Chng 3 CN BNG HA HC3.1. 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 :

Hng s cn bng tnh theo nng mol/l:

Hng s cn bng tnh theo phn mol:

Hng s cn bng tnh theo s mol:

Mi quan h ca cc hng s cn bng:

n l bin thin s mol kh ca h.n = (c + d) (a + b)Nu n = 0 ta c Kp = KC = Kx = Kn3.1.2. Phng trnh ng nhit Vant Hoff

Xt phn ng: aA(k) + bB(k) cC(k) + dD(k)Ti nhit khng i, ta c:

Vi

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

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 bngCh :

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:

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:

3.2.3. phn ly

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

n: lng cht phn ly

no: lng cht ban u3.3. 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

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

Nu phn ng thu nhit, H > 0 : nh vy khi nhit tng, gi tr Kp cng tng, phn ng dch chuyn theo chiu thun.

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

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

00

x x xx

() () xx

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

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) + H2bng 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

a0 0

xx x

(a x)x x

Tng s mol cc cht lc cn bng:

vi n = 1

x = 0,764a

Vy phn ly:

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 u5,380

Phn ngxx2x

Cn bng(5,3 x)(8 x)2x

Theo bi: 2x = 9,5 x = 4,75 (mol)

Hng s cn bng:

Hn hp 8 mol I2 v 3 mol H2.

H2 + I2 2HI

Ban u380

Phn ngyy2y

Cn bng(3 y)(8 y)2y

V nhit khng i nn hng s cn bng cng khng i:

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 ua00

Phn ngaaa

Cn bnga(1-)aa

Ta c

Vi n = 1, ni = a(1+)

3P2 = 1 - 2

Vi P = 1 atm

Vi P = 8 atm

b. p sut no phn ly l 10%Ta c

P = 33 atm

c. Lng Cl2 cn thm vo

Gi b l s mol Cl2 cn thm vo: PCl5(k) PCl3(k) + Cl2(k)

Ban u10b

Phn ng0,10,10,1

Cn bng0,90,1(b + 0,1)

Ta c:

b = 0,5 (mol)

V d 5. C th iu ch Cl2 bng phn ng

4HCl(k) + O2 = 2H2O(h) + 2Cl2Xc 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: ; n = -1

Theo bi ta c: 2x = 0,402 x = 0,201 (mol)

4HCl(k) + O2 2H2O(k) + 2Cl2(k)

10,48 00

4x x 2x2x

(1 - 4x)(0,48 - x) 2x2x

Hng s cn bng:

(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

1,3150

xx

(1,315 - x)x

Theo bi ta c: x = 1,255 (atm)

Hng s cn bng:

Gi x l p sut ring phn ca H2 lc cn bng: 3Fe + 4H2O(h) Fe3O4(r) + 4H2

30

xx

(3 x)x

V nhit khng i nn hng s cn bng cng khng i:

x = 2,863 (atm)

S mol kh H2 sinh ra:

(mol)

Khi lng kh H2 sinh ra: (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:

(atm2)

b. p sut tng cng:

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

0,60

xx

(0,6 + x)x

V nhit khng i nn hng s cn bng cng khng i:

x2 + 0,6x - 0,2025 = 0

x = 0,24 (atm)

p sut ca hn hp:

(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:

Mt khc:

(atm-3)

Suy ra: (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)

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

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

Suy ra:

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 u0,01x

Phn ng0,010,03

Cn bng0,00(x - 0,03)

Tng s mol ti thi im cn bng:

(mol)Hng s cn bng:

(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

(a x)

Tng s mol lc cn bng:

Ta c:

V n = 0, hng s cn bng:

b. Kp tng 10% khi tng 1oC ln cn 1000oC.

Hng s cn bng KP 1001oC: Kp = 4 + 0,04 = 4,04

Ta c:

(cal)

(cal)

Ta li c: G01000 = H01000 - TS01000

(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 l phn mol ca cc kh cn bng:

Ta c, hng s cn bng: vi n = 2 1 =1Suy ra:

M:

Vy

(1)

Vi Kp = 1,85 atm

EMBED Equation.3 Gii phng trnh ta c: xCO= 0,72 v = 0,28.

Ta tnh c:

KP,1200 = 58,28 atm

Thay vo phng trnh (1) c:

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

COCO2PbPbO

H0298,tt (KJ/mol)-110,43-393,130-219,03

G0298 (KJ/mol)-137,14-394,000-189,14

Cp,298 (J/mol.K)29,0536,6126,5046,27

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 = 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:

Kp = 7,4.1011 (atm)

b. Biu th di dng mt hm ca T.

Cp,298 = 36,61 + 26,50 29,05 46,27

= - 12,21 (J/K)

(J)

c. Tnh Kp 1270C

(atm)V d 14. Cho phn ng v cc s liu tng ng sau:

Ckc(r) + 2H2(k) = CH4(k)

H0298 (Kcal/mol)0,4530-7,093

S0298 (cal/mol.K)0,56831,2144,50

Cp

(cal/mol. K)2,186,524,170

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.

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

c.

Ta c

Ta c

d.

Vy (cal)

(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 = SO3Bit 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: ,

6. Tnh (G0 v hng s cn bng Kp 250C ca phn ng sau:

NO + O3 NO2 + O2 .

Cho bit cc s liu sau:

NO2O2NOO3

33,81090,25142,12

240,35240,82210,25237,42

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.

12. Cho phn ng: CO + Cl2 COCl2C 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) + 2Cl2Xc 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).1

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