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

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

Trng THPT chuyn Nguyn Hu - H NiI. M U

Trong qu trnh ging dy trng ph thng nhim v pht trin t duy cho hc sinh l nhim v rt quan trng, i hi tin hnh ng b cc mn, trong Ha hc l mn khoa hc thc nghim cp n 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 rng kin thc hc mt cch sinh ng, phong ph m cn thng qua n tp, rn luyn mt s k nng cn thit v ha hc, rn luyn tnh tch cc, t lc, tr thng minh sng to cho hc sinh, gip hc sinh hng th trong hc tp. Qua bi tp ha hc gio vin kim tra, nh gi vic nm vng kin thc v k nng ha hc ca hc sinh.

gio vin bi dng 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 li mt s cu hi v bi tp theo mt s chuyn , trong c phn dng luyn tp cho hc sinh phn Nhit ho hc

II. 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 tp phng 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)

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

chm 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 = - (Pn : p sut bn ngoi)

* Nu s bin i l BTN th Pn = Pkq = const

WBTN = - Pkq . = - 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 v cng b. Khi Pn mi lc thc t = P bn trong xi lanh = Pk Pn = Pk = n.RT/V

( WTN = - = - nRT . = - nRT .ln ( 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 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 ( = Q - ( (U = Q - * 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:

= P . = 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 b mol 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 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 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

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

* Tng t vi CV: (H = ; (U = CP, CV l hm ca nhit .

Vi 1 mol kh l tng: CP = ; CV =

M (U = (H - P. (V ( CP = = + = CV + R

Q, W: Khng phi l hm trng thi

QV = (U; QP = (H ( QV, QP l hm trng thi ( ch ph thuc vo trng thi u v trng thi cui ca 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 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:

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

(Ha = = - (Hb =

( (H2 = (H1 + = (H1 + - (H1 thng c xc nh iu kin chun: (HoT = (Ho298 + 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 = const

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

- Nu s bin i l bt thun nghch th dS > - 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 = ((STN = (SBTN)

12) NGUYN L II CA NHIT NG HC:

dS ( - 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 ti khi 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).

( S l 1 i lng dung .

13) S BIN THIN S TRONG QU TRNH BIN I TRNG THI CA CHT:

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

T = const ( (S = = (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 V1 ( 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. ln ( = -(- P. (V) = ).

T = const ((S = = nRln= n.R.ln15) 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 =

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

(S = * Trong khong nhit hp, coi CP = const ( (S = n.CP.ln- Qu trnh: V = const ( (S = n .CV.ln16) 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 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 = ( ST = 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 ( 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 mol

kh 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 0

a)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 = -+ iu kin t din bin ca h:

( (S c lp = (S 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 < 0

V 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 = - ( iu kin t din bin ca h trong qu trnh ng nhit, ng tch l

(F = (U T. (S < 0

V t trng thi cn bng khi (F = 0

Trong : 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 = Wmax

TM LI :

* Qu trnh ng p: P = const

- Cng: (WP = - P.dV = -n.R.dT ( WP = - P. (V = - nR(T

- Nhit: (QP = dH = n..dT ( QP = (H = n.

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

- Entropi: dS ( (S ( (STN = = Nu = const ( (STN = n. .ln* Qu trnh ng tch:

- Cng: (WV = - P.dV = 0 ( WV = 0

- Nhit: (QV = dUV = n. .dT ( QV = (UV = Nu = const (QV = n. .(T

- Ni nng: (UV = QV + W

- Entropi: (S = ( (S n. .ln ( = 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 = - ( WT = -- Nhit: (UT = QT + WT = 0 ( QT = - WT = nRT ln- Ni nng: (UT = 0

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

- Entropi: (S TN = hoc = * Vi qu trnh dn n kh l tng thun nghch

(S = Nu CV = const ( (S = n. + nRT lnV T = const ((S = nRT ln = nRT.ln* Qu trnh on nhit:

- Nhit: Q = 0

- Ni nng v cng: dU = (Q + (W = (W = -PdV = +Qu trnh bt thun nghch:

dUBTN = (WBTN = -Png .dV = -P2.dV

(UBTN = WBTN = -Png.(V2 V1) = n.CV. (T

* PT Poisson: (Dng cho qu trnh thun nghch)

T .= const

P.V( = const

* WBTN = -P2(V2 V1) = - P2.( ( T2 ( (U = W = .... ( V2

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

T1. V= T2 . V ( T2 = T1.()(-1- Entanpi: (H = n .CP(T2 T1)

- Entropi: (STN = = 0

* G = H TS = U + PV TS

- (S ; - (V

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

= - nF. = - (S

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

V (W = (W + (-PdV)

(Q T.dS ( dG (W + VdP SdT

Du = 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 max ( (G = W max

* i vi qu trnh BTN: W gim; Q tng khi hon ton BTN ( W = 0

20) 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 (= - S ( = - (S

( (G = (H T. (S = (H + T. ( T. - (G = -(H ( ( ( ( Nu coi (Ho khng ph thuc vo nhit th:

b) S ph thuc vo P:

Khi T = const (

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

- Vi cht kh l tng ( V =

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 = nRT.lnb) Trn ln ng nhit, ng p 2 kh l tng:

(G = nA.RTlnxA + nB.RTlnxBc) Qu trnh chuyn pha thun nghch (ti nhit chuyn pha): (Gcf = 0

d) Qu trnh chuyn pha thun nghch T ( TcfNguyn tc: p dng chu trnh nhit ng. V G l hm trng thi nn (G ch ph thuc trng thi u, 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)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 = 200l

d) Gin on nhit ti V = 200l

Chp nhn rng N2 l kh l tng v nhit dung ng p khng i trong qu trnh th nghim v bng 29,1J/mol.K

Giia) V = const ( W = (U = QV = n.(T = (- R).(T2 T1) .n

= (- R).(-1).T1.n = (29,1 - 8,314).(.273,15 = 14194,04(J)

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

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

QP = (H = .n .(T = = 29,1. (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 = - = - nRT.lnW = - .8,314 .273,15.ln= -7431,67(J)

(U = Q + W = 0

( Q = -W = 7431,67(J)

d) Q = 0 (S = const)

Theo PT poisson: T1. V= T2 . V ( T2 = T1.()(-1

Vi ( W = (U = n.(T2 T1) = (29,1-8,314).(189,33 -273,15) = 6222,4(J)

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

Bi 2:

Tnh ca Cl-(aq). Bit:

(a): H2 + Cl2(k) ( HCl(k) = -92,2(kJ)

(b): HCl(k) + aq ( H+(aq) + Cl-(aq) = -75,13(kJ)

(c): H2 + aq ( H+(aq) + e = 0

Gii:

Ly (a) + (b) (c) : Cl2 + e + aq = Cl-(aq) = - 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) Fe3+(aq) NO3-(aq) NO(k) H2O(l)(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) ( 3Fe3+(aq) + NO(k) + 2H2O (l)

(H=3.(Fe3+,aq)+(NO)+2.(H2O(l))3.(Fe2+,aq)- (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.RT

Phn ng trn c: (n = 1-2 = -1 ( (H = (U RT ( (H < (U

2) C6H12O6 + 6O2 ( 6CO2 + 6H2O

(U(() = (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)

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

( < ( Dng ( - glucoz c nh hn nn bn hn.

Bi 5:

1) Th no l entanpi sinh ca n cht? Tnh v v d on hot tnh ho hc ca chng t cc d kin sau:

(a): C than ch + O2(k) ( CO2(k) = -393,14(kJ)

(b): C kim cng + O2(k) ( CO2(k) = -395,03(kJ)

(c): 3As2O3 (r) + 3O2(k) ( 3As2O5(r) = -811,34(kJ)

(d): 3As2O3 (r) + 2O3(k) ( 3As2O5(r) = -1090,98(kJ)

2) T kt qu trn v cc d kin sau:

(H(O O) tnh t O2 = - 493,24kJ/mol; (H(O O) tnh t H2O2 = - 137,94kJ/mol

Chng minh rng: Khng th gn cho O3 cu trc vng kn.

Gii:

1)- Entanpi sinh ca cc n cht bn iu kin chun = 0

- i vi nhng nguyn t c nhiu dng th hnh th khi chuyn t dng ny sang dng khc bao gi cng km theo 1 hiu ng nhit. Hiu ng nhit ca qu trnh hnh thnh n cht (dng th hnh km bn) t dng th hnh bn iu kin nht nh c coi l nhit sinh ca n cht trong nhng iu kin .

( ca O3 v kim cng l hiu ng nhit quy v iu kin chun ca cc qu trnh:

O2 (O3(k) ; C than ch ( C kim cng - Ly (a) (b):

C than ch ( C kim cng = 1,89 kJ/mol

- Ly (c) - (d):

O2(k) (O3(k) = 139,82 kJ/mol

2) Nu coi O3 c cu trc vng kn:

( = -

= 3.(-137,94) - .(- 493,24) = 326,04 (kJ/mol) > 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 chun ca C4H10 (k). Bin lun v kt qu thu c. Cho bit entanpi thng hoa ca than ch v nng lng lin kt H- H ln lt bng: 710,6 v - 431,65 kJ/mol.

Gii:

* (1) C than ch + 2H2 (k) ( CH4(k) =-74,8kJ

(2) C than ch ( C (k) = 710,6 kJ

(3) H2 (k) ( 2H (k) = 431,65 kJ

Ly (1) [(2) + 2.(3)] ta c:

C(k) + 4H(k) ( CH4(k) = -1648,7(kJ/mol)

( Nng lng lin kt trung bnh ca lin kt C H l: (-1648,7) = - 412,175 (J/mol).

* (4) 2C than ch + 3H2 ( C2H6(k) = -84,6 (kJ/mol)

Ly (4) [2 .(2) + 3.(3)] ta c:

2C(k) + 6H (k) ( C2H6 (k) = -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:

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

* (5) 4C(k) + 10 H(k) ( C4H10 (k) = -5110,6 (kJ/mol)

Ly (2). 4 + (3).5 + (5) ta c:

4Cthan ch + 5H2(k) ( C4H10(k) = -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 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 = -822,2 kJ/mol; = -1676 (kJ/mol)

2) S(r) + O2(k) ( SO3(k) (2)

Bit (3) : S(r) + O2(k) ( SO2(k) = -296,6 kJ

(4): 2SO2(k) + O2(k) ( 2SO3(k) = -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:

1) = - = -1676 + 822,2 = - 853,8(kJ)

2) = + = -296,6 - .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) = 28,17 (kJ)

b) Cthan ch + O2(k) ( CO2 (k) = -393,1(kJ)

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

d) 2SO2(k) + O2(k) ( 2SO3(k) = -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 = 65,38

Gii:

1) (H = (U + (n.RT

Do 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. = -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 bit = - 74,8 (kJ/mol); = -46,1kJ/mol; = 91,1 kJ/mol

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

(a) : Cgr + 2H2(k) ( CH4 = -74,8 (kJ/mol)

(b) : N2(k) + H2(k) ( NH3(k) = - 46,1kJ/mol

(c) : 5Cgr + H2(k) + N2(k) ( C5H5N5(r) = 91,1 kJ.mol-1(d) : CH4(k) + NH3(k) (HCN(k) + 3H2(k) (Ho = 251,2 kJ.mol-1Ta 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 = -46,2 kJ.mol-1

= 24,7 + 37,48.10-3 T Jmol-1K-1= 27,8 + 4,184.10-3 T Jmol-1K-1= 286 + 1,17.10-3 T Jmol-1K-1Gii:

N2(k) + H2(k) ( NH3(k) = - 46,2kJ/mol

(CP = - - = - 24,7 + 37,48.10-3T - [27,8 + 4,184.10-3] - [28,6 + 1,17 .10-3T]

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

+ = + = + = - 46,2.103 +31,541 .10-3. (10002 -1982) 32,1(1000 298)= - 54364,183 (J/mol)

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

Q = .(-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/mol

Entanpi phn li ca Cl2: 238,26 kJ/mol

Entanpi thng hoa ca Ba: 192,28 kJ/mol

Nng lng ion ho th nht ca Ba: 500,76 kJ/mol

Nng lng ion ho th hai ca Ba: 961,40 kJ/mol

i 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/mol

Nhit hirat ho ion Cl- : - 363 kJ/mol

Trong cc kt qu thu c, kt qu no ng tin cy hn.

Gii:

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 ht cc qu trnh din ra trong dung dch, cc ion nht l cation t nhiu cn c tng tc ln nhau hoc 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 = R 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 = 0

(U = Q + W = 0 ( Q = -W

W = - = - nRTlnVi kh l tng: P1.V1 = P2 .V2 ( = ( V2 = . V1 = . 10 = 100(l)

( W = -(nRT).ln = -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.. (T = .R(T2 T1)

(U = W = (T2 T1)

Theo PT poisson: T.V(- 1 = const

M V = ( T.= const ( T( .P(- 1 = const

c) Q = 0 (U = W

( n.(T2 T1) = -Png .(V2 V1) = -P2 .( n. R(T2 T1) = -nR.1 ( T2 = 0,64T1

( (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 + Cl2 = 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 chuyn thnh 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 nhit th 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 cch no. V vy (U trong 2 trng hp trn ch l mt.

- V (U = Q + W = Q + W - P(V = Q + W - (n.RT

Do (nRT = const; (U = const

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

- (U = (H - (nRT = -126,566 + . 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 = -WTN = .8,314 .300. ln= -6201,39(J)

*WBTN = - Png . (V = -Png(V2 V1) = -Png= - P2.V1 = - P2. = -.8,314 .300 = -2993,04 (J)

b) WTN = nRTln= .8,314 .300.ln= 6201,39(J)

WBTN = - Png. (V= -Png(V2 V1) = -Png = -nRT.P2 = -nRT = -.8,314 .300 = 14965,2 (J)

KL: - Cng m h thc hin (sinh) trong qu trnh bin thin thun nghch t trng thi 1 n trng thi 2 bng cng m h nhn khi t trng thi 2 v trng thi 1. Cn trong qu trnh bin thin bt thun 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 + O2(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 = = 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.

= 30,2 + 10-2T(J/molK) ; = 35,56(J/molK); = 75,3(J/molK)

Gii:

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

+ 0,5.30,2 .(773 373) + .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 ca nc v nc lng ln lt bng 35,56 v 75,3J/molK; ca hi nc l (30,2 + 10-2T) J/molK

Gii:

= n.=0,5. = 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 c lp v nhit dung mol ca nc lng l 75,3 J/mol.K

Gii:

Gi T l nhit ca h sau khi pha trn.

Do Q thu = Q to nn:

. (T 288) = .(333 T)

T 288 = 2.333 2T ( T = = 318(K)

(S h = (S1 + (S2 = + = .75,3 ln+ .75,3 ln = 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 kh vo nhau bng cch ni 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) l = 0,2V; = 0,3V; = 0,5V.

Do %V = %n ( = 0,2 mol; = 0,5 mol; = 0,3mol.

- S bin thin entropi c tnh theo CT: (S = nRln= 0,2 .8,314.ln= 2,676J/K

= 0,5.8,314.ln= 2,881J/K

= 0,3.8,314.ln= 3,003J/K

( (S = + + = 8,56(J/K)

* Qu trnh khuch tn kh l tng l ng nhit nn (H = 0

( (G 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) + O2(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- < 0 ( (S 0

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

(kJ/mol) - 74,8 - 241,8 -110,5 0

(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:

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

= 3.(130,684) + 197,6 - 188,7 - 186,2 = 214,752 (J/K)

Do (H0, (S0 khng ph thuc vo T nn:

( = (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: < 0 ( (H0 - T. (S0 < 0

( T > = = 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: = -240000 + 6,95T + 12,9TlgT (J/mol)

Tnh (G0, (S0 v (H0 ca phn ng to thnh H2O 2000K

Gii:

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

dG = VdP - SdT ( = -S

(= -= 6,95 + 12,5.lgT + 12,9T.= 6,95 + 12,9lgT + = 6,95 + 12,9lg2000 + = 55,1357(J/molK)

(= ( + T. = -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 lun ny 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:

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) = - T. Vi phn ng ho hc T = const. Nu > 0 ( Phn ng t din bin theo chiu nghch.

= 0 : Phn ng trng thi cn bng.

< 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 > ( (S > 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 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.ln = -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 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 trong 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 pha long n nng ca dung dch th nht bng clorofom.

Gii:

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)

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; = 75,291 (J/K.mol); = 33,58 (J/molK)

Gii:

Xt chu trnh:

(S1 = = = nCP(l)ln= 3. 75,291.ln = 50,6822(J/K)

(S2 = = = 3. = 326,8605(J/K)

(S3 = = = nCP(h)ln= 3. 33,58.ln = - 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. = 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 bt thun nghch m ch ph thuc vo trng thi u v trng thi cui. V vy:

(S = nRln= 2. 8,314.ln= 26,76 (J/K)

T = const ( (H = 0; (U = 0

((G = (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 b vch ngn i.

Gii:

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

(S = (S(N2) + (S(O2) = .Rln+ Rln = Rln+ Rln = = 0,0174(l.at/K) = 0,0174 .101,325 = 1,763 (J/K)

- Qu trnh ng nhit ( (H = 0

( (G = (H - T. (S = - 298 .1,763 = - 525,374 (J)

Bi 38: Cho cc d liu sau y 298K

Cht (kJ/molK) S0(J/molK) V(m3/mol)

Cthan ch 0,00 5,696 5,31.106Ckim cng 1,90 2,427 3,416.10-61) 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?

Gii

1)

(Ho = (Ho than ch - (Ho kim cng = 0 - 1,9 = -1,9 (kJ)

(So = So than ch - Sokim cng = 5,696 - 2,427 = 3,269 (J/K)

( = (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 than ch.

2)

(V = VKC - VTC = 3,416.10-6 - 5,31.10-6 = 1,894.10-6 (m3/mol)

Ta c: dG = VdP - SdT

( = V ( = (V

( - = (V(P2 - P1)

iu ch c kim cng t than ch th: 0

( + (V(P2 - P1) 0

P2 - P1 - (Do (V < 0)

( P2 P1 - = 1 +P2 14977,65 (atm)

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

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. Cng phn 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 phn ng, (Smt v (Stp? Cho T = 300K

Gii:

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 = = -= 66,88(J/K)

(SmtBTN = - = = 769,12(J/K)

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

(Smt(TN) = - = -(Sh ( (Stp(TN) = 0

Bi 40:

- (Gp = WmaxXt 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 = = -( (S v tr = - + (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

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. (V

Q = (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 (Hmt ( (H 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) (kJ/mol) -152,4 0 0 64,39

(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? Tnh Epin?

Gii:

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

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

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

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

W = 0; qu trnh BTN; W = 0

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

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

Gii:

(G = - nEF ( = - nF. = - (S ( (S = nF .

( (H = (G + T. (S = nF.(T. - E)

( (H = 2. 96485 .(298,15.0,000429 - 1,09337) - - 187162,5(J)

Bi 43: Cho phn ng ho hc: Zn + Cu2+ ( Zn2+ + Cu

xy 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

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

a) Tnh hiu ng nhit Q, bin thin Gipx (G 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 = - = n.F . = 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 O2 c coi l l tung nhit 250 C khi:

a. Gin n t 10atm xung 1atm

b. Nn t 1atm n 10atm

p 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. N tng i ph hp vi yu cu v mc ch ging dy, bi dng hc sinh kh, gii trng chuyn chun b d thi hc sinh gii cc cp . N c th dng lm ti liu hc tp cho hc sinh cc lp chuyn Ho hc v ti liu 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 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 ngy cng c nhiu ti liu hc tp mt cch h thng hn.