C 6 Nhiet Dong Hoa Hoc

HUI 2006General Chemistry:Slide 1 of 48

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HA I CNG

Chapter 6: NHIT NG HA HC


Nhit ng lc ha hc

6.1 i tng nghin cu ca NLH6.2 Cc khi nim c bn6.2 Nguyn l 1 ca nhit ng hc6.4 nh lut Hess6.5 Nguyn l th 2 ca NLH v chiu qu

trnh HH6.6 Bi tp


6.1 i tng nghin cu ca NLH

i tng nghin cu ca nhit ng lc hc vnhit ng lc hc ho hc l:

Nhit ng lc hc l khoa hc nghin cu ccquy lut v s bin ha t dng nng lng nysang dng nng lng khc v thit lp cc nhlut ca s bin i . C s ca nhit ng lchc l l 2 nguyn l nhit ng lc hc

Nhit ng lc hc ha hc l khoa hc nghincu cc quy lut v s bin i qua li gia hanng v cc dng nng lng khc trong cc qutrnh ha hc.


6 .26 .2 Khai nim c bn s dng trong nhit nglc hc v nhit ho hc

6.2.1. H (nhit ng ): l mt vt thhay nhm vt th c nghin cu vtch bit vi mi trng xung quanh

Hoc pht biu cch khc: H l tp hpcc vt th xc nh trong khng gianno v phn cn li xung quanh gi lmi trng

1. H c lp: l h khng trao i cht v Evi mi trng bn ngoi


2.2. HH kknn ((hh ngng))

He kn la he ch co the trao oi E vi MT ngoai.

H kn

ChtCht Cht Cht

NhitNhit


3.3. HH anan nhinhitt

He oan nhiet la he khong trao oi chat va nhietnhng co the trao oi cong vi MT ngoai.

V2

V1


4. H ng th v h d th, pha, h cn bng

H ng th l h c cc tnh cht l ho hc gingnhau mi im ca h ngha l khng c s phn chiah thnh nhng phn c tnh cht ho l khc nhau

H d th l h c b mt phn chia thnh nhng phnc tnh cht ho l khc nhau

Pha l phn ng th ca h, c thnh phn, cu to vtnh cht nht nh. H ng th l h 1 pha, h d th lh nhiu pha

H cn bng l h c nhit , p sut, thnh phn gingnhau mi im ca h v khng thay i theo thi gian


6.2.2 Trng thi ca h v thng s ( tham s)trng thi, hm trng thi

Trng thi ca h l ton b cc tnh cht l, ho ca h. Thng s trng thi: Trng thi ca h c xc nhbng cc thng s (tham s) nhit ng l: nhit T, psut P, th tch V, nng C

Phng trnh trng thi m t tng quan gia ccthng s trng thi

C 2 loi thng s trng thi+ Thng s cng : Khng ph thuc vo lng cht :nh nhit , t khi, p sut+ Thng s khuych (dung ): l nhng thng sph thuc vo lng cht khi lng, s mol, th tch


Trng thi cn bng: l l trng thi tng ng vi hcn bng ( Khi cc thng s trng thi ging nhau miim v khng i theo thi gian)

Hm trng thi: i lng nhit ng c gi l hmtrng thi nu bin thin ca i lng ch ph thucv trng thi u v trng thi cui ca h, khng phthuc vo cch tin hnhNi cch khc Hm trng thi l i lng nhit ngc gi tr ch ph thuc vo cc thng s trng thi cah m khng ph thuc vo cch bin i ca h, hayni cch khc khng ph thuc vo con ng i ca h.(Nhit T, p sut P, Th tch V, Ni nng U, entanpiH, entropi S, th ng p Gl nhng hm trng thi)


6.2.3 Qu trnh Qu trnh l s bin i xy ra trong h gn lin vi

s thay i t nht 1 thng s trng thi Qu trnh xy ra p sut khng i (P= hng s) gi l

qu trnh ng p, th tch khng i gi l qu trnhng tch v nhit khng i gi l qu trnh ngnhit

Qu trnh thun nghch: l qu trnh bin i t trngthi ny qua trng thi khc ( t 12) c gi l thunnghch nu nh c th bin i theo chiu ngc li ( t21) i qua ng mi trng thi trung gian nh chiuthun sao cho khi h tr v trng thi ban u th khngcn tn ti mt bin i no trong chnh h cng nhmi trng

Qu trnh khng thun nghch l qu trnh m sau h v mi trng khng th quay tr li trng thi banu


6.2.4 Nng lng

Nng lng l i lng o thuc tnh vn ng cavt cht. N l thc o kh nng vn ng ca vt cht.

i vi h c hc th nng lng c c trng chokh nng sinh cng ca h

Cng c hc = lc x qung ng i Nhit v cng l hai hai hnh thc trao i ca h vi

mi trng+ Dng truyn nhit l dng truyn nng lng vhng, khng c trt t, c thc hin qua s chuynng hn lon+ Dng truyn cng l dng truyn nng lng c hng,c truyn t h thc hin cng sang h nhn cng


Cac dang nang lng

H thng H thng

V= hng s, U=Qv P= hng s, U=Qp + P V


Nang lng, nhiet va cong

Cong la thc o s chuyen ong co trat t, co hngcua cac tieu phan trong he. He nhan cong: A < 0. Hesinh cong: A > 0.

A = Pngoai(V2 V1)


Cc dng nng lng

ng nng: dng E c trng cho vt chuynng E=(mv2)/2

Th nng: l E ca h c do v tr ca n trongtrng lc Et= mgh

in nng: l E chuyn ng ca cc tiu phntch in

Nhit nng: nng lng c lin quan n schuyn ng hn lon ca cc tiu phn

Ho nng l nng lng gn lin vi s bin icht


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 gi l c nng Ni nng U l nng lng d tr bn trong h

gm ng nng ca cc phn t, lc ht y cacc tiu phn mang in, nng lng ca cc linkt ho hc, nng lng ht nhn

Nh vy E (h) = E + Et + U


6.3 Nguyen ly 1 cua nhit ng lc hoc.nh lut bo toan nng lng

6.3.1 Nguyen ly Nng lng khng t nhin sinh ra hay tbin mt m n ch c th chuyn t dng ny sang dngkhc theo t l tng ng nghim ngt

V du: H knHp th mt nng lng = QMt cng A lc bn ngoi tc dng vo h.Trng thi h: 1 sang trng thi 2Ni nng ca h t U1 sang U2

U = Q + A

Trong o: U = U2 U1 la bien thien noi nang cua he.

P

Q

k h i


6.3.2 Cac ai lng nhiet ong: Noi nang, entanpiva nhiet dung

1. Noi nang U va nhiet ng tchV dTc V1 = V2 A = 0 Do o: Qv = UV

Vy s tng hay gim ni nng ca h ng bng nhitlng h thu vo hay ta ra.


2. Entanpi va Nhiet ang ap

Tc Pngoai = Pkh = P trong o Ap = P(V2 V1) va QP = U +A

Do o QP = (U2 U1) + P (V2 V1)= (U2 + PV2) (U1 + PV1)at H = (U + PV) QP = H2 H1 Hay QP = HTrong o: H la entanpi , ham trang thai.

H = HSP - HTCVay: Lng nhiet thu vao (hay toa ra) ung bang s tang (hay giam)entanpi cua he


EntanpiEntanpi ccaa phphnn ngng1. Entanpi t l vi h s hp thc ptCH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H = -1604 kJ2. Khi i chiu phn ng th du ca entanpi i H:CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H = +802 kJ3. Entanpi ph thuc trng thiCH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H = -802 kJCH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = -890 kJ


Quan h gia Qp v Qv cua chat kh

Do H= U + PVH = (U + PV) ma PV = nRT nen PV = nRT

oi vi qt ang tch QV = Uoi vi qt ang ap Qp = H

Vay QP = QV + nRTTrong o: n = nSP - nC

Trong qa trnh cht kh ta c:H = U + nRT


Qui c dau

Qui c:

Q < 0 : He toa

nhiet,

Q > 0 : He thu

nhiet

A < 0 : He nhan

cong, A > 0 : He

sinh cong

To nhit Thu nhit


3. Nhit dung va nhiet dung mol

a. Nhiet dung: la nhiet lng can thiet e nangmot lng chat nao o len 1 o

b. Nhit dung ring : l nhit lng cn thit nng 1g cht ln 1

c. Nhit dung mol: Nhit lng cn thit nngnhit ca 1 mol cht ln 1o m khng c sbin i v trng thi


Nhiet dung

+ Trng hp ng p: C tr thnh Cp+ Trng hp ng tch: C tr thnh Cv+ Vi h khng c cht kh Cp= Cv+ H kh l tng (1mol): Cp= Cv +R+ oi vi he co 1mol va n mol th nhiet dung rieng trung

bnh la:

QTCnT

QTT

QC

.

12


4. Kh l tng v nguyn l 1 ng tch: cong Av= 0; nhiet Qv = Uv= nCv (T2-T1) ng p Cong Ap = -P(V2-V1) = - nR(T2-T1)

nhiet Qp= Hp= nCp(T2-T1) ng nhit: Cong AT do the tch kh ly tng t le nghch vi ap

suat nen AT=- nRTln (V2/V1)= nRTln (P1/P2)

Nhiet V noi nang phu thuoc vao nhiet o nenUT= 0 QT= -AT= nRTln(V2/V1) = nRTln (P1/P2)


- H = QPQh = -Qmt

- Mi trng xung quanh l nc v trong nhit lngk

Qh = - (qnc + qnhit lng k)

B qua Q nhit lng kQh = - Qnc

ChmhT = - cncmncTnc

oo nhinhitt dungdungP= hng s


NH4NO3(r) NH4+(aq) + NO3-(aq)Tnc = 18.4oC 23.0oC = -4.6oCmnc = 60.0gcnc = 4.184J/goCmmu = 3.88g

qmu = -qncqmu = -cncmncTnc

qmu = -(4.184J/goC)(60.0g+3.88g)(-4.6oC)qmu = 1229J

Tnh H bng kJ/molso mol NH4NO3 = 3.88g/80.032g/mol = 0.04848mol

H = q mu/moles

H = 1229J/0.04848molH = 25.4 kJ/mol


V d

15.5g hp kim c nungnng ti 98.90 C v thvo trong 25.0 g nctrong nhit lng k.Nhit tng t 22.50 Cn 25.7oC.

Tnh nhit dung ringca mu hp kim

Cho (H2O (l) C= 4.18 Jg-1 oC-1 )


1. Hieu ng nhiet phan ng: Nhit lng to ra hay thu voca mt phn ng ha hc e lam thay oi noi nang hay entanpicua he.

+Neu nhiet dung e bien oi noi nang goi la hieu ng nhiet angtch con e bien oi entanpi goi la hieu ng nhiet ang ap

6.3.3 Hiu ng nhit ca cc qu trnh ho hc(Nhiet hoa hoc)


+ oi vi nhng qua trnh xay ra trong chat ran valong th V thay oi khong ang ke khi o H = U

+ Vi chat kh th H = U + nRT + P khong oi. Hieu ng nhiet cua phan ng mot

nhiet o xac nh bang tong entanpi cua san pham tri tong entanpi cua chat au

H = Hsp - Hc Hieu ng nhiet cua phan ng c tnh bang kj/mol

hoac kcal/mol


Bt u Kt thcHiu ng nhit ca mt phn ng ha hc ph thuc vo

trng thi ca cc cht phn ng u v trng thi ca cc snphm cui m khng ph thuc vo cc giai oan trung gian

+ Phng trnh hoa hoc co ghi them nhiet lng toa rahay thu vao va trang thai cua cac chat goi la phngtrnh nhiet hoa hoc

C (r) + O2 (k) CO2 (k) , H= -395,50kJ


2. Hieu ng nhiet tieu chuan

ai lng nay c tnh oi vi 1mol chat va ng viieu kien chuan cua chat: la tat ca cac chat tham giavao phan ng phai trang thai ben, ap suat chuan101,325 kPa (1atm)

Hieu ng nhiet nay goi la hieu ng nhiet chuan va kyhieu H0298 hoac H0 neu khong chu y en nhiet o


3. Nhit tao thanh , nhiet phan huy va nhiet ot chay

a. Nhiet tao thanh (sinh nhiet) cua mot hp chat la hieung nhiet cua phan ng tao thanh 1 mol chat o t cacn chat ng vi trang thai t do ben vng nhat trongnhng ieu kien a cho ve ap suat va nhiet oV du: C (r) than ch + O2 (k) CO2(k)

H0tt (CO2,k) = - 393,51 kJ/mol (H0)Nhiet tao thanh chuan cua n chat bang 0: H0298 = 0.b. Nhiet phan huy cua mot hp chat la hieu ng nhiet cuaphan ng phan huy 1 mol chat o thanh cac n chat.V du: H2O (l) H2 (k) + 1/2O2 (k)

H0ph (H2O,l) = + 285,84 kJ/mol


Nhiet tao thanhva phan huy cuamot hp chat bangnhau ve tr sonhng trai dau

nh luat Lavoisier-Laplace


c. Nhiet ot chay (thieu nhiet)

- Nhiet ot chay la hieu ng nhiet cua phan ng otchay 1 mol chat bang oxy e tao thanh san pham chay ap suat khong oi.

V du: CH4 (k) + 2O2 (k) = CO2 (k) + H2O (l) H0c (CH4,k) = - 212,7 kcal/mol -Neu nhiet ot chay c qui ve ieu kien tieu chuan

c goi la nhiet ot chay tieu chuan, ky hieu H0c 298


6.4 nh luat Hess va nhng he qua l Hess

6.4.1 Noi dung Neu co nhieu cach e chuyen cacchat ban au nh nhau thanh nhng san phamcuoi cung nh nhau th hieu ng nhiet tong congtheo cach nao cung giong nhau

Hay noi cach khac Hiu ng nhit ca mtphn ng ha hc ch ph thuc vo ban chat vatrng thi ca cc cht phn ng u v ca ccsn phm cui m khng ph thuc vo cchthc hien phan ng


V du 1:C (than ch,r) + O2(kh) = CO2 (kh), H = -393,509 kJC (than ch,r) + O2 = CO (kh), H1 = -110,525 kJCO (kh) + O2 (kh) = CO2 (kh), H2 = - 282,984 kJ

H = H1 + H2 V du 2

CH4(k) + 2O2(k) CO2(k) + 2H2O(k) H = -802 kJ2H2O(k) 2H2O(l) H = -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ H = H1 + H2


Entanpi l hm trng thi vkhng ph thuc vo qu trnh


6.4.2 Mot so he qua

He qua 1: Hieu ng nhiet cua mot phan ng bangtong nhiet tao thanh (sinh nhit) cua cac san phamtr tong nhiet tao thanh cua cac tac chat

H0298 = H0tt (sp) H0tt(tc)


V du 1: Xc nh nhit phn ng cho phn ng sau:

4NH3(k) + 5O2(k) 4NO(k) + 6H2O(k)S dng h phn ng sau

N2(k) + O2(k) 2NO(k) H = 180.6 kJN2(k) + 3H2(k) 2NH3(k) H = -91.8 kJ2H2(k) + O2(k) 2H2O(k) H = -483.7 kJ

Gi : Ba phn ng trn phi c t hp ton hc thu c phn ng mong mun.V H ..


4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)S dung cac phan ng:

N2(g) + O2(g) 2NO(g) H1 = 180.6 kJN2(g) + 3H2(g) 2NH3(g) H2 = -91.8 kJ2H2(g) + O2(g) 2H2O(g) H3 = -483.7 kJ

Gii:

NH3:O2 :NO:H2O:

o ngc v x 2 4NH3 2N2 + 6H2 H = +183.6 kJ= 0

x 2 2N2 + 2O2 4NO H = 361.2 kJx 3 6H2 + 3O2 6H2O H = -1451.1 kJ


4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g):

NH3:O2 :NO:H2O:

o ngc v x2 4NH3 2N2 + 6H2 H = +183.6 kJ

x2 2N2 + 2O2 4NO H = 361.2 kJx3 6H2 + 3O2 6H2O H = -1451.1 kJ

Loi b phn t theo h s v ly tng

4NH3 + 5O2 4NO + 6H2O H = -906.3 kJ

Phn ng ta nhit hay thu nhit


V du 2 CaCO3 (r) CaO (r) + CO2 (k)H0298 = ? .Biet:

H0tt -1206,9 - 635,6 -393,5 (kJ)H0298 = - 635,6 -393,5 + 1206,9 = +177,8 kJ


He qua 2

He qua 2: Hieu ng nhiet cua mot phan ng bang tongnhit ot chay cua cac tac chat tr tong nhiet ot chay cuacac san pham

H0298 = H0c (tc) H0c(sp)

V du1: CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l)+ H2O (l)

H0c - 871,69 - 1366,91 - 2284,05 0 H0 = - 871,69 - 1366,91 + 2284,05 = + 45,45 kJ


V du 2 Xc nh nhit phn ng cho phn ng sau:

C2H4(k) + H2(k) C2H6(k), H ?S dng h phn ng:

C2H4(k) + 3O2(k) 2CO2(k) + 2H2O(l) H1 = -1401 kJH2(k) + 1/2O2(k) H2O(l) H2 = -286 kJC2H6(k) + 7/2O2(k) 2CO2(k) + 3H2O(l) H3 = -1550 kJ

Ans: C2H4(g) + H2(g) C2H6(g) H = -137 kJ

.H = H1 + H2 - H3 = -1401 - 286 + 1550 = - 137kJ


C2H4(k) + H2(k) + 7/2 O2 C2H6(k) + 7/2 O2

2CO2(k) + 3H2O(l)

H1H2

-H3

H= (H1 + H2) - H3

V d 3: C2H4(k) + H2(k) C2H6(k) , H?


He qua 3

He qua 3: Hieu ng nhiet cua mot phan ngbang tong nang lng lien ket co trong cac chatban au tr tong nang lng lien ket co trongcac san pham.

H0298 = Elk (tc) Elk(sp)


C2H4 + H2 C2H6 : H= ?Cho E C=C = 595.7 kj/molEC-H= 418.4 kj/molEH-H= 432.0 kj/molEC-C = 343.4 kj/mol

C2H4 + H2C2H6H

H=EC=C + 4EC-H + EH-H EC-C- 6EC-HEC=C 4 EC-H EH-H

-EC-C-6 EC-H


1.Tm hieu ng nhiet cua phan ng:V du: Xac nh hieu ng nhiet cua qua trnh.

C (than ch ran) C (kim cng ran) H = ?Biet: C(than ch ran) + O2 (kh) CO2 (kh)

H1 = -393,51kJC(kim cng ran) + O2 (kh) CO2 (kh)

H2 = -395,41kJ

C (ran)than ch

C (ran)kim cng

CO2 (kh)H1

H2H

6.4.3 Mot vai ng dung cua nh luat Hess


C (ran)than ch

C (ran)kim cng

CO2 (kh)H1

H2H

Theo nh luat Hess H= H1- H2 =1,9 kj


2 Xac nh nang lng lien ket hoa hoc V du: Xac nh E t/bnh cua cac lk O H trong pt

nc. Biet E lk HH = 435,9 kJ, EO=O = 498,7 kJ, khi t 2

mol H2 toa ra 483,68 kJ. Ta co p: 2H2 (k) + O2 (kh) 2H2O (l) H0298 = - 483,68kJ Theo he qua 3: H0298 = Elk (tc) Elk (sp) - 483,68 = 2 x 435,9 + 498,7 4Elk(O H)Elk(O H) = (2 x 435,9 + 498,7 + 483,68)=463,545 kJ


3.Tm nhiet tao thanh mot chat

V du xac nh nhiet tao thanh axetylen 2C(grafit,r) + H2 (k) C2H2(k) ; H0 298= ?Biet: C(grafit,r) + O2 (k) CO2(k) ; H0 298= -393,5 kJ H2 (k) + O2 (k) H2O(l) ; H0 298=-285,8 kJ C2H2(k) + 5/2O2 (k) CO2 (k) + H2 O(l) H0 298= -1299,4kJ H0 298=(-2 x 393,5 -285,8)-(-1299,4) = +226,6kJ Ket qua nhiet tao thanh cua axetylen bang H0tt (axetylen, kh) = + 226,6 kJ


6.5 Nguyen ly th 2 cua nhiet ong hoc vachiu t din bin ca mt qu trnh hoa hoc

6.5.1 Nguyn l th hai nhit ng lc hc: Nhit khng th t ng truyn t vt lnh sang vt

nng hn Khng c qu trnh trong nhit ly t mt vt c

chuyn thnh cng m khng c s n b

Khng t din bin

T din bin


6.5.2 Entropi va tnh chat cua no1. Khai niem

Entropi (S) la mot ai lng nhiet ong kho st s trao inhit ca h vi mi trng trong qu trnh h chuyn t trngthi nhit cao T1 sang nhiet o thap T2

+ Neu qua trnh trao oi nhiet thun nghchS = Q tn/T

+ Neu qua trnh trao oi nhiet bt thun nghch.S > Q btn/T

Tom lai S Q/T (j/mol, cal/mol) Trong : Qtn va Qbtn la nhiet lng trao oi trong qua trnh thuan nghch

va bat thuan nghch T: l nhit o tuyt i ti xy ra s trao i nhit

S = S2-S1S1,S2: entropi ca h cc trng thi nhit T1 v T2


2. Tnh chat cua entropi

+ Entropi S l i lng c trng cho mc hn nca phn t trong h cn xt

S= kblnW = R/N ln WTrong o W xac sut nhit ng

kb hng s Boltzmal kb=1,3860066.10-23+ Cung nh cac tnh chat khac cua he nh T,P, H, U,entropi la mot ai lng xac nh trang thai cua he

+S la ai lng dung o (khuyech o) gia tr cua nophu thuoc vao khoi lng

+ S la ham trang thai, bien thien cua no ch phu thuocvao trang thai au va cuoi, khong phu thuoc vao cacgiai oan trung gian


Tnh chat cua Entropi

+ Ve ban chat entropi (S) la thc o tnh hon loan cuahe, tnh hon loan cua cac tieu phan trong he cang caoth entropi cua he cang cao.

Do o He cang phc tap th entropi cang ln

V du S0 cua O = 160,95 j/molS0 cua O2 = 205,03 j/molS0 cua O3 = 238,82 j/mol


Tnh chat cua Entropioi vi cung mot chat th S tang khi chuyen t ransang long va sang chat khS0nc a = 41,31 j/molS0nc long = 63,31j/molS0hi nc = 185,6 j/mol

Nhiet o cang tang th entropi cang caoAp suat tang th entropi

cua he giamVay S phu thuoc vao

T, P, V


Tnh chat cua Entropi

Tnh hn lonEntropiEntropi

Rn Lng Kh


3. Entropy v s chuyn pha

S (kh) > S (lng)> S (rn)

So (J/Kmol)H2O(kh) 188.8H2O(lng) 69.9H2O (rn) 47.9 Nc

hi

S = Q/TS = Q/THH22O (l)O (l) ------> H> H22O(k)O(k)H = q = +40,700 J/molH = q = +40,700 J/mol

molJ/K109+=K373.15

J/mol40,700=

Tq

=S


4.Bien thien entropi cua mot so QT thuan nghch

i vi qu trnh ng pSSpp = n C= n CpplnTlnT22/T/T11 ( C( Cpp ll khngkhng ii))

ii vvii ququ trtrnhnh ngng ttchchSSvv = n C= n CvvlnTlnT22/T/T11 (( CCvv ll khngkhng ii))

ii vvii ququ trtrnhnh ngng nhietnhietSSTT = Q= QTT/ T/ T

NNuu CCpp vv CCvv phph thuthucc vvoo nhinhitt llcc phphii ttnhnh ssphph thuthucc CC vvoo nhinhitt

VV dd;; TTnhnh SS ccaa 1 mol1 mol nncc khikhi unun nnngng tt 273K273Knn 373K, C373K, Cpp = 75,5= 75,5 kjkj/mol/mol


6.5.3 Nguyen ly 3 cua nhiet ong hoc- Entropi tuyet oi va entropi tieu chuan

Nguyen ly: nhit khong tuyt i (0K) mi ncht cng nh hp cht dng tinh th hon ho u centropi bang 0 (S0 =0 ) hay gi nh lut Nernst

Entropi tuyt i: Entropi c xc nh nhiet o naoo tren c s i t nhiet o 0 tuyet oi

S= ST S0. T viec tnh S ta tnh c ST Entropi tieu chuan: la gia tr entropi tuyet oi cua chat

iu kin tiu chun: nhiet o 250C (298,15K), apsuat 1atm va ky hieu S0298. . oi vi chat kh o la khly tng , con dung dch la dd co nong o 1 mol/lit


V d; 2 H2(k) + O2(k) 2 H2O(long)So = 2 So (H2O) - [2 So (H2) + So (O2)]So = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1

mol (205.3 J/Kmol)]So = -326.9 J/K

S gim v ?

6.5.4 Bin thin S cua phn ng hoa

S = S (snphm) - S (cht tham gia)

So = So (snphm) - So (cht tham gia)


6.5.5 Bieu thc thong nhat cua 2 nguyen ly

Theo nguyen ly 1: Q = U + A Theo nguyen ly 2 cho qua trnh ang nhiet : S Q/T hay Q T.S Ket hp 2 bieu thc tren T.S U + A + oi vi qua trnh thuan nghch TS = U + ATN + oi vi qua trnh khong thuan nghch TS > U + AKTN


6.5.6 Th nhiet ong v chiu hng din bin ca ccqu trnh ho hc

1.1. KhaKhaii nieniemm veve thethe nhienhiett oongng aangng ttchch,, aangng aapp a)a) QuanQuan hehe cucuaa entanpientanpi vava entropientropiphan anh khuynh hng tap hp cua cac tieu phan S phan anh khuynh hng ngc lai, lam cho he phan

tan e co s phan bo hon loan+ Khuynh hng tap hp lam cho entanpi giam (

con khuynh hng phan tan lan tang entropi (S


b.The nhiet ong va phng trnh c ban cua nhiet ong hoc

bb11)) TheThe aangng ttchchTT nguyennguyen lyly II vava IIII tata coco momott aaii llngng mmii++ NeNeuu quaqua trtrnhnh xayxay rara TT vava VV khongkhong ooii tata coco

FF = U TSFF gogoii lala thethe aangng nhienhiett aangng ttchch ((gogoii tatatt lala thethe aangngttchch)) hoahoacc hahamm nangnang llngng tt dodo HelmholtzHelmholtz ((gogoii tatatt lalahahamm HelmholtzHelmholtz))

++ ooii vvii hehe kknn,, trongtrong ieieuu kiekienn aangng nhienhiett aangng ttchch,,quaqua trtrnhnh tt diendien biebienn kekemm theotheo ss giagiamm thethe aangng ttchch((F


b2) The ang ap++ NeNeuu quaqua trtrnhnh xayxay rara TT vava PP khongkhong ooii..

G = H - TS (1) GG gogoii lala thethe ngng nhinhitt ngng pp ((gogoii tatatt lala thethe aangng aapp) hay) hay hahamm nngnng llngng tt dodo

Gibbs (Gibbs ( gogoii tatatt lala hahamm GipGip))+ oi vi ieu kien ang nhiet ang ap T (1) ta co

G = G2-G1= (H2-H1)- T(S2-S1) =H TShay G = H TS (2)

Phng trnh (2) goi la phng trnh c ban cua nhietong hoa hoc


b3) The ang ap tieu chuan

The ang ap tieu chuan la the ang ap o iu kintiu chunKy hieu GGoo298298 hayhay GGooLng chat ung 1 mol, Ap suat 1 atm, Cac chat dang ben vngn v o kJ/mol

The ang ap tao thanh tieu chuan cua mot chat lao bien oi the ang ap cua phan ng tao thanh 1molchat o t cac n chat ng trang thai t do ben

1 atm va 250C. Ky hieu GGoo298298 tttt ooii vvii nn chachatt GGoo298298 tttt cc quiqui cc babangng 00


2. o bien oi the ang ap v chiu hng din bin cacc qu trnh ho hc

T nguyen ly I va II ngi ta chng minh c rangA -G (Ala cong co ch trong ieu kien ang nhiet, ang ap).Cong co ch cc ai khi qua trnh thuan nghch Am = - G

V vay oi vi qu trnh ng p, ng nhit

G > 0 Qu trnh khng t xy ra.G < 0 Qu trnh t xy ra.G = 0 h t trng thi cn bng.

a.ieu kien dien ra cua cac qua trnh hoa hoc


b. Bien thien the ang ap G va chieu dien ra cacqua trnh hoa hoc

G = H TS

H S -TS G = H - TS+

+ +

++

+---

-

--

-+ tt c T- Tt c T- T cao+ T thp

+ T cao- T thp


c. Bin thin th ng p tieu chuan G0 trongphn ng ha hc

Nng lng t do to thnh n cht Gtto = 0 kJ/mol i vi hp cht Gtt = Htt - TStt

V du Br2(l) Br2(k)H = 30.91 kJ/mol, S = 93.2 J/mol.K

G = kJ/mol - (298K)J/mol.K3.13 kJ/mol


Bin thin th ng p trong phn ng ha hc

Phn ng ha hc

GGoopp == GGootttt (sp)(sp) -- GGootttt ((cc))C2H4(k) + H2O(l) C2H5OH(l) Tnh Gr cho cc gi tr sau

Gtt(C2H5OH(l)) = -175 kJ/molGtt(C2H4(g)) = 68 kJ/molGtt(H2O (l)) = -237 kJ/mol


S ph thuc G v GG vvoo TT vv PP Ph thuc vo T

GG22/T/T22-- GG11/T/T11== H (1/TH (1/T22--1/T1/T11)) VV dd;;ii vvii ququ trtrnhnh chchyy llngng ccaa nncc H =6007 j/molH =6007 j/mol 273,15 K273,15 K thth GG00= 0= 0 ttnhnh GG00 T=272.15 KT=272.15 K

SS phph thuthucc vvoo pp susutt::GGpp= G= Goo ++ nRTlnPnRTlnPGGpp== GGoo ++ nRTlnPnRTlnPVV dd: CaCO: CaCO33 CaOCaO + CO+ CO22 GG298298oo = 129kj/mol= 129kj/molTTnhnh GG khikhi nhinhitt khngkhng ii P = 2P = 2 atmatm


6.5.8 i lng mol ring phn v ha th

i lng mol ring phnV= naVa + nbVbTh tch mol ring phn ca mt cu t l binthin th tch ca hn hp khi cho thm 1 molcu t vo hn hp trong iu kin cc thngs khc (T, P thnh phn ca cu t khc) khngi

Ha th;i= Gi : gi l ha th


HA TH Ha th ca mt hp cht i trong hn hp l bin

thin entanpi t do ca hn hp khi thm 1 molcht i vo hn hp trong iu kin cc thng skhc khc khng i (T, P v thnh phn mol cnli)

= o + RTlnPCu t th ii= io + RTlnPi


BI TP CHNG 6

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C 6 Nhiet Dong Hoa Hoc