10

20

h1

h2

V d 6. Mt dung dch l tng ca A v B cha 25% mol A. 250C, hi cn bng ca n cha 50% mol A. Nhit ha hi ca A v B ln lt l 5 v 7 Kcal/mol. Tnh t s p sut hi bo ha ca A v B khi nguyn cht 250C v 1000C.

Gii a. 250C:

Ta c: lA0AA x.PP v

Theo bi ta c: PA = PB lB0BlA0A x.Px.P 3

25,075,0

x

x

PP

lA

lB

0B

0A

b. 1000C: p dng cng thc: 1212 T1T1RPPln

i vi cht A: Ta c: 12A0 298A

0373A

T1

T1

RPP

ln (1)

Tng t i vi cht B:

12B0 298B0

373B

T1

T1

RPP

ln (2)

Ly phng trnh (1) (2), ta c:

12AB0 298A0 298B0 373B0 373A T1T1R1PPPPln 373298 37329850007000987,1 1PPPPln 0 298A0 298B0 373B0 373A 507,0PP31 0 373B

0373A

52,1PP0 373B0

373A

lB

0BB x.PP


32

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. Gii

a. Gin P - x (hnh 5.4)

lA

lA

0AA x100x.PP (1)

lB

lB

0BA x600x.PP (2) lAlA0B0A0Bt 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,.0xlA v 143,0x lB 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 dch cha 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 th p sut tng cng l 300 mmHg. Hy xc nh p sut hi bo ha ca A v B nguyn cht 500C.

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 ca bong bng hi u tin, bit rng 1000C p sut hi bo ha ca CCl4 v SnCl4 ln 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 ca dung 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 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 nc 17,5 mmHg.


33

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.


34

Chng 6

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.

xPP

PPP

00

0 Trong : P0: p sut hi ca dung mi nguyn cht P: 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. 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 nghim ng K

Hnh 6.1. Gii thch tng im si v h im kt tinh

1000.MR.TK

20

Trong : R: hng s kh T0: nhit chuyn pha M: phn t lng : nhit chuyn pha6.1.3. p sut thm thu p 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 thy tnh 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


35

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 ca cc 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). Vng nm gia ng lng v ng rn h tn ti cn bng ca hai pha: RA L 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

Hnh 6.3. Qu trnh a nhit ca h Q

H H = pha lng e + h rn chung RC

H rn chung RC = pha rnA + pha rn B

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 gm nhng 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


36

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 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

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 ae1 l ng kt tinh ca cht rn A. ng e1de2 l 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. Cho bit hng s nghim lnh ca nc l 1,86 v hng s nghim si ca nc l 0,513. p sut hi ca nc nguyn cht 250C l 23,76 mmHg.

Gii a. h im ng c ca dung dch: T = 1,50C

Ta c:

m .CKT 0,8061,861,5

KTC

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

23,420,806

181000

181000

23,76.xPP 0 (mmHg) V d 2. 200C p sut hi nc l 17,54 mmHg v p 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: 0,018

17,5417,2217,54

PP

x0


37

M 0,018

18m

60m

60m

xOHct

ct

2

(1) Gi s 100mm OHct 2 (g) (2) ctOH m100m 2 T (1) v (2), ta c: mct = 5,75 (g) Th tch ca dung dch: 99

1,01100

dmV dd (ml)

Nng ca dung dch:

0,96810009960

5,75

VnCM (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 - B c cho trong hnh sau.

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 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 h

Vng I: f = 1, c = k f + 1 = 2 Vng II: f = 2, c = k f + 1 = 1 Ti im eutecti: f = 3, c = k f + 1 = 0. b. Khi lng ca pha rn v pha lng

54

0,250,2

HeHR

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

0,20,8

RRRR

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 nh s 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 ti RC. Tnh lng rn A v rn B kt tinh v lng lng eutecti cn li.

c. Tnh lng lng eutecti ti a thu c t h trn.


38

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 h ta c: mRc = 80 (g), mle = 30 (g) Khi lng rn A v rn B. Ta c h pt:

173

0,850,15

RRRR

m

m

AC

BC

R

R

B

A mRa + mRb = 80

Gii h ta c: mRb = 68 (g); mRa = 12 (g).

c. Khi lng eutecti ti a khi RC trng vi RB

43

0,40,3

HeHR

m

m B

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.

c. Khi im h Q trng vi im H. Hy xc nh khi lng ca 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 2. V gin pha ca h Sb - Pb da vo cc d kin thc nghim sau:

Thnh phn hn hp lng (% khi lng) Nhit bt u Kt tinh (0C) Sb Pb

100 0 632 80 20 580 60 40 520


39

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 hnh di. 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 eutecti cn li. b. Tnh lng eutecti ti a thu c. c. Phi trn A v B vi thnh phn nh th no thu c hn

hp A v B ng thi kt tinh.

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 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. 81,360C; 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.


40

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 in gi 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 : : 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 s c trong dung dch v s tiu phn ban u: 11i Trong : : phn ly = m + n Vi m, n l h s ca phng trnh:

AmBn = mAn+ + nBm- 7.1.3. Hot , h s hot ca cc cht in ly

M+A- = +Mz+ + -Az-

Hot trung bnh ca ion: 1 .aaa Trong : = + + - Ta c: a+ = +.m+ v a- = -.m- .m.m 1 .ma Trong : 1 . m : molan trung bnh ca cc ion : h s hot trung bnh ca cc ion +, -: h s hat ca cc ion Lc ion: 2iim Zm21I hoc 2iiC ZC21I Trong : i: l k hiu ca tt c cc ion trong dung dch mi Ci: nng thc ca cc ion7.2. S chuyn vn in tch trong dung dch in ly 7.2.1. dn in ring dn in ring () l dn in ca mt dung dch c th tch V = 1 cm3, c t gia hai in cc phng song song c din tch nh nhau v cch nhau 1 cm.

1

Trong l in tr sut:

lS

s

l1

R1L

7.2.2. dn in ng lng ()


41

L dn in ca mt th tch tnh theo cm3 cha ng mt ng lng gam cht in ly nm gia hai in cc phng song song cch nhau 1 cm.

NC.1000

Trong : : dn in ring CN: nng ng lng Ngi ta kho st mi tng quan gia theo C v C = - A C 7.2.3. Mi quan h gia dn in v in ly Ta c: M: Trong : : dn in ng lng ca dung dch. : dn in ng lng gii hn ca dung dch.

:, dn in ng lng gii hn ca cation v anion hay cn gi l linh ion. 7.3. Pin v in cc 7.3.1. Th in cc Trn ranh gii phn chia 2 pha gm mt tm kim lai dng lm in cc (vt dn lai 1) v dung dch in phn (vt dn lai 2) xut hin mt hiu in th gi l th in cc c ln c xc nh bng phng trnh Nernst.

khoxhlnnFRT0 Trong : R: hng s kh T: nhit tuyt i F: hng s Faraday n: s electron trao i

7.3.2. Nhit ng hc ca pin v in cc 7.3.2.1. Cng ca pin Xt phn ng: aA + bB = cC + dD Ta c: G = - Amax Cng to c l: A = n.F.E G = -n.F.E 7.3.2.2. nh hng ca nng n sc in ng v in cc -

phng trnh Nernst. Sc in ng ph thuc vo nng cc cht c trong pin v nhit

ca pin. Phng trnh Nernst: ba dc0 B.A D.Clgn0,059EE

Trong : 000E : sut in ng tiu chun

00, : th in cc tiu chun ca in cc dng (+) v m (-)

7.3.2.3. Phng php kho st mt phn ng Ta c: G = - n.F.E

Nu E > 0 G < 0: phn ng xy ra theo chiu thun. Nu E < 0 G > 0: phn ng xy ra theo chiu ngc li. Nu E = 0 G = 0: phn ng cn bng. 7.3.2.4. H thc Luther H thc ny dng xc nh th in cc cho mt cp oxi ha - kh ca nguyn t c nhiu mc oxi ha khc nhau. V d, kim loi M c hai cation Mh+ v Mn+ (h > n). H ny ng vi ba qu trnh in cc: Mh+ + he = M (1) G1 = -hFh Mn+ + ne = M (2) G2 = -nFnMh+ + (h - n)e = Mn+ (3) G3 = -(h - n)Fh/ n V (3) = (1) - (2), nn: G3 = G1 - G2 Hay: (h - n)h/ n = hh - nn7.3.3. Cc loi in cc in cc loi 1: l mt h gm kim loi hoc kim c nhng vo dung dch cha ion ca kim lai hoc kim . K hiu Mn+/ M hoc An-/ A:

Mn+ + ne = M


42

Phng trnh Nernst: nnn M0 /MM/MM a1lnnFRT in cc loi 2: l mt h gm kim loi c ph mt hp cht kh tan (mui, oxit hay hydroxit) ca kim loi v nhng vo dung dch cha anion ca hp cht kh tan . K hiu An-/ MA/ M. MA + ne = M + An- Phng trnh Nernst : nnn A0 AMA/M,AMA/M, lnanFRT 7.4. Bi tp mu V d 1. Tnh nhit kt tinh ca dung dch cha 7,308g NaCl trong 250g nc cho bit 291K p sut thm thu ca dung dch l 2,1079.106 N/m2, khi lng ring ca dung dch l 1g/cm3, nhit nng chy ca nc nguyn cht l 333,48.103 J/kg.

Gii Ta c:T = i.k.Cm (1) Vi:

2 2o

dRT M 8,314.273 .18k 1,861000 1000.333,48.18

Trong : 0,5

0,2558,57,308Cm (mol/kg)

Ta li c: i.CRT Nng ca dung dch: 0,510007,30825058,5 7,308C (mol/l)

6

52,1079.10i 1,74

CRT 1,013.10 .0,082.291.0,5

Th vo cng thc (1), ta c: T = i.k.Cm = 0,5 1,74 1,86 = 1,62 Nhit kt tinh ca dung dch in ly l:

T = 0 - 1,62 = -1,620C

V d 2. h im kt tinh ca dung dch CH3COOH 0,1M l 0,1885 , hng s nghim lnh ca nc l 1,86. Tnh phn ly ca dung dch CH3COOH 0,1M v 0,05M.

Gii nng long th nng mol/l gn bng vi nng molan nn ta

c: Trng hp dung dch c nng 0,1M:

T = i.k.Cm 1,01340,11,860,1885i M ta li c: i 1 1,0134 1 0,0134

1 2 1

phn ly ca dung dch CH3COOH 0,1M l 1,34%. Trng hp dung dch c nng 0,05M: Ta c phng trnh phn ly:

CH3COOH = CH3COO- + H+ Hng s phn ly:

522 1,82.100,01341

0,10,01341

.CK i vi dung dch in ly yu, ta c: 11 5K 1,82.10 0,019

C 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 1,742

2910,0820,520,78

CRTi


43

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 NaCl 0,15M 370C bit phn ly ca dung dch l 95%.

Gii Ta c: iCRT M:

i 11

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 V d 5. Tnh nng ca dung dch ng sacaroza c gi tr p sut thm thu l 8,1134 atm 370C.

Gii Nng dung dch ng sacaroza l:

8,1134CRT 8,1134 C 0,32RT 0,082.310 mol/l

V d 6. Thit lp biu thc tnh sc in ng ca pin c nng sau: Ag/ AgCl/ HCl (C1)// HCl (C2)/ AgCl/ Ag

Cho bit C1 > C2. Gii

Cc m: Ag + Cl-(C1) = AgCl + e Cc dng: AgCl + e = Ag + Cl-(C2) Sc in ng ca pin c tnh nh sau:

1

2

CRTE ( ) lnF C

1

2

CRTE lnF C

V d 7. in tr ca dung dch KCl 0,02N 250C trong mt bnh o dn in o c l 457. Bit dn in ring ca dung dch l 0,0028 -1.cm-1. Dng bnh ny o dn in ca dung dch CaCl2 cha 0,555g

CaCl2 trong 1 lt dung dch c gi tr l 1050. Tnh hng s bnh in cc v dn in ng lng ca dung dch CaCl2.

Gii Vi dung dch KCl ta c:

k = R. 457 = 1,2796 (cm-1) Vi dung dch CaCl2 ta c:

0,001221050

1,2796Rk

( 1 .cm-1) Nng ca dung dch CaCl2 l:

0,011111

20,555Vm

CN (N) dn in ng lng ca dung dch CaCl2 c tnh theo cng

thc: 122

0,010,001221000

C1000 (cm2. 1 .lg-1)

V d 8. dn in ng lng gii hn ca axit propionic (C2H5COOH) 250C l 385,6 -1.lg-1.cm2. Hng s phn ly ca acid ny l 2,34.10-5.Tnh dn in ng lng ca dung dch acid propionic 0,05M cng nhit .

Gii Dung dch acid propionic l mt cht in ly yu nn:

5K 2,34.10 0,0216C 0,05

M: 2 1 10,0216.385,6 8,34cm . .dlg

Vy dn in ng lng ca dung dch l: = 8,34.cm2.-1.lg-1 V d 9. dn in ng lng ca NH4Cl trong dung dch v cng long l 149,7 -1.lg-1.cm2. Linh ion ca OH-1 l 198 ca Cl- l 76,3 -1.dlg-1.cm2. Tnh dn in gii hn ca dung dch NH4OH.

Gii Ta c:

4(NH Cl) ) Cl )


44

73,476,6149,7 )(NH4 (cm2.-1.lg-1) dn in ng lng gii hn ca NH4OH l:

4

2 1 1(NH OH) ) ) 73,4 + 198 = 271,4cm . .dlg

V d 10. in tr ca dung dch KNO3 0,01N l 423 . Hng s bnh in cc l 0,5 cm-1. Xc nh dn in ring, dn in ng lng v phn ly ca dung dch, bit linh ion ca NO3- v K+ ln lt l 71,4 v 73,4 -1.lg-1.cm2.

Gii dn in ring ca dung dch KNO3:

-3 -1 -1k 0,5 1,182.10 .cmR 423

dn in ng lng: 3

2 1 11000. 1000.1,182.10 118,2cm . .dlgC 0,01

Ta c:

3

2 1 1(KNO ) ) ) 73,4 + 71,4 =144,8cm . .dlg

in ly l:

144,8

Nh vy in ly ca dung dch l 81,63% V d 11. Xc nh nng ca dung dch HCl nu dng dung dch NaOH 8N chun 100ml dung dch HCl bng phng php chun dn in th th kt qu thu c l:

VNaOH (ml) 0,32 0,60 1,56 2,00 2,34 (-1.cm.10-2) 3,2 2,56 1,64 2,38 2,96 Gii

Ta xc nh im tng ng ca php chun in th bng s thay i t ngt ca dn in ring. T bng s liu th im tng ng chnh l im c th tch NaOH bng 1,56ml.

Ta tnh c nng ca dung dch HCl:

NaOH NaOHHCl

HCl

C .V 8.1,56C 0,125NV 100

V d 12. Tnh th in cc: Zn/ ZnCl2 (0,005N) 250C cho bit dn in ng lng ca dung dch l 89 -1.lg-1.cm2, dn in ng lng gii hn ca dung dch l 113,7 -1.dlg-1cm2 v in th tiu chun ca in cc Zn l -0,76V.

Gii phn ly ca dung dch ZnCl2 l:

89113,7

Nng ca Zn2+:

[Zn2+] = 0,783 0,005 = 0,004 N = 0,002 M Th in cc:

20,059 lg[Zn ]2

V 0,839lg0,0022

0.0590,76

V d 13. Cho pin Cd / Cd2+ // CuSO4 / Cu c sc in ng l 0,745V. Hy xc nh phn ly ca dung dch CuSO4 0,1N cho bit in th tiu chun ca in cc Cu l 0,34V, ca in cc Cd l -0,4V v nng ion Cd2+ trong dung dch l 0,05N.

Gii Phn ng xy ra trong pin:

Cd + Cu2+ = Cd2+ + Cu Sc in ng ca pin nh sau:

220 CuCdlg20,059EE Trong : 0,740,40,34E 000 (V) Th vo cng thc trn ta c: 20,059 [Cu ]lg

2 0,05


45

2[Cu ] = 0,074 (N) phn ly: 0,740,10,074CuSOCu 42 V d 14. Cho pin (Pt) Hg/ Hg2Cl2/ KCl 0,01N// H+ / Quinhydron. C sc in ng 250C l 0,0096V. Tnh pH ca dung dch bit in th in cc Calomen l 0,3338V v th in cc tiu chun ca in cc Quinhydron l 0,699V.

Gii p dng cng thc tnh pH ca dung dch:

oQuin Cal EpH

0,059

0,699 0,3338 0,0096pH 6,0270,059 V d 15. Cho pin, Cu/ CuCl2 (0,7M)// AgNO3 (1M)/ Ag. Cho bit phn ly ca dung dch CuCl2 l 80% v dung dch AgNO3 l 85%, in th tiu chun ca in cc Cu l 0,34V v in cc Ag l 0,8V. Tnh sc in ng ca pin v tnh lng AgNO3 cn thm vo sc in ng ca pin tng thm 0,02V, cho th tnh bnh l 1lt.

Gii CuCl2 = Cu2+ + 2Cl-

Nng ion Cu2+: 0,56100800,7Cu2

(M) AgNO3 = Ag+ + NO3-

Nng ion Ag+: 0,85100851Ag

(M) Sut in ng ca pin:

E0 = 0+ - 0- = 0,8 - 0,34 = 0,46 (V)

220 AgCulg20,059EE 0,4630,850,56lg20,0590,46E 2 (V) Thm vo sut in ng ca pin 0,02V E = 0,483 (V) 0,483 = 0,46 - 20,059 0,56lg2 Ag [Ag+] = 1,836 (M)

[AgNO3] = 1,836 2,16(M)0,85

3AgNOn = CM.V = 2,16 x 1 = 2,16 (mol)

Vy s mol AgNO3 thm vo l: 2,16 1 = 1,16 (mol) 3AgNO

m = 1,16 x 170 = 197,2 (g) V d 16. Vit cu trc pin trong cc m l in cc Hiro, cc dng l in cc Calomen. Cho bit in cc Calomen nhng vo dung dch KCl 0,1M v pH ca dung dch l 1,0. Tnh sc in ng ca pin.

Gii Pt, H2 /H+ //KCl 0,1M /Hg2Cl2/Hg, Pt

p dng cng thc:

0,059EpH cal

M: 0Cal = 0,268 (V) Ta c: 1

0,059E cal E = 0,059 + Cal

M: 0,8470,11lg20,0590,788 2cal (V) E = 0,059 + 0,847 = 0,906 (V)


46

V d 17. Cho in th tiu chun ca in cc Cu l 0,34V, ca in cc Ag l 0,799V. Chng minh phn ng sau khng xy ra:

2Ag + Cu2+ = 2Ag+ + Cu. Gii

Cc m: 2Ag 2e = 2Ag+ Cc dng: Cu2+ + 2e = 2Cu Vy pin c hnh thnh t phn ng trn l:

Ag/ Ag+// Cu2+/ Cu Sc in ng tiu chun ca pin tnh c:

E0 = 0,34 0,799 = - 0,459 (V) E0 < 0 nn phn ng khng t xy ra. V d 18. Vit cc phng trnh phn ng in cc v phn ng tng qut xy ra trong cc pin sau:

a. Zn / ZnSO4 // CuSO4 / Cu b. Cu / CuCl2 / AgCl / Ag c. (Pt) H2 / H2SO4 / Hg2SO4 / Hg (Pt) d. Cd/ CdSO4 / Hg2SO4 / Hg (Pt)

Gii a. Zn / ZnSO4 // CuSO4 / Cu

Cc m: Zn - 2e = Zn2+ Cc dng: Cu2+ + 2e = Cu Zn + CuSO4 = ZnSO4 + Cu

b. Cu / CuCl2 / AgCl / Ag Cc m: Cu - 2e = Cu2+ Cc dng: 2AgCl + 2e = 2Ag + 2Cl- Cu + 2AgCl = 2Ag + CuCl2

c. (Pt) H2 / H2SO4 / Hg2SO4 / Hg (Pt) Cc m: H2 - 2e = 2H+ Cc dng: Hg2SO4 + 2e = 2Hg + SO42- H2 + Hg2SO4 = 2Hg + H2SO4

d. Cd / CdSO4 / Hg2SO4 / Hg (Pt) Cc m: Cd - 2e = Cd2+ Cc dng: Hg2SO4 + 2e = 2Hg + SO42- Cd + Hg2SO4 = 2Hg + CdSO4

V d 19. Lp pin trong xy ra cc phn ng sau. a. Cd + CuSO4 = CdSO4 + Cu b. 2AgBr + H2 = 2Ag + 2HBr c. H2 + Cl2 = 2HCl d. Zn + 2Fe3+ = Zn2+ + 2Fe2+

Gii a. Cd + CuSO4 = CdSO4 + Cu

Cc m: Cd - 2e = Cd2+ Cc dng: Cu2+ + 2e = 2Cu Cd/ CdSO4// CuSO4/ Cu

b. 2AgBr + H2 = 2Ag + 2HBr Cc m: H2 - 2e = 2H+ Cc dng: 2AgBr + 2e = 2Ag + 2Br- Pt, H2 / HBr / AgBr / Ag

c. H2 + Cl2 = 2HCl Cc m: H2 - 2e = 2H+ Cc dng: Cl2 + 2e = 2Cl- Pt, H2/ HCl/ Cl2, Pt

d. Zn + 2Fe3+ = Zn2+ + 2Fe2+ Cc m: Zn - 2e = Zn2+ Cc dng: 2Fe3+ + 2e = 2Fe2+ Zn/ Zn2+// Fe3+, Fe2+/ Pt

V d 20. Cho phn ng ca pin l: Hg2+ + 2Fe2+ = 2Hg + 2Fe3+ c hng s cn bng 250C l 0,018 v 350C l 0,054. Tnh G0 v H0 ca phn ng 250C.

Gii Coi H khng thay i trong khong t 25 - 350C ta c:

12TT T1T1RHKKln 12 120

T1

T1

8,314H

0,0180,054ln

H0 = 83834,58 (J)G0 = -RTlnKp = -8,314 298 ln(0,018) = 9953,36 (J)

V d 21. Cho pin: Zn / ZnCl2 (0,5M) / AgCl / Ag. a. Vit phn ng in cc v phn ng trong pin. b. Tnh sc in ng tiu chun, bin thin th ng p tiu chun ca

pin. c. Tnh sc in ng v bin thin th ng p ca pin.

Cho bit in th tiu chun ca in cc Zn l -0,76V, ca in cc Ag/AgCl/Cl- l 0,2224V.

Bi tp ha l c s rt gn

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