I. OLYMPIC HA HC VIT NAM: OLYMPIC HA HC SINH VIN VIT NAM 2003 - BNG A

Tnh nhit ca ngn la CO chy trong hai trng hp sau: a) Chy trong oxy tinh khit (20% oxy v 80% nit theo th tch) b) Chy trong oxy tinh khit

Cho bit lng oxy va cho phn ng, nhit lc u l 25oC. Entanpi chy ca CO 25oC v 1atm l 283kJ.mol-1. Nhit dung mol chun ca cc cht nh

sau: Cop (CO2, k) = 30,5 + 2.10-2T Cop (N2, k) = 27,2 + 4,2.10-3T

BI GII:

a) KTdTCCHT

oP

oP

oNCO

25550)2(298

298 22==++

b) KTdTCHT

oP

oCO

40980298

298 2==+

OLYMPIC HA HC SINH VIN VIT NAM 2003 - BNG A Cho cc s liu sau 298K: Ag+(dd) N3-(dd) K+(dd) AgN3(r) KN3(r)

Gott(kJ.mol-1) 77 348 -283 378 77 1) Xc nh chiu xy ra ca cc qa trnh sau:

Ag+(dd) + N3-(dd) AgN3(r) K+(dd) + N3-(dd) KN3(r)

2) Tnh tch s tan ca cht in li t tan. 3) Hi phn ng g xy ra khi mui KN3 tc dng vi HCl c.

BI GII: 1) Ag+(dd) + N3-(dd) AgN3(r)

Go = 378 (77 + 348) = -47kJ: Chiu thun. K+(dd) + N3-(dd) KN3(r) Go = 77 (-283 + 348) = 12kJ: Chiu nghch.

2) AgN3 l cht t tan. Gi Ks l tch s tan ca n:

910.79,5

237,8298.314.8.303.2

47000lg

===

s

s

K

K

3) KN3 + HCl HN3 + KCl HN3 + 3HCl NH4Cl + N2 + Cl2 KN3 + 4HCl NH4Cl + N2 + Cl2 + KCl.

OLYMPIC HA HC SINH VIN VIT NAM 2005 - BNG B Mt phn ng dng luyn km theo phng php kh l:

ZnS(r) + 3/2O2(k) ZnO(r) + SO2(k) 1) Tnh Ho ca phn ng nhit 298K v 1350K, coi nhit dung ca cc cht khng ph

thuc vo nhit min nhit nghin cu. 2) Gi thit ZnS nguyn cht. Lng ZnS v khng kh (20% O2 v 80% N2 theo th tch) ly

ng t l hp thc bt u 298K s t n nhit no khi ch hp th lng nhit ta ra

do phn ng iu kin chum ti 1350K (lng nhit ny ch dng nng nhit cc cht u) Hi phn ng c duy tr c khng, ngha l khng cn cung cp nhit t bn ngoi, bit rng phn ng trn ch xy ra nhit khng thp hn 1350K?

3) Thc t trong qung sfalerit ngoi ZnS cn cha SiO2. Vy hm lng % ca ZnS trong qung ti thiu phi l bao nhiu phn ng c th t duy tr c?

Cho bit entanpi to thnh chun ca cc cht 25oC (kJ.mol-1) Hp cht: ZnO(r) ZnS(r) SO2(k) Hof -347,98 -202,92 -296,90 Nhit dung mol ng p ca cc cht (J.K-1.mol-1): Hp cht ZnS(r) ZnO(r) SO2(k) O2(k) N2(k) SiO2(r) Cop 58,05 51,64 51,10 34,24 30,65 72,65 Bit MZnS = 97,42g.mol-1; MSiO2 = 60,10g.mol-1

BI GII: 1) Ho298 = -347,98 296,90 + 202,92 = -441,96kJ

Cop = 51,64 + 51,10 58,05 3/2.34,24 = -6,67J.K-1H1350 = -448976,84J

2) KTdTH

JKCCCCT

o

oNp

oOp

oZnSp

oP

1829031,293

31,293623

2981350

1)()()( 22

==+

=++=

T = 1829K > 1350K nn phn ng t duy tr c. 3) Gi x l s mol SiO2 c trong 1 mol ZnS +=+++= )(65,7231,293623 1)()()()( 222 JKxxCCCCC o SiOpo Npo Opo ZnSpop ==++ 1350

298

1350

298

84,1065,7231,29384,448976 molxxdTdT

%ZnS = 47% K THI CHN HC SINH GII QUC GIA NM 2002 (Bng A) Kh NO kt hp vi hi Br2 to ra mt kh duy nht trong phn t c 3 nguyn t. 1. Vit phng trnh phn ng xy ra. 2. Bit phn ng trn thu nhit, ti 25oC c Kp = 116,6. Hy tnh Kp (ghi r n v) ti 0oC ; 50oC. Gi thit rng t s gia hai tr s hng s cn bng ti 0oC vi 25oC hay 25oC vi 50oC u bng 1,54. 3. Xt ti 25oC, cn bng ho hc c thit lp. Cn bng s chuyn dch nh th no? Nu: a) Tng lng kh NO. b) Gim lng hi Br2. c) Gim nhit . d) Thm kh N2 vo h m: - Th tch bnh phn ng khng i (V = const) - p sut chung ca h khng i (P = const). BI GII: 1. 2 NO(k) + Br2 (hi) 2 NOBr (k) ; H > 0 (1) Phn ng pha kh, c n = -1 n v Kp l atm-1 (2) 2. Do phn ng thu nhit nn c lin h Kp ti O2 < Kp ti 252 < Kp ti 502 (3)

Vy : Kp ti 250 = 1 / 1,54 x Kp ti 252 = 116,6 / 1,54 = 75,71 (atm-1) Kp ti 252 = 1,54 x Kp ti 252 = 116,6 x 1,54 179, 56 (atm-1) 3. Xt s chuyn di cn bng ho hc taji 25OC. Trng hp a v b: v nguyn tc cn xt t s: PNOBr Q = (4) (Khi thm NO hay Br (PNO) 2

) 2 Sau so snh tr s Kp vi Q kt lun. Tuy nhin, y khng c iu kin xt (4); do xt theo nguyn l L satlie. a. Nu tng lng NO, CBHH chuyn di sang phi. b. Nu gim lng Br2, CBHH chuyn di sang tri. c. Theo nguyn l Lsatlie, s gim nhit lm cho CBHH chuyn di sang tri, chng li s gim nhit . d. Thm N2 l kh tr. + Nu V = const: khng nh hng ti CBHH v N2 khng gy nh hng no ln h (theo nh ngha p sut ring phn). + Nu P = const ta xt lin h. Nu cha c N2: P = pNO + pBr2 + pNOBr (a) Nu c thm N2: P = pNO + pBr2 + pNOBr + Pn2 (b) V P = const nn pi' < pi Lc ta xt Q theo (4) lin h / tng quan vi Kp: 1. Nu Q = Kp: khng nh hng 2. Nu Q > Kp : CBHH chuyn di sang tri, Q gim ti tr s Kp. 3. Nu Q

Cho: Cl = 35,453 ; P : 30,974 ; H = 1,008 ; Cc kh u l kh l tng. BI GII: 1. Thit lp biu thc cho Kp, Kc PCl5 (k) PCl3 (k) + Cl2 (k) ban u a mol cn bng a x x x (mol) Tng s mol kh lc cn bng : a + x = n

= xa

; Khi lng mol: = 30,974 + 5 x 35,453 = 208,239 (g/mol) 5PCl

M

= 30,974 + 3 x 35,453 = 137,333 (g/mol) 3PCl

M

= 70,906 (g/mol) 2Cl

M

gam

208,239 gam/molm

= a mol PCl5 ban u

*p sut ring phn lc cn bng ca mi kh:

= 5PCl

P pa xa x

+ trong = = 3PClP 2ClP +

xP

a x

Kp =

2 3

5

Cl PCl

PCl

P P

P =

+ +

2

-

xp

a xa x

pa x

= ( ) +2

22

xp

a x +

a xa x

1p

Kp = + 2

( ) (

x pa x a x)

= 2

2 2x

a x p ; Kp =

2

2

2 2

2 2

xa p

a xa a

= 2

21p

* Kc = [PCl5] = (1 )a

V trong [PCl3] = [Cl2] =

aV

Kc = [ ] [ ]3 25

Cl

[ ]

PCl

PCl = ( ) 2

2

a

V ( ) 1

Va

= 2

(1 )

aV

= 2

208,239 (1 )

mV

Hoc: Kp = Kc (RT)V Vkh = 1

Kp = Kc (RT) pV = nRT = (a + x) RT RT = +pV

a x = +(1 )

pVa

Kp = Kc +pV

a x

2

1p = Kc +

pVa x

Thay x = a 2

21p = Kc +(1 )

pVa

Kc = +

2

2

(1 )

1

aV

Kc = ( )

+ +

2 (1 )

1 (1-

aV )

= 2

(1 )

aV

= 2

208,239 V (1 )

m

* Quan h Kp v Kc. T cch 1 : Kc = Kp 1RT

Thay RT = pVa(1 )+ Kc = Kp

a(1 )pV

+ = 2 a(1 )1 pp V

+ = 2a

V(1 )

2. Th nghim 1 : ban u = a = 5PCln83,30 g

208,239 g/mol = 0,400 mol

M ca hn hp cn bng: 68,826 2,016 = 138,753 g/mol Tng s mol kh lc cn bng: n1 = a (l + 1) = 83,30 g138,753 g/mol = 0,600 mol n1 = a (1 + 1) = 0,400 (1 + 1) = 0,600 1 = 0,500 * Tm Kp ti nhit T1 : Kp =

2

21

p = 2

2

(0,5)1 (0,5)

2,70 = 0,900 3. Th nghim 2: - Gi nguyn nhit Kp khng i. - Gi nguyn s mol PCl5 ban u: a = 0,400mol. - p sut cn bng P2 = 0,500 atm.

Ta c 22

221

p2 = Kp =

22

221

0,500 = 0,900 2

2 = 0,64286 2 = 0,802 Tng s mol kh lc cn bng: n2 = 0,400 + (1+ 2) 0,721 (mol). * Th tch bnh trong TN 2: V2 = 2 1

2

n RTp

so vi V1 = 1 11

n RTp

21

VV

= 21 2

n pn p

1 = 0,721 2,7000,600 0,500

= 6,486 (ln) 4. Th nghim 3: - Thay i nhit Kp thay i. - Gi nguyn s mol PCl5 ban u a = 0,400 mol v V1 - p sut cn bng P3 thay i do: nhit gim (T3 = 0,9 T1), tng s mol

kh thay i (n3 n1). P3 = 1,944 atm ; Tnh 3 : n3 = a (1+ 3) = 0,400 (1+ 3) ; p3V1 = n3RT3 = 0,9 n3RT1 ; P1V1 = n1RT1. 3 3

1 1

P 0,9nP n

= 30,400 (1 ) 0,91,9442,700 0,600

+ = 3 = 0,200 n3 = 0,48 mol

* KP (T3 ) = 23

323

p1

=

2

2

(0, 200)1 (0, 200)

1,944 = 0,081 * Khi h nhit , Kp gim cn bng chuyn dch theo chiu nghch. Chiu nghch l chiu pht nhit Chiu thun l chiu thu nhit. K THI CHN HC SINH GII QUC GIA NM 2004 (Bng A)

1. Ngi ta nung nng n 8000C mt bnh chn khng th tch 1 lt cha 10,0 gam canxi cacbonat v 5,6 gam canxi oxit. Hy tnh s mol kh cacbonic c trong bnh. Mun cho lng canxi cacbonat ban u phn hy ht th th tch ti thiu ca bnh phi bng bao nhiu? Bit ti nhit kh CO2 trong bnh c p sut l 0,903 atm . 2. Ti 200C, phn ng: H2 (k) + Br2 (lng) 2 HBr (k) (1) c hng s cn bng Kp = 9,0 .1016 . k hiu (k) ch trng thi kh. a) Hy tnh Kp ca phn ng: H2 (k) + Br2 (k) 2 HBr (k) (2) ti 20OC v p sut p = 0,25 atm. b) Hy cho bit s chuyn dch cn bng ha hc ca phn ng (2) nu gim th tch bnh phn ng hai trng hp:

Br2 (k)

*) Trong bnh khng c Br2 (lng) ; **) Trong bnh c Br2 (lng). BI GII: 1. a) Vi iu kin cho trong bnh c phn ng: CaCO3 CaO + CO2 (k) (*)

Trong bnh ch c kh CO2. Gi thit l kh l tng, ta c: n = = = 0,01 (mol). Vy n = 0,01 mol. PV RT

0,903 1,0 0,082054 1073,15 CO

Nhn xt: Theo bi, lng CaCO3 cho vo bnh chn khng l: n = = 0,1 mol

2

3

10 CaCO 100

Lng CaCO3 b phn tch ch l 0,01 mol. S c mt ca 5,6 gam CaO v lng CaCO3 cn li khng nh hng ti kt qu tnh v

cc cht ny trng thi rn chim th tch khng ng k. b) Gi thit lng CaCO3 cho vo bnh chn khng b phn tch ht p sut kh CO2 vn l 0,903 atm (v phn ng (*) t ti cn bng ho hc ).

Do : Vmin = n RT / P = 0,1 0,082054 1073,15 / 0,903 = 9,75 (lt)

2. a) Phn ng H2 (k) + Br2 (lng) 2 HBr (k) (1) c (Kp)1 = p2HBr / p H2 (a) cn phn ng: H2 (k) + Br2 (k) 2 HBr (k) (2) c (Kp)2 = p2HBr / p H2 p Br2 (b) Xt cn bng Br2 (lng) Br2 (k) (3) c (Kp)3 = pBr2 (k) (c) Khi t hp (1) vi (3) ta c cn bng (2):

H2 (k) + Br2 (lng) 2 HBr (k) (1) Br2 (l) Br2 (k) (3)

(1) (3): H2 (k) + Br2 (k) 2 HBr (k) (2) Vy (Kp)2 = = = 3,6 . 1017 (atm)

b) Khi gim th tch bnh phn ng ngha l tng p sut ring phn ca kh trong h. Xt: Q = p2HBr / p H2 p Br2 (d) Trng hp 1: Khng c brom lng trong bnh: Phn ng (2) c tng s mol kh trc v sau phn ng bng nhau (n = 0) nn s thay i p sut khng dn ti chuyn dch cn bng (2). Trng hp 2: C brom lng trong bnh: p sut ring phn ca cc kh H2 , HBr tng; trong lc p sut ring phn ca Br2 kh li khng i do cn Br2 lng. Theo (d), v s m ca pHBr ln hn s m ca pH2 nn s tng p sut ni trn dn n s tng Q v cn bng (2) chuyn dch theo chiu nghch.

II. OLYMPIC HA HC QUC T: OLYMPIC HA HC QUC T NM 1997:

a) iu kin ban u 300K v 1,01325.107Pa, kh clo c coi l kh l tng. Gin n mt mol kh Cl2 iu kin n p sut cui l 1,01325.105Pa. Trong qa trnh gin n p sut ngoi lun c gi khng i l 1,01325.105Pa. Kt qa ca s gin n l kh clo c lm lnh n 239K ( cng l im si thng thng ca Cl2 lng), thy c 0,1mol Cl2 lng c ngng t.

im si thng thng, entanpi ha hi ca Cl2 lng bng 20,42kJ.mol-1, nhit dung mol ca Cl2 kh iu kin ng tch l Cv = 28,66J.K-1mol-1 v t trng ca Cl2 lng l 1,56 cng ti 239K. Gi thit nhit dung mol iu kin ng p ca Cl2(k) l Cp=Cv+R. Bit 1atm = 1,01325.105Pa.

R = 8,314510J.K-1.mol-1 = 0,0820584L.atm.K-1.mol-1. Hy tnh bin thin ni nng (E) v bin thin entropy ca h (Ssys) trong cc bin i

m t trn. b) Vi cc phn ng sau 298K:

[Ni(H2O)6]2+ + 2NH3 [Ni(NH3)2(H2O)4]2+ + 2H2O (1) lnKc = 11,60 v Ho = -33,5kJ.mol-1 [Ni(H2O)6]2+ + en [Ni(en)2(H2O)4]2+ + 2H2O (2) lnKc = 17,78 v Ho = -37,2kJ.mol-1

Ghi ch: en l vit tt ca etylendiamin (phi t trung ha hai rng). Bit R = 8,314510J.K-1.mol-1 = 0,0820584L.atm.K-1.mol-1.

Tnh Go,So v Kc ca phn ng c (3) sau: [Ni(NH3)2(H2O)4]2+ + en [Ni(en)2(H2O)4]2+ + 2NH3 (3)

BI GII: a) Tm tt cc chuyn ha:

E1 E2 1,01325.105Pa (1atm)

Cl2(k)1mol 300K

Cl2(k)1mol 239K

Cl2(l)0,1mol 239K

Lm lnh 1,01325.107Pa (100atm)

(Kp) (Kp)

9,0 1 3

10160,25

Qa trnh chung l s gin n km theo s thay i trng thi ng p (kh sang lng) v do ni nng (E) l mt hm trng thi, s bin i ni nng tng cng l E = E1 + E2.

Qa trnh 1: E1 = nCvdT = -1748,3J Qa trnh 2: thun tin, cc s liu c i sang atm; th tc tng ng cho Pa i

hi cc h s thch hp. Theo quan im v nng lng, qa trnh hnh thnh cht lng 2 c th c chia thnh

hai giai on: . S ha hi mt nhit (gim ni nng, -) t h thng ra mi trng ngoi (v qa trnh

xy ra p sut khng i, nhit bng vi bin thin entanpy) . Cng thc hin bi mi trng ngoi nn h xung th tch nh hn (tng ni nng,

+). Th tch kh ngng t l V = nRT/P = 1,96L Th tch ca Cl2 lng = 4,54mL E2 = H2 - PextV (i pha) = H2 Pext(Vl Vk) Nhng Vl gn bng 0 nn c th b qua (th tch cht lng 4,5mL tnh ton so vi 17,6L, sai s tnh ton 0,03%) E2 = 0,1(-Hhi) + Pext.Vk = -1843,5J. E = E1 + E2 = -3591,8J Entropy S l mt hm trng thi hia bin s. Do trong qa trnh 1 cc bin s bit l T

v P, S c biu th S(T,P). Sh = S1 + S2. v Cp = Cv + R = 36,97JK-1mol-1. S1 = nCpln(T2/T1) nRln(P2/P1) = 29,89JK-1. Vi s chuyn pha (nhit khng i) v trong iu kin ng p th S2 = H2/T = -

8,54JK-1. Sh = 21,35JK-1.

b) Cn cnh gic vi vic b qua cc sai s s thay i n li gii bi ny: S khc bit l nh do chuyn i qua li vi ln nhng cch lp ln s nh hng nhiu hn. i du ca lnKc v Ho cho phn ng 1 khi o ngc. Cc hng s cn bng c nhn ln khi cc phng trnh c cng thm, nh vy s

phi thm cc lnK. Phn ng 3 = Phn ng 2 - phn ng 1. Vy S3 = S2 - S1 v G3 = G2 - G1Go1 = -RTlnKc1 = -28740J.mol-1 = -28,74kJ.mol-1. Ho1 = -33,5kJ.mol-1. So1 = (Ho1 - Go1)/T = -0,0161kJ.K-1.mol-1 = -16,1JK-1mol-1. Tng t: Go2 = -44,05kJ.mol-1. Ho2 = -37,2kJ.mol-1. So2 = -22,98JK-1mol-1. Vy cc s liu ca phn ng 3 l: Ho3 = -3,7kJ So3 = 39,01JK-1. Go3 =Ho3 - TSo3 = -15,35kJ.mol-1. Vy Kc3 = 4,90.10-2. Mt khc: Go3 = Go2 - Go1 = -15,31kJ.mol-1 Kc = 4,82.10-2.

So3 = (Ho3 - Go3)/T = 38,96JK-1. OLYMPIC HA HOC QUC T 1999:

Hp cht Q (khi lng mol phn t l 122,0g.mol-1) gm c cacbon, hydro v oxy. PHN A: Nhit sinh (entanpi to thnh) tiu chun ca CO2(k) v H2O(l) ti 25oC tng ng l

393,51 v 285,83kJ.mol-1. Hng s kh R = 8,314J.K-.mol-1. Dng lng d oxy t chy ht mt mu cht rn Q nng 0,6000g trong mt nhit lng

k ban u cha 710,0g nc ti 25,000oC. Sau khi phn ng xong, nhit ln ti 27,250oC v c 1,5144g CO2(k) v 0,2656g H2O c to thnh.

1. Hy xc nh cng thc phn t v vit, cn bng phng trnh phn ng t chy Q vi trng thi vt cht ng. Cho nhit dung ring ca nc l 4,184J.g-1.K-1 v bin thin ni nng ca phn ng trn

(Uo) l 3079kJ.mol-1. 2. Hy tnh nhit dung ca nhit lng k (khng k nc). 3. Hy tnh nhit sinh (entanpi to thnh) tiu chun (Hof) ca Q.

PHN B: S liu sau y thu c khi xt s phn b cht q gia benzen vi nc ti 6oC, CB v CW l

nng cn bng ca cht Q tng ng trong lp benzen v lp nc: Gi thit rng khng ph thuc vo nng v nhit , cht Q ch c mt dng trong benzen.

Nng (mol.L-1) CB CW

0,0118 0,00281 0,0478 0,00566 0,0981 0,00812 0,156 0,0102

4. Vi gi thit Q dng monome trong nc, bng s tnh ton hy ch ra liu trong benzen cht q dng monome hay dime.

gim nhit ng c ca mt dung dch long l tng c tnh theo biu thc sau:

f

sof

fof H

XTRTT =

2)(

Trong Tf l nhit ng c ca dung dch, Tof l nhit ng c ca dung mi, Hf l nhit nng chy ca dung mi cn Xs l phn mol ca cht tan. Khi lng mol phn t ca benzen l 78,0g.mol-1. 1atm benzen nguyn cht c nhit ng c l 5,40oC. Benzen c nhit nng chy l 9,89kJ.mol-1.

5. Hy tnh nhit ng c (Tf) ca dung dch gm 0,244g cht Q v 5,85g benzen ti 1atm.

BI GII:

1) 2:6:700984,0:0295,0:0344,0161575,0

0,182.2656,0

0,445144,1:: =====OHC

Khi lng phn t ca C7H6O2 = 122 ging nh khi lng phn t u bi cho. C7H6O2(r) + 15/2O2(k) 7CO2(k) + 3H2O(l)Hay 2C7H6O2(r) + 15O2(k) 14CO2(k) + 6H2O(l)

2) Tng nhit dung = 1.730,6 == KkJT

UnT

q oQv

Nhit dung ca nc = 710,0 . 4,184 = 2971J.K-1. Nhit dung ca nhit lng k = 6730 2971 = 3759J.K-1. 3) Ho = Uo - RTnk = -3080 kJ.mol-1. Hof (Q) = -532kJ.mol-1. 4)

CB(mol.L-1) 0,0118 0,0478 0,0981 0,156 CW(mol.L-1) 0,00281 0,00566 0,00812 0,0102 CB/Cw 4,20 8,44 12,1 15,3 CB/Cw2 1,49.103 1,49.103 1,49.103 1,50.103

T cc kt qa trn ta thy rng t s CB/Cw thay i ng k cn t s CB/Cw2 hu nh khng i cho php kt lun rng trong benzen Q dng dime.

5) Nu Q hon ton nh hp trong benzen th MQ = 244

CTT

X

off

Q

54,4861,0

10.32,1

0,7885,5

244244,0

244/244,0 22

==

=

+=

OLYMPIC HA HC QUC T 2000: Vo ngy 1thng 7 nm 2000, ng hm v cu ni gia an Mch v Thy in chnh thc

c m ca. N bao gm mt ng hm t Copenhagen n mt hn o nhn to v mt chisc cu t hn o n Malmo Thy in. Vt liu chnh dung xy dng l thp v btng. Bi ny s cp n vic sn xut v thoi ha ca tng vt liu.

Btng c hnh thnh t mt hn hp ca ximng, nc, ct v nh. Ximng cha ch yu canxi silicat v canxi aluminat c sinh ra bng cch un nng v nghin nh hn hp t st v vi. Bc tip theo trong vic sn xut ximng l thm mt lng nh thch cao CaSO4.2H2O lm tng cng cng ca btng. Bc cui cng ta nng nhit ln nhng c th nhn c sn phm khng mong mun hemihydrat CaSO4.0,5H2O theo phn ng:

CaSO4.2H2O(r) CaSO4.0,5H2O + 1,5H2O Cc ga tr nhit ng cho bng sau: (bit p = 1,00bar) Hosinh(kJ/mol) So(J.K-1.mol-1). CaSO4.2H2O9(r) -2021,0 194,0 CaSO4.0,5H2O(r) -1575,0 130,5 H2O(h) -24,1,8 188,6 R = 8,314J.mol-1K-1 = 0,08314bar.mol-1.K-1. 0oC = 273,15K

1. Tnh Ho (kJ) cho phn ng chuyn 1,00kg CaSO4.2H2O(r) thnh CaSO4.0,5H2O(r). Phn ng ny thu nhit hay ta nhit?

2. Tnh p sut cn bng (bar) ca hi nc trong bnh kn cha CaSO4.2H2O(r), CaSO4.0,5H2O(r) v H2O(h) 25oC.

3. Tnh nhit p(H2O)(cb) = 1,00 bar trong h cu 2.2. Gi s Ho v So l hng s. BI GII:

1. Ho = -1575,0 + 3/2(-241,8) - (-2021,0) = 83,3kJmol-1. n = m/M = 1000/172,18 = 5,808mol

nHo = 484kJ Vy phn ng thu nhit.

2. So = 130,5 + 3/2(188,6) 194,0 = 219,4J.K-1mol-1. Go = Ho - TSo = 17886J.mol-1. m Go = -RTlnK K = (p(H2O))3/2 = 7,35.10-4 p(H2O) = 8,15.10-3 bar.

3. p(H2O) = 1,00bar K = 0 Go = -RTlnK = 0 m Go = Ho - TSo = 0 T = 380K (hay 107oC).

OLYMPIC HA HC QUC T 2001: Nguyn l th hai ca nhit ng hc l nguyn l c bn ca khoa hc. Trong bi ny chng ta

s nghin cu nhit ng hc ca kh l tng, v cn bng ha hc. 3,00 mol CO2 gin n ng nhit (nhit ca mi trng l 15oC) chng li p sut bn ngoi

n nh p = 1,00bar. Th tch u v th tch cui tng ng l 10,0L v 30,0L. 1) Chn thong tin ng v bin thin entropy ca qa trnh gin n (Ssys) v mi trng (Ssur).

a) Ssys > 0; Ssur = 0. b) Ssys < 0; Ssur > 0. c) Ssys > 0; Ssur < 0. d) Ssys = 0; Ssur = 0

2) Tnh Ssys, gi s CO2 l kh l tng. 3) Tnh Ssur. 4) Tnh s chuyn i entropy ca h. Nguyn l hai c c nghim ng hay khng? 5) Kh CO c s dng rng ri trong tng hp hu c, c th thu c bng phn ng gia CO2

vi graphit. S dng cc d kin di y chng minh Kcb 298,15K b hn n v. 298,15K: CO2(k): Hos = -393,51kJ.mol-1; So = 213,79JK-1mol-1. CO(k): Hos = -110,53kJ.mol-1; So = 197,66JK-1mol-1. C(gr): So = 5,74JK-1mol-1.

6) Tnh nhit m phn ng t cn bng. B qua s ph thuc ca H, S vo nhit . 7) Phn ng cau 6 xy ra 800oC v p sut chung l 5,0 bar, Kp = 10. Tnh p sut ring phn

ca CO ti cn bng. BI GII: 1) c 2) Ssys = nRlnVc/V = 27,4JK-1.

3) 194,6 === JKT

VpTqS ext

pext: p sut ngoi. 4) Sh = Ssys + Ssur = 20,5JK-1.

Ngyn l hai vn c nghim ng. 5) Ho = 172,45kJ.mol-1.

So = 176JK-1mol-1. Go = Ho - TSo = 120kJ.mol-1 > 0 K < 1.

6) Go = 0 khi Ho =TSo T = 980K 7) CO2(k) + C(gr) 2CO(k).

1 - 2 5.

11

+ 5.

12

+

Kp = p2(CO)/p(CO2) pCO = 3,7 bar. OLYMPIC HA HC QUC T 2004:

Chun b cho ln sinh nht th 18 vo thng hai ca mnh, Peter c iinh bin ci tp lu ca cha m trong vn thnh mt b bi vi mt bi bin nhn to. c th c lng ga c ca vic cung nhit v nc cho nh. Peter nhn c d liu v cc kh thin nhin v gi c ca n: Cht ha hc phn mol (x) fHo(kJ.mol-1) So(Jmol-1K-1) Cop(Jmol-1K-1)

CO2(k) 0,0024 -393,5 213,8 37,1 N2(k) 0,0134 0,0 191,6 29,1

CH4(k) 0,9732 -74,6 186,3 35,7 C2H6(k) 0,0110 -84,0 229,2 52,5 H2O(l) - -285,8 70,0 75,3 H2O(k) - -241,8 188,8 33,6 O2(k) - 0,0 205,2 29,4

1) Vit phn ng chy ca kh thin nhin (ch yu l metan v etan) cho bit nit khng chy trong iu kin chn. Tnh H; S v G ca tng phn ng iu kin chun (1,013.105Pa v 25,0oC). Bit tt c cc sn phm dng kh v 0oC = 273,15K.

2) T khi ca kh thin nhin l 0,740gL-1 (1,013.105Pa; 25oC) a) Tnh hm lng ca metan v etan (mol) trong 1,00m3 kh thin nhin (CH4, C2H6 khng

phi kh l tng). b) Tnh thiu nhit khi t chy 1,00m3 kh thin nhin iu kin chun. Gi thit tt c cc

sn phm u dng hi (nu khng lm c cu a ta c th gi s rng trong 1,00m3 kh thin nhin ng 40,00mol).

c) Theo PUC (Public utility company) th nng lng thu c l 9981kWh/m3 khi ta t chy kh thin nhin (nu sn phm ch yu l kh). Sai s so vi cu b l bao nhiu.

B bi trong nh rng 3,00m, di 5,00m v su 1,50m (thp hn mt sn). Nhit nc trong vi l 8,00oC v nhit phng l 10,00oC. Gi s dnc = 1,00kg.L-1 v kh trong phng l kh l tng. Bit th tch phng l 480m3 v din tch phng l 228,16m2. 3) Tnh nng lng (MJ) khi a nhit ca nc ln 22,00oC v a khng kh trong phng ln

30,00oC. Cho bit trong khng kh cha 21,0%O2, 79%N2 v p = 1,013.105Pa. Vo thng hai, nhit bn ngoi min Bc c xp x 5oC. T khi tng b tong v mi nh

tr nn mng hn (20,0cm) th nng lng mt i nhiu hn. Nng lng c gii phng ra mi trng xung quanh (khng k n nc ngm). dn in ca tng v mi nh l 1,00WK-1m-1.

Nng lng c tnh theo cng thc: J = E(A.t)-1 = w.T.d-1. J: Nng lng gii phng. d: dy ca bc tng A: din tch t: chnh lch thi gian t thi im bt u nh thi im kt thc. T: chnh lch nhit gia bn trong v bn ngoi phng. E: Nng lng trong phng.

4) Tnh nng lng cn thit (MJ) gi cho nhit trong phng l 30,0oC. 1,00m3 kh thin nhin c gi 0,40euro v 1kWh in gi 0,137euro. Tin thu trang thit b t

nng kh l 150euro trong tin thu l si in l 100euro. 5) Ton b nng lng cn thit cho k hoch b bi ma ng ca Peter l bao nhiu?. Lng

kh thin nhin cn l bao nhiu bit hiu sut ca my t nng kh l 90%.

Peter phi tn bao nhiu tin khi dung my t nng kh v l si in bit hiu sut ca l si l 100%. BI GII: 1) Phng trnh phn ng:

a) Metan: CH4 + 2O2 CO2 + 2H2O Ho = -802,5kJ.mol-1. So = -5,3J.mol-1.K-1. Go = -800,9kJ.mol-1. b) Etan: 2C2H6 + 7O2 4CO2 + 6H2O Ho = -2856,8kJ.mol-1. So = +93,2J.mol-1.K-1. Go = -2884,6kJ.mol-1.

2) a) m = .V = 740g Mtb = x(i).M(i) =0,0024.44,01+0,0134.28,02+0,9732.16,05+0,011.30,08 = 16,43g.mol-1. ntng = 45,04mol. n(CH4) = 43,83mol. n(C2H6) = 0,495mol. b)Ethiu nhit (H2O(k)) = n(i).cHo(i) = -35881kJ. EPUC(H2O(k)) = 9,981.1.3600 = 35932kJ chnh lch = -0,14%.

3) * Vnc = 22,5m3. nnc = V..M-1 = 1,249.106mol. Enc = nnc.Cp.T = 1316MJ. * nkk = PV(RT)-1 = 2,065.104 mol Cp(kk) = 29,16JK-1mol-1. Ekk = nCpT = 12,05MJ

4) J = E(A.t)-1 = w.T.d-1. E = 1556MJ.

5) Etng = Enc + Ekk + Ephng = 2884MJ Th tch kh tng ng = 2884.106(3600.9981.0,9)-1 = 89,13m3. My t nng kh tn = 0,4.89,18 + 150 = 185,67euro. L si in tn = 2884.106.0,137.(3600.1)-1 + 100 = 209,75euro.

III. BI TP CHUN B CHO CC K THI OLYMPIC HA HC QUC T: BI TP CHUN B CHO IChO LN TH 30 :

Diliti, mt cht thit yu cho h thng y ca tu khng gian Enterprise, l mt tiu phn thc s bit (d n khng th hin mi tnh cht m Roddenberry v cng s gn cho n!). Diliti c to thnh do s kt ni hai nguyn t liti pha kh.

Li(k) + Li(k) Li2(k) (1) a) Entanpi to thnh ca diliti kh c th o c trc tip. Tuy nhin bit c cc tham s

nhit ho hc sau y: Hof(Li(k)) = 159,4kJ.mol-1. IE(Li(k)) = 5,392eV (1eV = 96,486kJ.mol-1) Do(Li2+(k)) = 129,8kJ.mol-1 (Do(Li2+(k)) l mnh lin kt ca Li2+(k)) IE(Li2+(k)) = 5,113eV. Dng cc ga tr trn, xc nh Hof(Li2(k)) v Do(Li2(k))

b) Nh ha hc ch huy l plasma xon trn tu Enterprise ang th nghim hot ng ca h thng. ng ta np 122,045g liti nguyn cht vo bung phn ng trng. Bung phn ng c th tch 5,9474.105m3 v c duy tr ti nhit hot ng l 610,25K. Mt thit b o p sut rt nhy cho thy p sut trong bung n nh ti 9,462.10-4Torr (1Torr = 0,133322kPa); phn tch bng phng php quang ph ca mt trong bung phn ng cho thy ton b liti ha hi (Bung phn ng lm bng hp kim durani c p sut hi bng khng ti 610,25K). Tnh p sut ring phn ca hi liti v diliti trong bung phn ng v tnh Kc ca phn ng (1) ti nhit ny?. Bit R = 8,31441JK-1mol-1; MLi = 6,9410gmol-1.

c) Sau , chuyn 265,834g liti vo bung phn ng trng khc ging nh trn (cng ti 610,25K). p k ca bung phn ng ny n nh ti 1,0455.10-3Torr. Tnh p sut hi ca diliti ti 610,25K?

d) Ga thit rng phn ng 2Li(k) Li2(k) l ngun nng lng y duy nht cho tu khng gian Enterprise, tnh khi lng liti ti thiu cn mang theo lm nhin liu nu tu Enterprise phi gia tc t trng thi dng n vn tc bng na vn tc nh sang. Khi lng tng cng ca tu Enterprise, mt tu khng gian ln vi phi hnh on gm vi trm ngi l 3,586.106kg (cha k nhin liu). tnh ton n gin, c th gi thit nh sau: Khng xt hiu ng tng i tnh. Cc bung phn ng c th c duy tr lien tc ti 610,25K m khng hao tn nng

lng. Liti rn c np vo bung phn ng, c th ha hi thnh cc nguyn t liti m cng

khng hao tn nng lng (o li, qa trnh ngc li cng khng phng thch nng lng).

Entanpi lin kt Li-Li c th c chuyn ha 100% thnh nng lng ca tu khng gian.

Lng liti mang theo lm nhin liu khng tnh vo khi lng thc t ca tu Enterprise.

(R rng mt s gi thit trn l khng hp l, nhng y ta khng quan tm n iu . D sao cng ch l chuyn khoa hc vin tng). e) Cui cng, hy nhn xt v cc thun li hin nhin (hay ngc li) ca s to thnh diliti dng

trong phi hnh vi tc cn nh sang. Liu s to thnh diberi Be2(k) s l mt phng n khc tt hn?.

BI GII: 1) C th xy dng cc i lng nhit ng vi cc i lng bit l:

A = 2Hof(Li(k)) B = IE(Li(k)) C = -Do(Li2+(k)) D = -IE(Li2+(k))

Li+(k) + Li(k) Li2+(k) Li(k) + Li(k) Li2(k) 2Li(r)Vi cc n l: E = A + B + C + D = Hof(Li2(k)) v F = A E = Do(Li2(k)) T tnh c Hof(Li2(k)) = 216,0kJ.mol-1 v Do(Li2(k)) = 102,8kJmol1

C

B F

E A

D

b) S mol Li trong mu th l: (122,045g/6,9410gmol-1) = 17,583mol. Nu ch c hi Li(k) th t PV = nRT P = nRT/P = 0,15000Pa Pd on = 1,1251.10-3Torr. T phn ng thun nghch: Li(k) + Li(k) Li2(k) c th thy c 2 mol Li(k) to thnh 1 mol Li2(k), v nh vy p sut ring phn ca diliti phi

bng hiu s ca p sut hi liti nguyn t d on vi p sut ton phn o c: P(Li2(k)) = 1,1251.10-3 9,462.10-4 = 1,789.10-4Torr. Tng t: P(Li(k)) = 9,462.10-4 1,789.10-4 = 7,673.10-4Torr. xc nh Kc phi i cc p sut ring phn thnh nng . Da trn phng trnh trnh thi

PV = nRT v cc hng s chuyn i khc cho thy 610,25K c th i Torr thnh mol.L-1 bng cch nhn cho 2,6276.10-5 mol.L-1.Torr-1, tm c [Li2(k)] = 4,701.10-9 mol.L-1 v [Li(k)] = 2,016.10-8mol.L-1. Vi Kc c nh ngha l:

Kc = [ ][ ] 72)(

)(2 10.156,1= ck

k KLi

Li

c) Trong bung phn ng ny, s mol Li = 265,384/6,9410 = 38,234mol. C th bit c l khng phi tt c liti u pha hi.

Nu tt c Li(r) chuyn thnh Li2(k) s c 38,234 mol Li2(k) trong th tch 5,9474.108 lt, tng ng vi nng Li2(k)] = 3,2143.10-8mol.L-1, ngha l vi p sut ton phn l 1,2233.10-3Torr. Kt qa ny cao hn p sut o c do phn ln hi liti dng nguyn t hn l dng hai nguyn t th p sut ton phn l ra phi ln hn 1,2233.10-3Torr nu ton b liti ho hi. C th kt lun, do p sut n nh, hi liti cn bng vi liti rn v lng: nh vy p sut o c tng ng vi p sut hi ca liti. p sut hi ca diliti c th c nh ngha l p sut ring phn ca hi Li2(k) trong pha kh ti cc iu kin ny.

C th gii c Kc bng cch thay vo [Li(k)] + [Li2(k)] = 2,7472.10-8mol.L-1 (t p sut o c v dng h s chuyn i Torr

mol.L-1 cu b) ta c [Li2(k)] = 5,553.10-9mol.L-1 tng ng vi Phi(Li2(k) = 2,113.10-4Torr ti 610,25K.

d) Trc ht, ta cn xc nh nng lng cn thit gia tc tu Enterprise n mt na tc nh sang. Nng lng ng hc l: E = mv2/2 vi m = 3,586.106kg v v = 1,48996.108m.s-1, tm c: E = 4,0286.1019kJ. V entanpi lien kt ca Li2(k) l 102,8kJ.mol-1 (xc nh cu a)

ccn 3,919.1017mol diliti ngha l 5,44.1015kg diliti! (Tht ra cn nhiu hn th v cn tn ti cn bng gia liti v diliti dng kh).

e) Cc tnh ton cu d ch ra rng diliti l cht y khng hiu qa lm nh mt s nh nghin cu ngh1. Tuy nhin diberi cn t hn: thuyt MO cho thy Li2(k) c bc lin kt l 1, trong khi Be2(k) coi nh khng c lin kt no.

BI TP CHUN B CHO IChO LN TH 30: Hai ng v thng c xem nh c hot tnh ho hc ging nhau: tuy nhin, iu ny khng

tht chnh xc. Hot tnh khc bit ca cc ng v khc nhau l do s ph thu ca nng lng dao ng khng gian trong cc phn t vi khi lng cc ht nhn nguyn t cu to nn phn t . y, ta khng cn quan tm n c ch chi tit m ch lu rng cc hp cht c cha cc ng v nh (nh 1H19F) c bn lin kt hi thp hn cc hp cht tng ng nhng c cha cc ng v nng hn (nh 2H19F).

Hiu ng ng v ny thng khng quan trng hoc t quang trng ti nhit phng, nhng li c ngha quyt nh trong ha hc ti nhit thp. Mt trong cc mi trng nghin cu c tin hnh rng ri l cc m my dy c gia cc v sao ti nhit rt thp (thng khong 10K 20K), l nhng m my ln cha kh v bi, ngun gc nguyn thy ca cc ngi sao. S chuyn ha ca deuteri trong cc m my lnh gia cc v sao xy ra vi nhiu qa trnh khc nhau, gm c ch sau:

H2 + D HD + H (1) HD + D D2 + H (2) Cc tham s nhit ho hc c lin quan n phn ng (1) l: Hof(H2(k)) = 0kJ.mol-1 So(H2(k)) = 130,57J.K-1.mol-1. Hof(HD(k)) = 0,33kJ.mol-1 So(HD(k)) = 143,69J.K-1.mol-1. Hof(H(k)) = 216,00kJ.mol-1 So(H(k)) = 114,60J.K-1.mol-1. Hof(D(k)) = 219,76kJ.mol-1 So(D(k)) = 123,24J.K-1.mol-1. (Entanpi to thnh c tr s ti 0K (thch hp hn so vi nhit gia cc v sao hn l o ti

298K), entropy c tr s ti 298K, nhng tha mn yu cu ca chng ta c th ga thit l khng ph thuc vo nhit ). a) Xc nh bin thin nng lng t do Go ca phn ng (1), ti T = 20K v T = 1000K. Phn

ng xy ra theo hng no, nu nng ban u ca mi cht tham gia v cht to thnh l bng nhau?

b) Du ca Ho (dng hoc m) s cho bit g v bn lin kt ca H2 v HD? Du ca So nh th no phn ng (1) xy ra theo chiu thun v c s vt l cho du ca So l g?

c) Nay xt phn ng (2). Du ca Ho, So phi nh th no phn ng xy ra theo chiu thun? Nu gi thit rng bin thin entanpi v entropy l bng nhau v ln (bin ) vi ga tr tm c cu a. Hy tnh bin thin nng lng t do v d on chiu phn ng xy ra ti 20K v ti 1000K.

d) Hydro phn t (vi cc dng nh v khc nhau) hin din nng cao hn nhiu so vi hydro nguyn t (v deuteri nguyn t) trong cc m my gia cc v sao. D on dng u i ca deuteri (D, HD hay D2) trong m my gia cc v sao, gi thit ti nhit 20K:

i) Khi hm lng trong v tr nD

BI GII: a) Bin thin nng lng t do tnh c nh Go = Ho - TSo

Ho = -3,43kJ.mol-1. So = +4,48J.mol-1.K-1. Ti 20K th Go = -2,52kJ.mol-1, ti 1000K th Go = -7,91kJ.mol-1.

Phn ng xy ra ti c hai nhit (Tht vy, v Ho m v So dng, phn ng thun phi c th t xy ra c mi nhit )

b) Ho m theo chiu thun: nh vy bn lin kt ca HD ln hn H2. iu ny ph hp vi khuynh hng tng qut nu trn l cc tiu phn cha ng v nh hn s c lin kt hi yu hn.

So dng theo chiu thun, ch ra rng sn phm km trt t hn so vi cht tham gia. S gia tng tnh khng trt t c th c hiu l do kh nng to c hai phn t HD ging nhau (cha ln lt mt trong hai nguyn t H ging nhau) t mt phn t H2.

c) S chuyn ha HD thnh D2 kt hp vi Ho m, do bn lien kt trong D2 ln hn HD (cc ng v nng hn c lien kt bn hn). Bin thin entropy theo chiu thun cng s m, do (gii thch tng t trn) HD c khng trt t ln hn D2. Gi s ln ca Ho v So bng vi ga tr ca cu a ta c: Go = -3,34kJ.mol-1 ti 20K, Go = +1,05kJ.mol-1 ti 1000K. Nh vy phn ng thun t xy ra ti 20K nhng ti 1000K th xy ra phn ng nghch.

d) i) Do cn bng (1) lun di theo chiu thun bt k nhit no, r rng l D nguyn t khng th l dng c hm lng ln nht ca deuteri. Thy ngay l HD s c u th hn D2 nu hm lng deuteri tng cng l thp, nhng vn c th xem xt k hn:

Nu cn bng (2) di theo chiu nghch, khi y kt hp vi (1) cho: H2 + D + D2 + H 2HD + H +D Thu gn: H2 + D2 2HD (3) Bin thin nng lng t do ca cn bng ny l Go(3) = Go(1) - Go(2). Cc tnh ton cu a) v cu c) c th tnh c Go(3) = -0,18kJ.mol-1. Do Go(3) m, cn bng s di chuyn v pha chiu thun v th HD s c nhiu hn D2. ii) Lp lun trn cng p dng cho trng hp nD = nH nn HD vn nhiu hn D2. Tng t, do cn bng (3) di theo chiu thun, HD vn l dng c nhiu hn.

BI TP CHUN B CHO IChO LN TH 30: Phn tch nguyn t ca mt hp cht hu c X cho bit thnh phn gm c C (40,02% theo

khi lng) v H (6,75% theo khi lng). X khng cha N hoc S c th phn tch tm c v c th gii thit rng phn cn li l oxy khng th phn tch trc tip. a) Tm cng thc thc nghim ca X. b) X l cht lng ti nhit phng, 10mL X (khi lng ring = 1,044g.mL-1) c thm vo xiclohexan c tng th tch l 500mL. Khi lng ring ca dung dch ny c xc nh l 0,777g.mL-1. im ng c ca dung dch l +2,02oC. Nht ng c (Tf) ca xiclohexan (x-C6H12) = 6,60oC; Hng s nghim ng Kf(x - C6H12) = 20,0oC.kg.mol-1. T cc s liu trn, tm khi lng phn t v cng thc phn t ca X.. c) X tan tt trong nc. Thm 50mL X vo nc siu tinh khit t tng th tch l 500mL. Dung dch ny c khi lng ring l 1,005g.mL-1 v im ng c l -3,54oC. Kf(H2O) = 1,86oC.kg.mol-1. Khi lng phn t ca X theo phng php ny l bao nhiu? d) Trong dung dch nc th X phn ng c vi baz. Chun 25,00mL dung dch trong nc ca X c iu ch cu c) vi dung dch NaOH 1,247M, im dng c xc nh bi pH k l 33,60mL

dung dch NaOH. Tng th tch ca h ti im dng l 58,50mL, vi khi lng ring l 1,003g.mL-1, ng c ti -2,78oC. Tm cu to ca sn phm (Na+)iY- to thnh t phn ng ca X vi dung dch nc NaOH? e) Xc nh cu to ca hp cht X? Nhn xt v cc thiu st hin nhin, nu c trong li gii b); c); d) BI GII: a) Trong 100g mu th: 40,02g l C: tng ng vi : (40,02/12,011) = 3,332 mol nguyn t C. 6,75g l H: tng ng vi (6,75/1,00797) = 6,697 mol nguyn t H. 53,23g l O: tng ng vi (53,23/15,9994) = 3,327 mol nguyn t O T l C : H : O = 1,001 : 2,013 : 1 nn cng thc thc nghim l CH2O b) Tf = Kf.M, trong Tf l h im ng c v M l nng mol ca dung dch. Bc th nht l xc nh nng mol ca dung dch: Tf = 6,60 - 2,02 = +4,58oC M = Tf /Kf nn M = 0,229mol.kg-1Bc k tip l xc nh khi lng dung mi: Khi lng dung dch = Khi lng ring x Th tch = 0,3885kg Khi lng cht tan = Khi lng ring x Th tch = 10,44g (hoc 0,01044kg) Khi lng dung mi = 0,3885 - 0,01044 = 0,3781kg Nng mol trong trng hp ny c nh ngha l s mol cht tan chia cho khi lng dung mi nn: S mol X = 0,229.0,3781 = 8,66.10-2 mol Khi lng X chia s mol X cho khi lng mol phn t: Mr(X) = 10,44/8,66.10-2 = 120,6g.mol-1T khi lng mol phn t s xc nh c cng thc ca X l (CH2O)4 hay C4H8O4. c) Cch gii cu ny phn ln tng t nh cu b): s khc bit ch yu l trong dung mi phn cc nh nc th phi c s ion ha. Kh nng ny (phn li thnh ion) cn n cng thc Tf = i.Kf.M, tng i cho bit s trung bnh ca cc tiu phn thu c do phn li vi mi phn t cht tan. Tf = 0,0 - (-3,54) = +3,54oC i.M = Tf /Kf = 3,54/1,86 = 1,903mol.kg-1Khi lng dung mi = (500.1,005) - (50.1,044) = 450,3g Vi Mr(X) c c cu b), ta xc nh c 50mL X tng ng vi 0,433mol M = 0,433/0,4503 = 0,961mol.kg-1Tnh h s phn li i: i = 1,903/0,961 = 1,980 Nh vy s phn li ca x thnh 2 tiu phn trong dung dch nc tnh c. Mt khc, nu ta gi thit rng i = 1 th khi lng mol phn t ca X tnh theo phng php ny l 120,6/11,980 = 60,9g.mol-1d) Trc tin tnh s mol hydroxit tiu th trong phn ng: S mol OH- = 1,247.33,60.10-3 = 4,190.10-2 mol Do theo di phn ng bng pH, c th gi nh th nghim ny gn nh l th nghim chun axit - baz. V vy s mol OH- phn ng vi cng s mol H+ to thnh nc (s lp lun ny l cn thit xc nh khi lng dung mi). Khi lng dung dch = th tch x khi lng ring = 58,50.1,003 = 58,68g Khi lng cht tan = Khi lng Yi- + khi lng Na+Khi lng Yi- = khi lng X - khi lng H+ x s mol OH- = (25,00.1,044.0,1*) - 1,00797.0,0419 = 2,57g

* 0,1: h s pha long (50mL X c pha long thnh 500mL cu c)

Khi lng Na+ = 22,990.0,0419 = 0,96g Vy khi lng dung mi = 58,68 - (2,57 + 0,96) = 55,15g Vi i.M = Tf /Kf ta c: i.M = 2,78/1,86 = 2,495mol.kg-1Tham s i.M biu th s mol cc tiu phn ho tan trong mt kg nc. Cc tiu phn ho tan ny l Na+ v Yi-. Vi khi lng dung mi bit ta c: S mol Na+ + s mol Yi- = 1,495.5,515.10-2 = 8,24.10-2 mol bit s mol Na+ = s mol OH- dng nn s mol Yi- = 8,24.10-2 - 4,19.10-2 = 4,05.10-2 mol V s mol Yi- : Na+ = 1 : 1,03; nh phn tch im ng c xc nh c mui to thnh c cng thc Na+Y-. So snh khi lng Yi- tm thy trn vi s mol Yi- ta c Mr(Yi-) = 64,4g.mol-1 so vi Mr(C2H3O2-) = 59,0g.mol-1 c th kt lun mui to thnh l C2H3O2Na. Ghi ch rng, nh nghi cu b), nu cng thc ca X l C4H8O4, phi c: nX = 25,00.1,044.0,1/120,6 = 2,16.10-2 mol Nh vy chuyn C4H8O4 thnh 2(C2H3O2-) hin nhin ph hp e) Cc kt qa ca cu b) v c) khng thng nht vi nhau: dung mi xiclohexan dn n khi lng mol phn t gp 2 ln so vi trng hp nc l dung mi. iu ny ch ph hp nu X phn li hon ton thnh 2 ion trong dung dch nc: nn c l X l mt axit mnh. Tuy nhin cc axit mnh tng i him gp trong ha hc hu c. Hn na, ta cng d on mt axit mnh cng khng th phn li hon ton trong mt dung mi khng phn cc nh xiclohexan! Kt qa l cu d) cho thy mt vn khc: c hai "sn phm in li" ca X trong dung dch nc u chuyn thnh C2H3O2-. iu ny khng th xy ra c v hai sn phm phi l ion c in tch tri du. Mt cch gii thch khc ph hp hn v phng din ho hc l "dng phn li" ca X trong H2O cu c) l cc tiu phn trung ho c kh nng phn ng nh mt axit trong cu d). Khi lng mol phn t tnh c ca cc tiu phn "phn li trung ho" l 60,9g.mol-1, so vi 60,1g.mol-1 d on cho C2H4O2. C vi ng phn ca C2H4O2 v ng phn phn ng nh mt axit l axit axetic (axit etanoic). Vy v sao X to thnh 2 phn t axit axetic trong dung dch nc? Cch tr li tt nht l X khng phn li trong dung dch nc m dime ha trong xiclohexan. Lin kt hydro trong dng nh phn ca axit axetic:

CH3C

O

O

H

C CH3

O

O

H Dng ny l i xng v vy tan c trong dung mi khng phn cc nh xiclohexan BI TP CHUN B CHO IChO LN TH 31:

a) Nhit chy (entanpi chy, Ho) v nhit to thnh tiu chun (entanpi to thnh tiu chun Hof) ca mt nhit liu (cht t) c th c xc nh bng cch o bin i nhit trong mt calo k khi mt lng xc nh nhin liu c t chy trong oxy.

i) Cho 0,542g iso-octan vo mt calo k c dung tch khng i (bom), m bao quanh bnh phn ng l 750g nc ti 25,000oC. Nhit dung ca chnh calo k (khng k nc) c o trc l 48JK-1. Sau khi iso-octan chy ht, nhit ca nc t 33,220oC. Bit nhit dung ring ca nc bng 4,184J.g-1.K-1, hy tnh bin thin ni nng Uo ca s t chy 0,542g iso-octan.

ii) Hy tnh Uo ca s t chy 1mol iso-octan. iii) Hy tnh Ho ca s t chy 1mol iso-octan. iv) Hy tnh Hof ca iso-octan.

Nhit to thnh tiu chun ca CO2(k) v H2O(l) ln lt bng 393,51 v -85,83kJ.mol-1. Hng s kh R bng 8,314JK-1mol-1.

b) Hng s cn bng Kc ca mt phn ng kt hp: A(k) + B(k) AB(k)

L 1,80.103L.mol-1 ti 25oC v 3,45.103L.mol-1 ti 40oC. i) Gi s Ho khng ty thuc vo nhit , hy tnh Ho, So. ii) Hy tnh cc hng s cn bng KP v Kx ti 298,15K v p sut ton phn l 1atm.

(Cc k hiu Kp, KC v Kx ln lt l hng s cn bng xt theo nng , p sut v theo phn s mol).

c) Mc d iot khng d tan trong nc nguyn cht nhng n d dng tan trong nc c cha ion I-(dd).

I2(dd) + I-(dd) I3-(dd) Hng s cn bng ca phn ng ny c o nh l mt hm nhit vi cc kt qa sau: Nhit (oC): 15,2 25,0 34,9 Hng s cn bng: 840 690 530 Hy c lng Ho ca phn ng ny.

BI GII: a) (i) C8H18(l) + 25/2O2(k) 8CO2(k) + 9H2O(l).

Nhit dung ca calo k v cc cht cha bn trong: Cs = 48 + (750.4,184) = 3186JK-1. Lng nhit phng thch th tch khng i bng: Qv = CsT = 26,19kJ. T ta c: Uo = -Qv = -26,19kJ. (ii) Xt mt mol iso-octan chy:

1.5520542,0

19,26.23,114 == molkJU o (iii) Bin i entanpi (Ho) quan h vi Uo nh sau: Ho = Uo + nkhRT = -5520 4,5.8,314.298,15 = -5531kJ.mol-1. (iv) Ho = 8Hof(CO2(k)) + 9Hof(H2O(l)) - Hof(C8H18(l)) = -190kJ.mol-1

b) (i) Vi Go = -RTlnK nn:

RS

RTH

RTGK

ooo +==ln K hiu nhit thp hn 298,15K bng T1:

RS

RTHK

oo +=1

1ln

Tng t cho nhit cao hn 313,15K (T2)

RS

RTHK

oo +=2

2ln

Vy 121

12

1

2 .67,33ln =

= molkJH

TTTT

RH

KK oo

Thay Ho va tnh c vo biu thc lnK2 ta tnh c So = 175,2JK-1.mol-1. (ii) T phng trnh cho ta c:

BA

ABp PP

PK

.=

V PV = nRT nn: [ ]( )[ ]( )[ ]( ) RT

KRTBRTA

RTABK CP == Ti 298,15K KP = 0,726atm-1. T P1 = X1P nn:

726,0.. 11 ==== PKKPKPXX

XK PXX

AB

ABP

c) Chn hai gi tr bt k ca K ti hai nhit khc nhau, v d nh ti 15,2oC (288,4K) v 34,9oC (308,2K):

kJmolkJHTT

TTRH

KK oo 2,17.10.72,1ln 14

21

12

1

2 ==

=

T TR

HRSK

o 1.ln = Gi s Ho v So khng thay i, th ca lnK theo 1/T d on l mt ng thng vi

dc bng -Ho/R:

6.26.36.46.56.66.76.8

3.2 3.25 3.3 3.35 3.4 3.45 3.5

1000/T

lnK

dc = -H/8,314=2,06.103. H = -1,71.104J = -17,1kJ.

BI TP CHUN B CHO K THI IChO LN TH 31:

a) Axeton (k hiu A) v clorofom (k hiu C) u tan v hn trong nc. Ti 35oC o c p sut ring phn ca axeton v clorofom ca cc dung dch sau: XC 0,00 0,20 0,40 0,60 0,80 1,00 PC(torr) 0,00 35 82 142 219 293 PA(torr) 347 270 185 102 37 0,00 Trong XC l phn s mol ca clorofom trong dung dch. (i) Hy chng t rng cc dung dch l khng l tng.

(ii) S sai lch vi tnh l tng c th dng hoc m. Dung dch trn th hin s sai lch nh th no?.

(iii)Tnh khng l tng c th c biu th mt cch nh lng da trn hot ca mi phn t trong dung dch. Hot (a) c th thy trong phng trnh sau (ly v d vi clorofom): ac = PC/PoC trong ac l hot ca clorofom v PPoC l p sut hi ca clorofom tinh khit. Hy tnh hot ca clorofom v axeton i vi mi dung dch.

b) (i) Tm gi tr ca Kf (hng s nghim lnh hay hng s h im ng c) ca dung mi p diclobenzen t cc s liu sau: KLPT im nng chy (K) Honc(kJ.mol-1)

p diclobenzen 147,01 326,28 17,88 (ii) Mt dung dch cha 1,50g cht tan khng bay hi trong 30,0g p-diclobenzen v c im

ng c l 323,78K. Hy tnh khi lng mol phn t ca cht tan. (iii) Hy tnh tan vi dung dch l tng ca p-diclobenzen ti 298,15K.

BI GII: a) (i) T nh lu Raoult Pi = Xi.Poi

Xc : 0,20 0,40 0,60 0,80 PC = XC.Poc(Torr): 59 117 176 234 Pc(o c): 35 82 142 219 (PoC = 293Torr) XA : 0,80 0,60 0,40 0,20 PA = XA.PoA(Torr): 277 208 139 69 PA(o c): 270 185 102 37 (PoA = 347Torr) C th thy rng p sut hi tnh ton ca axeton ln clorofom u cao hn gi tr o c tt

c cc thnh phn t l. Nh vy dung dch lch so vi dung dch l tng. Tuy nhin c th biu din bng cch v th tng quan gia p sut v t l thnh phn nh

sau:

050

100150200250300350400

0 0.2 0.4 0.6 0.8 1 1.2Xc

P(To

rr)

(ii) Cc dung dch th hin s lch m so vi tnh cht l tng. (iii) T cc s liu cho, ta c th tnh c hot ca clorofom v axeton: XC : 0,20 0,40 0,60 0,80 ac = PC/PoC : 0,12 0,28 0,48 0,75 XA : 0,80 0,60 0,40 0,20 aa = PA/PoA : 0,78 0,53 0,29 0,11 Cc hot ca c clorofom v axeton u b hn phn s mol cho thy s lch m so vi tnh

cht l tng:

b) (i) T phng trnh Gibbs Hemholtz:

21ln

RTH

dTXd ofus= (1)

Trong Xi l phn s mol ca dung mi lng v Hofus l nhit nng chy ca dung mi nguyn cht. Nu Hofus c lp vi T trn mt khong nhit va phi ta c th tch hp phng trnh (1) t Tof (l im ng c ca dung mi nguyn cht ti Xi = 1) n T (l nhit ti m cn bng gia dng rn vi dng lng ng vi phn s mol X1). Kt qa l:

=

TTRH

X of

ofus 11ln 1 (2)

Biu din X1 theo X2 l phn s mol ca cht tan:

=

TTRH

X of

ofus 11)1ln( 2 (3)

Nu X2 nh (so vi dung dch) th: ln(1 X2) -X2. h im ng c ( h bng im) l Tof T = Tf. Do Tf l nh so vi Tof ta c th t

tch TTof Tof2. Phng trnh (3) chuyn thnh.

22 )(. o

f

fofus

TT

RH

X= (4)

Trong dung dch long X2 = n2/(n1 + n2) n2/n1. Vi nng mol m2 tnh theo s mol cht tan n2: m2 = (n2/w1).1000. Trong w1 l khi lng

ca dung mi tnh theo gam. Vi dung mi n1 = w1/M1, trong M1 l khi lng phn t ca dung mi th X2 = m2.M1/1000. Xp t li phng trnh (4) v thay X2 cho:

Tf = 1000.

2

1

f

of

HRTM

.m2 (5) h im ng c hay hng s nghim lnh Kf c nh ngha l:

Kf = 1000.

2

1

f

of

HRTM

(6) Thay s vo biu thc (6) th ta tnh c Kf = 7,26K.kg.mol-1. (ii) Tf = 2,50K. Vi nh ngha Kf phng trnh (6) chuyn thnh.

Tf = Kf.m2

Do m2 = 21

2

1

2

.1000.

1000.Mw

wwn =

Trong M2 l khi lng mol phn t ca cht tan. Sau khi sp xp li ta c:

M2 = 11

2 .6,145.1000.. = molgwT

wK

f

f

(iii) S dng phng trnh (2) ta tnh c X = 0,537. Vy tan theo phn s mol ca p diclobenzen ti 298,15K trong dung dch l tng bng 0,537.

BI TP CHUN B CHO K THI IChO LN TH 31: Mt axit hu c yu c phn b trong dung dch nc v cacbon tetraclorua.

a) Tm t l phn b D theo hng s phn ly axit Ka v h s phn b D bit rng: HA(aq) H+(aq) + A-(aq). HA(aq) HA(CCl4)

b) Cc th nghim chit tch ti cc pH khc nhau cho cc kt qa sau:

pH T l phn b D: 1 5,200 3 5,180 4 5,190 6 2,605 7 0,470

8,0 0,052 8,5 0,016

Hy tnh Ka v Kd. BI GII: 1) Ta c:

[ ] [ ][ ]

[ ][ ]dd

od

dd

dddda

HAHA

K

HAAH

K

=

=+

T l phn b c th c c lng bng:

[ ][ ] [ ]

[ ][ ]

[ ][ ]

[ ][ ] [ ]+

+

=+

=+=HK

K

HAA

HAHA

HAHA

AHAHA

Da

d

dd

dd

dd

dd

dd

o

dddd

o

1

2) T cu (1) ta c: lgD = lgKd lg(1 + Ka/[H+]) Ti pH thp: [H+] >> Ka. lgD = lgKd = hng s. D = Kd; Kd = (5,200 + 5,180 + 5,190):3 = 5,190 Ti pH cao: [H+]

H2S(k) -20,4 205,6 a) Tnh Ho, So, Go ti 25oC ca phn ng trn. b) Hy tnh hng s cn bng Kp ti 25oC ca phn ng trn. c) Hy tnh hng s cn bng Kp ti 35oC ca phn ng trn gi thit rng c Ho v So khng ph

thuc nhit . d) Hy tnh p sut ton phn trong bnh cha nu phn ng phn hy t cn bng ti 25oC. B

qua th tch ca NH4HS(r). e) Nu dung tch bnh cha l 100,00L. Hy tnh li p sut ton phn trong th nghim trn.

BI GII: a) Ho = 90,6kJ.mol-1.

So = 284,8J.K-1.mol-1. Go = Ho - TSo = 5,7kJ.mol-1.

b) Go = -RTlnKa Thay s vo ta thu c Ka = 0,1008. Ka = p(NH3).p(H2S) = Kp Kp = 0,1008bar2.

c) Go = Ho - TSo = 2839J.mol-1. Ka = 0,3302 Ka = P(NH3).P(H2S) = Kp = 0,3302bar2.

d) p(ton phn) = p(NH3) + p(H2S) p(NH3) = p(H2S) = 0,5p(ton phn) (do c s mol bng nhau) Kp = P(NH3).P(H2S) = [0,5p(ton phn)]2 = 0,1008 Kp = 0,635bar. nkh = pV/RT = 0,64mol n(NH4HS) = 1,00 0,5.0,64 = 0,68mol ngha l vn cn cht rn.

e) nkh = pV/RT = 2,56mol. n(NH4HS) = 1,00 0,5.2,56 = -0,28mol ngha l khng cn cht rn. 1,00mol cht rn chuyn thnh 2,00mol kh. p(ton phn) = n(ton phn).RT/V = 0,50bar.

BI TP CHUN B CHO K THI IChO LN TH 35: Mt ngi nh c trang b ni hi c nc nng s dng trong ma ng. Nng lng cn

hot ng ca ni hi l 116kW.Cn nh cn c mt thng du c V=4m3. Entanpi ca vic t chy du (cha hu ht cc ankan c khi lng phn t ln th lng) l 4300J/kg v ddu = 0,73g/cm3. 1) Xc nh thi gian hot ng ca ni hi t khi y du n khi ht du: a) 5h b) 2,2 ngy c) 12 ngy d) 3,3 tun e) 2,1 thng. 2) Xc nh lng CO2 sinh ra trong mi gi khi ni hi hot ng: a) 300g b) 1kg c) 5kg d) 10kg e) 30kg BI GII: 1) Khi lng ca thng du: m = 2920kg

Nng lng cn hot ng ca ni hi P = 116kW

Vy tc tiu th nhin liu ca ni hi l: 113 .73,9.10.70,2 === hkgskgm

hP

tm

Vy thi gian hot ng ca ni hi l: 5,12300 === ht

mmt ngy

2) CnH2n+2 + (3n+1)O2 nCO2 + (n+1)H2O

nmm

ankan

CO

17

222

+= . T l ny khng ph thuc nhiu vo n. i vi n = 10 th t l ny bng 3:1

Do 1.73,9 = hkgt

mankan nn ta c th tnh c lng CO2 thot ra l:

1.2,303.73,9. 22

=== hkgmm

tm

mankan

COankanCO

BI TP CHUN B CHO K THI IChO LN TH 35: Ho tan 80g NH4NO3 vo 1kg H2O 0oC. Xc nh trng thi cui ca h. Cho bit Cnc(l) =

76J.mol-1.K-1; Hnc = 6,01kJ.mol-1; Hho tan(NH4NO3) = 25,69kJ.mol-1; Kd(H2O)=1,86K.kg.mol-1. 1) Nhit cui ca h l:

a) 1,86K b) 3,72K c) 3,72oC d) 1,86oC e) -3,72oC f) 3,72K g) -1,86K h) -1,86oC

2) Trng thi cui ca h gm: a) Mt pha rn v mt pha lng. b) Mt pha lng v hai pha rn. c) Mt pha lng. d) Mt pha rn. e) Hai pha lng. f) Hai pha rn. g) Hai pha lng v mt pha rn.

3) Qa trnh ho tan c th c din t bng cc t (nhiu hn mt t ng) a) Cm ng. b) T pht. c) Thun nghch. d) Bt thun nghch. e) S ng nht cc cu t. f) on nhit. g) Khng on nhit. h) ng nhit. i) ng p.

j) ng tch. k) ng entropy. l) ng nng lng.

4) S thay i entropy ca h l: a) Ln hn 0. b) Bng 0 c) Nh hn 0.

BI GII: Qa trnh ho tan l thu nhit v ng p, chnh v vy nhit c t sinh ra t chnh dung dch.

Do nc ang nhit ng nn n c xu hng ng c nhng dung dch to thnh c nhit ng c gim xung do c nhng ion ho tan trong . Lng nhit sinh ra do s solvat ha s buc phi lm cho mt s phn t nc ng cng li. A. Theo nh lut Hess ta phi lp mt chu trnh ba bc. B. Trn dung dch 0oC vi H1 > 0 C. H nhit ca hn hp n nhit t sao cho H2 < 0 D. Lm ng c mt lng m gam nc vi lng nhit H3 < 0

Nhit cui ca dung dch c tnh bi biu thc: s

f mmnK = 22 . Vi Kf l hng s

nghim ng ca nc, 2 l s phn t NH4NO3 v n l s mol ca NH4NO3; n = 1mol; ms l khi lng ca dung dch

H1 = Hho tan.n = 25,69kJ.

22 MmCH P= . Vi Cp l nhit dung mol ca nc v M l khi lng phn t ca n

(18g/mol).

Mm

HH sf .3 = Vi Hf l entanpy nng chy. H = H1 + H2 + H3 = 0 bi v khng c s trao i nhit ca h vi mi trng. T tt c nhng phng trnh trn ta c th dn n h thc sau cho ms:

f

Pf

f

s

f

ss H

mCnKnMHHmnM

HHmm +

+

+= 22

.22

.2

2

Chng ta cho biu thc trc du cn thc l du + (ms > m) v thay s vo ta thu c kt qa ms = 28,52g bng. Vy 2 = -3,83oC.

Nu chng ta gi thit ms

a) 2 bar. b) 5,185 bar. c) 20 bar. d) 73,8 bar. e) Khng xc nh.

2) Kh CO2 cng tn ti pha lng v pha kh th p sut trong bnh cu ho l bao nhiu? a) Khong 2 bar. b) 5,180 bar. c) Khong 20 bar. d) Khong 63 bar. e) 73,8 bar. f) Khong 100 bar. g) Khng xc nh.

BI GII:

Do nhit phng pha trn im ba nn chc chn nhit phng CO2 khng bao gi tn ti trng thi rn. 289K th p sut hi ca CO2 l 63,1 bar. Ga tr ny cn c th c tnh mt cch nh lng khi ta k mt ng thng ni gia im ba v im ti hn. Ngoi ra n cng c th

c tnh bng cch s dng phng trnh Antoine: K

TCBA

kPaP

+=lg vi A = 6,46212; B = 748,28

v C = -16,9. BI TP CHUN B CHO K THI IChO LN TH 35:

Chit l phng php phn tch thng dng nht v n c hnh thnh trong mt cn bng gia mt cht c phn b trong c hai dung mi khng trn ln vi nhau, c t trng khng khc nhau nhiu chng d dng phn lp khi pha trn.

Phng php chit thng dng nht l chit dung dch nc vi dung mi hu c. Sau th phn ion v c nm trong dung dch nc cn phn hu c th nm trong dung mi hu c. Ion v c cng c th phn ng c vi tc nhn thch hp to ra mt hp cht khng phn cc c phn b trong dung mi hu c.

Khi mt phn t S(cht tan) c phn b gia hai dung mi (1) v (2) th chng ta c cn bng:

S1 S2KD

Vi KD l h s phn b cho bi h thc:

1

2

)()(

s

sD a

aK = (1) Vi as2 v as1 l hot ca S trong hai pha 1 v 2. Trong mt h cho trc ch gm dung mi

v cht ta th KD ch ph thuc vo nhit . Cc th nghim phn lp bng chit thng s dng phu chit, n l mt dng c thy tinh rt

d s dng v c mt trong cc phng th nghim ha hc. Phng trnh (1) ch ng khi cht tan S tn ti cng mt dng trong c hai dung mi. Theo

cch khc, nu cc cht phn b dng khc nhau (dime, trime,) trong cc dung mi th ngi ta thay th KD bng t l phn b D cho bi phng trnh:

12

s

s

CC

D = (2) Vi Cs1 v Cs2 l nng phn tch ca S trong pha 1 v 2 (khc nng cn bng ca cc cu

t). Bnh thng, khi mt trong hai dung mi l nc th phng trnh (2) c vit vi nng ca

cht trong nc t s v trong dung mi hu c mu s. D l hng s iu kin ph thuc vo cc thng s th nghim nh nng cht tan, cc phn t khc c to cn bng vi S trong pha bt k v ph thuc nhiu nht vo phng trnh ca pha nc (V d: nu S tham gia vo cn bng axit baz).

Nu Wog S lc u xut hin trong V1 mL dung mi 1 v S c chit thnh cng vi mt lng V2 mL dung mi 2 th lng Wn ca S cn li trong pha 1 sau n ln chit s c tnh theo cng thc:

o

n

n WVDVV

W

+= 121 (3)

hay: n

o

nn VDV

VWW

f

+== 121 (4)

Vi fn l phn gam S tn ti trong dung mi 1 sau n ln chit: T (3) v (4) ta d dng nhn thy rng chit nhiu ln vi mt th tch nh dung mi chit s c

hiu qa hn so vi vic chit mt ln bng ton b th tch dung mi chit. 3) Chng minh phng trnh (1.3). 4) Cht S c phn b gia clorofom vi nc vi t l phn b D = 3,2. Nu 50cm3 dung dch

nc ca S c chit vi: a) 100cm3 CHCl3. b) Chit 4 ln vi mi ln 25cm3 CHCl3.

Tnh %S thu c mi phng php. 5) S ln chit nh nht s l bao nhiu c th chit c ti thiu 99% cht X t 100cm3 dung

dch nc cha 0,500g X nu mi ln chit ta s dng 25,0cm3 hexan v h s phn b D = 9,5. 6) Axit yu HA co hng s phn ly Ka (trong H2O) c phn b gia hai dung mi l nc v

dung mi hu c. Nu hp phn ch chu nh hng ca s chit l phn khng phn ly HA vi h s phn b KD v HA ch tn ti trong dung mi hu c. Hy vit phng trnh biu th s ph thuc ca t l phn b D vo [H+] ca dung dch nc v rt ra kt lun t phng trnh ny.

7) 8 hydroxyquinolin C9H6(OH)N (k hiu OxH) c bit n di ci tn oxin. N to ra trong dung dch axit cation C9H6(OH)NH+ (OxH2+) v trong dung dch kim n to ra anion C9H6(O-)N. Clorofom ch chit c dng trung ho ca oxin vi h s phn b KD = 720.

a) Vit phng trnh th hin s ph thuc KD vo pH ca dung dch nc. b) Khi D t cc i th pH ca nc l bao nhiu: Bit:

N

OHOxH

NH+

OHOxH2+

N

O-Ox-

K1 = 1.10-5 K2 = 2.10-10

+H+ -H+

BI GII: 1) Bt u vi mt lng Wo ca S pha 1. Sau kh chit th lng ny c phn b hai pha

nh sau: Wo = (Cs)1.V1 + (Cs)2.V2V D = (Cs)2/(Cs)1 nn ta c Wo = (Cs)1.V1 + (Cs)2.V2 = (DV2 +V1)(Cs)1Nh vy lng cht S cn li pha 1 s l:

W1 = (Cs)1V1 = Wo12

1.VDV

V+

Lp li qa trnh chit trn ln th hai th lng cht cn li pha 1 c tnh theo cng thc:

W2 = (Cs)1.V1 = W112

1.VDV

V+ = Wo

2

12

1 ).(VDV

V+

Nh vy sau n ln chit th lng cht S cn li pha 1 s l:

o

n

n WVDVV

W

+= 121

2) a) Phn s gam S cn li sau khi chit vi 100mL dung mi c tnh theo cng thc 3 thu c kt qa f1 = 0,135. Nh vy phn trm S c chit ra l 86,5%.

b) Phn s gam S cn li sau khi chit 4 ln vi mi ln 25mL dung mi c tnh theo cng thc 4 thu c kt qa f4 = 0,022. Nh vy phn trm S c chit ra l 97,8%.

Nhn xt: chit nhiu ln vi mt th tch nh dung mi chit s c hiu qa hn so vi vic chit mt ln bng ton b th tch dung mi chit. 3) S dng phng trnh 4 ta thu c kt qa l 3,78. Nh vy ta phi chit t nht 4 ln. 4) Ta c cc phng trnh sau:

D = ( )( )

[ ][ ] [ ]ww owHA oHA AHA

HACC

+= KD = [HA]o/[HA]wKa = [H+]w[A-]w/[HA]w

Kt hp ba phng trnh trn ta c: D = [ ][ ] aw wD KHHK

+++.

Phng trnh cui cng cho bit nu [H+]w >> Ka (axit mnh) th DKD (D nhn gi tr ln nht c th) v HA nm hon ton trong dung mi hu c. [H+]w

[ ][ ] [ ] [ ][ ][ ]

[ ] [ ][ ][ ] [ ][ ] 102

5

21

2

10.2

10.1

720

)()(

+

+

+

+

==

==

==++==

w

ww

ww

w

oD

w

o

w

o

OxHHOx

K

OxHHOxH

K

OxHOxH

K

OxOxHOxHOxH

OxHCOxHC

D

Kt hp 4 phng trnh trn ta c:

[ ] [ ]wwD

HK

KH

KD

+

+++

=2

1

1

b) Ta ly o hm cp 1 v cp 2 (theo [H+] mu s ca phng trnh cui cng: [ ] [ ][ ]

[ ]322

2

1

2

1

2"

1'

1

+

+

+

+

=

=

++=

H

Kf

H

KK

f

HK

KH

f w

Do f lun lun dng nn khi f = 0 th f t cc tiu di cc iu kin cho. Nh vy t s phn b t cc i khi f = 0. Thay s vo ta tnh c [H+] = 4,5.10-8M. BI TP CHUN B CHO K THI IChO LN TH 36: 3) Vit phn ng t chy kh propan v butan trong khng kh. Cho bit trng thi ca cc cht trong

phn ng iu kin chun. 4) Tnh thiu nhit ca phn ng t chy 1mol mi cht propan v butan. C th gi thit rng cht

phn ng v sn phm u nhn c iu kin chun. 5) Th tch kh c s dng trong phn ng l bao nhiu trong qa trnh ny?. Gi thit rng nit v

oxy l kh l tng v khng kh cha 21% oxy v 79% nit theo th tch. Sn phm thng khng nhn c iu kin chun nhng s nhn c khi ta tng nhit . Gi

thit rng i vi sn phm nhit nhn c l 100oC v p sut chun trong khi phn ng xy ra iu kin chun.

6) Tnh thiu nhit i vi s t chy 1 mol mi cht propan v butan trong khng kh cc iu kin cho.

7) So vi cu 2 th cu 4 c hiu sut bao nhiu % v cch tch tr nng lng khc nhau nh th no? 8) Tnh hiu sut ca qa trnh t chy nh l mt hm ca nhit khong nhit t 25oC n

300oC. Gi s rng nc khng ngng t. 9) So snh thiu nhit tch ly khi t chy 1L mi cht propan v butan lng. Gi thit rng nhit

ca sn phm l 100oC. Cho bit: dpropan(l) = 0,493g.cm-3; dbutan(l) = 0,573g.cm-3. Propan(k): fHo = -103,8kJ.mol-1 Cp = 73,6J.mol-1.K-1.

Butan(k): fHo = -125,7kJ.mol-1 Cp = 140,6J.mol-1.K-1. CO2(k): fHo = -393,5kJ.mol-1 Cp = 37,1J.mol-1.K-1. H2O(l): fHo = -285,8kJ.mol-1 Cp = 75,3J.mol-1.K-1. H2O(k): fHo = -241,8kJ.mol-1 Cp = 33,6J.mol-1.K-1. O2(k): fHo = 0kJ.mol-1 Cp = 29,4J.mol-1.K-1. N2(k): fHo = 0kJ.mol-1 Cp = 29,1J.mol-1.K-1.

BI GII: 1) 1C3H8(k) + 5O2(k) 3CO2(k) + 4H2O(l)

2C4H10(k) + 13O2(k) 8CO2(k) + 10H2O(k) 2) cHo(propan) = -2220kJ.mol-1.

cHo(butan) = -2877kJ.mol-1. 3. Do gi thit oxy v nit l kh l tng nn th tch ca cc kh t l vi s mol ca cc kh:

76,3..2

2

2

22 OO

NON nV

Vnn ==

5 mol O2 v 18 mol N2 cn t chy 1mol propan 6,5mol N2 v 24,4mol N2 cn t chy 1mol butan. Vi V = nRT/p th th tch ca cc kh s l: Propan: Vkk cn = 0,582m3. Butan: Vkk cn = 0,756m3.

4. Di cc iu kin trn th nc khng th lng m th hi. Thiu nhit thay i do s thay i entanpy ca s ho hi ca nc v nhit cao ca sn phm. Nng lng cn nc bc hi 25oC: vHo(H2O) = fHo(H2O(l)) - fHo(H2O(k)) = 44kJ.mol-1. Nng lng cn thit tng nhit ca sn phm ln 100oC c tnh theo cng thc: H(T)

= (T-To)i.ni.CP(i). Nng lng E gii phng khi t chy 1mol mi kh s l: Epropan = -1984,5kJ.mol-1. Ebutan = -2580,0kJ.mol-1.

5) %7,89

%4,89100.

)4tan(tan

)4(

==

==

oc

bubu

oc

propanpropan

HE

HE

Nng lng tch ly c dng t nng sn phm. 6) Da vo cu 1.4 ta c:

E(propan, T) = -2044 + (T - To).792,8. E(butan.T) = -2657 + (T To).1026,4 Hiu sut ca qa trnh t chy: (hm ca nhit ) Propan: propan(T) = 1 3,879.10-4(T To) Butan: butan(T) = 1 3,863.10-4(T To)

7) ni = i.Vi/Mi npropan = 11,18mol mbutan = 9,86mol Ei = ni.E(propan/butan, 373,15K) Epropan = -22,19MJ

Ebutan = -25,44MJ BI TP CHUN B CHO K THI IChO LN TH 36:

Khi xem xt thit k mt cn nh th vic tnh ton kh nng dn nhit ca tng, mi ngi v sn nh ng mt vai tr quan trng. 1) Tnh nhit truyn qua tng 150m2 (chun mc ca mt gia nh chu u) bao gm cc vin gch dy 24cm so vi mt bc tng tng t nhng lt bng gch c dy d = 36cm. Nhit trong nh l 25oC cn ngoi nh l 10oC. Cng thc tnh:

)( 12 TTdSPW =

S: din tch : dn nhit T2: nhit trong nh T1: nhit ngoi nh d: dy P: nhit lng 2) Nhit mt i c th nh hn khi ta s dng lp cch ly l bt polystiren. Tnh nhit lng mt i khi i qua 10cm lp cch ly bng bt polystiren. Bit Stng = 150m2

S li th hn nu chng ta s dng i lng nhit tr (heat resistance) -1 i vi kh nng dn nhit thng qua mt bc tng gm nhiu lp khc nhau:

...1

3

3

2

2

1

1 +++= ddd

i vi cc phn khc ca cn nh (ca s, tng) th h s thu nhit c tnh bi cng thc:

...332211 +++=SS

SS

SS

k

Nng lng tit kim c rt quan trng c th lm gim nng lng ca th gii. S cch li tt khng ch em li hiu qa tch cc cho nn kinh t (s kh CO2 phn hy) m cn tt cho c nn kinh t. Hin ti cn nh lu gi nng lng c h s thu nhit ln nht k = 0,50W.m-2.K-13) Tnh dy ca bc tng ch cha gch t kt qa cn thit trn? 4) dy ca bc tng c th gim i bng cch s dng cc lp cch ly. Mt bc tng cha gch c dy d1 = 15cm bn ngoi, lp btng c dy d2 = 10cm, lp cch ly (bt polystiren) c dy d3 v lp thch cao c dy d4 = 5cm bn trong bc tng. Tnh dy ca lp cch ly d3 v tng dy ca bc tng t c h s k trn. 5) Ca s lm tng lng nhit mt i. Gi s mt bc tng 15m2 c ch to nh cu 4 v c thm mt ca s c k = 0,70W.m-2.K-1. Hy tnh dy tng thm (%) ca lp bt cu 4 t c h s k trn. Bng s:

Vt liu (W.m-1.K-1) B tng 1,10 Gch 0,81 Bt polystiren 0,040 Linoleum (cht ph nn nh) 0,17 Thch cao 0,35

BI GII: 1) Nhit truyn qua:

PW =150.0,24-1.(25 -10).0,81 = 7,59kW PW =150.0,36-1.(25 -10).0,81 = 5,06kW 2) PW =150.0,1-1.(25 -10).0,040 = 0,90kW Mc d tng lc ny mng hn nhng nng lng mt i li t hn do dn nhit thp hn. 3) k = .d-1 d = .k-1 = 0,81.0,5-1 = 1,62m 4) k-1 = 0,50-1 =

4

4

3

3

2

2

1

11dddd +++= d3 = 6,3cm

Vy tng dy ca bc tng s l: 6,3 + 15 + 10 + 5 = 36,3cm 5) k = 1 .S1.S-1 + 2.S2.S-1

0,50 = 0,70.4.15-1. 2.11.15-1

2 = 0,427W.m-2.K-1Tnh ton tng t nh cu 4 ta thu c kt qa d3 = 7,7cm Vy tng dy lc ny s l: 37,7cm iu ny xy ra do kh nng dn nhit cao ca ca s. Chiu dy ca lp bt phi tng 22% BI TP CHUN B CHO K THI IChO LN TH 36:

Amoniac l mt trong nhng cht trung gian cht quan trng nht. Mt trong s nhng cng dng ca n l sn xut phn bn. Thng thng amoniac c iu ch t hydro v nit bng qa trnh Haber Bosch. 1. Vit phng trnh phn ng. 2. Tnh cc ga tr nhit ng ca phn ng iu kin chun (Ho, So v Go) bng cch s dng

nhng ga tr cho bng 1 v cho bit phn ng l to nhit hay thu nhit? 3. Hin tng g s xy ra nu nh ta trn hn hp nit v hydro nhit phng? Gii thch l

do. 4. Tnh cc ga tr nhit ng ca phn ng (H, S v G) 800K v 1300K p sut chun.

Phn ng lc ny l to nhit hay thu nhit. S ph thuc ca nhit dung v entropy vo nhit c miu t bi phng trnh Cp(T) = a +

bT + cT2 v S(T) = d + eT + fT2. Gi tr ca cc hng s t a f c cho bng 2. 5. Tnh phn mol ca NH3 c th c hnh thnh 298,15K; 800K v 1300K p sut chun.

Gi thit cc kh u l l tng v cc cht phn ng c ly ng theo h s t lng. Trong cng nghip th phn ng cn phi nhanh v cho hiu sut cao. Cu 3 cho ta thy nng

lng hot ho ca phn ng l rt ln v cu 5 cho thy hiu sut phn ng tng khi tng nhit . C hai cch gii quyt vn ny. 6. Phn ng c th c tin hnh nhit thp vi xc tc (v d st oxit). Cht xc tc nh

hng n tnh cht nhit ng v ng hc ca phn ng nh th no? 7. Ta cng c th tng p sut. p sut nh hng n tnh cht nhit ng v ng hc ca phn

ng nh th no? 8. iu kin tt nht cho phn ng ny l g? Bng 1:

Cht ho hc fHo(kJ.mol-1) So(kJ.mol-1) Cpo(kJ.mol-1) N2(k) 0,0 191,6 29,1

NH3(k) -45,9 192,8 35,1 H2(k) 0,0 130,7 28,8

Bng 2:

Cht ha hc

a (J.mol-1.K-1)

b (J.mol-1.K-2)

c (J.mol-1.K-3)

d (J.mol-1.K-1)

e (J.mol-1.K-2)

f (J.mol-1.K-3)

N2(k) 27,3 5,2.10-3 -1,7.10-8 170,5 8,1.10-2 -2,3.10-5

NH3(k) 24,2 4,0.10-2 -8,2.10-6 163,8 1,1.10-1 -2,4.10-5

H2(k) 28,9 -5,8.10-4 1,9.10-6 109,8 8,1.10-2 -2,4.10-5

BI GII: 1. N2(k) + 3H2(k) 2NH3(k) 2. Ho = -91,8kJ.mol-1

So = -198,1J.mol-1.K-1. Go = -32,7kJ.mol-1.

3. Amoniac s hnh thnh ngay lp tc nhng do nng lng hot ha cao nn hai cht kh khng th phn ng c vi nhau. Tc ca phn ng ny rt thp.

4. S dng phng trnh Kirchoff tnh entanpy v entropy 800K v 1300K ta thu c cc kt qa sau: N2(k) H2(k) NH3(k)fH(800K) 15,1kJ.mol-1 14,7kJ.mol-1 -24,1kJ.mol-1

fH(1300K) 31,5kJ.mol-1 29,9kJ.mol-1 4,4kJ.mol-1

S(800) 220,6J(mol.K)-1 252,9J(mol.K)-1 236,4J(mol.K)-1

S(1300K) 236,9J(mol.K)-1 174,5J(mol.K)-1 266,2J(mol.K)-1

fG(800K) = 72,9kJ.mol-1. fG(1300K) = 184,0kJ.mol-1.

5. Hng s cn bng c th tnh c t bin thin nng lng Gibbs bi biu thc: Ks(T) = exp(-G(RT)-1). Ta c cc kt qa tnh hng s cn bng sau: Ks(298,15K) = 5,36.105. Ks(800K) = 1,74.10-5. Ks(1300K) = 4,04.10-8.

S dng biu thc: 1;3;. 32222

22

3

3

2

=++== NHHNNHNH

NHX xxxxxxx

xK

Chng ta nhn c biu thc:

xxxN

xN

XN

N

NX

KKKx

Kx

Kx

xx

K

271

274

272

027

127

427

)41(

2

22

2

2 24

2

++=

=+=

Cc kt qa thu c cho bng sau: T(K) x(N2) x(H2) X(NH3) 298,15 0,01570 0,04710 0,03720 800 0,24966 0,74898 0,00136 1300 0,24998 0,74994 0,00008

6. Cht xc tc lm gim nng lng hot ho ca phn ng v lm tng tc phn ng. Cc gi tr nhit ng khng h thay i.

7. p sut cao dn n kt qa l phn mol ca NH3 tng ln do Kx=Kp.p2 tng ln. S tng p sut lm cn bng chuyn dch v pha to thnh sn phm nhng khng lm thay i tc phn ng.

8. iu kin tt nht ca phn ng l: p sut cao, nhit cng tp cng tt v c mt cht xc tc. Nhit cng phi ti u chuyn ha nhanh v hiu sut c th chp nhn c.

BI TP CHUN B CHO K THI IChO LN TH 37: Cho 10 lt kh l tung 0oC v 10 atm, ta nn kh v 1atm. Tnh th tch cui v cng gin n

ca kh di cc iu kin sau: 1. Gin n thun nghch ng nhit. 2. Gin n thun nghch on nhit. 3. Gin n khng thun nghch on nhit trong cc iu kin sau: p sut gim xung 1atm mt

cch t ngt v kh gin n on nhit ng p p sut ny. Cho bit nhit dung mol ng tch c cho bi phng trnh CV = 3/2RT vi R l hng s kh.

BI GII: 1. Do qa trnh gin n l thun nghch ng nhit nn:

LPVP

V 1002

112 ==

S mol kh iu kin ny: n = 100/22,41 = 4,481mol Vy cng gin n do kh thc hin s l: A = -Q = nRTln(V2/V1) = 23290J

2. i vi qa trnh gin n thun nghch on nhit th: = CP/CV = 5/3 Nh vy: V2 = (P1/P2)1/2.V1 = 39,8L. Nhit cui cng c tnh t cng thc: T2 = P2V2/nR = 108,8K i vi qa trnh on nhit: q = 0 E = q + A = A = nCVT = -9141J.

3. i vi qa trnh ng nhit khng thun nghch ta c: q = 0 E = q + A = A = nCV(T2 T1) A = -P(V2 V1) Vy ta c:

KTnRnRT

TnR 8,17410

2,273.1

)2,273(23

22

2 =

= T ta rt ra c E = -5474J.

BI TP CHUN B CHO K THI IChO LN TH 37: Gin pha l mt con ng thun tin ch ra pha ca mt cht nh l hm ca nhit v

p sut. Tr li cc cu hi sau da vo gin pha ca nc di y:

1. A, B, C tn ti nhng pha no? 2. Ti sao nc khng chm trong nc lng? 3. Khi nc ng bng, n n ra. Gii thch hin tng ny da vo phng trnh Clapeyron

Clausius. Phng trnh ny c dng: VT

HdTdP

= .

4. Mt ci ly ch cha mt phn nc c ni vi mt my ht chn khng. Chuyn g s xy ra khi my ht chn khng hot ng.

5. Mt ngi ang trt trn b mt ca mt tm bng khi p sut khng kh l 1atm v nhit l 0oC. Thay i g s xy ra trn b mt tm bng tip xc vi vn trt. Gi s rng tm bng khng v?

BI GII: 1. A: pha rn; B: c ba pha rn, lng, kh cng tn ti; C: pha lng v pha hi cng tn ti. 2. H s gc m ca ng cong rn/lng ch ra rng trng thi lng ca nc th c hn cho nn

nc phi ni trn nc lng. 3. T gin pha ta thy h s gc dP/dT ca ng cong lng - rn l m. iu c ngha l th

tch ca nc tng ln khi ng c. 4. Lc ta m van, th p sut gim nn pha lng chuyn trc tip sang pha hi cng nhit . Nh

vy nc s bc hi. cng mt thi im th qa trnh ha hi ca nc l thu nhit lm cho mi trng xung quanh lnh i v dn n phn nc cn li s ng c. Cht rn ny s thng hoa cho n khi khng cn g c.

5. B mt ca lp bng di vn trt s ho lng do p sut ln hn rt nhiu so vi 1atm. IV. OLYMPIC HA HC CC QUC GIA TRN TH GII:

OLYMPIC HA HC O 2004: OCW, mt cng ty chuyn sn xut H2O2 o theo phng php Antraquinon. iu quan trng

nht ca phng php ny l vic iu ch hydro, n c iu ch dng rt tinh khit t metan v hi nc (qa trnh cn bng). Trong qa trnh ny cacbon oxit c sinh ra v c kh nng phn ng vi nc bc tip theo.

1. Vit cc phng trnh phn ng xy ra v cn bng vi trng thi vt cht ng ca cc cht. By gi chng ta s nghin cu phn ng gia metan v hi nc. y l phn ng thun

nghch. 2. S dng cc s liu sau tnh Kp (n v p sut: bar) . Cho bit 100oC nc trng thi hi.

Gi s Ho v So khng i khong nhit t 298K n 373K. H2 H2O CO CH4

Ho(kJ/mol) 0 -242 -111 -75 So(kJ/mol.K) 0,131 0,189 0,198 0,186 Cp (kJ/mol.K) 0,029 0,034 0,029 0,036

Bnh phn ng (V = 3,00m3) cha 6,40kg CH4, 7,20kg H2O, 11,2kg CO v 2,4kg H2 100oC. 3. Tnh nng mol phn ca tng kh. 4. Tnh p sut cung v p sut ring phn ca tng kh. 5. Xc nh hng s cn bng vo lc ny v cho bit chiu dch chuyn cn bng ca phn ng.

Metan v hydro em trn vi t l 1:1 v cho vo mt bnh kn, un nng bnh n 900oC. Vi s c mt ca cht xc tc phn ng t nhanh n cn bng vi p sut chung l 20bar.

6. Tnh Kp 900oC (Gi s Cp khng ph thuc nhit ) 7. Tnh %CH4 phn ng (Nu khng lm c cu 4.6 th s dng gi tr Kp = 4000 tnh

ton). BI GII: 1. Phng trnh phn ng

CH4(k) + H2O(h) CO(k) + 3 H2(k) H2O(h) + CO(k) H2(k) + CO2(k)

2. Kp 100oC: RH298 = -111 + 242 + 75 = 206 kJ = RH373 RS298 = 30.131 + 0.198 0.186 0.189 = 0.216 kJ/K = RS373 RG298 = 206 3730.216 = 125 kJ

373314.810.125

P

3

eK = KP = 2.7410-18 (p: bar)

3. Phn mol ca tng kh: n(H2) = 1200 mol; n(CH4) = n(H2O) = n(CO) = 400 mol; n = 2400 mol x(H2) = 0.5 mol; n(CH4) = n(H2O) = n(CO) = 0.167 mol;

4. p sut ca h v p sut ring phn ca tng kh: 6

G 1048.23373314.82400

VTRn

p === pG = 24.8 bar p(H2) = 12.4 bar; p(CH4) = p(H2O) = p(CO) = 4.14 bar

5. Chiu phn ng v hng s cn bng iu kin ny:

QlnTRGG RR += KQ

lnTRGR = 5.46014.414.44.12

Q 23

== 5

18R 1055.11074.25.460

ln373314.8G == J Cn bng chuyn dch sang bn tri

6. KP 900C: CP = 30.029 + 0.029 0.036 0.034 = 0.046 kJ/K

RH1173 = 206 + (1173 298) 0.046 = 246.25 kJ RS1173 = 0.216 +0.046 ln(1173/298) = 0.279 kJ/K RG1173 = 246.25 11730279 = -81 kJ

1173314.881017

P eK = KP = 4054 (p : bar)

7. Lng chuyn ha ca metan: CH4 H2O H2 CO

n(trc) 1 1 0 0 n -x -x +3x +x neq 1-x 1-x 3x x

xeqx22

x1+

x22x1

+

x22x3

+ x22x

+

peq 20x22

x1 +

20x22

x1 +

20x22

x3 + 20x22x +

)OH(p).CH(p)H(p).CO(p

K24

23

P =

)x1)(x22(20x2.5

67.63

)x1()x22(

400x2720)x1)(x1()x22)(x22(

20)x22)(x22()x3(4054

2

22

4

23

43

+=

+=++

++=

3.1835(2+2x-2x-2x2) = 5.2 x2

6.367 6.367x2 = 5.2x2 x2 = 0.5504 x = 0.7419 Metan chuyn ha 74,2%.

OLYMPIC HA HC O 2001: A. Gin pha n gin:

1) Cc t hp trng thi no tng ng vi CO2 p sut thng? 2) p sut no l thp nht xy ra s ha lng CO2. 3) nhit ti thiu no th CO2 ngng t? 4) nhit no nc kh (dng rn) nm cn bng vi CO2 (dng kh). 5) Bnh cu ho cha CO2 dng lng. Vy p sut no l thp nht CO2 vn cn

hot tnh cu ho? B. Nhit si ca etanol:

p sut hi ca etanol 60oC l 46,7kPa. Trong khong t 60oC n nhit nng chy th nhit ha hi ca etanol l HV = 862Jg-1.

1) Tnh nhit si ca etanol 1013mbar. 2) Tnh entropy ha hi ca etanol SV nhit si: Gin pha ca CO2 cho hnh sau:

300 T(K)

0,5

200 250

1,0

5,0

10

50

P(bar)

BI GII:

A. Gin pha ca CO2: a) Rn v kh. b) 5,1bar c) 304K d) 78oC e) 55bar

B. Nhit si ca etanol 1) HV = 46.862 = 39652J.mol-1.

T phng trnh Clapeyron Clausius dng:

=

21

V

1

2

T1

T1

RH

ppln

Thay s ta c:

=

ST1

3331

314.839652

46700101300ln

TS = 352 K = 79.0C 2)

S

VV T

HS

= SV = 113 J/K OLYMPIC HA HC O 2001:

A. Sinh nhit. Tnh sinh nhit hnh thnh axit nitr trong dung dch nc trong iu kin ng p v ng tch.

S dng cc d kin sau: a. NH4NO2(s) N2(g) + 2H2O(l) H(kJ) = -300.4 b. 2H2(g) + O2(g) 2H2O(l) H(kJ) = -569.2 c. N2(g) + 3H2(g) + aq 2NH3(aq) H(kJ) = -170.8 d. NH3(aq) + HNO2(aq) NH4NO2(aq) H(kJ) = -38.08 e. NH4NO2(s) + aq NH4NO2(aq) H(kJ) = +19.88

B. Cn bng ha hc: Cc halogen c th to nn cc hp cht lin halogen (interhalides) bn hay km bn. Mt trong

s l BrCl, cht ny c kh nng phn hy thnh nguyn t 500oC. nhit ny th hng s cn bng ca phn ng KC = 32 nu ta phn hy 2 mol BrCl. Chng ta s xt mt h ban u cha 0,25mol/L BrCl.

a) Vit phng trnh phn ng cho s phn hy. b) S dng tnh ton chng minh hn hp trn cha vo trng thi cn bng. c) Phn ng s chuyn dch theo chiu no? d) Tnh nng lng t do ca phn ng. e) Tnh KP v KX ca phn ng. f) Tnh nng cn bng ca cc cht nu xut pht t hn hp ban u c thnh phn nh trn.

BI GII: A: Phn ng xy ra nh sau: H2(g) + N2(g) + O2(g) HNO2(aq) p dng nh lut Hess tnh c nhit hnh thnh ng p ca HNO2(aq) theo cng thc sau: HB = -H(4)+ H(5)- H(1)-0.5H(3)+ H(2) Ta tnh c nhit ng p ca HNO2(aq) = -125,4kJ.mol-1. Trong iu kin ng tch ta c: UB = HB - nBRT vi n = -2 thu c UB = -120.5 kJ/mol B. a) 2 BrCl Br2 + Cl2 b) C2 K125,0

25.025.0Q == c) Q < KC v bn phi d)

CKQlnRTG = = - 22 kJ

e) Khi n = 0 ta c: KP = KX = 32 Nng ca cc cht lc cn bng: n(BrCl) = 0.25-2y n(Br2) = n(Cl2) = 0.25+y

2

2

)y225.0()y25.0(32

+= . Gii phng trnh ta thu c y = 0,0943

[BrCl] = 0,061M; [Br2] = [Cl2] = 0,34M OLYMPIC HA HC O 2005:

6,0 mol hydro c un nng trong iu kin ng tch. Kh c nn li t p sut 100kPa 0,0oC n p sut gp 2,5 ln p sut ban u. Nhit dung mol ng tch ca hydro l CV = 20,56J/mol.K. 1. Tnh nhit sau khi nn. 2. Tnh s thay i entropy ca qa trnh ny. 3. Tnh nhit dung mol ng p ca qa trnh ny.

1,0L argon c lm lnh ng p t iu kin thng xung nhit thp hn 50K so vi iu kin ban u. 4. Tnh nhit thot ra khi lm lnh kh. 5. Tnh th tch cui cng ca qa trnh lm lnh. 6. Tnh bin thin ni nng ca qa trnh.

By gi chng ta kho st phn ng sau: CH3CHO(l) + O2(k) CH3COOH(l)

7. Tnh nhit phn ng 20oC. S dng cc d kin sau: (1) H2(g) + O2(g) H2O(l) H = -286.2 kJ (2) 2 C(s) + 2 H2(g) + O2(g) CH3COOH(l) H = -486.0 kJ (3) 2 C(s) + H2(g) C2H2(g) H = +229.6 kJ (4) C2H2(g) + H2O(l) CH3CHO(l) H = -138.1 kJ 8. Nng lng t do ca phn ng chuyn ho t borneol (C10H17OH) thnh isoborneol pha kh ti

503K l +9,4 kJ/mol. Lc u ta c mt hn hp gm 0,15mol borneol v 0,30mol isoborneol v un nng n n 503K. S dng tnh ton cho bit chiu chuyn dch cn bng ca phn ng.

9. xc nh s ph thuc ca nhit vo p sut hi ca n propanol th t thc nghim ta thu c cc kt qa sau y:

T (K) 313 333 353 373 P (kPa) 6.69 19.60 50.13 112.32

a) V th v da vo tnh entanpy ca s ha hi b) Nhit si ca n propanol 101,3mbar. c) p sut no th n propanol si 45oC? BI GII:

1. Nhit cui cng sau khi nn:

CTT

Tconst

Tp

===

==

4105.682100273250

250273100

2. S thay i entropy:

3. Nhit dung ring ng p. Cp = Cv + R = 28,9J/mol.K

K/J113S273

5.682ln56.206TTlnCnS

1

2V

===

4. Lng nhit thot ra:

5. Th tch cui:

J4.4650314.85.21046.4q

n

TCnq

2

2

P

===

===

1046.415.273314.8001.0101300Vp

TR

LV

VconstTV

82.0223273

0.1

===

6. Ni nng ca phn ng:

J9.271018.01013254.46UVpqU

3 =+=+=

7. Nhit phn ng c th tnh c t phng trnh: Hp = H(2) H(4) H(3) H(1) H = -486 + 138.1 -229.6 + 286.2 = - 291.3 kJ 8.

215.030.0

233.04.9 776314.89400

776

====+=

Q

eKkJG

Q > K Cn bng chuyn dch v bn tri 9. a)

5.0

4.0 lnp

3.0

2.0

1.0

2.7 2.8 2.9 3.0 3.11/T10-3

kJ9.44H10)68.283.2(

91.372.4RH

V3V =

=

b) Da vo gin ta thy:

C97TK4.370T

107.2T162.43.101ln 3

====

c) Da vo gin ta thy: OLYMPIC HA HC C 2001:

kPa6.8p15.2pln1014.3T1K318C45 3 ====

Cho phn ng sau: 2H2(k) + 2Cl2(k) 4HCl(k) Ho = -92,3kJ. Mnh no sau y l sai:

a) Ga tr Ho s bng 92,3kJ nu HCl sinh ra dng lng. b) Bn lin kt trong HCl mnh hn bn lin kt trong hai phn t H2 v Cl2. c) Nu phn ng xy ra theo chiu nghch th ga tr ca Ho s bng +92,3kJ. d) Gi tr Ho s bng 23,1kJ nu phn ng sinh ra 1 mol HCl(k). e) Cc cht phn ng u vo trng thi bn vng nht.

BI GII: Cu a. OLYMPIC HA HC C 2001:

Tnh nhit ca phn ng sau: FeO(r) + Fe2O3(r) Fe3O4(r)Cho bit cc thng tin sau: 2Fe(r) + O2(k) 2FeO(r) Ho = -544,0kJ. 4Fe(r) + 3O2(k) 2Fe2O3(r) Ho = -1648,4kJ. Fe3O4(r) 3Fe(r) + 2O2(k) Ho = +1118,4kJ.

a) 1074,0kJ b) 422,6kJ c) 22,2kJ d) +249,8kJ e) +2214,6kJ

BI GII: Cu c OLYMPIC HA HC C 2002:

Photpho triclorua v photpho pentaclorua l nhng ha cht hay s dng trong cng nghip ch to cc hp cht c photpho. Ta c th iu ch hai hp cht ny t photpho trng bng cc phng trnh sau:

P4(r) + 6Cl2(k) 4PCl3(l) Ho = -1280kJ. P4(r) + 10Cl2(k) 4PCl5(r) Ho = -1774kJ. Nh vy entanpy ca phn ng PCl3(l) + Cl2(k) PCl5(r) l bao nhiu:

a) 123,5 b) 494 c) 763,5 d) +123,5 e) 3054

BI GII: Cu a OLYMPIC HA HC C 2003:

Ta xt cn bng sau: 2NO(k) + O2(k) 2NO2(k) Ho = -113kJ.mol-1. L do no sau y lm thay i t l NO2/NO?

a) Thm O2(k) b) Thm cht xc tc. c) Tng p sut. d) Tng nhit e) Tt c u sai.

BI GII: Cu d OLYMPIC HA HC C 2003:

N2(k) + O2(k) 2NO(k) K12NO(k) + O2(k) 2NO2(k) K2NO(k) + NO2(k) N2O3(k) K3Tnh K ca phn ng sau: 2N2O3(k) N2(k) + 3O2(k) da vo cc hng s K1, K2, K3.

a) K1K2K3 b) 1/K1K2K3 c) K12K2K32 d) 1/K1K22K32 e) 1/K12K2K32

BI GII: Cu e OLYMPIC HA HC C 2004:

Cho bin thin entanpy ca cc phn ng sau: 2NH3(k) N2(k) + 3H2(k) Ho = 92kJ.mol-1. 2H2(k) + O2(k) 2H2O(k) Ho = -484kJ.mol-1. Tnh bin thin entanpy ca phn ng: 2N2 + 6H2O 3O2(k) + 4NH3(k)

d) 576kJ.mol-1. e) 392kJ.mol-1. f) 392kJ.mol-1. g) 1268kJ.mol-1. h) 1636kJ.mol-1.

BI GII: Cu d OLYMPIC HA HC C 2004:

Mt mu photpho pentaclorua t trong mt bnh kn. Trong iu kin photpho pentaclorua t phn hy theo cn bng:

PCl5 PCl3(k) + Cl2(k)Khi hn hp vo trng thi cn bng th ta thm mt lng nh heli ( p sut v nhit

khng i) vo v hn hp li quay tr v trng thi cn bng. Mnh no sau y m t tt nht mi lin h gia cn bng th hai vi cn bng th nht?

a) Cn bng th hai cng ging nh cn bng th nht bi v heli nh hn bt k phn t no khc c mt trong phn ng v khng gy c nh hng ng k ln phn ng.

b) Cn bng th hai cng ging nh cn bng th nht bi v heli khng h phn ng vi bt k cht no trong bnh knh.

c) Cn bng th hai dch chuyn theo chiu dng v c nhiu PCl3 hn. d) Cn bng th hai dch chuyn theo chiu dng v c nhiu PCl5 hn. e) Cha th kt lun g .

BI GII: Cu c OLYMPIC HA HC O 1999:

A. p sut hi bo ho ca hn hp: 20oC, p sut hi ca etyl etanoat tinh khit l 9706Pa v ca etyl propanoat l 3693Pa. Do

tnh cht ging nhau, hn hp hai cht ny c tnh cht l tng. Tnh p sut hi ca hn hp gm 25g etyl etanoat v 50g etyl propanoat 20oC.

B. Nhit ha hc: iu kin chun, entanpy phn ng v entropy ca cc cht c ga tr nh sau: 2NH3 + 3N2O = 4N2 + 3H2O Ho298 = -1011kJ (1) N2O + 3H2 = N2H4 + H2O Ho298 = -317kJ (2) 2NH3 + 0,5O2 = N2H4 + H2O Ho298 = -143kJ (3) H2 + 0,5O2 = H2O Ho298 = -286kJ (4) So298(N2H4) = 240J/mol.K; So298(H2O) = 66,6J/mol.K So298(N2) = 191J/mol.K; So298(O2) = 205J/mol.K.

a) Tnh entanpy to thnh Ho298, So298 ca N2H4; N2O v NH3. b) Vit phng trnh phn ng chy ca hidrazin to thnh nc v nit. c) Tnh nhit phn ng chy ng p ny 298K v tnh Go298 v hng s cn bng K. d) Nu hn hp ban u gm 2 mol NH3 v 0,5 mol O2 th nhit phn ng (3) th tch khng i

l bao nhiu? BI GII: A. p sut hi bo ho ca hn hp:

M(EtAc) = 88,11g/mol; M(Etpr) = 102,1g/mol. n(EtAc) = 0,284mol; n(Etpr) = 0,490mol; n = 0,774mol x(EtAc) = 0,367; x(Etpr) = 0,633. P(EtAc) = 3562Pa; P(Etpr) = 2338Pa Pg = 59hPa

B. Nhit ha hc: a) (1) + 3(2) + (3) (4) 4N2 + 8H2 4N2H4

Phn ng trn c: Ho298 = 203kJ. Phn ng N2 + 2H2 N2H4 c Ho298 = 203kJ/4mol = 50,8kJ/mol. T (2) ta suy ra Ho298 (N2O) = 81,8kJ/mol. T (3) ta suy ra Ho298 (NH3) = -45,6kJ/mol.

b) N2H4 + O2 = N2 + H2O c) Ho298 = -623kJ

So298 = -121J/K Go298 = -587kJ K = e-G/RT = 10103.

d) H = U + pV = U + nRT = -139kJ. OLYMPIC HA HC BUGARI 1999 (Vng II)

20oC, ho tan vo dung dch NaOH 0,16g/L mt lng iot phn ng:: 2NaOH + I2 NaI + NaOI +H2O tin hnh n cng. a) Tnh pH ca dung dch cui cng b) Tnh p sut thm thu ca dung dch ny. c) dng cao (tnh bng cm) ca dung dch trong bnh o thm thu l bao nhiu? d) V sao HOI l axit yu trong khi HI l mt trong nhng axit mnh nht?

D kin: Ka(HOI) = 2.10-11; R = 8,314; khi lng ring ca dung dch l 1g/mL: 1cm ct nc ng vi 98,07Pa BI GII: a) 2Na+ + 2OH- + I2 2Na+ + I- + IO- + H2O Co(Na+) = 0,016/40 = 4.10-4M Co(I-) = Co(IO-) = 0,5Co(Na+) = 2.10-4M IO- + H2O HIO + HO-C(IO-) = Co(IO- - C(HIO); C(HIO) = C(OH-) HIO H+ + IO-[ ][ ]

[ ] [ ] [ ][ ] [ ][ ] 185,10;10.53,6 010.210.110.2

;10.2

11

2514214

11

===

===

+

++

+

+

pHMHHH

HK

OHHIO

HIOK Wa

b) = CRT vi C = Co(Na+) + Co(I-) + C(IO-) + C(OH-) + C(HIO) C(OH-) = C(HIO) = 1,53.10-4M C(IO-) = 2.10-4 - 1,53.10-4 = 0,47.10-4M C = 9,53.10-4M = CRT = 2321,5Pa c) h = /98,07 = 23,67cm d) m in ca iot nh hn ca oxy v phn cc ca lin kt H - O trong HOI rt nh. OLYMPIC HA HC ITALY 1999:

293K phn ng sau y t xy ra: NH3(k) + HCl(k) = NH4Cl(r) + Q Ta c th ni rng:

a) Kh nng ny l do entropy. b) Phn ng t xy ra mi nhit . c) Kh nng ny l do entanpy. d) Entropy v entanpy cng du:

BI GII: Cu d. OLYMPIC HA HC ITALY 1999:

Khi hi ngng t ta lun lun c: a) Gim entropy ca h v tng entropy ca mi trng. b) Tng entropy ca h v gim entropy ca mi trng. c) Gim entropy ca h v ca mi trng. d) Entropy ca h v ca mi trng u khng i.

BI GII: Cu a. OLYMPIC HA HC ITALY 1999:

Nu hng s cn bng ca phn ng c ga tr l 4,16.10-3 25oC v 2,13.10-1 100oC th c th ni rng phn ng l:

a) Pht nhit. b) Thu nhit c) Thu nhit ch khi p sut tng. d) Pht nhit ch khi th tch tng.

BI GII: Cu b.

OLYMPIC HA HC ITALY 1999: C 3 phn t CH3Cl(k), CH3OH(k), CH4(k) P = 101325Pa. Th t tng entropy mol chun l:

a) So(CH4) < So(CH3OH) < So(CH3Cl) b) So(CH4) < So(CH3Cl) < So(CH3OH) c) So(CH3Cl) < So(CH3OH) < So(CH4) d) So(CH3Cl) < So(CH4) < So(CH3OH)

BI GII: Cu b. OLYMPIC HA HC UCRAINA 1999 (Lp 10): Hm nhit ng (298K)

H3PO4(dd) H2PO4-(dd) HPO42-(dd) PO43-(dd) H+ + OH-= H2O(l)

Ho (kJ/mol) -1288 -1269 -1292 -1277 -56 So (J/mol.K) 158 90 -33 -220 81

a) Tnh Ho v Go ca phn ng trung ho tng nc H3PO4 bng kim: OH- + HnPO4n-3 = Hn-1PO4n-4 + H2O

b) T nhng d kin bng trn, tnh cc hng s in ly ca H3PO4 25oC. c) Tnh th tch ca nhng dung dch 0,1M ca axit v kim m khi trn chng th thnh 25mL

dung dch v nhit pht ra l 90J. BI GII:

a) Ho v So ca H+ u bng 0: Ho(H2O) - Ho(OH-) = 56kJ/mol. Hon = Ho3-n - Ho3-n+1 56;So3 = So3-n - So4-n + 81

Go = Ho - TSo T ta tnh c: Ho1 = -64kJ/mol; So1 = 13J/mol.K; Go1 = -68kJ/mol Ho2 = -52kJ/mol; So2 = -42J/mol.K; Go2 = -39,5kJ/mol Ho3 = -41kJ/mol; So3 = -106J/mol.K; Go3 = -9,5kJ/mol

b) Ta c: Go(H2O) = 80kJ/mol H3PO4 = H+ + H2PO4-; Goa = Go1 - Gonc = 12kJ/mol; K1 = 7,9.10-3H3PO4 = 2H+ + HPO42-; Gob = 40,5kJ/mol; K2 = 8,0.10-8H3PO4 = 3H+ + PO43-; Goc = 70,5kJ/mol; K3 = 4,4.10-11

c) * Hn hp hai mui axit (n/OH- = 0,1V2; n(H3PO4) = 0,1V1) 64.0,1V1 + (0,1V2 0,1V1).52 = 0,09; V1 + V2 = 0,025 4V2 = 0,06; V2 = 0,015L; V1 = 0,01L

* Hn hp mui axit v mui thng: V1 = 0,0071L; V2 = 0,0175L OLYMPIC HA HC UCRAINA 1999 (Lp 11):

Hng s phn b ca axit benzoic HA trong h nc/benzen 10oC l K = [HA]w/[HA]B = 0,700. Hng s phn ly ca HA l Ka = 6,20.10-5. HA b dime ha mt phn trong benzen. trng thi cn bng 200mL lp nc cha 0,0429g HA v 200mL lp benzen cha 0,145g HA.

a) Vit cng thc tnh nng cn bng ca [H+] trong lp nc. b) Tnh nng cc phn t (k c H+) trong lp nc. c) Tnh nng cc phn t v hng s dime ha (KD) trong lp benzen. d) Nng cc dng ca HA thay i nh th no khi pH tng? e) Gii thch v sao c s dime ha HA.

BI GII:

a) [ ] [ ] [ ] [ ] [ ]

[ ] [ ] ( )[ ] 0...0

)(.

2

3 =++

=++=+

++

+

+++

AwwowA

A

w

A

owA

KKHKCKH

KH

HK

HKCK

HOHA

b) CoN = 1,76.10-3M. CoB = 5,94.10-3M. [H+] + K[H+] KCoN = 0 [H+] = 3,00.10-4; pH = 3,52 [A-] = 3,01.10-4; [HA] = 1,46.10-3; [OH-] = 3,33.10-11.

c) [HA]B = [HA]N/K = 2,09.10-3. {[HA]2}B = (CoB [HA]n).0,5 = 1,93.10-3 KD = 442. d) Lp nc: [A--] tng; lp benzen [HA], {[HA]2}gim.

e)

O

O O

OH

H ----: Lin kt hydro. OLYMPIC HA HC C 1999 (Vng 2):

a) Nitrosyl clorua l mt cht rt c, khi un nng s phn hy thnh nit monoxit v clo. Cc s liu nhit ng hc cho bng: NOCl NO Cl2Hof(298)(kJ/mol) 51,71 90,25 0 So298 (J/mol.K) 264 211 223 CP (J/mol.K) 44,7 29,8 33,9

b) Bn hy tnh KP i vi phn ng nhit 298K. Bn hy tnh ra kt qa bng n v atm v Pa.

c) Bn hy tnh gn ng Kp nhit 475K (ngha l khng s dng CP) bng cch p dng phng trnh Vant Hoff v s ph thuc ca nhng hng s cn bng vo nhit . Bn hy nu r trong nhng iu kin no th vic tnh ton nh vy l c ngha.

d) Bn hy cho bit i lng CP cung cp thng tin v vn g? Cc ga tr CP ca nitrosyl clorua v ca nit monoxit khc bit nhau nhiu. Hy l gii iu .

e) Bn hy kim tra xem liu nhng iu kin bin i i vi vic tnh ton phn c) c c tho mn hay khng? Bn hy tnh chnh xc Kp i vi nhit 475K v so snh vi kt qa cu c).

f) nhit -33,5oC, ngi ta cho 2,00g nitrosyl clorua vo mt bnh chn khng c th tch V = 2,00L khng thay i mi nhit .

(i) Bn hy tnh p sut trong bnh 298K (bng Pa v atm). (ii) Bn hy tnh p sut trong bnh 600K m khng k n s phn hu nhit. (iii) Bn hy tnh p sut trong bnh 600K khi c tnh n s phn hy nhit (KP(600K) =

28,1.103Pa).

i vi cc tnh ton ca mnh hy s dng cc khi lng mol sau (g/mol): N = 14,01; O = 16,01; Cl = 35,45. Hng s kh R = 8,314J/mol.K

BI GII: a) 2NOCl 2NO + Cl2 b) Hng s cn bng nhit ng lc hc c tnh theo phng trnh:

K = e-G/RT Vi G = H - TS = 42214 (J/mol) Thay vo phng trnh trn ta c:K= KP = 3,98.10-8atm = 4,04.10-3Pa

c) i vi trng hp gn ng. Phng trnh Vant Hoff c dng: ( )( )

=

212

1 11lnTTR

HTKTK

P

P

a cc s liu vo ta c: KP = 4,32.10-3atm = 437Pa. S tnh ton gn ng ny rt c ngha, nu cc nhit c xem xt, H v S gn nh

bng nhau. (iu c ngha l cc nhit c xt CP ca cc cht tham gia phn ng gn nh bng 0).

d) CP l nhit dung mol. Mt cch trc quan th ga tr CP.1K cho bit lng nhit cn phi cung cp cho 1mol cht nghin cu cho nhit ca n tng thm 1K. Nhit cao hn l do chuyn ng ca cc phn t gy ra. Nhng i vi cc phn t nhiu nguyn t th nhit lng cung cp khng ch c chuyn ha thnh nng lng chuyn ng ca cc phn t m cng cn thnh nng lng quay v nng lng dao ng. Hai dng nng lng ny khng h gp phn vo vic lm tng nhit khc vi cc cht kh hai nguyn t, i vi cc cht kh gp khc ba nguynh t cn c thm chuyn ng dao ng. Chuyn ng dao ng ny hp th thm nng lng m khng th gp phn lm tng nng lng ca chuyn ng. V vy nhit dung ca NOCl ln hn NO (Khng ngh ti vic gii thch bng cc bc t do dao ng).

e) Kim tra cc iu kin bin ca phn c) CP = 4,1J/mol.K H(475) = H(298) +CP(475 298) H(475) - H(298) = 726J/mol. t hn 1% so vi H(298)=77080J/mol. Tnh tng t cho S ta thu c S(475) - S(298) = 1,91J/mol.K, nh hn 1,7% so vi

S(298) = 117J/mol.K. Tnh chnh xc cc kt qa ta thu c: H(475) = 77806J/mol. S(475) = 118,9J/mol.K G(475) = 21329J/mol KP = 4,51.10-3atm. Ch khc bit cht t so vi c)