2011 -xÂy dỰng hỆ thỐng cÂu hỎi vÀ bÀi tẬp phẦn “nhỆt hoÁ hỌc”

7/31/2019 2011 -XY DNG H THNG CU HI V BI TP PHN NHT HO HC

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TT LUYN THI & BI DNG KIN THC NGY MI 18A/88INH VN T - TP. HI DNG

HOAHOC.ORG NG XUN QUNH09798.17.8.85 09367.17.8.85 - netthubuon Website: www.hoahoc.org - [email protected]

XY DNG H THNG CU HI V BI TP PHN NHT HO HCDNG CHO HC SINH CHUYN HO

TrngTHPT chuyn Nguyn Hu - H Ni

I. M U

Trong qu trnh ging dy trng ph thng nhim v pht trin t duy cho hc sinh l nhim vrt quan trng, i hi tin hnh ng b cc mn, trong Ha hc l mn khoa hc thc nghim cpn nhiu vn ca khoa hc, s gp phn rn luyn t duy cho hc sinh mi gc c bit l qua phn

bi tp ha hc. Bi tp ha hc khng nhng c tc dng rn luyn k nng vn dng, o su v m rngkin thc hc mt cch sinh ng, phong ph m cn thng qua n tp, rn luyn mt s k nngcn thit v ha hc, rn luyn tnh tch cc, t lc, tr thng minh sng to cho hc sinh, gip hc sinh hngth trong hc tp. Qua bi tp ha hc gio vin kim tra, nh gi vic nm vng kin thc v k nng hahc ca hc sinh.

gio vin bidng hc sinh kh, gii trng chuyn d thi hc sinh gii cp Tnh v cp

Quc gia c tt th nhu cu cp thit l cn c mt h thng cu hi v bi tp cho tt c cc chuyn nh : cu to cht, nhit ho hc, ng ho hc, cn bng ho hc,....V vy , trong qu trnh ging dy i tuyn hc sinh gii Tnh v Quc gia ti su tm v tp hp limt s cu hi v bi tp theo mt s chuyn , trong c phn dng luyn tp cho hc sinh phnNhit ho hcII. MC CH CA TI

Xy dng h thng cu hi v bi tp phn Nhit ho hcdng cho hc sinh lp chuyn Ho hc bc THPT gip hc tr hc tt hn v chun b tt hn cho cc k thi hc sinh gii Ha hc c v l thuyt bi tpphng php gii, gp phn nng cao cht lng ging dy v hc tp mn Ha hc.III. NI DUNG

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 tc gia 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 hp kh th: 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 cng


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TI LIU LUYN THI & BI DNG KIN THC NM 2011


chm qua lin tip cc trng thi cn bng th s bin i ny c gi l thun nghch. y l s bin i ltng 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 thiP1.V1 = n.RT1 ; P2.V2 = n.R.T2

6) 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 V1n V2 t

o = const trong 1 xilanh kn nh 1 pittngc 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 Pn nhng lng v cng b th tch kh tng nhng lng vcng 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 cngny ng bng lng cng cn thit a h v trng thi ban u mt cch thun nghch.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 t trong h.- U l i lng dung v l hm trng thi- 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 = 0U = QV QV l 1 hm trng thi.

* Nhit ng p:Nu h bin i P = const th:

dVP.

2

1= P . dV

2

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


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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 b mol Bto 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 p QP = H

- Nu phn ng c thc hin V = const th nhit phn ng c gi l nhit phn ng ng tch QV=U* Quan h gia QP v QV

QP = H = (U + PV)P = U + P. V H = U + P . V = U + n .RTQP = 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 1

otrong iu kin

ng p (m trong qu trnh khng c s bin i trng thi).

* Tng t vi CV: H = 2

1

.T

T

P dTC ; U = 2

1

.T

T

T dTC

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 thi cuica h m khng ph thuc vo qu trnh bin i l thun nghch hay khng thun nghch.9) NH LUT HESS:H (U) ca 1 qu trnh ch ph thuc vo trng thi u v trng thi cui ca hm khng ph thuc vo ng i.

Hp = Hs (sn phm) - Hs (cht u) = Hc (cht u) - Hc (sn phm)10) NH LUT KIRCHHOFF:

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

PP dTCnCn DC


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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

QVi 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.ln 1

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 ttin trn ta c th tnh c entropi tuyt i ca cc cht cc nhit khc nhau.

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

1iiS

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 c

gi l gi tr entropi chun, k hiu l S0T, thng T = 298K S0298

17) 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= S

0298(sp) - S

0298(t/g)

+ V S ca cht kh >> cht rn, lng nn nu s mol kh sn phm (sp) > s mol

kh tham gia th S > 0 v ngc li. Cn trong trng hp s mol kh 2 vbng 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 mt nhit lng

Hmt do h to ra Hmt = - H h = - H

S mt = -T

H

+ iu kin t din bin ca h:S c lp = S h -

T

H> 0 H T. S < 0


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+ 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 = 0Trong : F = U TS

V 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. dTCT

T

P.2

1

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

- Entropi: dS T

QTN S 2

1 T

QTN STN =T

dTCn

T

T

P..2

1

= TdCnT

T

P ln..2

1

Nu PC = const STN = n. PC .ln1

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. dTdE


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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)

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

2T

H

T

T

G

P

dTTH

TGd

T

T

T

G

T

G

T

T

..2

1

2

2

1

2

1

dTT

GTG

TG

T

T

TT .2

1

12

212

Nu coi Ho khng ph thuc vo nhit th:

298

11

298298

TH

G

T

G oo

T

b) Sph 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) = G

oT + 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.ln1

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


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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 thi u, trngthi cui, khng ph thuc vo qu trnh bin thin.

e) G ca phn ng ho hc: Gop = GoS(sn phm) - G

oS(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 0

oC, 1atm. Tnh Q, W, U, H trong cc bin i sau y c tin hnh thun nghchnhit 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 bng29,1J/mol.KGii

a) V = const W = 0.dVP U = 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

100l 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

2V

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. V1

1 = 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) = 28100

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

H = n PC .T = 28100

.29,1(189,33 273,15) = - 8711,3(J)

Bi 2:

Tnh oSH 298, ca Cl-(aq). Bit:

(a):

2

1H2 +

2

1Cl2(k) HCl(k) oSH 298, = -92,2(kJ)

(b): HCl(k) + aq H+

(aq) + Cl-(aq)

o

SH 298, = -75,13(kJ)


11/31



(c):2

1H2 + aq H

+(aq) + e

oSH 298, = 0

Gii:

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

1Cl2 + e + aq = Cl

-(aq)

o

SH 298, = - 167,33(kJ)

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) Fe

3+(aq) NO3

-(aq) NO(k) H2O(l)

o

SH 298, (kJ/mol) -87,86 - 47,7 -206,57 90,25 -285,6

Gii:Phng trnh ion ca phn ng:

3Fe2+(aq) + 4H+

(aq) + NO3-(aq) 3Fe

3+(aq) + NO(k) + 2H2O (l)

H=3. oSH 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 khng i,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 )( oH = 6.

oCOSH )( 2 + 6.

oOHSH )( 2 -

oSH )(

oSH )( -o

SH )( =oH -

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

oSH )( 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 entanpi tiu chunca C4H10(k). Bin lun v kt qu thu c. Cho bit entanpi thng hoa ca than ch v nng lng lin ktH- 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 kJ

Ly (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 C2H6nh nhau th:E C- C = =1800,75 6(- 412,175) = -327,7(kJ/mol)

* Coi EC-H; EC- C trong cc cht CH4, C2H6, C4H10u nh nhau th:o

HCtungSH104

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


13/31



* (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 Elk(C C), Elk(C- H) trong mi trng hp l nh nhau. V v vy

s khng tnh r c oSH 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 bit o 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) oH298 = -296,6 kJ

(4): 2SO2(k) + O2(k) 2SO3(k)o

H298 = -195,96 kJT kt qu thu c v kh nng din bin thc t ca 2 phn ng trn c th rt ra kt lun g?Gii:

1) opuH )1( =o

OAlS rH

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

2) opuH )2( =o

puH )3( + 21 o

puH )4( = -296,6 - 21

.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. Nh vy,

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) oH298 = -393,1(kJ)

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

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

2) Khi cho 32,69g Zn tc dng vi dung dch H2SO4 long d trong bom nhit lng k 25oC, 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 = HoPhn 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)


14/31



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

Cho bit o 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

1N2(k) +

2

3H2(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. Bit

o

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

),( 3 kNHPC = 24,7 + 37,48.10-3 T Jmol-1K-1

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

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

Gii:

2

1N2(k) +

2

3H2(k) NH3(k) o kNHSH ,, 3 = - 46,2kJ/mol

CP = ),( 3 kNHPC - 21

),( 2 kNPC -

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

oH1000oH298 + dTCP,

1000

298 = oH298 + dTT)10.541,311,32(

1000

298

3

= oH298 + )2

10.541,311,32(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


15/31



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/mol

Trong 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. ACl

Uml = 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+ -

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 m t htcc qu trnh din ra trong dung dch, cc ion nht l cation t nhiu cn c tng tc ln nhau hoc tngtc 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 ni trn?Gii:a) T = const U = 0; H = 0U = Q + W = 0 Q = -W

W = - 2

1

2

1

.VdVnRTdVP = - nRTln

1

2

VV


16/31



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 = 2

3.

.

.

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 .P2

P1

1-

101

1-

=CPCV

CV + R

CV

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

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.1 101

12 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 25o

C 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 chuyn thnhin nng v sn ra cng W = 109,622 kJ.


17/31


18/31



Bi 16:Tnh nhit lng cn thit nng nhit ca 0,5 mol H2O t -50

oC 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 100

oC 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 H2O 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;o

PC ca hi nc l

(30,2 + 10-2T) J/molKGii:

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(10373

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 h l clp 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:

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 + T

dT.3,75.

18

400318

333

=18

200.75,3 ln

288

318+

18

400.75,3 ln

333

318= 5,78 (J/K) > 0

Qu trnh san bng nhit ny t xy ra.


19/31



Bi 19: Tnh s bin thin entropi v G ca s hnh thnh 1 mol hn hp kh l tng gm 20% N2; 50%H2v 30%NH3 theo th tch. Bit rng hn hp kh c to thnh do s khuch tn 3 kh vo nhau bng cchni 3 bnh ng 3 kh thng vi nhau. Nhit v p sut ca cc kh lc u u 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) l2N

V = 0,2V;3NH

V =

0,3V;2H

V = 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

V

V

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


20/31



Gp- = Go - Go - Go

o

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

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

Gp-(298) = -8,13kJ < 0 Phn ng xy ra theo chiu thuno

c)

oS298,p- = S

o - So - So298(H2O)

298(C2H4)298(C

2H

5OH)

= 282 - 219,45 - 188,72 = - 126,17(J/K)

G = H - T. S

H298,p- = G298,p- + T. S298,p-

= -8,13 + 298(- 126,17 .10-3) = - 45,72866(kJ)

H298,p- < 0 phn ng to nhit

o

o

o o

Bi 22: Mt mol kh l tng n nguyn t 300K v 15atm gin n ti p sut 1atm. S gin nc thc hin bng con ng:a) ng nhit v thun nghch nhit ng.

b) ng nhit v khng thun nghch.c) on nhit v thun nghch.d) on nhit bt thun nghch.

Trong cc qu trnh bt thun nghch, s gin n chng li p sut 1atm. Tnh Q, W, U, H, Stp cho mitrng hp.Gii:a) T = const U = 0 ; H = 0

WTN = -

2

1PdV = - nRTln 1

2

V

V

= -nRTln 2

1

P

P

WTN = -1(mol).8,314 (J.mol-1K-1) .ln

1

15.300(K) = -6754,42(J)

Q = -W = 6754,42(J)

Qu trnh gin n thun nghch: Stp = Smt + Sh = 0b) T = const U = 0 ; H = 0

WBTN = -Png(V2 - V1) = -P2(2P

nRT-

1P

nRT) = nRT(

1

2

P

P- 1) = 1. 8,314.300.(

15

1- 1)= -2327,92(J)

QBTN = -W = 2327,92(J)Stp = Smt + Sh

Sh(BTN) = Sh(TN) =T

QTN =

T

WU2

1 T

dTnCV + dV

V

nR2

1

= nRln1

2

V

V= nRln

2

1

P

P

= 1.8,314 .ln1

15= 22,515(J/K)

Smt =T

Qmt = -T

Qhe =300

92,2327= -7,76(J/K) Stp = 22,515 - 7,76 = 14,755 (J/K)

( Qu trnh gin n ny t xy ra)

c) on nhit Q = 0on nhit thun nghch Theo Poisson T.V- 1 = const


21/31



M V =P

nRT T.

1

P

nRT= const T .P1- = const

T1 .P1 = T2 .P2

T1

T2= P2

P1

T1

T2

P2

P1=

1- 1-

1-1-

T2 = T1 .P2

P1

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)

535

3

15

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

WU2

1

T

T

VT

dTnC + dV

T

PV

V

2

1

= 2

1

lnT

T

V TdnC + dVV

nRV

V

2

1


22/31



= nCVln1

2

T

T+ nRln

1

2

V

V= 1.

2

3 .8,314.ln300

188 + 1. 8,314 .ln64,1

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

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 00298S (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:

0puH = 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 = H

0 - 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:

02000,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 = 02000G + T.

02000S = -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 hu c cho

rng: H = U + P. V. S t chy trong bom nhit lng k lm cho V = 0, do H = U. Kt lunny sai u?Gii:H = U + P.V H = U + (PV) = U + P. V + V. P

Hay 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:


23/31



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) 0TG =0TH - T.

0TS

Vi 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.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


24/31



2) Tnh nhit, cng th tch, cng phi th tch (tc l cng hu ch) m h trao i vi mi trng trong micch?

3) Tnh S ca mi trng v S tng cng ca v tr khi tin hnh qu trnh theo mi cch.4) Mt m hnh t bo in ho khc lm vic da trn phn ng oxi ho C3H8(k) bi O2(k) khi c mt dungdch KOH 5M vi in cc Pt. Cc loi phn t v ion (tr KOH) u trng thi tiu chun. Hy vit ccna phn ng catot v anot v phn ng tng cng trong t bo in ho. Nu t t bo in ho ,

25o

C, ta thu c dng in 100mA. Hy tnh cng sut cc i c th t c.Gii:

C3H8(k) + 5O2(k) 3CO2(k) + 4H2O(l)

1) Do cc hm H, U, S, G l hm trng thi nn d tin hnh theo cch no th cc gi tr U, H, S, Gcng nh nhau vi cng trng thi u v cui. Vy:

0puH = 3

0)(, 2 kCOS

H + 4. 0 )(, 2 lOHSH -0

)(, 83 kHCSH - 5. 0 )(, 2 kOSH

= -3. 393,51 - 285,83 .4 + 103,85 = -2220 (kJ)0puH = 213,74. 3 + 4.69,91 - 269,91 - 5. 205,138 = -374,74 (J/K)0

puG = H

0

- T. S

0

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


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

0puG =

0puH = T.

0puS = -2548,44 + 298,15.912,43.10

-3 = - 2276,399(KJ)

0puE = -nF

G 0=

96485.20

2276399= 1,18(V)

E = E0 -56

323

283 ..][

][lg

20

0592,0

OHC PPOH

CO

= 1,18 -20

0592,0lg(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 t0 = 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 =T

Q=

15,273150

4,418

= - 0,989(J/K)

2) H bin thin on nhit t 150oC n 50oC

S2 = 03) 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 vi H2O(l))

- 15oC, 1mol H2O1,428 Torr

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

1,215 Torr

(1), (3) l qu trnh chuyn pha thun nghch

G1 = 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 25

oC, 1atm1 mol O2 25

oC trong khng kh trn mt t (P = 1atm, O2 chim 21% V khng kh)

- So snh gi tr hm G ca 1 mol O2trong 3 trng hp trn hn km nhau bao nhiu J?. T rt ra ktlun: Khnng phn ng ca O2 trong mi trng hp trn cao hay thp hn so vi trng hp khc?Gii:


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* G0 l hm Gibb ca 1 mol O2 1atm

- 1 mol O2, 1atm, 25oC 1 mol O2, 2atm, 25

oC(G0) (G1)

G1 = G1 - G0 = nRTln

1

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, 25

oC, 1atm 1 mol O2, 25oC, 0,21atm

(G0) (G2)

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

1

21,0 = -3868,8(J)

G2 < G0.

Vy:G2(1mol O2, 25

oC, 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 cao nht 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 phenol trong148,69g clorofom l -172J.Tnh nhit pha long i vi dung dch c nng nh dung dch th 2 cha 1 mol phenol khi pha longn nng ca dung dch th nht bng clorofom.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)0,672

941,569

94

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 H298tnh c theo l thuyt sai khc vi gi tr TN l 0,48%.


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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 =TQ3 =

1

2

.. )(

T

T

hPTdTCn = nCP(h)ln

1

2

TT = 3. 33,58.ln

15,37315,298 = - 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

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 54oC.Gii:V S, G l cc hm trng thi nn S, G khng ph thuc vo qu trnh bin thin l thun nghch hay btthun 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 N2 298K v p

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


28/31



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

5(l)

S = S(N2) + S(O2) =2N

n .Rln1

2

V

V+

2On Rln

1

2

V

V=

RT

VP NN 22 . Rln5,2

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 298K

Cht 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 chc kim cng 298K?

Gii1)

Ckim c-ng 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)

G

o

< 0 (Tuy nhin G

o

khng qu m) Phn ng t xy ra theo chiu thun khng tn ti mt lng nh kim cng cng vi than ch.2)

G0 = +2874,162 (J)298

Cthan ch Ckim c-ng

V = VKC - VTC = 3,416.10-6 - 5,31.10-6 = 1,894.10-6 (m

3/mol)Ta c: dG = VdP - SdT

TP

G

= V TP

G

= V

2PG - 1PG = V(P2 - P1) iu ch c kim cng t than ch th:

2PG 0

1P

G + V(P2 - P1) 0

P2 - P1 -V

GP

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


29/31



15000 atm. p sut cao hn qu trnh chuyn than ch thnh kim cng l t din bin, mc du vi tc rt chm. Mun tng tc phi tng nhit v p sut, trong thc t qu trnh chuyn than ch thnh kim

cng c tin hnh khi c xc tc (Ni + Cr + ) nhit trn 1500oC v P 50000atm.

Bi 39: Phn ng gia Zn v dd CuSO4 xy ra trong ng nghim to ra lng nhit 230,736kJ. Cngphn ng trn cho xy ra trong pin in th mt phn ho nng chuyn thnh in nng. Cng in ca pin l

210,672kJ. Chng minh rng: U ca 2 qu trnh khng i, nhng nhit to ra thay i. Tnh S ca phnng, 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 = 0UBTN = 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 =T

Hmt = -T

Hhe S v tr = -T

Hhe + 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 = 0Hy 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 H

mt= - H

h

* Trong pin: Wmax = G < 0 nn HmtH h.Bi 41: Xt phn ng: Zn(r) + Cu

2+(aq) Zn

2+(aq) + Cu(r)


30/31



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) Cu

2+(aq)

0298,SH (kJ/mol) -152,4 0 0 64,39

0298S (J/mol.K) - 106,5 41,6 33,3 - 98,7

b) 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? TnhEpin?Gii:

a) 0puH =0

, 2

ZnSH + 0,CuSH -

0,ZnSH

0

, 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)

0puG =

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

Uo = QP = 0puH = -216,79 (kJ)

W = 0; qu trnh BTN; W = 0b) * 0puG = -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, U0khng thay i do H, S,G, U l cc hm trng thi nn khng ph thuc qu trnh bin i l thun nghch hay bt thun nghchnhng 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

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.


31/31

TT LUYN THI & BI DNG KIN THC NGY MI 18A/88INH VN T - TP. HI DNGSc 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 bnh cu

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 l thc hin

qu trnh mt cch bt thun nghch th G, S ca phn ng vn nh cu (a). Do G, S l 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 kh O2ccoi 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 = 3,3.10

4 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. N tngi ph hp vi yu cu v mc ch ging dy, bi dng hc sinh kh, gii trng chuyn chun b dthi hc sinh gii cc cp . N c th dng lm ti liu hc tp cho hc sinh cc lp chuyn Ho hc v tiliu tham kho cho cc thy c gio trong ging dy v bi dng hc sinh 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 gia vo cck 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 ngy cng c nhiu ti liu hctp mt cch h thng hn.

2011 -xÂy dỰng hỆ thỐng cÂu hỎi vÀ bÀi tẬp phẦn “nhỆt hoÁ hỌc”

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