HSG Nhiet Hoa Hoc

7/28/2019 HSG Nhiet Hoa Hoc

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A- C S L THUYT :Trc khi a ra h thng bi tp cho hc tr luyn tp th gio vin cn phi yu cu hc tr

nh li mt s khi nim v ni dung l thuyt c bn ca phn Nhit ho hc nh sau:1) KH L TNG:* Kh l tng l cht kh m khong cch gia cc phn t kh xa nhau, c th b qua tng tcgia chng.* Vi kh l tng th c th p dng :

- Phng trnh trng thi: P.V = nRT (R = 8,314 J/mol.K = 0,082 l.atm/mol.K)- Trong bnh c hn h p khth: P = Pi = V

ni .RT

cn Pi = Ni .P = .Pnini

2) H V MI TRNG:- H m: h trao i cht v nng lng vi mi trng.- H kn: H ch trao i nng lng vi mi trng.- H on nhit: H khng trao i nhit vi mi trng.* Quy c:

H nhn nng lng ca mi trng nng lng mang du +H nhng nng lng cho mi trng nng lng mang du -3) BIN I THUN NGHCH:

Nu h chuyn t trng thi cn bng ny sang trng thi cn bng khc mt cch v cngchm qua lin tip cc trng thi cn bng th s bin i ny c gi l thun nghch. y l s

bin i l tng khng c trong thc t.4) S BIN I BT THUN NGHCH: l s bin i c tin hnh vi vn tc ng k.

Nhng phn ng trong thc t u l bin i bt thun nghch.5) HM TRNG THI: l hm m gi tr ca n ch ph thuc vo cc thng s trng thi ca h,khng ph thuc vo nhng s bin i trc .V d: P.V = hm trng thi

P1.V1 = n.RT1 ; P2.V2 = n.R.T26) CNG (W) V NHIT (Q)- L 2 hnh thc trao i nng lng.- W, Q khng phi l hm trng thi v gi tr ca chng ph thuc vo cch bin i.V d: Cng ca s gin n kh l tng t th tch V1 n V2 to = const trong 1 xilanh kn nh 1

pittng c tnh bng cng thc:

W = - dVPn .2

1

(Pn : p sut bn ngoi)* Nu s bin i l BTN th Pn = Pkq = const

WBTN = - Pkq . dV2

1= - Pkq . V = - Pkq .(V2 - V1)

* Nu s bin i l thun nghch: Gim P n nhng lng v cng b th tch kh tng nhnglng v cng b. Khi Pn mi lc thc t = P bn trong xi lanh = Pk

Pn = Pk= n.RT/V

WTN = - dVPn .2

1

= - nRT . 2

1V

dV= - nRT .ln

1

2

V

V WBTN WTN

* Cc qu trnh thun nghch sinh cng ln nht khi h bin i t trng thi 1 sang trng thi 2.Lng cng ny ng bng lng cng cn thit a h v trng thi ban u mt cch thunnghch.

7) NI NNG U:- U ca mt cht hay mt h gm ng nng ca cc phn t v th nng tng tc gia cc phn ttrong h .- U l i lng dung v l hm trng thi


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- U ca n mol kh l tng ch ph thuc vo nhit .8) NGUYN L I CA NHIT NG HC: (S BIN I NI NNG CA H).

U = U2 - U1 = W + Q- i vi s bin i v cng nh: dU = W + Q (: Ch nhng hm khng phi l hm trng thi)- Thng gp cng c thc hin ch do s bin i th tch nn: W = -P.dV

dU = Q = P .dV dU = Q - dVP.2

1 U = Q - dVP.2

1* Nhit ng tch: Nu h bin i V = const dV = 0

U = QV QV l 1 hm trng thi.* Nhit ng p:Nu h bin i P = const th:

dVP.2

1

= P . dV2

1

= P. V2 - P. V1

U = U2 - U1 = QP - P. V2 + P .V1 QP = (U2 + P.V2) - (U1 + P .V1)t U + P.V = H = entanpi = hm trng thi QP = H2 - H1 = H = s bin thin entanpi ca h.* Nhit phn ng:Xt 1 h kn trong c phn ng: aA + bB cC + dD

Nhit phn ng ca phn ng ny l nhit lng trao i vi mi trng khi a mol A phn ng vi bmol B to ra c mol C v d mol D T = const.- Nu phn ng c thc hin P = const th nhit phn ng c gi l nhit phn ng ng pQP = H- Nu phn ng c thc hin V = const th nhit phn ng c gi l nhit phn ng ng tchQV=U* Quan h gia QP v QVQP = H = (U + PV)P = U + P. V H = U + P . V = U + n .RT

QP = QV + n .RT ( n = n kh sp - n kh p )Khi n = 0 QP = QV hay H = U

U = QV = n .CV . T

H = QP = n .CP . T * Nhit dung mol ng p (CP) l nhit lng cn cung cp lm 1 mol cht nng thm 1o trongiu kin ng p (m trong qu trnh khng c s bin i trng thi).

* Tng t vi CV: H = 2

1

.T

T

PdTC ; U =

2

1

.T

T

TdTC

CP, CV l hm ca nhit .

Vi 1 mol kh l tng: CP =T

H

; CV =T

U

M U = H - P. V CP =T

H

=T

U

+T

VP

.

= CV + R

Q, W: Khng phi l hm trng thiQV = U; QP = H QV, QP l hm trng thi ch ph thuc vo trng thi u v trng thicui ca h m khng ph thuc vo qu trnh bin i l thun nghch hay khng thunnghch.9) NH LUT HESS:H (U) ca 1 qu trnh ch ph thuc vo trng thi u v trng thi cui

ca h m khng ph thuc vo ng i.Hp = Hs (sn phm) - Hs (cht u) = Hc (cht u) - Hc (sn phm)10) NH LUT KIRCHHOFF:


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n1 A + n2 B n3C + n4 DT2

H2

HaHb

n1 A + n2 B n3C + n4 DT1

H1

Theo nh lut Hess: H2 = Ha + H1 + HbM:

Ha = +2

1

)...( 21

T

T

PP dTCnCn bA = - +2

1

)...( 21

T

T

PP dTCnCn BA

Hb = +2

1

)...( 43

T

T

PPdTCnCn

DC

H2 = H1 + ++2

1

)].()..[( 2143

T

T

PPPP dTCnCnCnCn BADC = H1 + 2

1

.

T

T

P dTC

- H1 thng c xc nh iu kin chun: HoT = Ho298 + T

o

P dTC298

.

Vi CoP = CoP(sp) - CoP(tham gia)CoP l nhit dung mol ng p iu kin chun (1atm).- Trong khong hp ca nhit c th coi CoP = constTh: H2 = H1 + CP.(T2 -T1) HoT = Ho298 + CoP (T - 298)11) ENTROPI (S)- Trong s bin i thun nghch v cng nh T = const h trao i vi mi trng mt lng

nhit QTN th s bin thin entropi trong qu trnh ny l: dS =TQTN

S l hm trng thi (J/mol.K)

- Nu s bin i l bt thun nghch th dS >T

QTN

- V l hm trng thi nn khi chuyn t trng thi 1 sang trng thi 2 bng bin thin thun nghch

hay bt thun nghch th S2 - S1 = S = 2

1T

QTN

(STN = SBTN)12) NGUYN L II CA NHIT NG HC:

dS T

Q

- Trong h c lp Q = 0. nn:+ dS = 0: trong h c lp entropi ca h khng i nu xy ra qu trnh thun nghch.+ dS > 0 : trong h c lp, qu trnh t xy ra (BTN) theo chiu tng entropi ca h v tng cho tikhi t gi tr max th h s t trng thi cn bng.* Entropi l thc o hn n ca h: hn n ca 1 h hay 1 cht cng ln khi h hay cht gm nhng ht v s dao ng ca cc ht cng mnh (khi lin kt gia cc ht cng yu).VD: S < S < S

S < S < SH2(k) O2(k) O3 (k)

H2O(r) H2O (l) H2O (h)

S l 1 i lng dung .13) S BIN THIN S TRONG QU TRNH BIN I TRNG THI CA CHT:


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Khi cht nguyn cht nng chy hoc si P = const th:

T = const S = 2

1T

Q=

T

H

H = nhit bin thin trng thi = Ln/c hoc Lh14) S TRONG QU TRNH GIN N NG NHIT KH L TNG:Xt n mol kh l tng gin n th tch t V 1 V2 to = const. V ni nng ca kh l tng ch ph

thuc nhit nn trong s bin i ny:U = QTN + WTN = QBTN + WBTN = 0

QTN = - WTN = nRT. ln1

2

V

V( = -(- P. V) = dV

V

nRT.

2

1

).

T = const S =T

QTN = nRln1

2

V

V= n.R.ln

2

1

P

P

15) S BIN THIN ENTROPI CA CHT NGUYN CHT THEO NHIT .- Qu trnh P = const: un nng 1 cht nguyn cht t T1 T2, khng c s chuyn pha:

S =

2

1

T

T

TN

T

Q

Vi Q = QP = dH = n.CP.dT

S =T

dTCn

T

T

P..

2

1

* Trong khong nhit hp, coi CP = const S = n.CP.ln1

2

T

T

- Qu trnh: V = const S = n .CV.ln1

2

T

T

16) ENTROPI TUYT I

* Nguyn l III ca nhit ng hc:- Entropi ca cht nguyn cht di dng tinh th hon chnh 0(K) bng 0: S(T = 0) = 0* Xut pht t tin trn ta c th tnh c entropi tuyt i ca cc cht cc nhit khcnhau.VD: Tnh S ca 1 cht nhit T no , ta hnh dung cht c un nng t 0(K) T(K) xt P=const. Nu trong qu trnh un nng c s chuyn pha th:

S = ST - S(T = 0) = ST = =

5

1i

iS

ST =T

dTCn

T

Ln

T

dTCn

T

Ln

T

dTCn

T

T

hP

S

S

T

T

lP

nc

nc

T

rP

S

S

nc

nc

........ )()(0

)(

1

++++

Gi tr entropi c xc nh P = 1 atm = const v nhit T no cgi l gi tr entropi chun, k hiu l S0T, thng T = 298K S029817) S BIN THIN ENTROPI TRONG PHN NG HO HC:+ Khi phn ng thc hin P = const, T = const th: S = S(sp) - S(t/g)+ Nu iu kin chun v 250C th: S0298= S0298(sp) - S0298(t/g)+ V S ca cht kh >> cht rn, lng nn nu s mol kh sn phm (sp) > s molkh tham gia th S > 0 v ngc li. Cn trong trng hp s mol kh 2 v

bng nhau hoc phn ng khng c cht kh th S c gi tr nh.18) TH NHIT NG

Sc lp = S h + S mt 0a)Th ng p G:Xt h xy ra s bin i P, T u khng i trong qu trnh ny mi trng nhn ca h mtnhit lng Hmt do h to ra Hmt = - H h = - H


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S mt = -T

H

+ iu kin t din bin ca h:

S c lp = S h -T

H> 0 H T. S < 0

+ H trng thi cn bng khi H T. S = 0+ t G = H TS nhit , P khng i th qu trnh xy ra theo chiu c

G = H T. S < 0V t ti trng thi cn bng khi G = 0.b) Th ng tch: (Nng lng Helmholtz)

Nu h bin i iu kin T, V khng i nhit ng tch m mi trng nhn ca cc h l

Umt Smt = -T

Umt

iu kin t din bin ca h trong qu trnh ng nhit, ng tch lF = U T. S < 0

V t trng thi cn bng khi F = 0

Trong : F = U TSV H = U + PV G = H TS = U TS + PV G = F + PV+ i vi qu trnh T,P = const G = Wmax+ i vi qu trnh T, V = const S = WmaxTM LI :* Qu trnh ng p: P = const- Cng:WP = - P.dV = -n.R.dT WP = - P. V = - nRT

- Nhit:QP = dH = n.PC .dT QP = H = n. dTC

T

T

P.

2

1

- Ni nng: dU = Q + W U = H P. V = H n.R. T- Entropi: dS

T

QTN S 2

1T

QTN STN =T

dTCn

T

T

P..

2

1

= TdCnT

T

P ln..2

1

NuPC = const STN = n. PC .ln

1

2

T

T

* Qu trnh ng tch:- Cng: WV = - P.dV = 0 WV = 0

- Nhit: QV = dUV = n. VC .dT QV = UV = dTnCT

T

V ..2

1

Nu VC = const QV = n. VC .T

- Ni nng: UV = QV + W

- Entropi: S T

QV = = TdT

Cn

T

T

V..2

1

TdCn

T

T

V ln..2

1

S n. VC .ln1

2

T

T( VC = const)

- Entanpi: H = U + PV dH = dU + P.dV + V.dP = dU + V.dP (dV = 0)H = U + V . P

* Qu trnh ng nhit:

- Cng:WT = - PdV = - dVV

nRT.

WT = -1

2

2

1

1

2 lnlnln..2

1P

PnRT

V

VnRT

V

VnRT

V

dVRTn

V

V

===


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- Nhit:UT = QT + WT = 0 QT = - WT = nRT ln1

2

V

V

- Ni nng:UT = 0- Entanpi:HT = UT + (PV)T = UT + nR. T = 0

- Entropi: S TN =nc

ncTN

T

L

T

Q= hoc =

S

h

T

L

* Vi qu trnh dn n kh l tng thun nghch

S = =

=T

WU

T

QTN + TdT

Cn V

T

T

..2

1

dVV

nRTV

V

2

1

Nu CV = const S = n.1

2lnT

TCV + nRT ln

1

2

V

V

V T = const S = nRT ln1

2

V

V= nRT.ln

2

1

P

P

* Qu trnh on nhit:- Nhit: Q = 0

- Ni nng v cng: dU = Q + W = W = -PdV =T

dTCn

V

T

T

..2

1

+Qu trnh bt thun nghch:dUBTN = WBTN = -Png .dV = -P2.dVUBTN = WBTN = -Png.(V2 V1) = n.CV. T* PT Poisson: (Dng cho qu trnh thun nghch)

T . 1V = constP.V = const

V

P

C

C=

* WBTN = -P2(V2 V1) = - P2.( )() 121

1

2

2 TTnCP

nRT

P

nRTV = T2U = W = .... V2

* Qu trnh thun nghch: W = U = n.CV(T2- T1)

T1. V1

1 = T2 . V

12 T2 = T1.(

2

1

V

V)-1

- Entanpi: H = n .CP(T2 T1)

- Entropi: STN = T

QTN

= 0* G = H TS = U + PV TS

=

PT

G- S ; =

TP

G- V

Vi phn ng oxi ho kh c th din ra trong pin in: G = - nEF

dT

Gd= - nF.

dT

dE= - S S = nF.

dT

dE

H = G + T. S = nF( T.dT

dE- E)

19) NGHA VT L CA G:G = H TS = U + PV TS dG = dU + P.dV + V.dP T.dS SdT = (W + Q) + PdV + VdP T.dS SdTV W = W + (-PdV)


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Q T.dS dG W + VdP SdTDu = ng vi qu trnh thun nghch v cng ln nht.dG = Wmax + VdP SdT* i vi qu trnh ng nhit, ng p dP = dT = 0 dGT,P = W maxG = W max* i vi qu trnh BTN: W gim; Q tng khi hon ton BTN W = 020) MT S TNH CHT CA HM G:dG = V.dP SdT ( coi W = 0)a) S ph thuc ca G vo T:- Khi P = const

PT

G

= - S PT

G

= - S

G = H T. S = H + T.PT

G

T.PT

G

- G = -H 22

.

T

H

T

GT

GT

P =

2

T

H

T

T

G

P =

dTT

H

T

Gd

T

T

T

G

T

G

T

T

..2

1

2

2

1

2

1

=

dTT

G

T

G

T

G T

T

TT.

2

1

12

2

12

=

=

Nu coi Ho khng ph thuc vo nhit th:

=

298

11

298

298

TH

G

T

G oo

T

b) S ph thuc vo P:

Khi T = const VP

G

T

=

( ) ( ) ==2

1

1

2

1

2..

2

1

P

P

PT

P

P

PT dPVGGdPVdG

- Vi cht rn, lng coi V = const khi P bin thin (tr min p sut ln) th:( ) ( ) )( 1212 PPVGG PTPT +=

- Vi cht kh l tng V =P

nRT ( ) ( )

1

2ln.12 P

PnRTGG PTPT +=

Nu p sut bnh thng: P1 = Po = 1bar (1 atm) GT(P) = GoT + nRT.lnP(P tnh bng bar (atm)).

21) TNH G CA MT S QU TRNH:a) Gin nn ng nhit kh l tng

G = nRT.ln 1

2

P

P

= nRT.ln 2

1

V

V

b) Trn ln ng nhit, ng p 2 kh l tng:G = nA.RTlnxA + nB.RTlnxB

c) Qu trnh chuyn pha thun nghch (ti nhit chuyn pha): Gcf= 0d) Qu trnh chuyn pha thun nghch T Tcf

Nguyn tc: p dng chu trnh nhit ng. V G l hm trng thi nn G ch ph thuc trng thiu, trng thi cui, khng ph thuc vo qu trnh bin thin.e) G ca phn ng ho hc: Gop = GoS(sn phm) - GoS(tham gia)


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B - H THNG CC CU HI V BI TP PHN NHIT HA HC :Bi 1:Cho 100 g N2 0oC, 1atm. Tnh Q, W, U, H trong cc bin i sau y c tin hnh thun

nghch nhit ng:a) Nung nng ng tch ti P = 1,5atm.

b) Gin ng p ti V = 2V ban u.

c) Gin ng nhit ti V = 200ld) Gin on nhit ti V = 200lChp nhn rng N2 l kh l tng v nhit dung ng p khng i trong qu trnh th nghim v

bng 29,1J/mol.KGii

a) V = const W = = 0.dVPU = QV = n VC .T = ( PC - R).(T2 T1) .n

= ( PC - R).(1

2

P

P-1).T1.n = (29,1 - 8,314).( )1

1

5,1 .273,15 = 14194,04(J)

b) Vo = )(804,22.28

100 l= V= 2Vo = 160 (l)

W = -P. V = -1(160 80) = -80 (l.at) = -80 .101,33 = -8106,4(J)

QP = H = PC .n .T =

11

1

2 ..1,29.28

100TT

V

V= 29,1.

28

100(2.273,15 273,15) = 28388,1(J)

U = Q + W = 28388,1 = 8106,4 = 20281,7(J)c) T = const U = 0; H = 0

W = - 2

1

.V

dVnRT = - nRT.ln

1

2

V

V

W = -28

100.8,314 .273,15.ln

80

200= -7431,67(J)

U = Q + W = 0 Q = -W = 7431,67(J)d) Q = 0 (S = const)Theo PT poisson: T1. V

11 = T2 . V

12

T2 = T1.(2

1

V

V)-1 Vi 4,1

314,81,29

1,29

=

+==

P

P

V

P

CR

C

C

C

W = U = n. VC (T2 T1) =28

100(29,1-8,314).(189,33 -273,15) = 6222,4(J)

H = nPC .T = 28

100.29,1(189,33 273,15)= - 8711,3(J)

Bi 2:

Tnho

SH 298, ca Cl-(aq). Bit:

(a):2

1H2 +

2

1Cl2(k) HCl(k)

o

SH 298, = -92,2(kJ)

(b): HCl(k) + aq H+(aq) + Cl-(aq)o

SH 298, = -75,13(kJ)

(c): 2

1

H2 + aq H+

(aq) + e

o

SH 298, = 0Gii:

Ly (a) + (b) (c) :2

1Cl2 + e + aq = Cl-(aq)

o

SH 298, = - 167,33(kJ)


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Bi 3:Tnh hiu ng nhit ca phn ng:

3Fe(NO3)2(aq) + 4HNO3(aq) 3Fe(NO3)3(aq) + NO(k) + 2H2O (l)Din ra trong nc 25oC. Cho bit:

Fe2+(aq) Fe3+(aq) NO3-(aq) NO(k) H2O(l)o

SH 298, (kJ/mol) -87,86 - 47,7 -206,57 90,25 -285,6Gii:Phng trnh ion ca phn ng:3Fe2+(aq) + 4H+(aq) + NO3-(aq) 3Fe3+(aq) + NO(k) + 2H2O (l)H=3.

o

SH 298, (Fe3+,aq)+o

SH 298, (NO)+2.o

SH 298, (H2O(l))3.o

SH 298, (Fe2+,aq)-o

SH 298, (NO3-, aq)= 3.(-47,7) + 90,25 + 2.(-285,6) + 3.87,6 + 206,57 = -153,9(kJ)Bi 4:1) So snh H, U ca cc phn ng: CnH2n + H2 CnH2n+2

2) Khi t chy hon ton 2 anome v ca D glucoz mi th 1 mol p sut khngi, ngi ta o c hiu ng nhit ca cc phn ng 500K ln lt bng:-2790,0kJ v - 2805,1kJ

a) Tnh U i vi mi phn ng.b) Trong 2 dng glucoz, dng no bn hn?Gii:1) H = U + P. V = U + n.RTPhn ng trn c: n = 1-2 = -1 H = U RT H < U2) C6H12O6 + 6O2 6CO2 + 6H2OU() = H() - n.RT = - 2799 6.8,314.10-3.500 = -2824(kJ)U() = H() - n.RT = - 2805,1 6.8,314.10-3 .500 = -2830 (kJ)

oH = 6.o

COSH )( 2 + 6.o

OHSH )( 2 -o

SH )(o

H = 6.o

COSH )( 2 + 6.o

OHSH )( 2 -o

SH )(

o

SH )( -o

SH )( =oH -

oH = -2805,1 + 2799 = -6,1(kJ)

o

SH )( 139,82(kJ/mol)

O3 c cu trc vng kn rt khng bn cu trc ny khng chp nhn c.

Bi 6:Entanpi sinh tiu chun ca CH4(k) v C2H6(k) ln lt bng -74,80 v -84,60 kJ/mol. Tnh entanpitiu chun ca C4H10(k). Bin lun v kt qu thu c. Cho bit entanpi thng hoa ca than ch vnng lng lin kt H- H ln lt bng: 710,6 v - 431,65 kJ/mol.Gii:

* (1) C than ch + 2H2 (k) CH4(k)o

CHSH 4, =-74,8kJ

(2) C than ch C (k)o

thH = 710,6 kJ

(3) H2 (k) 2H (k) lkH = 431,65 kJLy (1) [(2) + 2.(3)] ta c:

C(k) + 4H(k) CH4(k)o

CHtungSH 4,/, = -1648,7(kJ/mol)

Nng lng lin kt trung bnh ca lin kt C H l:4

1(-1648,7) = - 412,175 (J/mol).

* (4) 2C than ch + 3H2 C2H6(k)o

KHCSH ),( 62 = -84,6 (kJ/mol)

Ly (4) [2 .(2) + 3.(3)] ta c:2C(k) + 6H (k) C2H6(k)

o

HCtungSH62

,/, = -2800,75 (kJ/mol)Coi EC H trong CH4 v C2H6 nh nhau th:

E C- C = =1800,75 6(- 412,175) = -327,7(kJ/mol)* Coi EC-H; EC- C trong cc cht CH4, C2H6, C4H10 u nh nhau th:

o

HCtungSH104

,/, = 3. EC- C + 10.EC- H = 3.(- 327,7) + 10( -412,75) = -5110,6 (kJ/mol)

* (5) 4C(k) + 10 H(k) C4H10(k)o

HCtungSH104

,/, = -5110,6 (kJ/mol)Ly (2). 4 + (3).5 + (5) ta c:

4Cthan ch + 5H2(k) C4H10(k)o

HCSH104

, = -109,95(kJ/mol)* Kt qu thu c ch l gn ng do coi E lk(C C), Elk(C- H) trong mi trng hp l nh nhau. V

v vy s khng tnh r co

SH ca cc ng phn khc nhau.

Bi 7: Tnh Ho ca cc phn ng sau:1) Fe2O3(r) + 2Al(r) 2Fe(r) + Al2O3(r) ( 1)

Cho bito

OFeS rH

)(32, = -822,2 kJ/mol;

o

OAlS rH

)(32, = -1676 (kJ/mol)

2) S(r) +2

3O2(k) SO3(k) (2)

Bit (3) : S(r) + O2(k) SO2(k)o

H298 = -296,6 kJ

(4): 2SO2(k) + O2(k) 2SO3(k)oH298 = -195,96 kJ

T kt qu thu c v kh nng din bin thc t ca 2 phn ng trn c th rt ra kt lun g?Gii:


11/27

1) opuH )1( =o

OAlS rH

)(32, -

o

OFeS rH

)(32, = -1676 + 822,2 = - 853,8(kJ)

2) opuH )2( =o

puH )3( +2

1 opuH )4( = -296,6 -

2

1.195,96 = -394,58 (kJ)

KL: Hai phn ng (1), (2) u to nhit mnh. Song trn thc t 2 phn ng khng t xy ra. Nhvy, ch da vo H khng khng nh chiu ca 1 qu trnh ho hc (tuy nhin trong nhiu

trng hp, d on theo tiu chun ny l ng).

Bi 8:1) Tnh hiu ng nhit ng tch tiu chun ca cc phn ng sau 25oC.

a) Fe2O3(r) + 3CO(k) 2Fe(r) + 3CO2(k)oH298 = 28,17 (kJ)

b) Cthan ch + O2(k) CO2 (k)o

H298 = -393,1(kJ)

c) Zn(r) + S(r) ZnS(r)oH298 = -202,9(kJ)

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

H298 = -195,96 (kJ)

2) Khi cho 32,69g Zn tc dng vi dung dch H 2SO4 long d trong bom nhit lng k 25o

C,ngi ta thy c thot ra mt nhit lng l 71,48 kJ. Tnh hiu ng nhit nhit . Cho Zn =65,38Gii:1) H = U + n.RTDo cc phn ng a), b), c) c n = 0 nn Uo = Ho

Phn ng d): Uo = Ho - n.RT = -195,96 + 1.8,314. 298,15. 10-3 = -193,5 (kJ)2) Zn(r) + H2SO4(dd) H2(k) + ZnSO4(dd)Trong bom nhit lng k c V = const.

U = - 71,48.38,65/69,32

1= -142,96 (kJ/mol)

H = U + n.RT = - 142,96 + 1. 8,314 .298,15 .10-3 = - 140,5 (kJ/mol)

Bi 9: Tnh Ho ca phn ng tng hp 1 mol adenine C5H5N5(r) t 5 mol HCN(k).

Cho bito

kCHSH ),, 4 = - 74,8 (kJ/mol);o

kNHSH ,, 3 = -46,1kJ/mol;o

radeninSH )(, = 91,1 kJ/mol

V CH4(k) + NH3(k)HCN(k) + 3H2(k) Ho = 251,2 kJ.mol-1

Gii:

(a) : Cgr + 2H2(k) CH4o

kCHSH ),, 4 = -74,8 (kJ/mol)

(b) : 2

1

N2(k) + 2

3

H2(k) NH3(k)

o

kNHSH

,, 3 = - 46,1kJ/mol

(c) : 5Cgr+2

5H2(k) +

2

5N2(k) C5H5N5(r)

o

radeninSH )(, = 91,1 kJ.mol-1

(d) : CH4(k) + NH3(k)HCN(k) + 3H2(k) Ho = 251,2 kJ.mol-1

Ta ly: -5 .(a) + [-5 .(b)] + (c) + [-5.(d)] ta c:5HCN(k) C5H5N5(r)Ho(4) = 251,2 kJ/mol

Bi 10:

Tnh nhit thot ra khi tng hp 17kg NH3 1000K. Bito

kNHSH ),(298, 3 = -46,2 kJ.mol-1

),( 3 kNHPC = 24,7 + 37,48.10-3

T Jmol-1

K-1

),( 2 kNPC = 27,8 + 4,184.10-3 T Jmol-1K-1

),( 2 kHPC = 286 + 1,17.10-3 T Jmol-1K-1


12/27

Gii:

2

1N2(k) +

2

3H2(k) NH3(k)

o

kNHSH ,, 3 = - 46,2kJ/mol

CP = ),( 3 kNHPC -2

1),( 2 kNP

C -2

3),( 2 kHP

C

= - 24,7 + 37,48.10-3T -2

1[27,8 + 4,184.10-3] -

2

3[28,6 + 1,17 .10-3T]

= - 32,1 + 31,541.10-3 T

= oH1000o

H298 + dTCP,1000

298

=o

H298 + dTT)10.541,311,32(1000

298

3 +

=oH298 + )

210.541,311,32(

21000

298

3 TT +

= - 46,2.103 +31,541 .10-3.2

1(10002 -1982) 32,1(1000 298)= - 54364,183 (J/mol)

Khi tng hp 17 kg NH3 th nhit lng to ra l:

Q =17

17000 .(-54364,183 .10-3) = -54364,183 (kJ)

Bi 11:Tnh nng lng mng li tinh th BaCl2 t 2 t hp d kin sau:

1) Entanpi sinh ca BaCl2 tinh th: - 859,41 kJ/molEntanpi phn li ca Cl2: 238,26 kJ/molEntanpi thng hoa ca Ba: 192,28 kJ/mol

Nng lng ion ho th nht ca Ba: 500,76 kJ/molNng lng ion ho th hai ca Ba: 961,40 kJ/moli lc electron ca Cl : - 363,66 kJ/mol

2) Hiu ng nhit ca qu trnh ho tan 1 mol BaCl2 vo mol H2O l: -10,16kJ/mol.Nhit hirat ho ion Ba2+ : - 1344 kJ/molNhit hirat ho ion Cl- : - 363 kJ/molTrong cc kt qu thu c, kt qu no ng tin cy hn.Gii:

Ba(r) + Cl2(k) BaCl2(tt)

Hth(Ba) Uml

Ba(k) +2Cl (k) Ba2+

+2Cl-I1(Ba) +I2(Ba)

2. AClUml = H - Hth (Ba) - H - I1(Ba) - I2(Ba) - 2ACl

=- 859,41 - 192,28 - 238,26 - 500,76 - 961,40 +2 .363,66=- 2024,79 (kJ /mol)

HS(BaCl2, tt)

Hpl(Cl2)

o

S(BaCl2, tt) pl(Cl2) o

2) BaCl2 (tt) Ba(aq) +2Cl(aq)

- Uml +H2O

H1 H2

Ba2+ +2Cl-

Hht(BaCl2)

2+ -


13/27

Uml = H1 + H2 - H

=-1344 - 2.363 +10,16 =-2059,84 (kJ/mol) ht(BaCl2)

Kt qu 1) ng tin cy hn, kt qu tnh theo m hnh 2) ch l gn ng do m hnh ny khng mt ht cc qu trnh din ra trong dung dch, cc ion nht l cation t nhiu cn c tng tc ln nhauhoc tng tc vi H2O.

Bi 12:Cho gin n 10 lt kh He 0oC, 10atm n p sut l 1atm theo 3 qu trnh sau:

a) Gin ng nhit thun nghch.b) Gin on nhit thun nghch.c) Gin on nhit khng thun nghch.

Cho nhit dung ng tch ca He CV =2

3R v chp nhn khng i trong iu kin cho ca bi

ton.Tnh th tch cui cng ca h, nhit Q, bin thin ni nng U v cng W trong mi qu trnh nitrn?Gii:

a) T = const U = 0; H = 0U = Q + W = 0 Q = -W

W = - =2

1

2

1

.V

dVnRTdVP = - nRTln

1

2

V

V

Vi kh l tng: P1.V1 = P2 .V2 1

2

V

V=

2

1

P

P V2 =

2

1

P

P. V1 =

1

10. 10 = 100(l)

W = -(nRT).ln2

1

P

P= -10.10 .ln 10 = 230,259 (l.at)

W = 230,259 .101,33 .10-3 = 23,332 (kJ)Q = - W = -23,332 (kJ)

b) Q = 0

U = W = n. VC . T = 23

..

.

1

11

TR

VP.R(T2 T1)

U = W =2

3.

.

1

11

T

VP(T2 T1)

Theo PT poisson: T.V- 1 = const

M V = P

nRT

T.

1

P

nRT

= const T

.P

- 1

= const

T1 .P1 =T2 .P2

T1T2

= P2P1

T1T2

P2P1

=

1- 1-

1-1-

T2 =T1 . =273,15 .P2P1

1-

101

1-

=CPCV

CV +RCV

32 R +R

R=5

3=

1 -=- 0,4 = = 3

2

1-53

53

32

T2=273,15 .(10)-0,4=108,74 (K)

U =W = . (108,74 - 273,15) .101,33 =9148,6(J)273,1610 .10


14/27

V2 = = 39,81 (l)1 .273,1510.10.108,74

P2 .T1

P1 .V1.T2 ~~

c) Q = 0 U = W

n. VC (T2 T1) = -Png .(V2 V1) = -P2 .

1

1

2

2

P

nRT

P

nRT

n. 23

R(T2 T1) = -nR.1

10112 TT

T2 = 0,64T1

V2= = 64(l)T1 . 110.10.0,64T1

P2 .T1

P1 .V1.T2=

U = W = -Png(V2 V1) = -1(64 10) = -54(l.atm)= -54(l.atm) .101,33 .J/l.atm = - 5471,82 (J)

Bi 13 :

Phn ng sau: Ag +2

1Cl2 = AgCl

Xy ra di p sut 1 atm v 25oC to ra 1 nhit lng l 126,566 kJ.

Nu cho phn ng xy ra trong 1 nguyn t ganvani P, T = const th ho nng s c chuynthnh in nng v sn ra cng W = 109,622 kJ.Hy chng t rng trong c 2 trng hp trn, bin thin ni nng ca h vn ch l mt, cn nhitth khc nhau v tnh gi tr bin thin ni nng .Gii:- Do U l hm trng thi nn U = U2 U1 = const, cho d s bin i c thc hin bng cchno. V vy U trong 2 trng hp trn ch l mt.- V U = Q + W = Q + W - PV = Q + W - n.RTDo nRT = const; U = const

Nn khi W (cng c ch) thay i th Q cng thay i

- U = H - nRT = -126,566 + 21 . 8,314 .298,15.10-3 = - 125,327 (kJ)

Bi 14:Tnh cng ca s bin i ng nhit thun nghch v bt thun nghch 42g kh N2 300K khi:a) Gin n t 5atm n 1atm.

b) Nn t 1atm n 5atm.(Kh c coi l l tng). So snh cc kt qu v rt ra kt lun.Gii:

a) * WTN = -1

2

1

2

2

1

2

1

lnlnP

PnRT

V

VnRT

V

dVnRTPdV ===

WTN =2842 .8,314 .300. ln

51 = -6201,39(J)

*WBTN = - Png . V = -Png(V2 V1) = -Png

1

2

11 VP

VP= - P2.V1

1

2

1

P

P

= - P2.

=

1

2

2

1

1

11.P

PnRT

P

P

P

nRT= -

28

42.8,314 .300

5

11 = -2993,04 (J)

b) WTN = nRTln1

2

P

P=

28

42.8,314 .300.ln

1

5= 6201,39(J)

WBTN = - Png. V= -Png(V2 V1) = -Png

12 P

nRT

P

nRT


15/27

= -nRT.P2

12

11

PP= -nRT

1

21P

P= -

28

42.8,314 .300

1

51 = 14965,2 (J)

KL: - Cng m h thc hin (sinh) trong qu trnh bin thin thun nghch t trng thi 1 n trngthi 2 bng cng m h nhn khi t trng thi 2 v trng thi 1. Cn trong qu trnh bin thin btthun nghch th cng h sinh nh hn cng h nhn.- Trong s bin thin thun nghch th h sinh cng ln hn trong qu trnh bin thin bt thun

nghch.

Bi 15: Phn ng: C6H6 +2

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

300K c QP QV = 1245(J). Hi C6H6 v H2O trong phn ng trng thi lng hay hi?Gii:

QP QV =nRT = 1245(J) n =300.314,8

1245= 0,5

H2O v C6H6 phi th hi th n = 0,5

Bi 16:Tnh nhit lng cn thit nng nhit ca 0,5 mol H2O t -50oC n 500oC P = 1atm. Bit nhit nng chy ca nc 273K l Lnc = 6004J/mol,

nhit bay hi ca nc 373K l Lh = 40660 J/mol.o

hOHPC ),( 2 = 30,2 + 10-2T(J/molK) ;

o

rOHPC ),( 2 = 35,56(J/molK);o

lOHPC ),( 2 = 75,3(J/molK)

Gii:

H2O(r) H2O(r) H2O(l) H2O(l) H2O(h) H2O(h)(500oC)

H1 H2 H3 H4 H5

-50oC 0oC 0oC 100oC 100oC

Ho = ho

lPnc

o

rP LndTCnLndTCnH ......

373

273

)(

273

223

)(

5

1

+++=

+

773

373

)( .. dTCno

hP

= 0,5 .35,56(273 223) + 0,5 .6004 + 0,5 .75,3 .(373 273) + 0,5 .40660 +

+ 0,5.30,2 .(773 373) +2

10 2.0,5 (7732 3732) = 35172(J)

Bi 17: Tnh s bin thin entropi ca qu trnh un nng 0,5 mol H 2O t 50oC n 500oC P =1atm. Bit nhit nng chy ca nc 273K = 6004J/mol; nhit bay hi ca nc 273K =40660J/mol. Nhit dung mol ng p oPC ca nc v nc lng ln lt bng 35,56 v

75,3J/molK; oPC ca hi nc l (30,2 + 10-2T) J/molK

Gii:

H2O(r) H2O(r) H2O(l) H2O(l) H2O(h) H2O(h)S1 S2 S3 S4 S5 773K373K373K273K273K223K

o o o o o

S2 S3 S4 S5 o o o o

= + + + +So

S1o

= n.

++++

273

223

373

273

773

373

)()()( .373

.273

.T

dTC

L

T

dTC

L

T

dTC hP

h

lP

nc

rP

=0,5.

+++++ )373773(10

373

773ln.2,30

373

40660

273

373ln.3,75

273

6004

223

273ln.56,35

2 = 93,85(J/K)

Bi 18:Tnh s bin thin entropi khi trn ln 200g nc 15oC vi 400g nc 60oC. Bit rng hl c lp v nhit dung mol ca nc lng l 75,3 J/mol.KGii:Gi T l nhit ca h sau khi pha trn.Do Q thu = Q to nn:


16/27

18

200.

PC (T 288) =18

400.

PC (333 T)

T 288 = 2.333 2T T =3

288333.2 += 318(K)

S h = S1 + S2 =T

dT.3,75.

18

200318

288

+ TdT

.3,75.18

400318

333

=18

200 .75,3 ln288

318 +18

400 .75,3 ln333318 = 5,78 (J/K) > 0

Qu trnh san bng nhit ny t xy ra.Bi 19: Tnh s bin thin entropi v G ca s hnh thnh 1 mol hn hp kh l tng gm 20%

N2; 50%H2 v 30%NH3 theo th tch. Bit rng hn hp kh c to thnh do s khuch tn 3 khvo nhau bng cch ni 3 bnh ng 3 kh thng vi nhau. Nhit v p sut ca cc kh lc uu kc (273K, 1atm).Gii:V kh l tng khuch tn vo nhau nn qu trnh l ng nhit.Gi th tch ca 1 mol hn hp kh l V th tch mi kh ban u ( cng iu kin) l

2NV =

0,2V; 3NHV = 0,3V; 2HV = 0,5V.

Do %V = %n 2N

n = 0,2 mol;2H

n = 0,5 mol;3NH

n = 0,3mol.

- S bin thin entropi c tnh theo CT: S = nRln1

2

V

V

2NS = 0,2 .8,314.ln

V

V

2,0= 2,676J/K

2HS = 0,5.8,314.ln

V

V

5,0= 2,881J/K

3NHS = 0,3.8,314.ln VV

3,0 = 3,003J/K

S =2N

S +2H

S +3NH

S = 8,56(J/K)* Qu trnh khuch tn kh l tng l ng nhit nn H = 0G 273 = H T. S = -273.8,56 = -2336,88(J)Bi 20: Trong cc phn ng sau, nhng phn ng no c S > 0; S < 0 v S 0 t.

C(r) + CO2(k) 2CO(k) (1)

CO(k) +2

1O2(k) CO2(k) (2)

H2(k) + Cl2(k) 2HCl(k) (3)S(r) + O2(k) SO2(k) (4)

Gii:Phn ng (1) c n kh = 2 -1 = 1 > 0 S > 0

Phn ng (2) c n kh = 1 -1-2

1< 0 S


17/27

b) iu kin chun v 25oC phn ng i theo chiu no?c) Tnh oH298 ca phn ng. Phn ng to nhit hay thu nhit?Gii:a) iu kin chun: )(42 kHCP = )(2 hOHP = )(52 hOHHCP = 1atm v phn ng c thc hin t

o, P khngi.

b)

Gp = Go

- Go

- Goo

S,298(C2H5OHh) S,298(C2H4k) S,298(H2Oh)

=168,6 - 68,12 +228,59 = - 8,13 (kJ)Gp (298) =-8,13kJ


18/27

on nhit thun nghch Theo Poisson T.V- 1 = const

M V =P

nRT T.

1

P

nRT= const T .P1-= const

T1 .P1 =T2 .P2

T1T2 =

P2P1

T1T2

P2P1=

1- 1-

1-1-

T2=T1 .P2P1

1-

Vi kh l tng n nguyn t th CV =2

3R; CP =

2

5R

= =CPCV

5

3

1 -=

1 -

= -0,4

T2 = 300.-0,4

= 101,55(K)

53

5

315

1

U = W = nCV(T2 - T1) = 1.2

3.8,314.(101,55 - 300) = -2474,87(J)

H = nCP(T2 - T1) = 1.

2

5.8,314 .(101,55 - 300)= - 4124,78(J)

STN =T

Q= 0

d) on nhit Q = 0on nhit, khng thun nghch khng p dng c PT poissonU = W nCV. T = -Png. V

n.2

3.R(T2 - T1) = -P2(

2

2

P

nRT-

1

1

P

nRT)

2

3(T2 - 300) = -( T2 -

1

2

P

P.T1)

2

3T2 - 450 = -T2 +

15

1.300

V2 =2

2

P

nRT=

1

188.082,0.1= 15,416(l)

V1 =2

2

P

nRT= 1,64(l)

U = W = 1.2

3.8,314.(188- 300) = -1396,752(J)

Stp = Sh = T

QTN = =

T

WU

2

1

T

T

VT

dTnC + dVT

PV

V

2

1

= 2

1

ln

T

T

V TdnC + dVV

nRV

V

2

1

= nCVln1

2

T

T+ nRln

1

2

V

V= 1.

2

3.8,314.ln

300

188+ 1. 8,314 .ln

64,1

416,15= 12,801(J/K) > 0


19/27

Bi 23: Tnh 0273G ca phn ng: CH4(k) + H2O (k) CO(k) + 3H2(k)Bit: CH4(k) H2O (k) CO(k) H2(k)

0

298,SH (kJ/mol) - 74,8 - 241,8 -110,5 00

298S (J/molK) 186,2 188,7 197,6 130,684

a) T gi tr G0 tm c c th kt lun g v kh nng t din bin ca kh nng phn ng 373oK?

b) Ti nhit no th phn ng cho t xy ra iu kin chun?(Coi H0, S0 khng ph thuc T)Gii:

0

puH = 3.0 + 1(-110,5) -(-74,8) -(-241,8) = 206,1(kJ)0

puS = 3.(130,684) + 197,6 - 188,7 - 186,2 = 214,752 (J/K)Do H0, S0 khng ph thuc vo T nn:

0273G = H0 - T. S0 = 206,1 = 373.214,752.10-3 =125,9975(kJ) > 0 kc v T = 373K Phn ng khng th t din bin.

b) phn ng t din bin nhit T(K) th: 0TG < 0 H0 - T. S0 < 0

T >0

0

S

H

=752,214

10.1,206 3

= 959,71(K)

Bi 24: Entanpi t do chun ca phn ng to thnh H2O t cc n cht ph thuc vo T theo

phng trnh sau:0

,TSG = -240000 + 6,95T + 12,9TlgT (J/mol)Tnh G0, S0 v H0 ca phn ng to thnh H2O 2000KGii:

0

2000,SG = -240000 + 6,95.2000 + 129.2000lg2000= -140933,426(J/mol)

dG = VdP - SdT PT

G

= -S

02000S = -P

T

G

0

= 6,95 + 12,5.lgT + 12,9T.10ln.

1

T= 6,95 + 12,9lgT +

10ln

9,12

= 6,95 + 12,9lg2000 +10ln

9,12= 55,1357(J/molK)

02000H = 0

2000G + T.0

2000S = -140933,426 + 2000. 55,1357 = -30662,054 (J/mol)

Bi 25: Mt Hc sinh khi lm bi tng trnh th nghim o nhit t chy mt hp cht huc cho rng: H = U + P. V. S t chy trong bom nhit lng k lm cho V = 0, do H =

U. Kt lun ny sai u?Gii:H = U + P.V H = U + (PV) = U + P. V + V. PHay H = U + (nRT)Trong bom nhit lng k th: V = 0 nn: H = U + V. P = U + (nRT)

Bi 26: Hy ch ra nhng mnh sai:a) i vi 1 h kn, qu trnh gin n kh l on nhit h l c lp Q = 0; S = 0.

b) Mt h bt k c th t din bin ti trng thi c entanpi thp hn (H < 0) v entropi ln hn(S > 0). Hay h c th din bin theo chiu gim entanpi t do (G < 0).

c)0

TG =0

TH - T.0

TSVi phn ng ho hc T = const. Nu

0G > 0 Phn ng t din bin theo chiu nghch.0

G = 0 : Phn ng trng thi cn bng.


20/27

0G < 0 : Phn ng t xy ra theo chiu thun.Gii:a) Sai . Do S = 0 ch khi qu trnh bin i thun nghch.

Cn vi qu trnh bin i bt thun nghch th S >T

QS > 0.

b) Sai . Do mnh ny ch ng trong iu kin T, P = const.

Cn vi qu trnh bin i m V, T = const th phi xt F.c) Sai. Do vi qu trnh ho hc th phi xt gi tr:G = G0 + RTlnQ ch khng phi da vo G0.

(Tuy nhin, c th coi rng 0TG


21/27

0

puH = 213,74. 3 + 4.69,91 - 269,91 - 5. 205,138 = -374,74 (J/K)0

puG = H0 - T. S0 = -2220 + 298,15 .374,74.10-3 = -2108,27 (kJ)U0 = H0 - (PV) = H0 - nRT = -2220 - (-3).8,314.298,15.10-3 = -2212,56(kJ)2) a) Qu trnh bt thun nghch:- Nhit m h trao i vi mi trng l QBTN = H0 = -2220 (kJ)

- Wtt = -

2

1.dVP = -P. V = -n(k) .RT

= 3. 8,3145.298,15 = 7436,9(J)- W = 0

b) Qu trnh thun nghch:- QTN = T. S = 298,15 (-374,74) = - 111728,731(J)- Wmax = G = -2108,27(kJ) < 0 : H sinh cng- Wtt = - n(k) .RT = 7436,9(J) > 0: h nhn cng3) a) Qu trnh bt thun nghch:

Smt =T

Qmt = -

T

QBTN = -T

H0= -

15,298

10.2220 3= 7445, 916 (J/K)

S v tr = Smt + S h = 7445,916 - 374,74 = 7071,176(J/K)b) Qu trnh thun nghch:

Smt =T

Qmt = -T

QTN = )/(74,37415,298

731,111728KJ=

S v tr = Smt + S h = 04) Cc na phn ng:Anot: C3H8 + 26OH- 3

23CO + 17H2O + 20e

Catot: O2 + 2H2O + 4e 4OH-

Phn ng tng cng:

C3H8(k) + 5O2(k) + 6OH-(aq) 3 2

)(3 aqCO + 7H2O(l)

S pin:(-) Pt, C3H8(1atm)/KOH(5M), K2CO3(1M)/ O2(1atm), Pt (+)

0

puH = 3(-677,14) + 7.(-285,83) + 103,85 - 5.0 - 6(-229,99) = -2548,44(KJ)0

puS = 3.(-56,9) + 7.69,91 - 269,91 - 5.205,138- 6(-10,74) = -912,43(KJ)0

puG =0

puH = T.0

puS = -2548,44 + 298,15.912,43.10-3 = - 2276,399(KJ)

0puE = -nF

G 0=

96485.20

2276399= 1,18(V)

E = E0 - 56323

283..][

][lg200592,0

OHC PPOH

CO

= 1,18 -

20

0592,0 lg(5)-6 = 1,19(V)

P = E .I = 1,19 .0,1 = 0,119(W)

Bi 29: Tnh bin thin entropi khi chuyn 418,4J nhit t vt c t 0 = 150oC n vt c t0 =50oC.Gii:Qu trnh bin i trn l khng thun nghch c coi nh gm 3 qu trnh bin thin thun nghch:1) Vt 150oC truyn nhit thun nghch T = const.

S1 = TQ

= 15,273150

4,418

+

= - 0,989(J/K)2) H bin thin on nhit t 150oC n 50oCS2 = 0


22/27

3) Vt 50oC nhn nhit thun nghch T = const

S3 = -T

Q=

15,27350

4,418

+ = 1,295(J/K)

Do S l hm trng thi nn:SBTN = STN = S1 + S2 + S3 = 0,306(J/K)

Bi 30: Bit -15oC, Phi(H2O, l) = 1,428 (torr) -15oC, Phi (H2O,r) = 1,215(torr)

Hy tnh G ca qu trnh ng c 1 mol H2O(l) thnh nc -15oC v 1atm.Gii:

15oC, 1 mol H2O l -15oC, 1mol H2O(r)

(3)

GBTN =?

(1)(Qu trnh TN doH2O hi, bo honm cn bng v i H2O(l))

- 15oC, 1mol H2O1,428 Torr

(2) -15oC, 1mol H2O (h)1,215 Torr

(1), (3) l qu trnh chuyn pha thun nghchG1 = G3 = 0

G = G2 = nRTln1

2

P

P= 1.8,314. 258,15 ln

428,1

215,1= -346,687(J)

Bi 31: C 1 mol O2 nguyn cht 25oC, 2atm, 1 mol O2 nguyn cht 25oC, 1atm1 mol O2 25oC trong khng kh trn mt t (P = 1atm, O2 chim 21% V khng kh)- So snh gi tr hm G ca 1 mol O2 trong 3 trng hp trn hn km nhau bao nhiu J?. T rt

ra kt lun: Kh nng phn ng ca O2 trong mi trng hp trn cao hay thp hn so vi trnghp khc?Gii:* G0 l hm Gibb ca 1 mol O2 1atm- 1 mol O2, 1atm, 25oC 1 mol O2, 2atm, 25oC

(G0) (G1)

G1 = G1 - G0 = nRTln1

2

P

P= 1. 8,3145 .298,15.ln

1

2= 1718,29(J)

G1 > G0.- Gi G2 l hm Gibb ca 1mol O2 25oC trong khng kh (0,21 atm)

1mol O2, 25oC, 1atm 1 mol O2, 25oC, 0,21atm(G0) (G2)

G2 = G2 - G0 = 1. 8,3145 .298,15.ln1

21,0= -3868,8(J)

G2 < G0.Vy:G2(1mol O2, 25oC, 0,21atm) < G0(1 mol O2, 25oC, 1atm) < G1(1 mol H2O, 25oC, 2atm)- 1 cht c hm G cng cao th cng km bn 1 mol O2 25oC, 2atm c kh nng phn ng caonht cn 1 mol O2 nm trong khng kh th b nht c kh nng phn ng km nht.

Bi 32: Nhit ho tan (Hht) 0,672g phenol trong 135,9g clorofom l -88J v ca 1,56g phenoltrong 148,69g clorofom l -172J.Tnh nhit pha long i vi dung dch c nng nh dung dch th 2 cha 1 mol phenol khi phalong n nng ca dung dch th nht bng clorofom.


23/27

Gii:

94g phenol +CHCl3 dd 2 dd 1H pha long

+CHCl3Hht (2)

Hht(1)

H pha long= Hht(1) - Hht(2)

=- .(-172) + (-88) =- 2004,87(J)0,672941,56994

Bi 33: Nhit ho tan 1 mol KCl trong 200 ml nc di p sut P = 1amt l:t0C 21 23

H 18,154 17,824 (kJ)Xc nh H298 v so snh vi gi tr thc nghim l 17,578 (kJ)Gii:Theo nh lut Kirchhoff:H294 = H298 + CP.(294 - 298) = 18,454 (kJ)

H286 = H298 + CP.(296 - 298) = 17,824(kJ)

H298 = 17,494 (kJ)

CP = -0,165 (kJ/K)

H298(LT) - H298(TN)H298(TN)

0,48%~~

Vy H298 tnh c theo l thuyt sai khc vi gi tr TN l 0,48%.Bi 34: Tnh S ca qu trnh ho hi 3 mol H2O (l) 25oC, 1atm.

Cho: Hhh, H2O(l) = 40,656 kJ/mol; )(, 2 lOHPC = 75,291 (J/K.mol); )(, 2 hOHPC = 33,58 (J/molK)Gii:Xt chu trnh:

25oC, 3 mol H2O (l), 1atm 25oC, 3 mol H2O(r), 1atm

100oC, 3mol H2O(l),1atm 100oC, 3mol H2O (h),1atm

S

S1

S2

S3

S1 =T

Q1 = 2

1

.. )(

T

T

lPT

dTCn = nCP(l)ln

1

2

T

T= 3. 75,291.ln

15,298

15,373= 50,6822(J/K)

S2 =T

Q2 =T

Hn lhh.. = 3.15,373

10,656,40 3= 326,8605(J/K)

S3 =T

Q3 = 1

2

.. )(

T

T

hPT

dTCn = nCP(h)ln

1

2

T

T= 3. 33,58.ln

15,373

15,298= - 22,6044(J/K)

S = S1 + S2 + S3 = 354,9383 (J/K)

Bi 35:a) Tnh cng trong qu trnh t chy 1 mol ru etylic kc v 25oC.

b) Nu H2O dng hi th cng km theo qu trnh ny l bao nhiu?

Gii:a) C2H5OH(l) + 3O2 (k) 2CO2 (k) + 3H2O (l)n = -1


24/27

W = -Png . V = -Png.ngP

RTn.= R.T = 8,314.29815 = 2478,82 (J)

b) Nu H2O dng hi th: n = 2. W = - n. RT = -2 .8,314 .298,15 = - 4957,64(J)

Bi 36: Tnh S, G trong qu trnh gin khng thun nghch 2 mol kh l tng t 4lt n 20

lt 54o

C.Gii:V S, G l cc hm trng thi nn S, G khng ph thuc vo qu trnh bin thin l thun nghchhay bt thun nghch m ch ph thuc vo trng thi u v trng thi cui. V vy:

S = nRln1

2

V

V= 2. 8,314.ln

4

20= 26,76 (J/K)

T = const H = 0; U = 0G = H - T. S = 0 -(273,15 + 54) .26,76 = - 8755,1 (J)

Bi 37: Mt bnh c th tch V = 5(l) c ngn lm 2 phn bng nhau. Phn 1 cha N 2 298K

v p sut 2atm, phn 2 298K v p sut 1atm. Tnh G, H, S ca qu trnh trn ln 2 kh khingi ta b vch ngn i.Gii:

T = 298K ; Vb (N2) = Vb(O2) =2

5(l)

S = S(N2) + S(O2) = 2Nn .Rln1

2

V

V+

2On Rln

1

2

V

V=

RT

VPNN 22

.Rln

5,2

5+

RT

VP OO 22 . Rln5,2

5

= 2ln..

2222

T

VPVP OONN + = 0,0174(l.at/K) = 0,0174 .101,325 = 1,763 (J/K)

- Qu trnh ng nhit H = 0G = H - T. S = - 298 .1,763 = - 525,374 (J)

Bi 38: Cho cc d liu sau y 298KCht 0SH (kJ/molK) S0(J/molK) V(m3/mol)Cthan ch 0,00 5,696 5,31.106

Ckim cng 1,90 2,427 3,416.10-6

1) 298K c th c mt phn rt nh kim cng cng tn ti vi than ch c khng?2) Tnh p sut ti thiu phi dng c th iu ch c kim cng 298K?

Gii1)

Ckim cng Cthan ch G0 = ?298

Ho = Hothan ch - Hokim cng = 0 - 1,9 = -1,9 (kJ)So = Sothan ch - Sokim cng = 5,696 - 2,427 = 3,269 (J/K)

0

,298 puG = Ho - T. So = -1900 - 298.3,269 = -2874,162(J)Go < 0 (Tuy nhin Go khng qu m) Phn ng t xy ra theo chiu thun khng tn ti mt lng nh kim cng cng vi thanch.2)

G0 =+2874,162 (J)

298Cthan ch Ckim cng

V = VKC - VTC = 3,416.10-6 - 5,31.10-6 = 1,894.10-6 (m3/mol)Ta c: dG = VdP - SdT


25/27

TP

G

= V TP

G

= V

2P

G -1P

G = V(P2 - P1)

iu ch c kim cng t than ch th:2P

G 0

1P

G + V(P2 - P1) 0

P2 - P1 -VGP

1

(Do V < 0)

P2 P1 -V

GP

1 = 1 +

325,101.10.894,1

162,28743

P2 14977,65 (atm)Vy p sut ti thiu phi dng iu ch c kim cng t than ch l 14977,65atm.

Nh vy 25oC, cn bng than ch kim c ng tn ti p sut khong

15000 atm. p sut cao hn qu trnh chuyn than ch thnh kim cng l t din bin, mc duvi tc rt chm. Mun tng tc phi tng nhit v p sut, trong thc t qu trnh chuynthan ch thnh kim cng c tin hnh khi c xc tc (Ni + Cr + ) nhit trn 1500 oC v P

50000atm.

Bi 39: Phn ng gia Zn v dd CuSO4 xy ra trong ng nghim to ra lng nhit 230,736kJ.Cng phn ng trn cho xy ra trong pin in th mt phn ho nng chuyn thnh in nng. Cngin ca pin l 210,672kJ. Chng minh rng: U ca 2 qu trnh khng i, nhng nhit to ra thayi. Tnh S ca phn ng, Smt v Stp? Cho T = 300KGii:

Zn + CuSO4 = ZnSO4 + Cu- Khi thc hin trong ng nghim: (Tin hnh bt thun nghch)VZn VCu Wtt = 0 ; W = 0

UBTN = QBTN = H = -230,736kJ- Khi thc hin phn ng trong pin in (qu trnh thun nghch)Wmax = - 210,672 (kJ) G = Wmax = -210,672(kJ)HTN = HBTN = - 230,736(kJ) QTN = T. S = H - G = -230,736 + 210,672 = -20,064(kJ)UTN = Q + W + P. V = -20,064 = 210,672 + 0 = -230,736 (kJ) = UBTN

- Sh =T

QTN = -300

10.064,20 3= 66,88(J/K)

SmtBTN = -

T

QBTN =

300

10.736,230 3= 769,12(J/K)

Stp(BTN) = 702,24(J/K)

Smt(TN) = -T

QTN = -ShStp(TN) = 0

Bi 40:- Gp = Wmax

Xt 1 phn ng thun nghch trong pin in th Gp = Wmax < 0- Nhng mt hc sinh vit rng:

Trong mi qu trnh lun c: S v tr = Smt + S h (1) Hmt = - H h (2)

Smt =THmt = -

THhe S v tr = -

THhe + S h

T. S v tr = - H h + T. S h = -G hVi qu trnh thun nghch th S v tr = 0 G h = 0 Gp = 0


26/27

Hy gii thch mu thun ny.Gii:(2) ch ng khi ngoi cng gin n h khng thc hin cng no khc:H = U + P. V U = H - P. VQ = U - W = (H - P. V) - (-P. V + W) Q h = H h - W = - H mt Ch khi W = 0 th Hmt = - H h* Trong pin: Wmax = G < 0 nn HmtH h.Bi 41: Xt phn ng: Zn(r) + Cu2+(aq) Zn2+(aq) + Cu(r)din ra trong ktc 25oC.a) Tnh W, Q, U, H, G, S ca phn ng iu kin trn.Bit: Zn2+(aq) Zn(r) Cu(r) Cu2+(aq)

0

298,SH (kJ/mol) -152,4 0 0 64,390

298S (J/mol.K) - 106,5 41,6 33,3 - 98,7b) Xt kh nng t din bin ca phn ng theo 2 cch khc nhau.c) Nu thc hin phn ng trn 1 cch thun nghch trong pin in th cc kt qu trn c g thay

i? Tnh Epin?Gii:

a)0

puH =0

, 2+

ZnSH +

0

,CuSH - +0

,ZnSH0

, 2+

CuSH = -152,4 - 64,39 = -216,79 (kJ)

0puS =0

)(2 aqZnS + +

0

)(rCuS -0

)(rZnS -0

)(2 aqCuS + = -106,5 + 33,3 - 41,6 + 98,7 = -16,1 (J/K)

0

puG =0H - T. 0S = -216,79 + 298,15 .16,1.10-3= -211,99(kJ)

Uo = QP =0

puH = -216,79 (kJ)W = 0; qu trnh BTN; W = 0

b) *0

puG = -211,99 (kJ) 0 Qu trnh l bt thun nghch phn ng t xy ra.c) Khi thc hin phn ng trn TN trong pin in th cc gi tr H0, S0, G0, U0 khng thay ido H, S, G, U l cc hm trng thi nn khng ph thuc qu trnh bin i l thun nghch hay btthun nghch nhng cc gi tr Q, W th thay i.C th: Wtt = 0; Wmax = G0 = -211,99(kJ)Q = T. S = 298,15 .(-16,1) = - 4800,215 (J)

Smt =T

Qmt =T

Qhe = 16,1 (J/K) S v tr = Smt + Sh = 0

Epin = -nF

G 0=

96485.2

211990 1,1(V)

Bi 42:i vi nguyn t anien 15oC ngi ta xc nh c sc in ng E = 1,09337V v h s

nhit ca sc in ngT

E

= 0,000429 V/K. Hy tnh hiu ng nhit ca phn ng ho hc?

Gii:G = - nEF

T

G

= - nF.T

E

= - S S = nF .T

E


27/27

H = G + T. S = nF.(T.T

E

- E)

H = 2. 96485 .(298,15.0,000429 - 1,09337) - - 187162,5(J)Bi 43: Cho phn ng ho hc: Zn + Cu2+ Zn2+ + Cuxy ra mt cch thun nghch ng nhit, ng p 25oC trong nguyn t Ganvani.Sc in ng ca nguyn t o c l 1,1V v h s nhit ca sc in ng l

PT

E

= 3,3.10-5 (V/K).

a) Tnh hiu ng nhit Q, bin thin GipxG v bin thin entropi S ca phn ng ho hc cho.

b) Tnh Qtn ca qu trnh?c) Nu cng phn ng ho hc trn thc hin cng nhit v cng p sut nhng trong mt bnhcu thng th cc gi tr ca G, S s l bao nhiu?Gii:a) G = - nEF = - 2 .1,1 .96485 = - 212267(J)

S = -T

G

= n.F .

T

E

= 2 .96485 .3,3 .10-5 = 6,368 (J/K)

H = G + T. S = 212267 + 298,15 .6,368 = -210368,4(J)b) Qtn = T . S = 298,15 .6,368 = 1898,62 (J)c) Nu phn ng ho hc thc hin cng nhit , p sut nhng trong 1 bnh cu thng tc lthc hin qu trnh mt cch bt thun nghch th G, S ca phn ng vn nh cu (a). Do G, Sl cc hm trng thi gi tr ca G, S khng ph thuc vo qu trnh bin thin.Bi 44: Tnh cng ca s bin i ng nhit thun nghch v bt thun nghch ca 48 gam khO2 c coi l l tung nhit 250 C khi:a. Gin n t 10atm xung 1atm

b. Nn t 1atm n 10atmp s: Wtn = -8,6.10-3J; Wbtn =-3,3.103J

b. Wtn = 8,6.10-3; Wbtn = 3,3.104 J.

IV- KT LUN - KIN NGH :Trn y l h thng cu hi v bi tp phn Nhit ho hc m ti p dng trong ging dy. Ntng i ph hp vi yu cu v mc ch ging dy, bi dng hc sinh kh, gii trngchuyn chun b d thi hc sinh gii cc cp . N c th dng lm ti liu hc tp cho hc sinh cclp chuyn Ho hc v ti liu tham kho cho cc thy c gio trong ging dy v bi dng hcsinh gii Ho hc bc THPT gp phn nng cao cht lng ging dy v hc tp mn Ho hc.Tuy nhin, y ch l mt phn rt nh trong chng trnh n luyn cho hc sinh chun b tham giavo cc k thi hc sinh gii cc cp. V vy, ti rt mong cc Thy , C ng nghip gp kin cho

ti v chuyn ny v cng nhau pht trin sang cc chuyn khc hc tr chuyn Ho ngycng c nhiu ti liu hc tp mt cch h thng hn.

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