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NHIT NG LC HC

5.1. nh lut bo ton v chuyn ha nng lng5.1.1. H

a/ nh nghaH l mt phn v tr c nghin cu, xem xt, phn cn li l mi trng.

b/ Phn loi- H c lp: L h khng trao i cht v trao i nng lng vi mi trng.- H kn: L h khng c trao i cht, song c th trao i nng lng vi mi trng.- H h: L h c trao i cht v trao i nng lng vi mi trng.

5.1.2. Trng thi- Tham s trng thi ca h: l bt k mt thuc tnh no (khch hoc cng ) c s dng

m t trng thi ca h.+ Thuc tnh khch : l nhng thuc tnh ph thuc vo khi lng, c tnh cng tnh i vih ng nht. VD: V, d, m

+ Thuc tnh cng : l nhng thuc tnh khng ph thuc vo khi lng v khng c cngtnhVD: nhit , p sut, t khi, nng .

- Trng thi cn bng nhit ng: l trng thi c c khi cc thuc tnh ca h khng thay i theothi gian.

5.1.3. Hm trng thiMt hm F(p,v,T) c gi l mt 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. iu ny c ngha rng nu hchuyn t trng thi 1 (P1, V1, T1) sang trng thi 2 (P2, V2, T2) th F = F2 F1 ch ph thuc vogi tr (P1, V1, T1) v (P2, V2, T2) ch khng ph thuc vo tnh cht ca qu trnh bin i (thunnghch hay bt thun nghch).

5.1.4. Qu trnhQu trnh nhit ng: l mi bin i xy ra trong h m c lin quan vi s bin thin d ch mttham s trng thi ca h.

5.1.5. Qu trnh t din bin v khng t din bin

- Qu trnh t din bin l qu trnh m t bn thn n c th xy ra ch khng cn tiu th nnglng bn ngoi. VD: S t khuch cc cht kh.- Ngc li qu trnh khng t din bin. cho qu trnh ny xy ra th ta phi cung cp nng lngcho h.

5.1.6. Qu trnh cn bng- L qu trnh m trong sut thi gian din bin ca n, trong h lc nocng ch c nhng sai lch v cng nh so vi trng thi cn bng.

5.1.7. Qua trnh thun nghch v khng thun nghch

- L qu trnh c th thc hin theo chiu thun v chiu nghch v khi theochiu nghch h cng nh mi trng ngoi u tr v ng nh trng thiban u, khng c s thay i nh no. Nu ngc li, l qu trnhkhng thun nghch.

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- Trong t nhin, thng gp cc qu trnh khng thun nghch, cn qu trnh thun nghch hon tonkhng c theo mt chiu.

5.1.8. Nng lnga/ nh ngha

Nng lng l o vn ng cc vt cht trong mi bin i ca n t dng ny sang dng khc.b/ Phn loiC nhiu dng nng lng:

+ ng nng: Dng nng lng c trng cho mt vt ang chuyn ng

E = 2mv2

1

+ Th nng: Nng lng m h c c do v tr ca n trong trng lcEt = mgh

+ in nng: L nng lng chuyn ng ca cc tiu phn tch in ( electron, ion.)+ Ha nng: L nng lng gn lin vi qu trnh bin i cht.

Nng lng ton phn ca mt h gm: ng nng ca ton b h Th nng do v tr ca h trong trng lc ngoiTng ng nng v th nng ca h c gi ngoi nng

Nng lng d tr bn trong ca h (ni nng) gm : ng nng cc phn t, nng lng hty ca cc tiu phn cu to nn h, nng lng ha hc, nng lng ht nhn.

5.1.9. nh lut bo ton v chuyn ha nng lng. S tng ng gia nhit v cnga/ nh lut bo ton v chuyn ha nng lng

- Nng lng v tr khng i. Nu mt h no gim nng lng th nng lng mi trng xungquanh phi tng tng ng. Khi mt dng nng lng no chuyn thnh dng khc th phi c mtquan h nh lng nghim ngt. VD : ng lng c hc : 1cal = 4,184J- Khng th sng to ra nng lng, khng th hy dit c nng lng m ch c th chuyn nnglng t dng ny sang dng khc.

b/ S tng ng gia nhit v cng- S tng ng gia nhit v cng trong cc chu trnh c th pht biu nh sau : Khi mt h nhitng thc hin mt chu trnh trong n ch trao i nng lng vi bn ngoi di dng nhit vcng th : Nu n nhn nhit (Q>0)th n sn cng (A0) th n nhng nhit (Q

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Nu A (Jun)Q(Calo)

Nu A v Q o cng 1 n v th J = 1 -A = Q Khng th c ng c vnh cu loi 1 ( L loi my lun sinh cng m khng cn cung cp nng

lng / nhn nhit)

5.1.11. Nguyn l th nht nhit ng lc hc- Nguyn l mt chnh l s p dng s bo ton v chuyn ha nng lng vo cc h v m, c linquan n s trao i nhit v cng vi mi trng ngoi.

- Nu h trao i nng lng vi bn ngoi di dng nhit v cng th tng ni nng ca h ( tng nng lng ca h) ph bng ng phn nng lng chuyn t ngoi vo h di dng nhit Q,tr phn nng lng chuyn t h ra ngoi di dng cng A (Sinh cng)

*Biu thc ton hc ca nguyn l I:U= Q + A

* Biu thc vi phn ca nguyn l th nht:i vi mt qu trnh v cng nh (qu trnh nguyn t). Khi h trao i vi mi trng lng to vlng cng v cng nh, ta c: dU = AQ +

5.1.12. Nhit v hiu ng nhit ca qu trnh- Trong trng hp chung, cng do h thc hin gm:

+ Cng gin n th tch: A = -pdV+ Cng c tch khc: A

Th khi biu thc vi phn ca nguyn l 1 s c vit di dng:dU = 'ApdVQ

- Nu h khng thc hin cng c ch th A = 0 dU = pdVQ (I)

a/ Qu trnh ng tchV = const dV = 0 dUv = vQUv = Qv

Nu phn ng ha hc tin hnh trong iu kin V = const th ton b lng nhit do h thu vo

hay ta ra trong qu trnh ng tch dung lm tng ni nng ca h.+ Nu phn ng thu nhit th ni nng ca h tng: 0U >+ Nu phn ng ta nhit th ni nng ca h gim: 0U

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Trong iu kin p = const, lng nhit thu vo (hay ta ra) bng bin thin entanpi ca h:+ Nu phn ng thu nhit, entanpi ca h tng : H > 0+ Nu phn ng ta nhit, entanpi ca h gim: H< 0

-Hiu ng nhitca mt qu trnh ha hc: L lng nhit km theo phn ng khi tin hnh

trong qu trnh khng thun nghch nhit ng, sao cho cng c ch khng c sinh ra. Khi hiung nhit ca qu trnh ha hc c xc nh bng thay i ca U v H.+ Trong phn ng ch c mt cht lng v cht rn, s thay i th tch l khng ng k

th UH + Trong phn ng c cht kh th: nRTUH +=

n : bin thin s mol khR = 1,987 cal/mol.

= 8,314 J/mol.5.1.13. Nhit dung- Nhit dung ca mt cht l lng nhit cn dng nng nhit ca cht ln thm 10

- Nhit dung ring: l nhit dung ca 1 gam cht- Nhit dung mol: l nhit dung ca 1 mol cht

+ Nhit dung mol p sut khng i: Cp+ Nhit dung mol th tch khng i: Cv

i vi kh l tng: Cp Cv = R i vi cht rn v cht lng: Cp Cv

* Biu thc:

pp T

HC

=

v vv T

UC

=

i vi 1 mol kh l tng: dU = CvdT; dH = CpdT i vi n mol kh l tng: dU = nCvdT; dH = nCpdT

5.1.14. Cng v nhit trong mt s qu trnh (i vi kh l tng)a/ Qu trnh ng tchV = const dV = 0* Cng gin n th tch: vA = -pdv = 0 Av = 0* Nhit: vQ = dUv + pdV

= CvdTTrong qu trnh hu hn:

=2

1

T

T

vv dTCQ

Nu Cv = const Qv = Cv(T2 T1) (1mol)

b/ Qu trnh ng pp = const* Cng : pdVAp =

Qu trnh hu hn : =

2

1

V

Vp

pdVA

Ap = -p(V2 V1)Ap = -p V

+ i vi 1 mol kh l tng : TRAp =

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+ i vi n mol kh : TnRAp =

* Nhit :dTCdHQ ppp ==

Qu trnh hu hn : =2

1

T

Tpp dTCQ

Nu Cp = const Qp = Cp(T2-T1) (1 mol kh)

c/ Qu trnh ng nhitT = const* Cng : pdVAT =

i vi n mol kh : VnRTdV

dVnRTAT ln==

Qu trnh hu hn : =

2

1

ln

V

VT

VnRTdA

= -nRT(lnV2 lnV1)

1

2lnV

VnRTAT =

T = const ta c th tch ca kh l tng t l nghch vi p sut hoc nng mol nn :

2

1

2

1

1

2

C

C

P

P

V

V==

21

2

1

1

2 lnlnlnC

CnRT

P

PnRT

V

VnRTA

T ===

* Nhit : T = const th ni nng ca kh l tng khng i QT = TAU

QT = -AT2

1

2

1

1

2 lnlnlnC

CnRT

P

PnRT

V

VnRT ===

d/ Qu trnh va ng p , va ng nhitT, p = const

* Cng:pdVA

PT

=,

Qu trnh hu hn

AT,p = VppdVV

V

= 2

1

nRTA PT =,*Nhit:

QT, p = PTAU ,QT, p = -AT, p

e/ Qu trnh on nhit( Q = 0)* Cng: Q,n = nn AU ,,

A,n = n,U

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M dU = CvdT =2

1

T

T

VdTCU

A,n = 2

1

T

T

vdTC

Nu Cv = const A,n = Cv(T2 T1)Hay A,n =- Cv(T1 T2) (1mol kh)

5.2. Nhit ha hc5.2.1. Hiu ng nhit phn ng-Hiu ng nhit phn ng: l nhit lng pht ra hay hp th ca mt phn ng ha hc. VD : Khit than, nhit lng ta ra, tri li nung vi l mt phn ng thu nhit.

Ngnh ha hc nghin cu hiu ng nhit cc phn ng, s ph thuc hit ng nghit vo thnhphn, cu to cc tc cht cng nh iu kin tin hnh phn ng c gi l nhit ha hc.-Phng trnh nhit ha hc : l phng trnh phn ng c ghi nhit lng pht ra hay thu c.

C + O2 = CO2 - 393,6kJN2 + O2 = 2NO + 22,16kcal

Nhit lng c n v kJ hay kcal (1kcal = 4,184kJ)- Thng thng vic o nhit phn ng c thc hin trong iu kin ng p, khi ta c hiung nhit ng p k hiu (bin thin entanpi).- Ch mt s trng hp mi thc hin c trong iu kin ng tch: ta c hiu ng nhit ng tch(bin thin ni nng).- Gia hiu ng nhit ng p vi hiu ng nhit ng tch c mi lin h sau:

nRTUH +=( n : bin thin s mol kh)

5.2.2. Nhit to thnh Nhit phn hy Nhit t chy* Nhit to thnh mt hp cht l hiu ng nhit ca phn ng to thnh 1 mol cht t cc ncht ng vi trng thi t do bn vng nht.

VD: C(graphit) + O2(k) = CO2(k) ttH (CO2) = -94,05kcal

Nhit to thnh ca cc n cht bn iu kin chun c chp nhn bng 0

* Nhit phn hy l nhit phn ng phn hy 1 mol cht to thnh cc n cht.VD

: H2O(l) = H2(k) + 1/2O2(k)phH

(H2O) = 68,3 kcal nh lut Lavoissier Laplace Nhit to thnh v nhit phn hy ca cng mt cht bng gi tr v ngc du

VD: 1/2H2(k) + 1/2I2(r) = HI(k) ttH (HI) = +6,2kcalHI(k) = 1/2H2(k) + 1/2I2(r) phH (HI) -6,2kcal

* Nhit t chy l hiu ng nhit ca phn ng t chy bng O2 mt mol cht hu c to thnhkh CO2 v nc lng ( v mt vi sn phm khc).

VD: C2H6(k) + 7/2O2(k) 2CO2(k) + 2H2O(l) H = -372,82kcal

Ta c: cH (C2H6) = -372,82kcal

5.2.3.nh lut HessNm 1840 G.I.Hess pht minh nh lut cn bn ca nhit ha hc.

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Ni dung: Nu c nhiu cch chuyn nhng cht ban u thnh nhng sn phm cui cngging nhau th hiu ng nhit tng cng theo cch no cng nh nhauNi khc i: Hiu ng nhit ca qu trnh ha hc ch ph thuc vo bn cht v trng thi ca cccht u v sn phm ch khng ph thuc vo ng i

VD: C vi cch chuyn 1 mol Na 0,5 mol H2, 0,5 mol O2, thnh 1 mol NaOH v hiu ng nhittng cng ca cch no cng bng = -102,0kcal.Cch 1:

Na(r) + 1/2O2(k) = 1/2Na2O2(r) H1= -60,3kcal1/2H2 (k) + 1/4O2(k) = 1/2H2O (l) H2= -34,1kcal1/2Na2O2 + 1/2H2O (l) = NaOH(r) + 1/4O2(k) H3= -7,6kcal

Tng cng: Na(r) + 1/2H2(k) + 1/2O2(k) = NaOH (r) H= ?Ta c: H = -60,3 - 34,1 7,6 = -102kcal

Cch 2 :H2(k) + 1/2O2(k) = H2O(l) H1 = -68,3kcal

Na(r) + H2O(l) = NaOH (r) + 1/2H2(k) H2 = -33,7kcal

Tng cng: Na( r) + 1/2H2(k) + 1/2O2(k) = NaOH( r) H = ?Ta c: H = -68,3 37,7 = -102kcal

CH : Entanpi ca mt cht c tnh i vi mt mol cht.Bin thin entanpi tnh c tentanpi ca cc cht iu kin chun c gi l bin thin entanpi chun v c k hiu l H0

hoc khi ch c nhit na th c k hiu l 0298H

+ i vi cc kh, trng thi chun l trng thi ca kh l tng p sut 1atm.+ i vi cc cht rn v cht lng, trng thi chun l trng thi ca cht tinh khit.+ Nhit thng c ly l : 250C = 2980K.

5.2.4. H qu ca nh lut Hess(1) Hiu ng nhit mt phn ng bng tng nhit to thnh ca cc sn phm tr tng nhit tothnh ca cc cht u ( c k cc h s).

H =Htt(sn phm) - Htt(tc cht)

VD : CaCO3(r) = CaO (r) + CO2(k) H =?

Gii:H = Htt(CaO) + Htt(CO2) - Htt(CaCO3)

= -151,9 - 94,1 + 288,5 = 42,5kcal Nung vi l qu trnh thu nhit.

(2) Hiu ng nhit phn ng bng tng nhit t chy ca cc cht u tr tng nhit t chyca cc sn phm ( c k cc h s).

H = Hc(tc cht) - Hc(sn phm)VD: CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) H = ?H = Hc(CH3COOH) + Hc(C2H5OH) - Hc(CH3COOC2H5)

= -208,2 - 326,7 + 545,9 = 11kcal

5.2.5. ng dng ca nh lut Hess

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(1) Tm hiu ng nhit ca mt s phn ng khng th xc nh bng thc nghimVD: C( r) + 1/2O2(k) = CO(k) H1 = ?

Than chBit:

C( r) + O2 (k) = CO2 (k) H = -94,1kcal

Than chCO(k) + 1/2O2(k) = CO2(k) H2 = -67,7kcalGii:Theo nh lut Hess ta c:

H - H2 = H1Vy H1 = -94,1 + 67,7 = -26,4kcal

(2) Tm nhit to thnh ca mt chtVD: Tm nhit to thnh ca ru etylic t cc d kin:C2H5OH(l) + 3O2(k) 2CO2(k) + 3H2O(l) H = -327kcal

Htt=? 0 -94kcal -68,3kcalGii:p dng h qu (1) ca nh lut Hess ta c:H = 2(-94) + 3(-68,3) - Htt(C2H5OH) = -327kcalRt ra: Htt(C2H5OH) = 2(-94) + 3(-68,3) - (-327) = -65,9kcal

(3) nh nng lng lin ktH = Hlk(tc cht) Hlk(sn phm)

VD: nh nng lng trung bnh ca cc lin kt O-H trong phn t nc, bit nng lng lin kt H-H v O=O tng ng 435,9 v 498kJ. Khi t chy 2 mol H 2 ta ra 483 kJ

Gii:2H2(k) + O2(k) 2H2O(k) H = -483,68kJ -483,68 = 2(+435,9) + 498,7 - 4*Hlk(O-H)

Hlk(O-H) = (2.435,9 + 498,7 + 483,68) = 463,545kJ

(4) Xc nh nng lng mng li ca tinh th.Nng lng mng ion l nng lng to thnh mng tinh th hp cht t cc ion ca trng thi kh.VD: Nng lng mng ion ca tinh th NaCl chnh l hiu ng nhit ca phn ng:

Na+(k) + Cl-(k) = NaCl( r) H = ?

T cc d kin sau, ta c th tnh c nng lng mng ion ca tinh th NaCl:Nhit thng hoa ca Na:

Na( r) = Na(k) H1 = 20,64kcalNng lng lin kt Cl2:

1/2Cl2(k) = Cl(k) H2 = *58kcali lc vi electron ca clo:

Cl(k) - e- = Cl-(k) H3 = -83,17kcalNng lng ion ha natri:

Na(k) - e- = Na+(k) H4 = +119,98kcalNng lng mng ion:

Na+(k) + Cl-(k) + NaCl( r) H0 = ?Hiu ng nhit ca phn ng:

Na( r) + 1/2Cl2(k) = NaCl( r) H = -98,23kcal

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Theo nh lut Hess ta c:H1 + H2 + H3 + H0 = H20,64 + *58 + (-83,17) + 119,98 + H0 = -98,23T ta c nng lng mng tinh th ion mui n :

H0 = -184,68kcal

5.3. Chiu t din bin ca cc qu trnh5.3.1. Entropia/ Biu thc

KTN

TN

T

QdS

T

QdS

>

=

T

QdS

T

QS

Q: Nhit lng m h pht ra hay thu vo trong qu trnh nhit T S: bin thin entropi ca h

b/ Tnh cht- Entropi S l mt thuc tnh khch ca h, tng t nh ni nng, tc l n c cng tnh, gi trca n ph thuc vo lng cht.- Entropi S l mt hm trng thi n tr, lin tc v hu hn ca h. iu ny c ngha l bin thinentropi ca h trong mi qu trnh bt k ch ph thuc v trng thi u v cui ca h, khng phthuc vo ng i.c/ ngha* ngha vt l ca entropi:Bin thin ca entropi l o tnh khng thun nghch ca qu trnh trong nhng h c lp.

* ngha thng k ca Entropi:- Entropi S ca h ti mi trng thi cn bng c trng cho xc sut nhit ng W ca trng thi :

S = klnW : Cng thc Boltzman

N

Rk= : Hng s Boltzman

* R : Hng s kh* N : S Av gar

W : xc sut nhit ng S : Entropi

ngha thng k ca entrpi l: Tnh cht bin thin mt chiu ca entrpi trong h c lp gnlin vi vic chuyn h t trng thi t xc sut sang trng thi nhiu xc sut hn ( hay S l otnh hn lon, mt trt t ca h)

5.3.6. Entrpi ca h c lp

- i vi h c lp th Q = 0 Q = 0 dS 0 Sh c lp 0

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T hc Ha i cng

- H c lp gm: vt kho st v ngun nhitS h c lp = Sh kho st + Sngun 0

i vi qu trnh thun nghch: Sh c lp = S2 S1 = 0 S2 = S1 = const

i vi qu trnh khng thun nghch nhit ng:Sh c lp = S2- S1 > 0 S2 > S1

Vy Entrpi trong h c lp ch c th khng i hoc ch c th tng ch khng th gim hayEntropi l o tnh khng thun nghch ca mt qu trnh

VD: * Khi ta b vch ngn th kh s khch tn, chuyn ng hn lonra (khng thun nghch). Lc ny hn lon tng ln nn S tng.* Khi cc phn t kh phn b u trong ton h th khi h t

n trng thi cn bng Smax = const

- Cc qu trnh t nhin u khng thun nghch nn S l tiu chun vchiu din bin ca qu trnh v iu kin cn bng ca h (ang xt lh c lp)

+ iu kin t din bin: Tng S+ iu kin cn bng: Smax = const

5.3.7. Bin thin Entrpi ca mt s qu trnh thun nghch

TNT

QS

=

a/ Qu trnh thun nghch ng nhit

T

QS=

M: Q = dU A= CvdT + RTdlnV

dS = TQ

= Cv TdT

+ RdlnVng nhit : dS = RdlnV

S = 2

1

V

V

RdlnV

S = Rln1

2

V

V

= Rln2

1

p

p(1 mol kh)

b/ Qu trnh thun nghch on nhitQ = 0

Sn = 0

c/ Qu trnh thun nghch ng p (p = const)

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KhChn

khng

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T hc Ha i cng

T

dTC

T

QdS

pp =

=

( Qp = Cp.dT)

dS: bin thin entrpi trong mt qu trnh v cng nh

==2

1

2

1

lnT

T

T

T

ppTdC

T

dTCS

Nu Cp = const 1

2p

T

TlnCS= (1mol kh)

d/ Qu trnh thun nghch ng tch

T

dTC

T

QdS vv =

=

==2

1

2

1

lnT

T

T

T

vv TdCT

dTCS

Nu Cv = const 1

2v

T

TlnCS= (1 mol kh)

( Qv = Cv.dT)

5.3.8. Nguyn l th ba nhit ng lc hc Entrpi tuyt i Entrpi chuna/ Pht biu nguyn l III( W.Nernst, 1864-1941)

Entrpi ca cc cht nguyn cht di dng tinh th hon ho khng tuyt i bng khng S 0T 0

b/ Entrpi tuyt i- Nguyn l III cho php tnh entrpi tuyt i ca cc cht nguyn cht bt k nhit no.- Gi s nng 1 mol ca cht nguyn cht dng tinh th hon ho t 0K ln nhit T di p sutkhng i

0K Tnc Ts TNng chy Si

S bin i entrpi trong qu trnh ny s l :S = ST S0

= +

++

+nc

S

S

nc

T

0

T

T

p

S

S

T

T

p

nc

ncp dTT

C

T

HdT

T

C

T

HdT

T

C

Trong qu nng chy v si th nhit s khng thay i* 0K entrpi ca cht nguyn cht di dng tinh th hon ho S0 = 0

ST = +

++

+nc

S

S

nc

T

0

T

T

p

S

S

T

T

p

nc

ncp dTT

C

T

HdT

T

C

T

HdT

T

C

ST : c gi l entrpi tuyt i ca cht nguyn cht nhit T v di p sut p

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(rn) (lng) (hi)

(rn) (lng) (hi)

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- Gi tr ca entrpi tuyt i trong cc s tay ha hc l entrpi tuyt i ca 1 mol cht nguyn cht 2980K v di p sut 101,325kPa tnh bng J.K-1.mol-1, c k hiu l 0298S

c/ Entrpi chunEntrpi S ca cht thng c xc nh trong nhng iu kin nht nh thng ngi ta ly gi tr

ca S nhit 250C = 2980K v p sut 1 atm, trong kh c coi l kh l tng, cn dung dchc ly nng bng n v, v c k hiu l 0298S hoc vit gn l S.

5.3.9. Bin thin Entrpi ca mt phn ng ha hc- Bin thin entrpi S ca mt phn ng : Tng entrpi ca cc sn phm phn ng tr i tngentrpi ca cht phn ng

S = S(sn phm) - S(cht phn ng)VD1: Phn ng:

C(than ch) + CO2(k) = 2CO vi 0298S ca cc cht l:

5,74 213,68 197,54 (J/mol.) 0298S = 2.197,54 - (5,74 + 213,68)= 175,56(J/mol.)

y l mt phn ng lm tng th tch ca h nn entrpi tng ln.

VD2: Phn ngH2(k) + 1/2O2(k) = H2O(k) vi 0298S ca cc cht l:

130,52 205,04 188,72 (J/mol.)

0298S = 188,72 - (130,52 +2

04,205)

= -44,32(J/mol.) y l phn ng lm gim th tch ca h nn entrpi gim xung

Ch : Entrpi tuyt i S ca mt cht khng bng bin thin entrpi to thnh S ca cht

5.3.10. Chiu t din bin ca cc qu trnh Th ng p hay nng lng t do Gibbs- Cho n cui th k 19, khi khi qut ha cc d kin thc nghim v hiu ng nhit phn ng hahc, ngi ta thy iu kin nhit khng i, ch nhng phn ng ta nhit (H0) ch xy ra khi c cung cp nng lng t

bn ngoi cho h. T Marcelin Berthelot pht biu qui tc: Cc phn ng ha hc ch t din

bin theo chiu ta nhit- Cn thng qua i lng S li th hin xu hng t din bin ca qu trnh l xu hng phn bhn lon ca cc ht (khuynh hng t n trng thi c xc sut ln nht)

Vy chiu t din bin ca cc phn ng ha hc c xc nh bng s tc ng tng hpca hai yu t: khuynh hng chuyn n trng thi c nng lng nh nht (gim entanpi) vkhuynh hng t n trng thi c xc sut ln nht (tng entrpi). Xu hng trong ccqu trnh ha hc xy ra nhit v p sut khng i c th hin trong s bin i mti lng gi l entanpi t do hoc nng lng Gibbs (G) (ly tn nh vt l ngi M lW.Gibbs,1839-1903):

G = H - TSH = G + TS

Nhit H ca bt k qu trnh no u gm c hai phn: G l phn nhit dng sinh cng(thc hin qu trnh) v TS l phn nhit khng th sinh cng m dng lm bin i entrpi ca

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T hc Ha i cng

h (iu ny lm sang t nguyn l II ca nhit ng lc hc l nhit khng th bin hon ton thnhnng lng c, in, ha hcm lun lun cn li mt i lng khng th bin thnh dng nnglng khc)* Nu tt c cc cht phn ng u trng thi chun th nng lng Gibbs to thnh ca cht trngthi chun v c k hiu l G0 hoc 0298G khi ch n nhit .

* Nng lng Gibbs to to thnh chun ca cc n cht bng khng* Bin thin nng lng Gibbs ca phn ng bng tng nng lng Gibbs to thn ca cc sn phm

phn ng tr i tng nng lng Gibbs to thnh ca cc cht phn ng:G = G(sn phm) G(cht phn ng)

- Ngi ta chng minh c rng: Trong iu kin nhit v p sut khng i phn ng tdin bin theo chiu gim nng lng t do Gibbs. S gim nng lng t do Gibbs cng ln thqu trnh din ra cng mnh. Qu trnh s t din ra cho n khi h t trng thi cn bng, lc nng lng t do t gi tr cc tiu v khng thay i na (ngha lG = 0)* nhit thp (T nh), TS c th b qua

G HG< 0 hay H < 0 th phn pht nhit t din bin. Khi ta c qui tcBerthelot : Phn ng t din bin theo chiu ta nhit

* nhit cao (T ln), ta c :

STG

STH