21 co so lt hoa hoc dhbk hn.pdf

7/30/2019 21 Co So LT Hoa hoc DHBK HN.pdf

1/75

Bi ging mn Csl thuyt Ha hc

PHN I: NHIT NG HO HC

Mun xt mt phn ng ho hc c thc hin c hay khng cn bit:

- iu kin no th phn ng xy ra v xy ra n mc no?

- Phn ng xy ra nh th no? Nhanh hay chm? Nhng yu t no nh hngn tc phn ng?

Khi tr li c c hai cu hi ny, ngi ta c thiu khin c phn ng,

tm c iu kin ti u thc hin phn ng, nhm t hiu qu cao nht.

Cu hi th nht l i tng ca nhitng ho hc, cn cu hi th hai l i

tng ca ca ng ho hc.

Nhitng hc l b phn ca vt l hc, nghin cu cc hin tng cv nhit,

cn nhitngho hc l b phn ca nhit ng hc nghin cu nhng quan h nng

lng trong cc qu trnh ho hc.

http://hhud.tvu.edu.vn


2/75


CHNG I: P DNG NGUYN L THNHT

CA NHIT NG HC VO HO HC

I. MT S KHI NIM MU

1. Kh l tng:- Cht kh c coi l l tng khi m khong cch gia cc phn t kh xa nhau,

c th b qua s tng tc gia chng v coi th tch ring ca cc phn t kh l

khng ng k (kh c p sut thp).

- Phng trnh trng thi ca kh l tng: nu c n mol kh p sut P, nhit Tv chim th tch V th: PV = nRT = RT

M

m(1.1)

trong : m- khi lng ca kh, g

M: Khi lng mol ca kh, g

T Nhit tuyt i, K ( T = t0C +273)

R: Hng s kh l tng, ty theo n v ca P v V m c ga tr

khc nhau:

- Nu P (atm), V(dm3=l) R = 0,082 atm.l.K-1.mol1

- Nu P (Pa=N/m2), V(m3) R = 8,314 J.K-1.mol-1

1atm = 1,013. 105 Pa= 1,013. 105N/m2= 760 mmHg

- Nu trong bnh c mt hn hp kh th mi kh gy nn mt p sut gi l p sut ringphn ca kh v c k hiu l Pi .Tng tt c cc p sut ring phn bng p sut

chung P ca hn hp.Nu gi V l th tch chung ca hn hp kh ( bng dung tch bnh

ng th phng trnh kh kh l tng c dng:

V

RTnPP ii

== (1.2)

in : Tng s mol kh trong hn hp.

p sut ring phn Pi ca kh i trong hn hp c th tnh:

V

RTnP ii = hoc Pi= NiP vi Ni =

i

i

n

n

(1.3)

2. H v mi trng

-H: H l i tng cn nghin cu cc tnh cht nhit ng hc. i km vi khi nim

h l khi nim mi trng xung quanh, l ton b phn cn li ca v tr bao quanh h.

Hc phn cch vi mi trng xung quanh bng mt mt thc hay tng tng.

- C 4 loi h:

+ H c lp: l h khng trao i cht v nng lng vi mi trng+ H m: l h trao i cht v nng lng vi mi trng.



3/75


+ H kn l h ch trao i nng lng vi mi trng

+ H khng trao i nhit vi mi trng c gi l hon nhit.

3.Quy c v du trong qu trnh trao i nng lng

Nng lng trao i gia h v mi trng c th l cng , nhit, nng lng in...

-

H nhn nng lng: du (+)- H nhng nng lng du ()

4.Trng thi ca h v cc thng strng thi:

- Trng thi vm ca mt hc c trng bng nhng i lng xc nh nh: t0C, P,

V, C...Cc thng s ny c tho c, gi l cc thng s trng thi.

v d: gia s mol kh n, nhit T v p sut P ca mt h kh (gi s l kh l tng)

c mi quan h cht ch, c biu din bng phng trnh trng thi ca kh l tng

PV=nRT.

- C hai loi thng s trng thi: dung v cng

+ Thng s trng thi dung l nhng thng s trng thi t l vi lng cht, th d

th tch, khi lng.

+ Thng s trng thi cng khng t l vi lng cht, v d nhit p sut, nng

, nht.

5. Trng thi cn bng ca h

L trng thi ti cc thng s trng thi ca h khng i theo thi gian. VD phn ng

thun nghch CH3COOH + C2H5OH CH3COOC2H5 + H2O t trng thi cn bngkhi nng ca 4 cht khng bin i .

6. Bin i thun nghch v bin i bt thun nghch

- Nu h chuyn t mt trng thi cn bng ny sang mt trng thi cn bng khc v

cng chm qua lin tip cc trng thi cn bng th s bin i c gi l thun

nghch.y l s bin i l tng khng c trong thc t.

- Khc vi s bin i thun nghch l s bin i bt thun nghch. l nhng bin i

c tin hnh vi vn tc ng k. Nhng bin i xy ra trong thc tu l bt thun

nghch.

7.Hm trng thi

- Mt hm F( P,V,T...) c gi l hm trng thi nu gi tr ca n ch ph thuc vo

cc thng s trng thi ca h m khng ph thuc vo cch bin i ca h.

- V d: n mol kh l tng:

+ trng thi 1 c c trng bng P1V1=nRT1

+ trng thi 1 c c trng bng P2V2=nRT2

PV l mt hm trng thi, n khng ph thuc vo cch bin i t trng thi 1sang trng thi 2.



4/75


8.Cng v nhit: L hai hnh thc trao i nng lng.

Cng W (J, kJ)

Nhit Q (J, kJ)

Cng v nhit ni chung khng phi l nhng hm trng thi v gi tr ca chng ph

thuc vo cch bin i.* Cng gin n( cng chuyn dch)

W = - Pngoi.dV =-PndV (1.4)

W ph thuc vo Pn ( v h sinh cng nn c du -).

Nu qu trnh l hu hn => W = 2

1dVPn (1.5)

Nu gin ntrong chn khng Pn =0W=0.

Nu gin nbt thun nghch: gin nchng li Pn khng i:

Pn= const (Pn=Pkq)Wbtn = -Pn(V2-V1) (1.6)Nu gin n thun nghch: tc l Pn =Ph

Wtn= 2

1

V

VndVP (1.7)

Nu kh l l tng v gin nng nhit c :

Pn = Ph =V

nRT=>

1

2ln2

1 V

VnRT

V

dVnRTW

V

VTN ==

Vy WTN =- nRT ln1

2

V

V=- nRT ln

2

1

P

P(1.8)

II. NGUYN L I P DNG VO HA HC

1. Khi nim ni nng (U)Nng lng ca h gm 3 phn

- ng nng chuyn ng ca ton h- Th nng ca h do h nm trong trng ngoi- Ni nng ca hTrong nhit ng ho hc nghin cu ch yu ni nng.

Ni nng ca h gm:

- ng nng chuyn ng ca cc phn t, nguyn t, ht nhn v electron (tinhtin, quay..)

- Th nng tng tc (ht v y) ca cc phn t, nguyn t, ht nhn v electron.Nh th ni nng (U) ca h l mt i lng dung , gi tr ca n ch ph thuc

vo trng thi vt l m khng ph thuc vo cch chuyn cht ti trng thi . N l

mt hm trng thi.



5/75


6/75


t H=U+PV

Ta c: Qp= H2-H1 = H (1.11)

H c gi l entapi, n l hm trng thi v U v PV u l nhng hm trng thi.

III. NHIT PHN NG HO HC.

1. Nhit phn ngL nhit lng thot ra hay thu vo khi phn ng xy ra theo ng h s t lng,

cht tham gia v sn phm cng mt nhit T.

c th so snh nhit ca cc phn ng cn ch r iu kin phn ng xy ra:

- Lng cc cht tham gia v sn phm to thnh theo h s t lng.- Trng thi vt l ca cc chtVi mc ch ny ngi ta a ra khi nim trng thi chun. Trng thi chun ca

mt cht nguyn cht l trng thi l hc di p sut 101,325kPa(1atm) v nhit

kho st n bn nht.

V d: Cacbon tn ti hai dng th hnh l graphit v kim cng. 298K v di p

sut 101,325kPa, graphit l bin i th hnh bn nht do trng thi chun 298K ca

cacbon l graphit.

- Nu phn ng c thc hin P=const th nhit phn ng c gi l nhit phnng ng p Qp= H .

- Nu phn ng c c thc hin V=const th nhit phn ng c gi lnhit phn ng ng tch Qv= U .

Phn ng ta nhit v phn ng thu nhit- Phn ng ta nhit: l phn ng nhng nhit lng cho mi trng. Khi

PQH = 0. V d phn ng nung vi..

Quan h gia nhit ng tch v nhit ng p:( ) VpUpVUH p +=+=

Qp= Qv+ nRT (1.12)

Trong : n = s mol sn phm kh s mol cht kh tham gia phn ng.

R = 8.314 J/mol.K: hng s kh l tng

T: K

V d: C6H6 (l) +2

15O2(k) = 6CO2(k) + 3H2O(l)

n= 6-7,5=-1,5.

C(r) + O2(k) = CO2(k)

n= 1- 1= 0



7/75


2. Nhit sinh chun ca mt cht:

L nhit ca phn ng to thnh 1 mol cht t cc n cht bn iu kin chun

(cht sn phm v cht phn ng phi l cc cht nguyn cht 1atm v gi P, T=const,

cc s liu nhit ng chun trong cc ti liu thng c xc nh nhit T=298

K).K hiu 0 sTH , (kJ.mol

-1)

Nu T =298 => 0 ,298sH

V d: 0 ,298 sH (CO2)=-393,51(kJmol-1). N l nhit phn ng ca phn ng sau 250C

khi atmpp COO 122 ==

Cgr + O2(k) = CO2(k).

C graphit l n cht bn nht ca cacbon 1 atm v 298K.

- Tnh ngha trn ta suy ra nhit sinh chun ca n cht bn bng khng.3. Nhit chy chun ca mt cht:

L nhit ca qu trnh t chy han ton 1 mol cht bng O2 to thnh cc xit bn

nht ( vi ha tr cao nht ca cc nguyn t), khi cc cht trong phn ng u nguyn

cht P=1atm v gi T, P khng i (thng T=298K).0

,cTH (kJ.mol-1)

V d: )( 40

,298 CHH c =-890,34kJ.mol-1ng vi nhit ca phn ng sau 250C v p=const

khi atmPPP COOCH 1224 === .

CH4 (k)+ 2O2 (k) CO2 (k) + 2H2O(l)

Tt c cc xit bn vi ha tr cao nht ca cc nguyn tu khng c nhit chy.

IV.NH LUT HESS V CC H QU

1.Pht biu: Hiu ng nhit ca mt phn ng ch ph thuc vo trng thi u v trng

thi cui ca cc cht tham gia v cc cht to thnh ch khng ph thuc vo cc giai

on trung gian, nu khng thc hin cng no khc ngoi cng gin n.

V d:

Cgr + O2(k) CO2(k)

Theo nh lut Hess: 21 HHH += (1.13)

2.Cc h qu

CO(k) + 1/2 O2(k)

H

H2H



8/75


H qu 1: Hiu ng nhit ca phn ng thun bng hiu ng nhit ca phn ng nghch

nhng ngc du.: nt HH = (1.14)

H qu 2: Hiu ng nhit ca mt phn ng bng tng nhit sinh ca cc cht cui tri

tng nhit sinh ca cc cht u.

= )()( thamgiaHmnphsHHss (1.15)

Nu iu kin chun v T=298K th

= )()( ,,, thamgiaHnphmsHH sspu 298029802980 (1.16)

Tnh ngha ny suy ra: nhit sinh ca mt n cht bn vng iu kin chun

bng khng: 0 sTH , (n cht) = 0.

V d: Tnh H0 ca phn ng:

C2H4(k) + H2 (k) --> C2H6 298K?

Cho bit 0298 sH , ca cc cht (kJ.mol-1) nh sau:

C2H4(k): +52,30

C2H6(k): -84,68

Gii:

Ta c:0298H =

0298 sH , (C2H6(k)) - [

0298 sH , (C2H4(k)) +

0298 sH , (H2(k))]

=-84,68-52,30-0

=-136,98kJ.mol-1H qu 3: Hiu ng nhit ca mt phn ng bng tng nhit chy ca cc cht u tri

tng nhit chy ca cc cht cui.

= )()( spHtgHH ccp (1.17)

Nu iu kin chun v T=298K th

= )()( ,,, spHtgHH ccp 298029802980 (1.18)

3.Cc ng dng

* nh lut Hess v cc h qu ca n c mt ng dng rt ln trong Ho hc, n chophp tnh hiu ng nhit ca nhiu phn ng trong thc t khng tho c.

V d1: khng tho c nhit ca phn ng Cgr + 1/2 O2(k) =CO(k) v khi t chy

Cgr ngoi CO (k) ra cn to thnh CO2(k) nhng nhit ca cc phn ng sau y o

c:

Cgr + O2(k) = CO2(k)0298H =-393513,57 J.mol

-1

CO(k) + O2(k) = CO2(k)0298H =-282989,02 J.mol

-1

tnh c nhit ca phn ng trn ta hnh dung s sau:



9/75


Cgr O2(k)

1/2O2(k)+

+ CO2(k)

CO(k)

x=?

Trng thi u (Cgr+O2) v trng thi cui (CO2(k)) ca c hai cch bin i l nh nhau,do theo nh lut Hess:

-393.513,57 = x - 282.989,02

x=-110507,81J.mol-1V d 2: Xc nh nng lng mng lui tinh th ca NaCl(r) bit

+ Nhit nguyn t ha Na(r)

Na(r) Na(h) 11 724108+= molJH ..

+ Nhit phn ly Cl2(k)Cl2(k) 2Cl(k)

12 672242

+= molJH ..

+ Nng lng ion ha Na(h)

Na(h) Na+(h) + e 13 528489+= molJH ..

+i lc i vi electron ca Cl(k)

Cl(k) + e Cl-(k) 14 192368= molJH ..

+Nhit ca phn ng

Na(r) + 1/2 Cl2(k) NaCl1

5 216414

= molJH .. xc nh nng lng mng li tinh th NaCl ta dng chu trnh nhit ng Born

Haber:

Na(r) + 1/2 Cl2(k)

Na(h) + Cl(k)

NaCl(r)

Na+

(h) + Cl-

(k)

Trngthi u

Trngthi cui

1H

3H

4H

5H

x=?1/2

Theo nh lut Hess ta c:

xHHHHH ++++= 43215 21/

x= )/( 43215 21 HHHHH +++ x= -765.612J.mol-1

V. SPH THUC HIU NG NHIT VO NHIT .NH LUT KIRCHHOFF

1. Nhit dung mol ca 1 cht

L nhit lng cn thit nng nhit ca 1 mol cht ln 1K v trong sut qu trnhny khng c s bin i trng thi(nng chy, si, bin i th hnh...)



10/75


- n v thng dng ca C l: J.K-1mol-1

- Nhit dung mol ng p. Qu trnh c thc hin P=const.

dT

dH

T

HC

p

P =

= => dH=CpdT => =

2

1

2

1

dTCdH P

==> =2

1

dTCH P

-Nhit dung mol ng tch. Qu trnh c thc hin V=const.

dT

dU

T

UC

v

v =

= => dU=CvdT => =

2

1

dTCU v

==> =2

1

dTCU v

2.Nhit chuyn pha-Chuyn pha: bay hi ,nng chy, ng c, thng hoa...

- cfH l nhit lng trao i vi mi trng khi 1 mol cht chuyn pha. P=const, khi

mt cht nguyn cht chuyn pha th trong sut qu trnh chuyn pha, nhit khng

thay i.

3.nh lut Kirchhoff

Xt 1 h kn, P=const.. Xt phn ng sau thc hin bng hai con ng:

n1A + n2B

n1A + n2B

n3C + n4D

n3C + n4D1H

aH

2H

bH

T1

T2

Theo nh lut Hess ta c ba HHHH ++= 12

+=+=

2

1

1

2

2121

T

T

PP

T

T

PPa dTCnCndTCnCnHBABA

)()(

+=2

1

43

T

T

PPb dTCnCnH DC )(

T dTCnCnCnCnHHBADC PP

T

T

PP )]()[( 214312

2

1

+++=

=> +=2

1

22

T

T

PTT dTCHH => Cng thc nh lut Kirchhoff

Vi: = )()( tgCspCC PPP



11/75


iu kin chun(P=1atm) v T1=298 K c:

+=T

PT dTCHH298

00298

0

Nu trong khong hp ca nhit => coi constCP =0 th

)( 298002980 += TCHH PT

4.Mi quan h gia nng lng lin kt v nhit phn ng

C th quy uc nng lng lin kt (Elk) tng ng vi nng lng ph vlin kt hoc

hnh thnh lin kt.

y ta qui c Elkng vi qu trnh ph v lin kt: nng lng lin kt l nng

lng ng vi qu trnh ph v lin kt do nng lng lin kt cng ln th lin kt

cng bn.

- Mt phn ng ho hc bt k v bn cht l ph vlin kt c v hnh thnh cc linkt mi do Hp c thc tnh qua Elk ca cc lin kt ho hc.

V d1: Ph v1 mol thnh cc nguyn t c lp:

H2(k,cb) --> H(k,cb) + H (k,cb)

298K, p= 1atm => EH-H = +432kJ.mol-1 = 0298H

Trong trng hp ny Elk coi nh hiu ng nhit ca qu trnh.

V d2: Xt phn ng N2(k) + 3H2(k) => NH3(k). Thc hin bng 2 con ng

N2(k) + 3H2(k) 2NH3(k)H

2H(k) + 6H(k)

EN-N3EH-H

-6EN-H

HNHHNN EEEH += 63

Ti liu tham kho:

1. Nguyn nh Chi, CSL Thuyt Ha Hc, NXB GD, 2004.2. Nguyn Hnh, , CSL Thuyt Ha Hc, Tp 2, NXB GD 1997.3. L Mu Quyn, CSL Thuyt Ha Hc - Phn Bi Tp, NXB KHKT, 2000.



12/75


CHNG II: NGUYN L II CA NHIT NG HC CHIU V GII HN

TDIN BIN CA QU TRNH

MU

Trong t nhin, cc qu trnh l hc v ho hc xy ra theo chiu hon ton xc nh.- Nhit t truyn t vt nng sang vt lnh hn- Kh t truyn t ni c p sut cao n ni c p sut thp- Cc phn ng ho hc t xy ra, v d: Zn + HCl --> ZnCl2 + H2Cn cc qu trnh ngc li th khng t xy ra c.

Nguyn l I cho php tnh nhit ca cc phn ng nhng khng cho php tin

on chiu v gii hn ca qu trnh

Nguyn l II cho php gii quyt cc vn ny.

I.NGUYN L II. HM ENTROPY

1.Nguyn l II (Tiu chun xt chiu ca qu trnh)

- Tn ti mt hm trng thi gi l entropi (S).

- nhit T khng i, trong s bin i v cng nh, h trao i vi mi trng mt

nhit lng Q th bin thin entropi ca qu trnh c xc nh:

Nu l bin i thun nghch:T

QdS TN

=

Nu l bin i bt thun nghch: TQdS bTN> Tng qut

T

QdS

Du > : qu trnh bt thun nghch

2

1 T

QS

Du = : qu trnh thun nghch

* Ch :

V S l hm trng thi --> S ch ph thuc vo trng thi u v trng thi cui, tc

l: ==2

1 T

QSS TN

TNBTN

>=2

1 T

QSS btnBTNTN

==> QTn> QBTN : Nhit qu trnh thun nghch ln hn nhit qu trnh bt thun nghch.

+ xc nh Sbtn , trc ht hnh dung mt qu trnh thun nghch c cng trng

thi

u v trng thi cu

i v

i qu trnh b

t thu

n ngh

ch, sau

tnh

Stheo cng th

c:

=2

1 T

QS TN

(khng xc nh c trc tip Sbtn)



13/75


2. Nguyn l II p dng trong h c lp

i vi h c lp:

Qtn= 0 --> 0=S

Qbtn=0 --> 0>S

Nh vy i vi h c lp:- Trong qu trnh thun nghch (cn bng), entropi ca h l khng i.- Trong qu trnh bt thun nghch ngha l t xy ra, entropi ca h tng.iu ny c ngha rng trong cc h c lp, entropy ca h tng cho ti khi t ti gi

tr cc i th ht ti trng thi cn bng.o li ta c th ni:

Trong h c lp:

- Nu dS >0 ( S tng) h t din bin

- Nu dS=0, d2S 2 kh khuch tn vo nhau cho n khi c s phn bng u trong ton b

th tch ca 2 bnh.

S khuch tn cc kh l tng vo nhau l qu trnh c T=const(Q=0) --> 0>S

(S2> S1) --> hn n ca trng thi cui (hn hp 2 kh) c trng bng S2 ln hn hn n ca trng thi u ( mi kh 1 bnh ring bit) c trng bng S1.

Vy trong h c lp, qu trnh t xy ra theo chiu tng hn n ca h (tng

entropi, 0>S ). Qu trnh ngc li: Mi kh t tch ra khi hn hp kh trli trng

thi u khng th t xy ra.

* Kt lun:

- Entropi c trng cho hn n: hn n ca h cng ln th S cng ln.- Nu s ht trong h cng ln--> hn n cng ln--> Sln- Lin kt gia cc ht trong h cng yu --> hn n cng ln--> S ln. V d:

SH2O(r) ,SH2O(l)< SH2O(k) .

- S l hm trng thi v l i lng dung .b. ngha thng k ca S

Trng thi ca mt tp hp bt k c thc c trng bng 2 cch:

- Bng gi tr ca cc tnh cht o c : T, P,C...--> c gi l cc thng s trngthi vm.

-

Nhng c trng nht thi ca cc phn t to nn hc gi l cc thng s vim.



14/75


* S thng s trng thi vi m ng vi mt trng thi vm c gi l xc sut nhit

ng

Nu s phn t trong h tng th S tng--> tng. Gia S v c quan h vi nhau

thng qua h thc Bolzomann.

H thc Boltzmann (l csca nguyn l III)S=kln

k: hng s Boltzmann

Nhn xt: Trong h c lp, qu trnh t din bin theo chiu tng xc sut nhit ng .

4.Bin thin entropi ca mt squ trnh

a. Bin thin entropi ca qu trnh bin i trng thi ca cht nguyn cht

Trong sut qu trnh ny T=const ==> S ca mt mol cht nguyn cht trong qu

trnh bin i trng thi xy ra P=const l

==2

1 cf

cf

T

H

T

QS

cfH nhit chuyn trng thi

b. Bin thin entropi ca qu trnh gin nng nhit kh l tng

T=const, dn nn mol kh l tng t V1-->V2

==2

1 T

Q

T

QS TNTN

v T=const

V T=const --> 0=

U 12

V

V

nRTWTN ln=

Theo nguyn l I: 0=+= TNTN WQU -->1

2

V

VnRTWQ TNTN ln+==

==>2

1

1

2

P

PnR

V

VnRS lnln ==

Nu P1>P2 --> 0>S : qu trnh gin nny t din bin

==> Cch pht biu khc ca nguyn l II: Cc cht kh c th t chuyn di t ni c p

sut cao n ni c p sut thp.c. Bin thin entropi ca cht nguyn cht theo nhit:

un nng n mol 1 cht nguyn cht t nhit T1-->T2 vi iu kin trong khong

nhit cht ny khng thay i trng thi

- Trong iu kin P = const: ===2

1

2

1

2

1 T

dTnC

T

dH

T

QS p

p

p

Vy =

2

1 T

dT

nCS pp

- Trong iu kin V= consthttp://hhud.tvu.edu.vn


15/75


===2

1

2

1

2

1 T

dTnC

T

dU

T

QS v

v

v

==> Vy =

2

1 T

dTnCS vv

Nu coi Cp hoc Cvkhng i theo T th:

1

2

T

TnCS pp ln=

1

2

T

TnCS vv ln=

II. Nguyn l III ca nhit ng hc

Nhn xt: dng tinh th hon ho ca mt cht nguyn cht OKng vi 1 trng thi

vm ch c 1 trng thi vi m ==> OK th 1=

1. Nguyn l III (tin Nernst)Entropi ca mt cht nguyn cht di dng tinh th hon ho OK bng khng:

( ) == ln00 kS K ( 1= )

2.Entropi tuyti ca cc cht nguyn chtcc nhit T

V d: un nng n mol 1 cht nguyn cht 0K -->TK, trong khong ny xy ra cc qu

trnh bin i trng thi v iu kin P=const. Tnh ST?

0K---> Tnc--->Ts--->T

== ===

T

T

pTTT T

dT

nCSSSS0

0

+

++

+== =

nc S

nc S

T

K

T

T

T

T

hP

S

s

lp

nc

nc

rpTTT

dTnC

T

Hn

T

dTnC

T

Hn

T

dTnCSSS

0

0 )()()(

thng th P=1atm, T=298K, n=1mol ==> )..( 110298

molKJS

-->Bng entropi chun ca cc cht 25oC

* Nhn xt: Gi tr S cht nguyn cht lun > 0, tr khi xt cho ion trong dung dch, c

th c Sion>Slng,Srn ==> c th cn c vo s mol kh 2 v ca phn ng nh

gi ln cng nh l du ca S ca phn ng.

0=n ==> S nh0>n ==> S > 0 ==> phn ng tng S



16/75


0 S nh ==> phn ng gim S

V d: SO2(k) + 1/2 O2(k) --> SO3(k) c 0 S CO2(k) c 0=n ==> 0S

III. HM TH NHIT NG. TIU CHUN XT CHIU CA QU TRNH

- H c lp: 0S --> tiu chun t din bin v gii hn ca qu trnh- H khng c lp: gm h + Mi trng --> av 1 h c lp mi bng cch gp

h v mi trng thnh 1 h c lp.

==> tiu chun t din bin v gii hn ca h mi l : 0+ mtSS

mtS cha xc nh nhng c tha v cc thng s ca h bng cch tm 1

hm thay th cho c ( mtSS + ), hm thay th ny gi l hm th nhit ng. Thng

gp h:

+ ng nhit, ng p ==> c hm thng nhit ng p+ ng nhit, ng tch==> c hm thng nhit ng tch

1.Hm thng nhitng p

a.nh ngha: Xt 1 h: T, P = const

H thc hin mt bin i no

T

HS

T

HSSS mtmt

=

+=+

H Mi trng

H _Nhit lng trao i vi mi trng HHH hmt ==

S _Bin thin entropi ca h.T

H

T

HS mtmt

=

=

==> tiu chun t din bin v gii hn ca qu trnh l:

0+ mtSS

0

T

HS

0 STH ( ) 0 TSH t H-TS =G => G l hm trng thi

0G

0= STHG : P,T=const ==> qu trnh t xy ra theo chiu 0 WQdU += , m 'WpdVdW += ( 'W : cng hu ch)



17/75


=> SdTTdSPdVVdPWPdVQdG +++= '

SdTTdSVdPWQdG ++= '

Nguyn l II => TdSQ ; 'WPdVTdSdU +

=> SdTTdSVdPWTdSdG ++ '

hay: SdTVdPWdG + '

l phng trnh cbn ca nhit ng hc.

- Nu qu trnh l thun nghch-->cng l ln nht --> du =SdTVdPWdG += max'

- T v P =const => dT=0 v dP=0 c:'

max,WdG PT =

max,'WG PT =

ngha ca G: G biu th cng hu ch ca qu trnh thun nghch ng nhit ng

p.

2. Hm thng nhitng tch (lm tng t G)

a.nh ngha: Xt 1 hT, V =const, h thc hin 1 bin i no

H Mi trng

U _Nng lng trao i vi mi trng UUU hmt ==

S _Bin thin entropi ca qu trnh

T

U

T

US Mt

=

=

Tiu chun t din bin v gii hn ca qu trnh l:

0+ mth SS

0

T

US

0 STU

( ) 0 TSU t U - TS = A => A l thng tch ( nng lng Helmholtz).

0A

b. ngha A A = U TS

dA = dU - TdS SdT ; 'WPdVQWQdU +=+=

TdSQ => SdTTdSWPdVTdSdA + '

=> SdTPdVWdA '

- Nu qu trnh l thun nghch-->cng l ln nht --> du =SdTPdVWdA = max'

- T v V =const => dT=0 v dV=0 c:max'WdA =



18/75


max'WA =

ngha ca A : Bin thin thng tch A biu th cng c ch ca qa trnhthun nghch ng nhit ng tch.

Tm li tiu chun t din bin v gii hn ca qu trnh lT,P = const => 0dG

T,V = const => 0dA

Mi lin h gia G v AG = H- TS U +PV TS =(U-TS) + PV =A + PV

Vy G = A+PV

3.Bin thin thng p:

SdTVdPWdG + '

0dG --> iu kin t din bin v gii hn=> Qu trnh t din bin theo chiu lm gim G cho ti khi t gi tr Gmin: dG=0

(G=0)

d2G>0 (G>0)

a. Thng p sinh chun ca 1 chtnhit T:

L s bin thin thng p ca phn ng to thnh 1 mol cht t cc n cht

bn iu kin chun v nhit T ca phn ng.

K hiu: 0 sTG , (J.mol-1 hoc kJ.mol-1)

Thng T=298K => 0298 sG , --> c bng thng p sinh chun ca cc cht 250C.

VD:

)(, 30298 NHG s =-16,65kJ.mol

-1ng vi qu trnh )()()( kNHkNkH 322 21

2

3+

)(, HClG s0298 =-95,5kJ.mol

-1ng vi qu trnh )()()( kHClkClkH + 22 21

2

1

=> HCl(k) bn hn NH3(k) v nng lng to ra nh hn.

-

Tnh ngha =>

0

,sTG

(n cht) =0b. Tnh bin thin thng p ca phn ng ho hc

V G l hm trng thi v l i lng dung nn c:

- = )()( tgGspGG ssp

Nu iu kin chun v 250C c:

- = )(,)(, tgGspGG ssp 29802980

G thun=- G nghch

G


19/75


G >0; phn ng xy ra theo chiu nghch

- TTT STHG =

0298

0298

0298 298 SHG = .

- =

TT GG (cc qu trnh trung gian)IV. CC YU TNH HNG N THNG P

1. nh hng ca nhit

Xt h: ch c bin i thun nghch, khng sinh cng c ch, c P=const.

SdTVdPWdG += ' , v 0='W (khng sinh cng hu ch) v P=const

nn ST

G

P

=

=> S

T

G

P

=

)(, thay vo biu thc STHG = ta c:

PT

G

THG

+=

)(

=> HGT

GT

P

=

)(

Chia c hai v cho T2 ta c:

22T

H

T

GT

GT

p =

)(

=> 2THTGT p

=

hoc l2

T

H

T

T

G

p

=

Phng trnh Gibbs- Helmholtz cho php xc nh G theo T:

dT

T

H

T

Gd

2

=

)(

thng xt iu kin chun (p= 1atm) v T=298K:

dTT

H

T

Gd

2

00 =

)( -> ly tch phn t 298-->T v thng thng bit gi tr 0298G v

)(TfHT =0

C:

= T T

T

dTT

H

T

Gd

2982

0

298

0

)(

==> =T

TT dTTHG

TG

2982

00

298

0

298 .....)( ++== bTaTfHT0

Phng trnh Gibbs-Helmholtz



20/75


=> )(TfG = 0

2. nh hng ca p sut:

Xt h: bin i thun nghch, khng sinh cng hu ch, T=const.

T biu thc: SdTVdPWdG += ' , 0='W , T= const nn c

VP

G

T

=

=> =

2

1

2

1

P

P

VdpdG

=> =2

1

12

P

P

TT VdpGG PP

- i vi cht rn v cht lng --> coi V=const khi P bin i ( tr min P ln) nn: )( 1212 PPVGG PP TT =

)( 1212

PPVGGPP

TT += hay )( 12 PPVG =

- i vi cht kh (coi l kh l tng) -->P

nRTV =

1

22

1

12 P

PnRT

P

dPnRTGG

P

P

TT PPln==

Vy:1

2

P

PnRTG ln= hay

1

2

12 P

PnRTGG

PP TTln+=

Nu ban u P1=1atm (iu kin chun) th0

2 TTGG

P=

PnRTGG TT ln+=0

VD: Nn 0,5 mol kh l tng t P=1atm n P= 2atm 250C. Hi qu trnh nn c t

xy ra c khng?

01

2

1

2 >== lnln nRTP

PnRTG

==> qu trnh nn khng t xy ra.

3. nh hng ca thnh phn cc cht. Khi nim thho

Xt h gm i cht: ni ,1= vi s mol tng ng l n1, n2, ...ni.G=G(T, P, n1, n2...ni)

i

inPTinPTnPTNT

NP dnn

Gdn

n

Gdn

n

GdP

P

GdT

T

GdG

jjj

++

+

+

+

=

,,,,,,,,,,,,,

, ...)( 221

1

11

'WVdpSdTdG ++=

Ch s N ch ra rng n1,n2, n3...ni l khng i, ch s nj#i ch ra rng ch c ni l bin i.

t: iinPT

G

ni

G

j

=

,,,,

= i

Trong : iG thng p mol ring ca cht i trong h



21/75


i Th ho ca cht i

==> = ii dnW ' => ii dn l 1 dng cng hu ch ==> gi l cng ho

T: += ii dnSdTVdPdG

* ngha vt l ca i :- Th ho ca cht i ( i ) l thng p mol ring phn ca i trong hn hp- i l tng kh nng sinh cng hu ch ca h khi thm 1 lng v cng nh

cht i vo h trong iu kin P,T v thnh phn (s mol) ca cc cht khc l khng i.

ic tnh cho 1 mol cht.

- i l i lng cng nhng dni l i lng dung

- ii dn c th lm tiu chun xt chiu v gii hn trong iu kin T,P khng

i:

+ ii dn i ph thuc vo T, P (ging Gi)

- Nu h ch gm 1 cht kh th th ho chnh l thng p ca 1 mol cht: G= .do : V

PP

G

TT

=

=

T lm tng t nh hm G, s thu c phng trnh ca T (ging GT):

PRTTT ln+=0 P(atm) (i vi 1 mol kh).

0T

: th ho chun ca cht kh nhit T, P=1atm v tnh cho 1 mol.

- Nu h gm 1 hn hp kh c p sut chung l P th p sut ring phn ca kh itrong hn hp l Pi=Ni.P ( vi = i

iin

nN )

)ln(ln)()()(

PNRTPRTiTiiTiTi

+=+= 00

iTiTiNRTPRT lnln

)()(++= 0

=> iPTiTi NRT ln),()( +=0

- i vi phn ng ho hc: aA + bB --> cC + dD

= )()( tgspG iip

iu kin chun: = )()( tgspG iip000

Vy trong iu kin ng nhit, ng p:



22/75


+ Nu < )()( tgsp ii : phn ng t din bin (t tri qua phi)

+Nu = )()( tgsp ii phn ng trng thi cn bng

=> th ho cng l tiu chun xt chiu v gii hn ca cc qu trnh (phn ng xy ra

theo chiu gim th ho)4.Mi quan h gia du G v ln SH , v T:

T=const --> TTT STHG =

0>= Sn --> phn ng ch xy ra nhit cao

- phn ng: SO2(k) + 1/2 O2 (k) --> SO3(k) kJHp 12990 , =

050 = Sn , --> phn ng xy ra nhit thp, nhng nu thp qu th vn

tc khng ln --> phn ng khng xy ra ngay c

==> iu kin: nhit khng qu thp

Ti liu tham kho:




23/75


CHNG III: CN BNG HO HC

I.PHN NG THUN NGHCH V PHN NG 1 CHIU

1.Phn ng thun nghch

L phn ng xy ra theo hai chiu tri ngc nhau trong cng mt iu kin (cim ca phn ng thun nghch: khng tin hnh n cng m dn n cn bng)

V d: N2O4(k) 2NO2(k)

Khi ly kh NO2 (hoc N2O4) nghin cu --> lun thu c ng thi c kh N2O4

(hoc NO2) trong bnh ngay nhit thng do tn ti s chuyn ho gia hai kh trn -

-> gi phn ng trn l phn ng thun nghch.

2.Phn ng 1 chiu: L phn ng ch xy ra theo mt chiu xc nh

V d: Phn ng phn hu KClO3(xc tc MnO2) : KClO3 --> KCl + 3/2O2

c im: tin hnh ti cng3.Trng thi cn bng ho hc

Mt phn ng khi t trng thi cn bng th:

- Xt v mt ng hc: vt=vn- Xt v mt nhit ng: 0= pG

* Vy trng thi cn bng ho hc: l trng thi ca phn ng thun nghch khi tc

phn ng thun bng tc phn ng nghch (hoc khi bin thin th ng p bng

khng).

*c im ca trng thi cn bng ho hc:

- L cn bng ng (phn ng vn xy ra theo 2 chiu ngc nhau nhng vt=vn).

-Ti trng thi cn bng thnh phn ca cc cht khng thay i: cht tham gia

phn ng mt i bao nhiu theo phn ng thun th li c sinh ra by nhiu theo phn

ng nghch).

- Trng thi cn bng ch tn ti khi cc iu kin thc hin phn ng (C, t0,p)

khng i.

II.PHNG TRNH NG NHIT VANT HOFF. HNG S CN BNG K1.Thit lp phng trnh ng nhit Vant Hoff

Xt phn ng: aA + bB cC + dD

T=const v A,B,C,D l cc kh l tng

Ta c: ++== )()()()( BADCiiT badctgspG

M: iTiTi PRT ln),()( +=0

(i vi 1 mol)

=> cCTCCTCTC PRTcPcRTcc lnln )()()( +=+=00

(cho c mol)

Tng t: bBTBTB PRTbb ln)()( += 0 : dDTDTD PRTdd ln)()( += 0



24/75


a

ATATA PRTaa ln)()( +=0

=>

)]ln()ln[()]ln()ln[( )()()()(b

BTB

a

ATA

d

DTD

c

CTCTPRTbPRTaPRTdPRTcG ++++++= 0000

bd

b

B

a

A

d

D

c

CTTTTT

PPPPRTBbAaDdCcG

+++=

.

.ln)]()(()()([( 0000

bd

b

B

a

A

d

D

c

C

TTPP

PPRTGG

+=

.

.ln0

Pi: p sut ring phn ca cc cht kh i= A,B,C,D)

t Pbd

b

B

a

A

d

D

c

C

PP

PP=

.

.

=> PTT RTGG ln+=0

(*)

- Ti trng thi cn bng==> 0= TG =>cb

b

B

a

A

d

D

c

C

TPP

PPRTG

=

.

.ln

0

t: constKPP

PPp

cb

b

B

a

A

d

D

c

C ==

.

.T=const (v trng thi cn bng, thnh phn cc

cht khng bin i na)

==> PT KRTG ln=0 (**)

vP

PT

KRTG

ln= (***)

Trong : 0TT GG ,, (J)

R=8,314J.K-1mol-1

P (atm)

Cc phng trnh (*) (**) v (***) c gi l cc dng khc nhau ca phng trnh

ng nhit Vant Hoff

2.Phng trnh hng s cn bng K:a.Hng scn bng Kp

cb

b

B

a

A

d

D

c

C

pPP

PPK

=

.

.Pi: p sut cc kh trng thi CB

- KP khng c n v.- Kp ch ph thuc vo bn cht phn ng v nhit .- i vi 1 phn ng xc nh, T=const --> Kp=const --> gi l hng s cn

bng ca phn ng.

b.Cc hng scn bng khc



25/75


*cb

b

B

a

A

d

D

c

C

CCC

CCK

=

.

.Ci: nng mol/lit cc cht trng thi cn bng

- KC ph thuc vo bn cht phn ng v nhit .- i vi 1 phn ng xc nh --> KC =f(T): KC ch ph thuc vo nhit

*cb

b

B

a

A

d

D

c

C

nnn

nnK

=

.

.ni; s mol kh i trng thi cn bng

Knph thuc vo bn cht cc cht , T v P chung ca h khi cn bng v tng s mol

kh ca h khi cn bng

*cb

b

B

a

A

d

D

c

C

NNN

NNK

=

.

.Ni: nng phn mol ca kh i:

=

i

i

in

nN

KNph thuc vo bn cht cc cht , T v P chung ca h khi cn bng

c.Mi quan h gia cc hng scn bng

- Ta c PiV=niRT (V th tch hn hp kh (lt))

=> RTCV

RTnP i

i

i == thay vo phng trnh Kp ta c

( ) ( )

( ) ( )( ) ( )[ ] n

c

abdc

a

A

b

B

d

D

c

C

a

A

b

B

d

D

c

C

P RTKRTCC

CC

RTCRTC

RTCRTCK

++ === ).(.

.

n = s mol kh sn phm- s mol kh tham gia (da vo phng trnh phn ng)

Vyn

cP RTKK

= ).( (vi R= 0,082 atm.l.mol-1

K-1

)-Mt khc ta c:

Pn

nPNP

i

i

ii .. == --> thay vo Kp ta c:

n

cbNP PKK= . v

n

cbi

nPn

PKK

=

.

( in )cb tng s mol kh c mt trong h phn ng khi cn bng.

Vy c:n

cbi

n

n

cbN

n

CPn

PKPKRTKK

===

..)(

*Nhn xt:

- Khi 0=n (tng s mol kh 2 v phn ng bng nhau) --> Kp=KC=Kn=KN=K=f(T)

- i vi 1 phn ng cho th:

+ Kp,KC ch ph thuc vo nhit

+ KN ph thuc vo nhit , P chung ca h khi cn bng



26/75


+Kn ph thuc vo T,P chung ca h khi cn bng, tng s mol kh ca h khi cn

bng.

* Ch :

- Gi tr hng s cn bng K ca phn ng u phi gn vi 1 phn ng c th no .

V d: SO2(k) + 1/2O2(k) SO3(k)2

1

22

3

OSO

SO

P

PP

PK

.

=

2SO2(k) + O2(k) 2SO3(k)2

2

2

22

3

P

OSO

SO

P KPP

PK ==

.'

SO3(k) SO2(k) + 1/2O2(k)1

21

3

22 == PSO

OSO

P KP

PPK

.''

- Nu K c gi tr kh ln --> coi phn ng xy ra hon ton, K nh -> phn ng thunnghch.

- Nu phn ng c cht rn hoc cht lng tham gia (v khng tan ln vo cc cht khc)

th chng khng c mt trong phng trnh hng s cn bng ( v 0 )()( TiTi = trong

sut qu trnh phn ng--> khng cn RTlnPi)

V d: Fe2O3(r) + 3 CO(k) 2Fe(r) + 3 CO2(k) Kp= 3

3

2

CO

CO

P

P

Hg(l) + 1/2 O2(k) HgO(r) 212

1

O

P

PK =

3. Cc phng php xc nh hng scn bng

a.Xc nh theo thnh phn cc cht ti thiim cn bng

V d: CaCO3(r) CaO(r) + CO2(k)

Nung CaCO3nhit T , khi cn bng: mmHgPCO 7402 = .Tnh Kp

Gii:

760

7402 == COP PK

b.Xc nh thng qua 1 si lng nhit ng

PT KRTG ln=0

==>

=RT

GK

T

P

0

exp

V d: Tnh Kp ca phn ng sau 250C:

2NH3(k) N2(k) + 3H2(k) bit1

30298 6516

= molkJNHG s .,)(,

Gii:



27/75


13

0298

0298 333651622

+=== molkJNHGG s .,),)(()(,

630

298 104512983148

10333

298=

=

= .,

.,

.,exp

.expR

GKP

c. Xc nh theo phng php gin tip: Phn tch qu trnh theo 1 chu trnh kn

V d:

Cgr O2(k) CO2(k)

CO(k) O2(k)1/2

KP=?

K1 K2

02

01

0GGG +=

)ln()ln(ln 21 KRTKRTKRT P +=

lnKP=lnK1+lnK2 ==> K=K1.K2

III.SCHUYN DCH CN BNG. NGUYN L LE CHATIELIER1.Schuyn dch cn bng

- Xt v mt nhit ng, khi phn ng t trng thi cn bng--> 0=G --> 1=P

P

K

khi

hang trng thi cn bng nu ta thay i mt trong cc thng s trng thi (P,T,C)

ca h --> 0G => cn bng b ph v, qu trnh s tin hnh theo chiu 0 vt=vn => v=0. Khi h

t trng thi cn bng, ta thay i 1 trong cc thng s trng thi (P,C,T) => vt vn ==>

qu trnh tin hnh theo chiu v > 0 t trng thi cn bng mi ng vi cc thng

s mi.

==> gi l s chuyn dch cn bng.

nh ngha: S chuyn dch cn bng l s chuyn t trng thi cn bng ny sang trng

thi cn bng khc di nh hng ca cc tc ng bn ngoi (P,T,C) ln h.

2.nh hng ca nhit ti schuyn dch cn bng. Phng trnh ng p VantHoff

Xt phn ng: aA + bB cC + dD P=const

Ta c:RT

GK TP

0=ln

Mt khc:2

00

T

H

T

G

T

T

P

T =

do : => 2

0

RTH

TK T

P

P=

ln => phng trnh ng p Vant Hoff



28/75


- Nu 00 > TH (phn ng thu nhit) --> hm ng bin

+ Khi nhit tng --> Kp tng --> cn bng chuyn dch theo chiu thun ( chiu

thu nhit)

+ Nu T gim --> Kp gim ==> cn bng chuyn dch theo chiu nghch ( chiu

to nhit)

- Nu 00 hm nghch bin:

+Khi nhit tng --> Kp gim --> cn bng chuyn dch theo chiu nghch (

chiu thu nhit)

+Nu T gim --> Kp gim ==> cn bng chuyn dch theo chiu thun ( chiu to

nhit)

* Nhn xt: Kt qu ca s chuyn dch cn bng l chng li s thay i bn ngoi:

+ Khi nhit tng th cn bng chuyn dch theo chiu lm gim nhit ca hl chiu thu nhit c 0> TH

+ Khi nhit gim th cn bng chuyn dch theo chiu lm tng nhit ca h

l chiu to nhit c 0T2 hp -->c th coi constHT =0

th:

=2

1

2

1

2

0T

T

T

T

PT

dT

R

HKdln

=>

=

21

011

ln1

2

TTR

H

K

K

T

T

P

P

vi 0H (J) v R=8,314 J.K-1mol-1

3. nh hng ca p sut n schuyn dch cn bng

Ta c:n

cbNP PKK

= . V Kp khng ph thuc vo P--> thay i P th Kp=const nn:

- Nu 0>n : khi tng P--> KN gim ( gi Kp=const) => chuyn dch cn bng theo

chiu nghch ( lm gim s mol kh) v ngc li

- Nu 0 KN tng ( gi Kp=const) => chuyn dch cn bng theo

chiu thun ( lm gim s mol kh) v ngc li khi P gim --> KN gim --> cn bng

dch chuyn theo chiu nghch ( lm tng s mol kh)

- Nu 0=n => P khng nh hng ti s chuyn dch cn bng

* Nhn xt: Kt qu ca s chuyn dch cn bng l chng li s thay i bn ngoi:



29/75


- Nu P tng => cn bng chuyn dch theo chiu P gim (gim s mol kh0 cn bng chuyn dch theo chiu tng P (tng s mol kh0>n )

4. nh hng ca nng

Xt phn ng: aA + bB cC + dD c constCC

CCK

cb

b

B

a

A

d

D

c

C

C =

=

.

.T=const

- Nu tng CA, CB => cn bng chuyn dch theo chiu tng CC,CD ( gi KC=const) =>

cn bng chuyn dch theo chiu thun lm gim CA,CB- Nu gim CA, CB => cn bng chuyn dch theo chiu gim CC,CD ( gi KC=const)

=> cn bng chuyn dch theo chiu nghch lm tng CA,CB

* Nhn xt: Kt qu ca s chuyn dch cn bng l chng li thay i bn ngoi.Nutng Ci th cn bng chuyn dch theo chiu lm gim Ci v ngc li.

5. Nguyn l chuyn dch cn bng Le Chatelier

Khi mt hang trng thi cn bng, nu ta thay i 1 trong cc thng s trng

thi ca h ( T, P hoc C) th cn bng s chuyn dch theo chiu chng li s thay i

.

Ti liu tham kho:




30/75

Chng IV: Cn bng pha

I. Mt s khi nim1.Pha ( ) l phn ng th ca h c thnh phn, tnh cht l hc , tnh cht ho hc

ging nhau mi im ca phn ng th v c b mt phn chia vi cc phn khc

ca h.- Pha ch gm 1 cht gi l pha nguyn cht (pha n) cn pha gm 2 cht tr ln--> gi

l pha phc tp.

- H gm 1 pha --> h ng th.

- H 2 pha -> h d th.

V d: H gm H2O + H2O lng + H2O hi => gm 3 pha: rn, lng, hi.

H gm CaCO3(r), CaO(r),CO2(k) --> 3 pha: 2 fa rn + 1 pha kh

2. Cu t: L phn hp thnh ca h c th c tch ra khi h v tn ti c bn ngoi

h.

S cu t trong h k hiu l R

V d: dung dch NaCl gm 2 cu t l NaCl v H2O --> R=2

3.S cu t c lp (K): L s ti thiu cc cu t xc nh thnh phn ca tt c

cc pha trong h.

- Nu cc cu t khng phn ng vi nhau v nu pha c thnh phn khc nhau th K=R

(trong h khng c phng trnh lin h nng cc cu t)

V d: dung dch NaCl => R=K=2.

-Nu cc cu t tng tc vi nhau v nm cn bngvi nhau--> chng khng cn c lp

vi nhau na--> K=R-q

q: s h thc lin h gia cc nng ( q c th l phng trnh hng s cn bng, iu

kin u v nng ca cc cu t)

V d: H gm 3 cu t HCl, Cl2, H2 u l cc cht kh c tng tc,nm cn bng vi

nhau: 2HCl(k) H2(k) + Cl2(k)

[ ][ ][ ]2

22

HCl

ClHKC = => bit c nng ca 2 cu t s bit c nng

ca cu t cn li.

Vy h c: R=3, q=1, ==> K= R-q=2

Nu gi thit ban u h ch c HCl ( hoc cho t l mol H2:Cl2 ban u) => q=2 => K=1

4.Bc t do ca h(C): L s ti thiu cc thng s trng thi cng (P,T,C) xc

nh trng thi cn bng ca 1 h ( l s thng s trng thi cng c th thay i

1cch c lp m khng lm bin i s pha ca h)

V d: H2O(l) H2O(k)

==> cn bng c 2 pha==> C=1 v

+ C th thay i 1 trong 2 thng s P hoc T m khng lm thay i s pha ca

h.



31/75

+ Hoc: mt nhit xc nh th P hi H2O nm cn bng vi H2O lng l xc

nh, tc l ch cn bit 1 trong 2 thng s T hoc P th xc nh c trng thi cn bng

ca h.

5.Cn bng pha: Cn bng trong cc h d th, cc cu t khng phn ng ho hc

vi nhau nhng xy ra cc qu trnh bin i pha ca cc cu t => cn bng pha.II. Quy tc pha Gibbs.

Xt h gm R cu t 1,2,....R c phn b trong pha ( ,...,,, pha)

1.iu kin cc pha nm cn bng vi nhau: m bo cc cn bng sau:

- Cn bng nhit: nhit cc pha bng nhau

TTTT ==== ...

-Cn bng c: p sut cc pha bng nhau

PPPP ==== ...

-Cn bng ho: th ho ca mi cu t trong cc pha bng nhau:iiii

==== ...

2.Qui tc pha Gibbs

- Cc thng s trng thi cng xc nh trng thi ca h l T,P, C

Gi Ni l nng mol phn ca cu t i trong 1 pha th N 1+N2+N3+...+Ni=1

=> Vy xc nh nng ca R cu t trong 1 pha cn bit nng ca (R-1) cu

t.

V c pha => xc nh nng ca R cu t trong pha th s nng cnbit l (R-1).

T s thng s trng thi cng xc nh trng thi ca h l

(R-1)+ 2

trong s 2: biu th 2 thng s bn ngoi l T v P xc nh trng thi ca h

V cc pha nm cn bng vi nhau => cc thng s khng c lp vi nhau na: c lin

h vi nng m khi cn bng th ca mi cu t trong cc pha phi bng nhau ( iu

kin cn bng ho)

)(...)()( 111 ===

)(...)()( 222 ===

)(...)()( RRR ===

=> Mi cu t c (-1) phng trnh lin h ==> R cu t c c R(-1) phng trnh

lin h gia cc thng s.

Nu c thm q phng trnh lin h nng cc cu t, v d: khi c phn ng ho hc

gia cc cu t th s phng trnh lin h cc thng s trng thi cng ca h l:

R(-1) + q



32/75

Bc t do ca h = Cc thng s trng thi s phng trnh lin h gia cc thng s

C= [(R-1)+2]-[R(-1)+q] C=R-q-+2 C= K - q + 2 => Biu thc ton hc ca quy tc pha Gibbs

* Nhn xt:+ Khi K tng, => C tng, tng v C gim.

+ Bc t do 20 + KC

+Nu trong iu kin ng nhit hoc ng p th: C =K - + 1 (Nu phng trnh

c 0=n => P khng nh hng ti phn ng --> dng phng trnh ny)

+Nu h va ng nhit va ng p th C=K-

V d1: Xt h 1 cu t (R=K=1), v d nc nguyn cht

- Nu trng thi hi => =1 => C= K-+2= 1-1+2=2 => trng thi ca hi n

c

cxc nh bi 2 thng s trng thi cng l T v P

- Nu hi nc nm cn bng vi nc lng th =2=> C=1-2+2=1 => trng thi ca h

gm H2O lng v hi c xc nh bi 1 trong 2 thng s l T hoc P ( v 1nhit

xc nh th P ca hi nc l xc nh)

V d2: Xt h gm: Mg(OH)2 (r) MgO (r) + H2O(k)

=2 pha rn + 1 pha kh =3 pha

C=R-q+2=3-1-3+2=1

=> c php thay i 1 trong 2 thng s l T hoc P m khng lm thay i s pha cah hoc trng thi cn bng c xc nh bng 1 trong 2 thng s T hoc )(hOHP 2

III.Cn bng pha trong h 1 cu t1.Cn bng pha trong h 1 cu t

Xt h gm 1 cht nguyn cht, khi trong h c 2 pha nm cn bng nhau:

Rn(R) Lng(L)

Lng(L)Hi (H)

Rn (R)Hi (H)

( ( ) ( ) RR ) => v h 1 cu t, s pha 3 (3 2+ K )

=> C= K-+2 =1-2+2 =1 (R=K-1) trng thi cn bng gia hai pha c c trng

bi hoc T hoc P, tc l nu 1 trong 2 thng s trng thi l P hoc T bin i th thng

s kia phi bin i theo: p=f(T) hoc T=f(P). C th l :

- P=const=> cht nguyn cht nng chy, si hoc chuyn trng thi tinh th 1 nhit

nht nh, c gi l nhit chuyn phaTcf, nhit ny khng b bin i trong

sut qu trnh chuyn pha. Khi p sut thay i => Tcf thay i theo.

Vd: P=1atm, nc nguyn cht ng c 00

C v si 1000

C P=2atm, nc nguyn cht ng c 0,00760C v si 1200C



33/75

- T=const, hi nm cn bng vi lng v rn c P nht nh gi l P hi bo ho (hi

c goi l hi bo ho)

Cc ng cong biu th s ph thuc ca Phi bo ho ca pha rn vo nhit , ca pha lng

vo nhit v nhit nng chy vo P ct nhau ti 1 im gi l im ba, im ba

ny ba pha rn lng hi (R, L, H) nm cn bng vi nhau: R L

H Khi C=1-3+2 =0 => v tr im ba khng ph thuc vo T v P m ch ph thuc vo

bn cht cht nghin cu.

2. nh hng ca p sut n nhit nng chy, si v chuyn dng tinh th ca

cht nguyn chtV h 1 cu t nn th ha ng nht vi th ng p mol (Gi= i ). Khi T, P khng i

iu kin cn bng gia hai pha v l:)()( GG =

V h c C=1 nn nu mt thng s bin i, v d, p sut bin i mt lng dP th

mun hai pha tn ti cn bng, nhit cng phi bin i mt lng dT. Khi th

ng p mol phi bin i:)()()(

dGGG

+> )()()( dGGG +>

Sao cho: )()()()( dGGdGG +=+ => )()( dGdG = Thay vo cng thc: dG= VdP SdT ta c:

dTSdPVdTSdPV )()()()( =

=>S

V

SS

VV

dP

dT)()(

)()(

=

=

C T

HS = suy ra:

cfH

V

dP

dT

Tcf= phng trnh Clapeyron

Trong H c tnh bng J th V tnh bng m3, T bng K v P bng Pa.

- Khi mt cht si th V =Vh- Vl >0 v H hh>0 (hh:ha hi), nn p sut bnngoi tng th nhit si tng theo.

- Khi nng chy H nc >0 v a s trng hp =V Vl-Vr >0, do P tng thnhit nng chy tng. i vi nc Vl


34/75

3.nh hng ca nhit n p sut hi bo ho ca cht nguyn cht

Xt cc trng hp: L H

R H

V Vr,Vl hlh VVVV = v hrh VVVV =

Nu hi c coi l kh l tng,xt i vi 1 mol c:

P

RTVh = thay vo phng trnh Clayperon c:

PRT

H

VT

H

VT

H

dT

dP

cf

cf

cf

cf

cf

cf.

... 2

=

=

=

=> dTT

H

P

dP2

= (v Pd

P

dPln= ) nn c:

2RT

H

dT

Pd

=

ln

-> phng trnhClaypeyron-Clausius

Trong khong nhit hp -> c th coi constH= khi c

=

211

2 11

TTR

H

P

Pln (*)

Biu thc (*) cho bit c th:

-

Tnh p sut hi bo ho

nhit T2(hoc T1) khi bit P nhit T1 v cfH

- Tnh nhit si P bt k khi bit nhit si mt p sut no v H bay hi.Tnh H bng cch o P1 v P2 2 nhit khc nhau.

Ti liu tham kho:


P1,P2 : cng n v

R=8,314J.K-1.mol-1

H : J



35/75

Chng V Dung dchI.H phn tn1.nh ngha: L h gm 2 hay nhiu cht trong cht ny c phn b trong cht kia

di dng nhng ht rt nh.

- Cht phn b c gi l cht phn tn, cht kia l mi trng phn tn. Cht phn tn

v mi trng phn tn c th 1 trong 3 trng thi: rn, lng hay hi.

Vd: ng tan trong nc => ng l cht phn tn, H2O l mi trng phn tn

- Da vo kch thc ca ht phn tn, chia lm 3 h phn tn:

a.H phn tn th: Kch thc ht t 10-7-10-4m

-c im: Khngbn,cht phn tn d tch ra khi mi trng phn tn.

-C 2 dng:

+ Huyn ph: cht phn tn l rn, pha phn tn l lng. v d: nc ph sa

+ Nh tng: cht phn tn l lng, mi trng phn tn cng l lng v d: sa c

ln nhng ht m l lngb.Dung dch keo: (H keo): Kch thc ht t 10-7 10-9m

- c im: tng i bn

c. Dung dch thc (dung dch): Kch thc ht 10-10m (bng kch thc phn t hoc

ion), trong cht phn tn v mi trng phn tn c phn b vo nhau di dng

phn t hoc ion, gia chng khng cn b mt phn chia, to thnh 1 khi ng th gi

l dung dch thc

Cht phn tn c gi l cht tan, mi trng phn tn c gi l dung mi

-c im: H ny rt bn

2.Cc loi nng -Nng % (C%): l t l % khi lng cht tan so vi khi lng dung dch (C%chnh l

lng cht tan c trong 100g(100kg) dung dch)

100.%ba

aC

+= % a,b lng cht tan v lng dung mi, tnh bng g (kg)

-Nng mol/lit (CM): l s mol cht tan c trong 1 lit dung dch

-Nng ng lng (CN):S mol ng lng cht tan c trong 1 lt dung dch

-Nng molan ( ) l s mol cht tan c trong 1000gam dung mi

-Nng phn mol:

=i

ii

nnN

II. ng lng ()Trong cc phn ng ho hc, cc cht tc dng va vi nhau theo nhng s phn khi

lng tng ng gi l ng lng.

Chn ng lng ca H lm n v H=1

1.nh ngha nglng

- ng lng ca mt cht hoc ca 1 nguyn t l phn khi lng ca cht hoc

nguyn t tc dng va vi 1 ng lng ca H.

V khng phi mi cht u phn ng vi H => nh ngha c m rng nhsau:



36/75

ng lng ca 1 nguyn t hay 1 hp cht l s phn khi lng ca n tc dng va

vi 1 ng lng ca 1 cht bt k

V d: Cl2 + H2 = 2HCl

Cl2 + Zn =ZnCl2

=> 71 phn khi lng ca Cl2 tc dng vi 2 phn khi lng ca H

Vy kt hp vi 1 ng lng ca H ch cn 1 khi lng ca Cl2 bng khi lngnguyn t ca n => Cl=35,5, Zn= 32,5

-Mol ng lng ca mt cht: l khi lng tnh ra g c gi tr ng bng ng lng

V d: O=8g

2.Cch tnh ng lng

a.ng lng ca nguyn t

n

A=

Vi nguyn t c nhiu ho tr

khc nhau s c nhiu ng lng v ng lng ca chng gn vi 1 phn ng c th

m chng tham gia.

b.ng lng ca hp cht

n

M=

Cch xc nh n

- i vi phn ng oxi ho kh: n l s e trao i ng vi 1 phn t cht (n l se m 1 phn t trao i trong phn ng)

- i vi phn ng trao i: n l s in tch (+) hoc (-) m 1 phn t cht traoi trong phn ng

V d: H2SO4 + NaOH = NaHSO4 + H2O (1)

981

98

1

42

42 1===

SOH

SOH

M)(

401

40

11 ===

NaOHNaOH

M)(

H2SO4 + 2NaOH = Na2SO4 + H2O (2)

492

98

2

42

42 2===

SOH

SOH

M)(

401

40

12 ===

NaOHNaOH

M)(

- i vi trng hp tnh ng lng ca mt cht khng c phn ng c th th:n

M=

+ i vi axit: n l s H axit trong phn t+ i vi baz: n l s nhm OH- baz trong phn t

A: khi lng nguyn t nguyn tn: ho tr nguyn t

M: khi lng phnn: tu tng trng hp



37/75

+ i vi mui: n l s in tch (+) hoc in tch (-) m cc ion mang

trong 1 phn t.

V d:2

42

42

SONa

SONa

M= ;

6

342

342

)(

)(SOAl

SOAl

M=

+Mi quan h gia CM v

CN: CM= n

CN

3.nh lut ng lng

Cc cht tc dng va vi nhau theo cc khi lng t l vi ng lng ca chng.

=>B

A

B

A

m

m

= mA l khi lng cht A tc dng va vi khi lng mB cht B

=>B

B

A

A mm

= => n ng lng cht A phn ng va vi n ng lng cht B ( S

ng lng cht A bng s ng lng cht B)

=> Nu c VA(l) cht A nng mol ng lng l ANC phn ng va vi VB(l) cht

B nng mol ng lng lBN

C th

VA. ANC =VB. BNC -> ng dng nhiu trong chun th tch

V d: trung ho 25ml NaOH cn 28ml dung dch axit 0,1N => tnh lng NaOH c

trong 1l dung dch ?

Gii

Gi x l nng ng lng mol ca NaOH

=> x.25=0,1.28 =>x=28.0,1/25

NaOH=40=> s g NaOH c trong 1 lt dung dch lm=40.x=40.28.0,1/25=4,48g

III. ho tan1.Qu trnh ho tan. Nhit ho tan ca mt cht

Qu trnh ho tan (khng phi l qu trnh trn ln) gm qu trnh

+ Qu trnh phn tn cht tan( di dng nguyn t, phn t,ion) vo trong khp

th tch dung mi.

+Qu trnh tng tc gia cc phn t ca dung mi vi cc phn t ca cht tan

=> to thnh dung dch (hp cht ho hc)Hp cht ho hc to thnh gi l hp cht sonvat, nu dung mi l nc th gi l hp

cht hydrat.

=> Qu trnh ho tan c s ph v lin kt cng loi to lin kt khc loi v c th

biu din bng s :

T_T +

dm

T_d

T+T

dm

+

htH

0> ptH

0


38/75

dm dung mi

ptH nng lng phn tn ( cn tiu tn -> 0> ptH )

svH nng lng qu trnh sovat (Qu trnh l to nhit --> 0 = svH hH nhit hidrat ho)

htH nhit ho tan

Theo s trn ta c:

svptht HHH += (do 0 ptH --> htH c th m hoc dng)

+Nu 0>> htsvpt HHH : qu trnh ho tan thu nhit, l qu trnh ho tan

ca a s cht rn vo trong nc.

+Nu 0


39/75

+ Khi nhit tng -> cn bng chuyn dch theo chiu thun tan

tng.

+ Khi nhit gim -> cn bng chuyn dch theo chiu nghch tan

gim.

Trong mt s trng hp c th iu ch dung dch qu bo ha c nng ln hn

ha tan s nhit T dung dch qu bo ha khng bn.- Nu htH cn bng chuyn dch theo chiu nghch tan

gim.

+ Khi nhit gim -> cn bng chuyn dch theo chiu thun tan

tng.

c.nh hng ca P (i vi cht kh). nh lut Henry

CT(kh) + dm dd (*)

qu trnh ha tan cht kh lm gim n nn:+ Khi P tng => ha tan s tng+Khi P gim => ha tan s gim

T cn bng (*) thy rng: S tng P s dn n s chuyn dch cn bng sang phi. Nu

tng P ln n ln th ha tan ca cht kh cng tng ln n ln.

nh lut Henry: mt nhit khng i, khi lng cht kh ha tan trong 1 thtch cht lng xc nh t l thun vi P ca n trn b mt cht lng.

m= k.P

trong : k- h s t l ph thuc vo bn cht ca cht kh, dung mi v

nhit -> gi l h s Henry.

P p sut ring phn ca cht kh trn mt cht lng.

Nu trn mt cht lng c 1 hn hp kh th s ca mi kh t l vi P ring phn ca tng

kh. nh lut Henry ch ng cho cht kh c s nh, P khng ln v kh khng tc dng

ha hc vi dung mi.

IV. Tnh cht ca dung dch.- Ch xt dung dch: + long

+ cht tan khng bay hi

+ cht tan khng to dung dch rn vi dung mi.VD: dd mui hoc dd ng khi un ch c H2O bay hi.

- Dung dch l tng: l dung dch m ca n tun theo nh lut tng t nhi vi kh l tng. Cc dung dch rt long c coi nh dung dch l tng.

ca cu t i trong dung dch l tng c tnh theo cng thc ging nh i vi

kh l tng: iiTi NRTPT ln),()( +=0

.

Tuy nhin v P nh hng rt t n tnh cht ca cht lng nn y 0i v hu

nhch ph thuc vo T.

dung dch xt y l (rt) long-> coi l dung dch l tng.1. p sut hi bo ha ca dung dch. nh lut Raun I.



40/75

Kho st h 2 cu t: dung mi ch cha 1 cht tan ( cc kt lun rt ra cng ng cho

h nhiu cu t).

K hiu: 2 l cht tan, 1 l dung mi0

1 : th ha ca dung mi nguyncht.

i : th ha ca dung mi trong dung dch.

1

0

11 NRT ln+=

V N1 khi c mt cht tan , ca dungdch s gim i 1 lng l (RTlnNi).

a. p sut hi bo ha ca dung mi nguyn cht ( 01P )

L H(dm) ( vi cht lng L l dung mi nguyn cht)

Hi nm cn bng vi lng gi l hi bo ha, hi bo ha gy P hi bo ha C=1-

2+2=1: p sut hi bo ha ca dung mi nguyn cht ch ph thuc vo nhit . cng

1 nhit , cht no cng d bay hi th P hi bo ha cng ln

Qu trnh bay hi l qu trnh c 0>H => khi nhit tng th Phi bo ha

cng tng.

b. p sut hi bo ha ca dung dch ( 1P )

Ch xt dung dch cha 1cht tan v 1 dung mi: dd (L) H

C=2-2+2 =2

Phi bo ha ca dung dch ph thuc vo c T v C.iu kin cn bng pha (ca cht lng v hi ca n) t0C, P xc nh l:

li

hi =

m ih

i

h

i PRT ln+=0 (tnh cho 1 mol).

i

l

i

l

i NRT ln+=0

li

hi = => 0=

i

ili

ohi

N

PRT ln+ 0

constRTN

PR

hi

li

i

i =

=00

ln t0C xc nh.

Ngha l constkN

Pi

i

i == => Pi= ki.Ni

Khi Ni =1 (cht nguyn cht) th ki= Pi0 l p sut hi ca cu t i nguyn cht. T c:

Pi= Pi0.NiKt lun:

- Khi Ni Pi < Pi0: dung dch cha cht tan khngbay hi th Pp sut hi bo ha ca dung dch lun nh hn Pp sut hi bo ha ca dung mi

nguyn cht cng nhit .

- Nu nng dung dch cng ln => Pp sut hi bo ha ca dung dch cng nh.c. nh lut Rault I

Gi N1 l nng phn mol ca dung dch

P1

v P1,0

ln lt l Pp sut hi bo ha

ca dung dch v dung mi nguyn cht cng

mt t0C.

Th: P1 =N1.P1,0



41/75

V N1= 1-N2

=> 20101

101N

P

P

P

PP=

=

,,

, =>21

22

01 nn

nN

P

P

+==

,

Trong : P : gim p sut hi bo ho ca dung dch so vi dung mi.

01,P

P

l gim p sut hi bo ho tng i ca dung dch

n2: s mol cht ho tan

n1: s mol dung mi.

Nu dung dch long(N1-> 1)=> n1>>n2 th:

1

2

01 n

n

P

P=

,

=> Biu thc ca nh lut Rault I

Pht biu: gim tng i ca dung dch cha cht ho tan khng bay hi t l vi s

mol cht tan c trong 1 lng dung mi xc nh.

2. Nhit si v nhit ng c ca dung dch. nh lut Rault IIa.Nhit si ca dung dch

* Nhit si ca cht lng: L nhit p sut hi bo ha ca cht lng bng p

sut bn ngoi.

Xt cn bng L H

C=2+2-2 =2

Vy nhit si ca dung dch ngoi s ph thuc vo p sut bn ngai cn ph thuc

vo nng cht ha tan.

- p sut bn ngoi nhnhau, khi dung mi nguyn cht si th dung dch cha cht tankhng bay hi s cha si v p sut hi bo ha ca dung dch lun lun nh hn p sut

hi bo ha ca dung mi nguyn cht cng 1 nhit . Vy dung dch cha cht tan

khng bay hi c nhit si cao hn dung mi nguyn cht.

- tng nhit si ca dung dch so vi dung mi nguyn cht st c tnh theo cng

thc Rault 2:

M

mkkt sss ==

trong st =ts,dd-ts,dm (ts,dd: nhit si dung dch, ts,dm: nhit si dung mi nguyn

cht, nng molan; ks: hng s nghim si (ch ph thuc vo bn cht dung mi).

- Khi dung dch si th hi bay ra l ca dung mi nn nng dung dch tng dn, do

nhit si ca dung dch tng dn, nhng khi t n dung dch bo ha th hi dung

mi bay ra lm cho cht ha tan kt tinh li, lc ny nng dung dch khng thay i v

nhit si ca dung dch khng bin i na. Vn dng quy tc pha ta c: C=2-3+2 =1.

Ngha l khi xut hin tinh th cht tan th nhit si ca dung dch ch cn ph thuc

vo p sut bn ngoi.

b. Nhit ng c ca dung dch cha cht tan khng bay hi

Nhit ng c ca cht lng l nhit c cn bng sau: R L



42/75

i vi dung dch ta c: C=2+2-2 =2

Vy ng c ca dung dch ngoi s ph thuc vo p sut bn ngai cn ph thuc

vo nng cht ha tan.

Nhit ng c ca dung dch cha cht tan khng bay hi lun thp hn dung mi

nguyn cht. V tun theo nh lut Rault 2:

M

mkkt ddd ==

trong st =ts,dd-ts,dm ,ks: hng s nghim ng (ch ph thuc vo bn cht dung mi).

3. S thm thu v p sut thm thu:

a. S thm thu

L s khuch tn mt chiu ca cc phn t dung mi qua mng bn thm (mng bn

thm l mng ch cho cc phn t dung mi i qua m khng cho cc phn t cht ha tan

lt qua). Hin tng ny thy rt r khi hai bn ca mng bn thm cha dung dch c

nng khc nhau hoc 1 bn l dung dch cn bn kia l dung mi nguyn cht; khi

cc phn t dung mi s khuch tn t dung dch long hoc t dung mi nguyn cht

sang pha bn kia nhiu hn s khuch tn theo qu trnh ngc li, do lm tng th

tch ca dung dch pha bn kia.

b. p sut thm thu

Hin tng thm thu lm cho mc dung dch mt pha ca mng bn thm dng ln

cao. Chiu cao ca ct dung dch ny to nn mt p sut lm cho hin tng thm thu

ngng li. p sut c to ra bi ct dung dch ny c trng nh lng cho s thmthu v c gi l p sut thm thu P. N c tnh theo cng thc:

PV=nRT= RTM

m

Trong : V l th tch ca dung dch

M

mn = : s mol cht ha tan.

R: Hng s kh l tng.



43/75

Chng IX: Dung dch cht in ly

I.Tnh cht ca dung dch in li1.Cht in ly

L cht khi ha tan trong nc, cc phn t ca n phn ly nhiu hay t thnh cc ion.Nguyn nhn c bn ca s phn li phn t thnh ion l do tng tc gia cc cht in li

v cc phn t dung mi to thnh cc ion b hidrat ha.

V d: NaCl + mH2O = Na+.nH2O + Cl

-(m-n)H2O.

Cc gi tr m, n thng khng xc nh c v ph thuc vo nng v nhit nn

thng c vit:

NaCl + aq = Na+.aq + Cl-.aq

aq: Lng nc khng xc nh

2. Cht in li mnh: L cht khi tan trong nc, tt c cc phn t ca n phn ly thnhion.

V d: cht in ly mnh bao gm:

+ cc mui trung tnh: NaCl, NaBr, Na2SO4, NaNO3,...;

+ Cc axit mnh: HCl, HBr, HI, HNO3, H2SO4, HClO4;

+ Cc baz mnh: NaOH, KOH,...

ch s in li mnh, trong phng trnh in li c ghi bng du

V d: NaCl + aq Na+.aq + Cl-.aq

Ngi ta thng vit phng trnh ny mt cch n gin nhsau:

NaCl Na+ + Cl-

3. Cht in li yu: cc axit hu c HCOOH, CH3COOH, mt s axit v c: H2S,

H2CO3, H2SO3, HClO, HClO2, HClO3, H2SiO3, HF..., cc baz yu v baz t tan: NH3,

Mg(OH)2, Fe(OH)3, cc mui t tan: HgCl2, Hg(CN)2, CdCl2,...

L cht khi ho tan ch c mt phn cc phn t b phn ly thnh ion, trong dung dch

cht in ly yu tn ti cn bng ng gia cc cc ion v cc phn t khng b phn ly.

biu th s in ly khng hon ton, trong phng trnh in ly ca cht in ly yu

dng du

V d: Trong dung dch axit acetic tn ti cn bng

CH3COOH H+ + CH3COO

-

4. Tnh cht bt thng ca dung dch cht in li so vi dung dch cht khng in

li

- Dung dch cht in li dn in tt, l do trong dung dch c cc phn t mangin l ion.



44/75

- Dung dch cht in li c gim p sut hi 'P , tng nhit si ''st , gim nhit ng c dt' v p sut thm thu P u ln hn so vi dung dch

cht khng in li cng nng :

1>==

=

i

P

P

t

t

P

P '''

Tnh cht bt thng ny do cc phn t cht in li phn li thnh cc ion lm s phn

t trong dung dch tng ln. Cc tnh cht trn ca dung dch ph thuc vo nng

cc phn t ny. T c th tnh i bng cch sau:

S phn t c trong dung dch=i --------------------------------------

S phn t ho tanS phn t trong dung dch bng s ion c to thnh cng vi s phn t cha b

phn li thnh ion.II. in ly

nh gi mc in ly ca tng cht a ra khi nim in ly .

1.nh ngha:

S phn t b phn ly

= -------------------------------- (0 b qua lc tng tc tnh in gia cc ion.

- i vi cht in ly mnh:+ Trong dung dch long -> khong cch gia cc ion ln => b qua tng tc tnh

in gia cc ion.

+ Trong dung dch c -> s ion c mt trong dung dch nhiu => khong cch

gia cc ion nh =>Khng b qua tng tc tnh in gia cc ion, cc ion ht

hoc y nhau dn n hin tng bao quanh 1 ion c th c nhiu ion tri du to

kh quyn ion => c hin tng nh1 s phn t khng in ly.

2.Cch xc nh

Gi N l s phn t ho tan, q l s ion m 1 phn t phn ly ra.

S phn t b phn ly lN, do s ion to thnh l.N.q

S phn t cn li khng phn ly l N - .N

S phn t c trong dung dch l N= S ion + s fn t khng phn ly= q N + (N -N)



45/75

q. .N + N - .N q. - + 1

=> i = -------------------------- = -----------------

N 1

1q

1i

=

Da vo cng thc ny ta c th tnh c

I c th xc nh bng thc nghim t vic o gim nhit ng c hoc gim

Phi bo ha hoc t st

P

P

t

t

t

ti

d

d

s

s

=

=

=

'''

nhng i o c t vic o st hoc P cho kt qa km chnh xc so vi vic o 'dt

nn thng xc nh i t vic o 'dt :d

d

tti=

'

. Bit i tnh c v ngc li.

V d: Mt dung dch cha 8g NaOH ho tan trong 1000g nc ng c -0,677oC.

Hy xc nh .

Gii:

V q= 2

t

i = ----- t = t(dm) - t(dd) = 0- (-0,677) = 0,677oC

t

t = k.C =40

8861 ., = 0,372 =

12

1821

,= 0,82 hay 82%.

III. Cn bng trong dung dch cht in ly yu1. Hng s in ly

Trong dung dch cht in ly yu AB tn ti cn bng sau:

AB A+ + B-

Hng s in ly ca phn ng [ ]AB

B.AK

+

=

K ph thuc vo bn cht ca cht AB v nhit . i vi mi cht nht nh T=

const th K l hng s. K c trng cho kh nng in ly ca cht in ly yu, K cng ln

th kh nng in ly ca cht in ly cng mnh v ngc li.

2. Mi lin h gia K v . nh lut pha long Ostwald

AB A+ + B-

Ban u: C 0 0

Cn bng C- C C C



46/75

( )CC

CK

2

=

1

.CK

2

= Cng thc ton hc ca nh lut pha long

Ostwald

T cng thc ny nhn thy rng C cng nh th cng ln, c ngha l dung dch

cng long th in ly cng ln.

Khi long 1/C th i vi cht in ly yu nng v cng long th

c th coi qu trnh in ly l hon ton.

C th xc nh c cc C khc nhau nu bit K v ngc li

Nu < 0,05 th 1- 1 2.C = K

C

K =

3. Dch chuyn cn bng trong dung dch cht in ly yu

- S in ly ca cht in ly yu l phn ng thun nghch => cn bng trong dung dch

cht in ly yu cng tun theo mi qui lut ca cn bng ho hc.

VD: xt cn bng trong dung dch CH3COOH

CH3COOH CH3COO- + H+

Nu tng thm nng ca ion axetat bng cch thm mt vi tinh th mui axetat natri

th theo nguyn l dch chuyn cn bng cn bng s dch chuyn t phi sang tri, sao

cho nng ca ion axetat gim i, ngha l lm gim in ly ca axit.* Kt lun: Khi tng nng ca ion ng loi vi ion ca cht in ly th in ly ca

cht in ly yu gim i.

IV. Thuyt axit- baz ca Bronsted1.nh ngha axit-baz

Axit l tiu phn (ion hay phn t) c kh nng cho H+ (proton). Baz l cht c kh nng nhn H+.V d 1: Trong dung dch HCl

HCl + H2O = Cl-

+ H3O+

(1)axit1 baz2 baz1 axit2

ax1 - bz1: HCl/Cl-

1

1/C



47/75

ax2 - bz2: H3O+/H2O l 2 cp axit baz lin hp

V d 2: Trong dung dch NH3

NH3+ H2O NH4+ + OH-

bz1 ax2 ax1 bz2

ax1- bz1- NH4+/NH3ax2- bz2- H2O/OH

-

Nhn xt:

- nh ngha axit baz ch c tnh cht tng i: H2O (1) l 1 baz nhng (2)th H2O li l 1 axit.

- mnh ca 1 cp axit baz ngoi s ph thuc vo bn cht ca cp cnph thuc vo mnh ca cp cng tn ti vi n trong dung dch.

- i vi mt cp axit baz lin hp nu axit cng mnh th baz lin hp ca nc

ng yu v

ng

c li.2. Tch s ion ca nc - ch s hydro (pH)

Nc l cht in ly rt yu, va c tnh axit v va c tnh baz:

H2O + H2O OH- + H3O

+

=>[ ]OH

OHHK

2

3C

+

=O

V H2O l cht in ly rt yu => [H2O]cb OHC 2 ban u.

=> [ ]+

= OHHOHK 32C O = const= OHK 2 = Tch s ion ca H2O.=> Trong nc nguyn cht : [ ] [ ] OH3 2OHH KO ==

+

OHK 2 ch ph thuc vo nhit : OHK 2 =10-14 250C.

Trong nc nguyn cht(mi trng trung tnh) : [ ] [ ] OH3 2OHH KO ==+

250C [ ] [ ] )/(OHH OH3 2 lmolKO714 1010 + ====

Vy: - Mi trng trung tnh l mi trng trong + = OHO3H v 250C

710+ =O3H M.

- Mi trng axit l mi trng trong + > OHO3H v 250C 710+ >O3H M.- Mi trng baz l mi trng trong + < OHO3H v 250C 710+


48/75

Ka l hng s in ly ca axit trong nc, c trng cho mnh ca axit. Ka cng cao

th axit cng mnh. Ka ph thuc vo bn cht ca axit v ph thuc vo nhit .

- Axit mnh : khng dng Ka v coi 1 .

- Axit yu ( 10 pH= 7 250C

- Mi trng axit c 710+ >O3H M => pH 7 250C

Cch xc nh pH: + Xc nh bng my o pH.+ S dng giy o pH.

- Cht ch th mu: L cht thay i mu theo gi tr pH. Thng gp cc ch th mutrong phng th nghim.

+Qu tm.

+Metyl da cam: ch th mi trng axit yu.

+Phenol phatalein: ch th mi trng kim.



49/75

- Khong chuyn mu: l khong pH trong mu ca ch th bin i c. ivi ch th mu c 1 khong chuyn mu xc nh.

VD: phenolphtalein: + pH =0 8: khng mu.

+ pH =8-10 : hng

+pH =10-14 thm

Metyl da cam: + pH =0 3: hng+ pH =3-4,4 : da cam

+ pH =4,4-14 vng.

V. Tnh pH ca cc dung dch axit- baz- mui1. Tnh pH ca dung dch axit mnh 1 bc

Xt dung dch axit mnh HA, nng Ca, trong dung dch tn ti cc cn bng:

HA + H2O A- + H3O

+ (1)

2H2O H3O+ + OH- (2)

Trong dung dch tn ti cc ion H3O+

, OH-

, A-

.Phng trnh bo ton in tch: + += AOHOH 3 . Suy ra, ta c h phng trnh

sau

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

+=+=

=

+

+

a3

OH3

COHAOHOH

KOH.OH2

=> [ ] [ ][ ] [ ] 0KOHCOHC

OH

KOH OH3a

2

3a

3

OH

3 2

2 =+= +++

+

=> [ ]2

4 22

3

OHaa KCCOH ++=+

Nu dung dch c Ca> 3,16.10-7 (M) (pH pH = -lg Ca

V d: Tnh pH ca dungdch HCl 0,01M. HCl l axit mnh v Ca> 3,16.10-7 (M)

pH =-lg Ca

=-lg10-2=2.

2. Tnh pH ca dung dch baz mnh 1 bc

Trong dung dch tn ti cn bng:

BOH = B+ + OH-

2H2O H3O+ + OH-

Nu dung dch khng qu long Cb > 3,17.10-7M th b qua [OH-] do nc in ly.

[OH-] = Cb p(OH) = - lgCb

pH + p(OH) = 14 pH= 14 + lgCb



50/75

Nu dung dch qu long Cb < 3,17.10-7M th phi tnh n [OH-] do nc in ly ra,

do gii da vo tnh trung ho v in tch v tch s ion ca nc

[ ][ ]

[ ] [ ] [ ] [ ]

+=+=

=

+++

+

b33

OH3

COHNaOHOH

KOH.OH2

[ ] [ ][ ] [ ] 0KOH.COHC

OH

OKHOH OHb

2

b2

2=+=

Gii phng trnh tm c OH- pH

VD: Tnh pH ca dung dch NaOH 0,01M => pH = 14+ lgC b =14-2=12.

3. Tnh pH ca axit yu 1 bc

Axit HA c nng ban u l Ca

HA + H2O A- + H3O+ (1)2H2O H3O

+ + OH- (2)

=> ta c h:

[ ][ ][ ][ ]

[ ]

[ ] [ ] [ ][ ] [ ]

=

+=

=

=

+

+

+

ACHA

OHAOH

HA

AOHK

KOHOH

a

a

OH

3

3

3 2

.

Gii phng trnh bc 3 i vi [H3O+] tm c pH

Tuy nhin khng phi gii phng trnh bc 3 m p dng cc phng php gn ng

vi sai s < 5%

* Nu Ca.Ka 10-12 v 0,1 10-12

1001081

10

5

1

>=

.,K

C

a

a

[H3O+

] < < Ca

CH3COOH + H2O CH3COO- + H3O

+

[ ] 15 101081 + == ..,.CKOH aa3 8721081 6 ,.,lgpH == 4. Tnh pH ca baz yu 1 bc

Xt cn bng trong dung dch baz yu B (A l axit lin hp ca B)

B + H2O A + OH-

2H2O H3O+ + OH-

L lun t

ng t nh

tr

ng hp axit yu thy rng tm

c

OH phi giiphng bc 3 . i n cc php gn ng:

* Nu Cb.Kb 10-12 v 10010

b

b

K

Cbaz yu x


52/75

Nu ion ca axit yu hoc baz c nhiu in tch th s in lythnh nhiu nc, nc sau yu hn nc trc. Nu dung dch khng

qu long th cc nc sau c th b qua. Nu dung dch qu long th

phn ng thu phn c th chuyn dch n mc hon ton.

* Cc loi mui c th thu phn:

- Mui to bi anion ca axit yu v cation ca baz mnh (to bi axit yu v bazmnh), khi ho tan trong nc to thnh mi trng kim. V d: NaCH3COO,

Na2S, Na2CO3, NaClO

NaCH3COO Na+ + CH3COO

-

CH3COO- + H2O CH3COOH + OH

-

- Mui to bi anion ca axit mnh vi cation ca baz yu, khi ho tan trong nc

to mi trng axit. V d: NH4Cl, (NH4)2SO4, NH4NO3...

NH4Cl NH4+ + Cl-

NH4+

+ H2O NH3 + H3O+

- Mui to bi anion ca axit yu v cation ca baz yu, khi ho tan c 2 gc u

thu phn, pH ca mi trng cn ph thuc vo mc thu phn ca c 2 gc.

Nu Ka(cation baz yu) > Kb(anion ca axit yu) th mi trng c pH < 7

Nu Ka(cation baz yu) < Kb(anion ca axit yu) th mi trng c pH > 7

Nu Ka(cation baz yu) = Kb(anion ca axit yu) th mi trng c pH = 7

V d: NH4CH3COO,

NH4CH3COO NH4+ + CH3COO

-

NH4+ + H2O NH3 + H3O

+ Ka(NH4+) = 5,6.10-10

CH3COO- + H2O CH3COOH + OH- Kb(CH3COO-) = 5,7.10-10

C Ka(NH4+) = Kb(CH3COO

-) nn pH = 7

NH4 NO2 NH4+ + NO2

NH4+ + H2O NH3 + H3O

+ Ka(NH4+) = 5,6.10-10

NO2- + H2O HNO2 + OH

- Kb(NO2-) = 2.10-11

C Ka(NH4+) > Kb(NO2

-) nn pH < 7

- Mui to bi axit mnh v baz mnh khng thu phn: NaCl, NaClO4

- Mui axit l cht lng tnh. pH ca mi trng ca dung dch cha n ph thuc

vo mnh ca tnh axit hay tnh baz. V d: Dung dch NaHCO3 l lng tnh v:

HCO3- + H2O H2CO3 + OH

- Kb(HCO3-) = 2,4.10-8

HCO3- + H2O CO3

2- + H3O+ Ka2(H2CO3) = 4,8.10

-11

* Tnh pH ca dung dch mui:

Nguyn tc: Ging nhcch tnh pH ca dung dch axit yu hoc baz yu

V d 1: Tnh pH ca dung dch mui MCl3 0,1M. Bit rng ion M3+ c tnh axit, c

hng s in ly Ka = 2.10-3

cc nc thy phn sau c th b qua.

Ka.Ca = 2.10-3.10-1 > 10-12

3102

10

=.

,

K

C

a

a< 100 nn b qua [H3O

+] do nc in ly



53/75

M3+.H2O + 2H2O MOH2+ + H3O

+

[ ]3102

+

++

== .M

OH.MOHK

3

32

a

t x= [H3O+]

32

10210

=

= .,

Kax

x x pH

V d 2: Cho dung dch K2CO3 0,2M, bit rng Ka2(H2CO3) = 5,6.10-11. Tnh pH ca

dung dch, b qua nc in ly th 2 ca CO32-.

K2CO3 2K+ + CO3

2-

CO32- + HOH HCO3

- + OH-

CO32- l baz lin hp ca HCO3

-

Ka.Kb = 10-14 Kb(CO3

2-) = 411

1414

107811065

1010

== .,.,K a

Kb.Cb = 0,2.1,78.10-4 > 10-12 nn b qua s in ly ca nc

10010781

204>=

.,

,

K

C

b

b

[ ] 7112010781142010781 44 ,,..,lgpH,..,.CKOH bb =+=== V d 3: Tnh pH ca NH4NO2 10

-2 M, bit rng Ka ca NH4+ l 6,3. 10-10 v Kb ca

NO2- l 2.10-11.

V KaC v KbC u ln hn 10-14 rt nhiu nn phn ng ch yu trong dung dch l:

NH4- + NO2

- NH3 + HNO2

Ban u: 10-2 M 10-2 M 0 0

Cn bng: 10-2 -x 10-2 -x x x

Vy [NH4+]=[NO2

-] v [NH3]=[HNO2]

Ta bit rng1a

K ca NH4+ l:

][

]][[+

+

=4

33

1 NH

OHNHKa

v

2aK ca NO2-

l

: ][

]][[

2

23

2 HNO

NOOH

Ka

+

=

=>2121 3

2

3 aaaa KKOHOHKK .][][. ==>=++

=> 411

14

105102

102

== ..

aK ; MOH7410

3 10651051036+ == .,...,][

=> pH =6,25.

6.Tnh pH ca dung dch axit nhiu bc:

- i vi axit nhiu bc nh H2S, H2SO3, H3PO4 s in ly theo nhiu nc, nc utin mnh nht cn cc nc sau yu dn.

V d: Nhs in ly ca axit H2S

H2S + H2O HS- + H3O

+ Ka1 = 10-7



54/75

HS- + H2O S2- + H3O

+ Ka2 = 10-13

Do nc 2 yu hn nhiu so vi nc 1 nn c th b qua s in ly ca nc 2 v a v

bi ton tnh pH ca axit 1 bc.

VI. Cn bng trong dung dch cht in ly t tan

1.Tch s ha tan ca cht in ly t tanXt dung dch bo ha cht in ly t tan AmBn: Trong dung dch lun tn ti cn bng

gia phn rn khng tan v ion ca n trong dung dch:

AmBn (r) mAn+ + nBm-( thc cht l AmBn (r) AmBn dd (tan) --> mA

n+ + nBm- ).

Kc = [An+ ]m[Bm-]n

Trong trng hp ny Kc c trng cho tnh tan ca cht in ly t tan v c gi l tch

s ha tan ca cht in ly t tan Ks.* nh ngha: Tch s ha tan ca 1 cht in ly t tan l tch s nng ca cc ion trong

dung dch bo ho cht in ly t tan (vi s m l h s tng ng trong phng trnhin ly).

VD: CaSO4 Ca2+ + SO4

2- ]][[,

+= 242

4SOCaK CaSos

Nhvy Ks l 1 trng hp ca hng s cn bng Kc, bn cht ca Ks l hng s cn bng

K, do mi tnh cht ca K u p dng c i vi Ks.

- Ks ph thuc vo bn cht tng cht v nhit .2. Mi quan h gia tch s tan Ks v ho tan s

Gi s l nng ca dung dch bo ho, v d i vi dung dch

Ag2SO4 2Ag

+

+ SO42-

2s s

4s,CaSOK = [ ] [ ]

bh

2

4bh SO.Ag+ 2 = (2s)2.s = 4s3 3

s

4

Ks =

ho tan ca cht in ly t tan s gim i, nu thm vo dung dch mt lng ion

ng loi.

V d: Cho AgIs,K = 1,5.10-16 t = 250C. Tnh ho tan ca AgI trong ncnguyn cht v trong dung dch KI 0,1M.

Gii:

AgI Ag+ + I- (*) ho tan ca AgI l s

s s

bhbhAgIs, I.AgK+= = s2 M816 102211051 === .,.,,Ks AgIs

Trong dung dch KI 0,1M KI = K+ + I-

0,1 0,1

Nng ca ion I- tng ln lm cho cn bng (*) dch chuyn theo chiu nghch lm

cho ho tan ca AgI gim xung. Gi s l ho tan ca AgI trong dung dch KI



55/75

AgI Ag+ + I-

s s + 0,116105110 + =+== .,),'(']][[, ssIAgK AgIs V s


56/75

Chng VII: ng ho hc

Nhit ng ho hc nghin cu phn trc ch mi cho php xt on chiu hng

t din bin ca mt phn ng ho hc v ch kho st h trng thi cn bng, nn

khng h cho bit mt tn hiu no v tc , ngha l s bin i cc tham s ca h theothi gian.

V d: Phn ng gia H2(K) + 1/2 O2(K) = H2O(l) c Go298 = -237,2 kJ/mol, G

o298 ca

phn ng rt m, c ngha l v mt nhit ng hc phn ng c th xy ra mt cch hon

ton T= 298K v P = 1atm (K= 1041), song thc t cho thy phn ng hu nhkhng

xy ra iu kin cho, bi v tc ca phn ng cc k nh, do iu kin

thng ngi ta tng phn ng ny khng xy ra.

ng ho hc l mn khoa hc nghin cu v tc v c ch ca cc qu trnh

ho hc.I.khi nim v vn tc phn ng1. nh ngha vn tc phn ng

Vn tc trung bnh ca phn ng: c o bng bin thin nng ca mt

trong cc cht tham gia phn ng hay to thnh sau phn ng trong mt n v thi gian.

Xt phn ng: aA + bB cC + dD

Gi s thi im t1 nng ca cc cht l CA1, CB1, CC1, CD1, thi im t2 th

nng tng ng l CA2, CB2, CC2, CD2. Khi vn tc trung bnh ca phn ng l:

12

AA

tbA,tt

CCV 12

= ,

12

BB

tbB,tt

CCV 12

= ,

12

CC

tbC,tt

CCV 12

= ,

12

DD

tbD,tt

CCV 12

=

t

CVtb =

(+)- ng vi cht kho st l sn phm

(-) - ng vi cht kho st l cht tham gia

Vn tc tc thi ca phn ng:

limV0

=t dt

dC=

t

C

vn tc ca 1 phn ng l n tr:

dt

dC

a

1V A= =

dt

dC

b

1 B =dt

dC

c

1 C =dt

dC

d

1 D

2.Cc yu t nh hng ln vn tc

- Nng cc cht.

- Nhit

- Cht xc tc.



57/75

II. Thuyt va chm hot ng1.Ni dung:

Gi s xt phn ng A(K) + B(K) AB(K). phn ng xy ra th A v B phi va chm

vi nhau.

C 2 loi va chm:+ Va chm gy phn ng: gi l va chm c hiu qu(s va chm ny nh).

+ Va chm khng gy phn ng: gi l va chm khng hiu qu(s va chm ny ln).

gy va chm c hiu qu => cc phn t phi c nng lng ln hn nng lng

trung bnh ca h => gi l cc phn t hot ng => vn tc phn ng t l vi tn s va

chm gia cc phn t hot ng.

2.Phn b Boltzman:

C kh l tng A vi tng s mol l N, trong c s phn t hot ng l N* th:

RT

EA

eN

N =*

=> Biu thc nh lut phn b Boltzman.

Trong : EA- c gi l nng lng hot ho, n v J.mol-1

R - l hng s kh l tng, R = 8,314 J.K-1.mol-1

III. nh hng ca nng cc cht tham gia phn ng n vn

tc v nh lut tc dng khi lng.1. nh lut tc dng khi lng

a. i vi h ng th( Cc cht phn ng cng 1 pha).

* nh lut: Vn tc phn ng t l thun vi tch nng cc cht tham gia phn ng(vi s m thch hp).

V d: aA +bB -> cC (1)

=> v =k[A]n[B]m => gi l phng trnh ng hc ca phn ng.

trong [A], [B]: Nng mol/l ca A, B thi im xt.

v : Vn tc tc thi thi im xt.

n,m: Bc phn ng i vi cht A, B -> Xc nh bng thc nghim.

(n+m): Bc chung ca phn ng.

k: H s t l ph thuc vo bn cht ca cht tham gia v nhit.Vi 1 phn ng c th T =const -> k=const-> gi l hng s vn tc.

Khi [A]=[B]=1mol/l-> v=k -> gi l v ring ca phn ng.

b.i vi phn ng d th: Nu phn ng c cht rn tham gia -> coi nng cht rn =

const v a vo hng s vn tc => cht rn khng c mt trong phng trnh ng hc

ca phn ng.

V d 1: C(gr) + O2(K) CO2(K)v = nOk ][]Ok'.const.[ n2 2=

C. Gii thch: Khi nng tng th vn tc tng: Theo thuyt va chm hat ng: Khinng cc cht tham gia phn ng tng th s phn t hot ng c trong 1 n v th

tch tng -> dn n s va chm c hiu qu tng -> vn tc tng.



58/75

2.Bc phn ng:

* Bc phn ng c xc nh bng tng s m trong phng trnh ng hc (m+n). Bc

phn ng c th nguyn, hoc l s thp phn hoc bng 0.

- Nu (m+n)=1: phn ng bc 1.



* Cch xc nh bc phn ng:

- Xc nh theo tng cht ri cng li:

Dng phng php c lp: Coi nng cc cht # bng const ( ch c nng cht kho

st bc thay i theo thi gian) bng cch cho nng cc cht ln hn rt nhiu nng

cht xt.

Mt phn ng ha hc l phn ng tng cng ca nhiu giai on trung gian. Mi giai

on trung gian gi l

1 giai on s cp. Vn tc ca giai on s cp n

o chm cht squyt nh vn tc ca c phn ng.

S phn t tham gia vo 1 giai on s cp gi l phn t s ca giai on s cp .

Phn t s ca giai on s cp chm nht xc nh bc chung ca phn ng.

VD: 2HI + H2O2= 2H2O + I2 (a)

V=k[H2O2]n[HI]m

Phn ng (a) xy ra theo theo 2 giai on s cp:

HI + H2O2 -> HIO + H2O (1) xy ra chm

HIO + HI -> I2

+ H2

O (2) xy ra nhanh.

giai on (1) quyt nh bc phn ng -> phn t s ca (1) quyt nh bc ca

phn ng (a).

Phng trnh ng hc ca (a) cng l ca (1): v= k[H2O2][HI]

Phn t s ca (1) v (2) u l 1+1=2.

Bc ca p (a) l 1+1=2.

* Ch : nu phn ng n gin ch xy ra theo 1 giai on th n=a, m=b (a, b l cc

h s t lng trong phng trnh phn ng) => Bc phn ng (m+n) =(b+a).

IV. nh hng ca nhit ln vn tc phn ng

1.Quy tc Vant HoffBng thc nghim Vant Hoff cho thy rng nhit c tng thm 10oC th vn tc

ca phn ng tng ln ln, trong khong t 2- 4.

=+t

10t

V

V

- l h s nhit cho bit vn tc tng ln bao nhiu ln khi nhit tng thm10oC.

Tng qut: nhit t1 vn tc ca phn ng l v1, nhit t2 vn tc ca phn ng l v2, ta c:

110

tt

2 VV12

= l biu thc ton hc ca quy tc Vant Hoff



59/75

- Quy tc Vant Hoff ch gn ng trong khong nhit khng cao lm.2. Phng trnh Arrhenius:

lnT

Alnk += => T

A

.ek =

Trong : A v l nhng hng s c trng cho phn ng xc nh bng thc

nghim.* Theo th lnk- 1/A vi tg = A* Da vo gi tr K hai nhit khc nhau:

=

+=

+=

121

2

1

1

2

2

T

1

T

1A

K

Kln

lnT

AlnK

lnT

AlnK

* Gii thch nh hng ca nhit ti v theo thuyt va chm:

Khi nhit thay i -> c s phn b li nng lng trong h-> s phn t hot ng

thay i -> v thay i.

C th: khi nhit tng vn tc tng do: chuyn ng nhit ca cc phn t tng ln

tn s va chm ca cc cht tham gia tng v khi nhit cao th cc phn t km bn

d phn ng vi nhau.

* ngha ca A trong phng trnhArrheniusXt phn ng: A(k) + B(k) -> AB(k) c bc i vi A v B u bng 1.

Phng trnh ng hc: =kCA.C

B(a)

RT

E

ii

i

eCC

=* (Ci mol/l ca phn t hat ng i).

Theo thuyt va chm hot ng v ch ph thuc vo Ci*

=> RTE

B

RT

E

ABA

AAi

eCeCCCv

== ** vi l h s t l.

=> RTEE

BA

BA

eCCv

)( +

= .

t EA + EB = Ea gi l nng lng hot ha ca phn ng. Ea l nng lng cn thit

a mt mol cc cht tham gia phn ng c nng lng trung bnh tr thnh hot

ng.

=> RTE

BA

a

eCCv

= (b)

=> So snh (a) v (b), c: RTEa

ek

= => lnRT

Elnk a += => ln

T

Alnk +=

A=-Ea/R

* ngha ca Ea:

hiu r ngha ca Ea ta xt gin nng lng ca phn ng:

I2(K) + H2(K) 2HI(K)

Ea

KH2I2

Ea

E



60/75

Cc cht tham gia c nng lng ng vi mc I, mun phn ng c vi nhau phivt qua hng ro th nng c cao l K. Hiu gia mc nng lng K v I chnh lnng lng hot ho ca phn ng thun Ea. Hiu gia mc nng lng K v II l nng

lng ca phn ng nghch. Hiu gia mc I v II c gi l hiu ng nhit ca phnng thun.Vy nng lng hot ho Ea: Chnh l hng ro th nng m cc cht tham gia phi

vt qua hnh thnh cc sn phm phn ng. Nh vy, nu cc lin kt trong cc chttham gia cng bn th nng lng hot ho ca phn ng cng ln.V.nh hng ca xc tc ln vn tc phn ng1. nh ngha:

Cht xc tc l cht lm tng vn tc phn ng, nhng n khng b bin i v tiu tn

do phn ng xy ra.

Nu cht xc tc v cc cht tham gia phn ng trong cng mt pha th c gi l xc

tc ng th.

V d: 2SO2(K) + O2(K) 2SO3(K)

Nu cht xc tc khc pha vi cc cht tham gia phn ng th c xc tc d th.

V2O5(r)

V d: 2SO2(K) + O2(K) 2SO3(K)

Xc tc men ng vai

21 co so lt hoa hoc dhbk hn.pdf

Documents