Nhiet Dong

  • Published on
    04-Jul-2015

  • View
    156

  • Download
    13

Embed Size (px)

Transcript

I.

OLYMPIC HA HC VIT NAM: OLYMPIC HA HC SINH VIN VIT NAM 2003 - BNG A Tnh nhit ca ngn la CO chy trong hai trng hp sau: a) Chy trong oxy tinh khit (20% oxy v 80% nit theo th tch) b) Chy trong oxy tinh khit Cho bit lng oxy va cho phn ng, nhit lc u l 25oC. Entanpi chy ca CO 25oC v 1atm l 283kJ.mol-1. Nhit dung mol chun ca cc cht nh sau: Cop (CO2, k) = 30,5 + 2.10-2T Cop (N2, k) = 27,2 + 4,2.10-3T BI GII:T o a) H 298 + 298 T

(C

o PCO 2

o + 2C PN )dT = 0 T = 2555 K2

o o b) H 298 + C P 298

CO 2

dT = 0 T = 4098 K

OLYMPIC HA HC SINH VIN VIT NAM 2003 - BNG A Cho cc s liu sau 298K: Ag+(dd) N3-(dd) K+(dd) AgN3(r) o -1 G tt(kJ.mol ) 77 348 -283 378 1) Xc nh chiu xy ra ca cc qa trnh sau: Ag+(dd) + N3-(dd) AgN3(r) K+(dd) + N3-(dd) KN3(r) 2) Tnh tch s tan ca cht in li t tan. 3) Hi phn ng g xy ra khi mui KN3 tc dng vi HCl c. BI GII: 1) Ag+(dd) + N3-(dd) AgN3(r) Go = 378 (77 + 348) = -47kJ: Chiu thun. K+(dd) + N3-(dd) KN3(r) Go = 77 (-283 + 348) = 12kJ: Chiu nghch. 2) AgN3 l cht t tan. Gi Ks l tch s tan ca n: 47000 = 8,237 2.303 .8.314 .298 K s = 5,79 .10 9 lg K s =

KN3(r) 77

3)

KN3 + HCl HN3 + KCl HN3 + 3HCl NH4Cl + N2 + Cl2 KN3 + 4HCl NH4Cl + N2 + Cl2 + KCl. OLYMPIC HA HC SINH VIN VIT NAM 2005 - BNG B Mt phn ng dng luyn km theo phng php kh l: ZnS(r) + 3/2O2(k) ZnO(r) + SO2(k) o 1) Tnh H ca phn ng nhit 298K v 1350K, coi nhit dung ca cc cht khng ph thuc vo nhit min nhit nghin cu. 2) Gi thit ZnS nguyn cht. Lng ZnS v khng kh (20% O2 v 80% N2 theo th tch) ly ng t l hp thc bt u 298K s t n nhit no khi ch hp th lng nhit ta ra

do phn ng iu kin chum ti 1350K (lng nhit ny ch dng nng nhit cc cht u) Hi phn ng c duy tr c khng, ngha l khng cn cung cp nhit t bn ngoi, bit rng phn ng trn ch xy ra nhit khng thp hn 1350K? 3) Thc t trong qung sfalerit ngoi ZnS cn cha SiO2. Vy hm lng % ca ZnS trong qung ti thiu phi l bao nhiu phn ng c th t duy tr c? Cho bit entanpi to thnh chun ca cc cht 25oC (kJ.mol-1) Hp cht: ZnO(r) ZnS(r) SO2(k) Hof -347,98 -202,92 -296,90 -1 Nhit dung mol ng p ca cc cht (J.K .mol-1): Hp cht ZnS(r) ZnO(r) SO2(k) O2(k) N2(k) SiO2(r) o Cp 58,05 51,64 51,10 34,24 30,65 72,65 Bit MZnS = 97,42g.mol-1; MSiO2 = 60,10g.mol-1 BI GII: 1) Ho298 = -347,98 296,90 + 202,92 = -441,96kJ Cop = 51,64 + 51,10 58,05 3/2.34,24 = -6,67J.K-1 H1350 = -448976,84J C Po = C po ( ZnS ) + 3 2 C po(O2 ) + 6C op ( N 2 ) = 293,31JK 1 T 2) o H 1350 + 293,31dT = 0 T = 1829 K298

T = 1829K > 1350K nn phn ng t duy tr c. 3) Gi x l s mol SiO2 c trong 1 mol ZnS

C

o p

o o = C o ( ZnS) + 3 C p (O2 ) + 6C p ( N 2 ) + xC o ( SiO2 ) = 293,31+ 72,65x( JK 1 ) p p 21350

448976 ,84 +

298

1350

293 ,31dT +

298

72 ,65 xdT

= 0 x = 1,84 mol

%ZnS = 47% K THI CHN HC SINH GII QUC GIA NM 2002 (Bng A) Kh NO kt hp vi hi Br2 to ra mt kh duy nht trong phn t c 3 nguyn t. 1. Vit phng trnh phn ng xy ra. 2. Bit phn ng trn thu nhit, ti 25oC c Kp = 116,6. Hy tnh Kp (ghi r n v) ti 0oC ; 50oC. Gi thit rng t s gia hai tr s hng s cn bng ti 0oC vi 25oC hay 25oC vi 50oC u bng 1,54. 3. Xt ti 25oC, cn bng ho hc c thit lp. Cn bng s chuyn dch nh th no? Nu: a) Tng lng kh NO. b) Gim lng hi Br2. c) Gim nhit . d) Thm kh N2 vo h m: - Th tch bnh phn ng khng i (V = const) - p sut chung ca h khng i (P = const). BI GII:

2 NO(k) + Br2 (hi) 2 NOBr (k) ; H > 0 Phn ng pha kh, c n = -1 n v Kp l atm-1 2. Do phn ng thu nhit nn c lin h 1.

(1) (2)

Kp ti O2 < Kp ti 252 < Kp ti 502 (3) -1 Vy : Kp ti 250 = 1 / 1,54 x Kp ti 252 = 116,6 / 1,54 = 75,71 (atm ) Kp ti 252 = 1,54 x Kp ti 252 = 116,6 x 1,54 179, 56 (atm-1) 3. Xt s chuyn di cn bng ho hc taji 25OC.Trng hp a v b: v nguyn tc cn xt t s:

Q =

Sau so snh tr s Kp vi Q kt lun.

PNOBr (PNO)2

(4)

(Khi thm NO hay Br2)

Tuy nhin, y khng c iu kin xt (4); do xt theo nguyn l L satlie. a. Nu tng lng NO, CBHH chuyn di sang phi. b. Nu gim lng Br2, CBHH chuyn di sang tri. c. Theo nguyn l Lsatlie, s gim nhit lm cho CBHH chuyn di sang tri, chng li s gim nhit . d. Thm N2 l kh tr. + Nu V = const: khng nh hng ti CBHH v N2 khng gy nh hng no ln h (theo nh ngha p sut ring phn).+ Nu P = const ta xt lin h.

Nu cha c N2: P = pNO + pBr2 + pNOBr (a) Nu c thm N2: P = pNO + pBr2 + pNOBr + Pn2 (b) V P = const nn pi' < pi Lc ta xt Q theo (4) lin h / tng quan vi Kp: 1. Nu Q = Kp: khng nh hng 2. Nu Q > Kp : CBHH chuyn di sang tri, Q gim ti tr s Kp. 3. Nu Q 0; Ssur = 0. b) Ssys < 0; Ssur > 0. c) Ssys > 0; Ssur < 0. d) Ssys = 0; Ssur = 0 Tnh Ssys, gi s CO2 l kh l tng. Tnh Ssur. 4) Tnh s chuyn i entropy ca h. Nguyn l hai c c nghim ng hay khng? 5) Kh CO c s dng rng ri trong tng hp hu c, c th thu c bng phn ng gia CO2 vi graphit. S dng cc d kin di y chng minh Kcb 298,15K b hn n v. 298,15K: CO2(k): Hos = -393,51kJ.mol-1; So = 213,79JK-1mol-1. CO(k): Hos = -110,53kJ.mol-1; So = 197,66JK-1mol-1. C(gr): So = 5,74JK-1mol-1. 6) Tnh nhit m phn ng t cn bng. B qua s ph thuc ca H, S vo nhit . 7) Phn ng cau 6 xy ra 800oC v p sut chung l 5,0 bar, Kp = 10. Tnh p sut ring phn ca CO ti cn bng. BI GII: 1) c 2) Ssys = nRlnVc/V = 27,4JK-1. p V q = 6,94 JK 1 3) S = = ext T T pext: p sut ngoi. 4) Sh = Ssys + Ssur = 20,5JK-1. Ngyn l hai vn c nghim ng. Ho = 172,45kJ.mol-1. So = 176JK-1mol-1. Go = Ho - TSo = 120kJ.mol-1 > 0 K < 1. o G = 0 khi Ho =TSo T = 980K CO2(k) + C(gr) 2CO(k). 1- 2 1 2 .5 .5 1 + 1 + Kp = p2(CO)/p(CO2) pCO = 3,7 bar. OLYMPIC HA HC QUC T 2004: Chun b cho ln sinh nht th 18 vo thng hai ca mnh, Peter c iinh bin ci tp lu ca cha m trong vn thnh mt b bi vi mt bi bin nhn to. c th c lng ga c ca vic cung nhit v nc cho nh. Peter nhn c d liu v cc kh thin nhin v gi c ca n: Cht ha hc phn mol (x) fHo(kJ.mol-1) So(Jmol-1K-1) Cop(Jmol-1K-1)

a)

CO2(k) 0,0024 -393,5 213,8 37,1 N2(k) 0,0134 0,0 191,6 29,1 CH4(k) 0,9732 -74,6 186,3 35,7 C2H6(k) 0,0110 -84,0 229,2 52,5 H2O(l) -285,8 70,0 75,3 H2O(k) -241,8 188,8 33,6 O2(k) 0,0 205,2 29,4 1) Vit phn ng chy ca kh thin nhin (ch yu l metan v etan) cho bit nit khng chy trong iu kin chn. Tnh H; S v G ca tng phn ng iu kin chun (1,013.105Pa v 25,0oC). Bit tt c cc sn phm dng kh v 0oC = 273,15K. 2) T khi ca kh thin nhin l 0,740gL-1 (1,013.105Pa; 25oC) a) Tnh hm lng ca metan v etan (mol) trong 1,00m3 kh thin nhin (CH4, C2H6 khng phi kh l tng). b) Tnh thiu nhit khi t chy 1,00m3 kh thin nhin iu kin chun. Gi thit tt c cc sn phm u dng hi (nu khng lm c cu a ta c th gi s rng trong 1,00m3 kh thin nhin ng 40,00mol). c) Theo PUC (Public utility company) th nng lng thu c l 9981kWh/m3 khi ta t chy kh thin nhin (nu sn phm ch yu l kh). Sai s so vi cu b l bao nhiu. B bi trong nh rng 3,00m, di 5,00m v su 1,50m (thp hn mt sn). Nhit nc trong vi l 8,00oC v nhit phng l 10,00oC. Gi s dnc = 1,00kg.L-1 v kh trong phng l kh l tng. Bit th tch phng l 480m3 v din tch phng l 228,16m2. 3) Tnh nng lng (MJ) khi a nhit ca nc ln 22,00oC v a khng kh trong phng ln 30,00oC. Cho bit trong khng kh cha 21,0%O2, 79%N2 v p = 1,013.105Pa. Vo thng hai, nhit bn ngoi min Bc c xp x 5oC. T khi tng b tong v mi nh tr nn mng hn (20,0cm) th nng lng mt i nhiu hn. Nng lng c gii phng ra mi trng xung quanh (khng k n nc ngm). dn in ca tng v mi nh l 1,00WK-1m-1. Nng lng c tnh theo cng thc: J = E(A.t)-1 = w.T.d-1. J: Nng lng gii phng. d: dy ca bc tng A: din tch t: chnh lch thi gian t thi im bt u nh thi im kt thc. T: chnh lch nhit gia bn trong v bn ngoi phng. E: Nng lng trong phng. 4) Tnh nng lng cn thit (MJ) gi cho nhit trong phng l 30,0oC. 1,00m3 kh thin nhin c gi 0,40euro v 1kWh in gi 0,137euro. Tin thu trang thit b t nng kh l 150euro trong tin thu l si in l 100euro. 5) Ton b nng lng cn thit cho k hoch b bi ma ng ca Peter l bao nhiu?. Lng kh thin nhin cn l bao nhiu bit hiu sut ca my t nng kh l 90%. Peter phi tn bao nhiu tin khi dung my t nng kh v l si in bit hiu sut ca l si l 100%. BI GII: 1) Phng trnh phn ng: Metan: CH4 + 2O2 CO2 + 2H2O Ho = -802,5kJ.mol-1.

b)

2)

4)

So = -5,3J.mol-1.K-1. Go = -800,9kJ.mol-1. Etan: 2C2H6 + 7O2 4CO2 + 6H2O Ho = -2856,8kJ.mol-1. So = +93,2J.mol-1.K-1. Go = -2884,6kJ.mol-1. a) m = .V = 740g Mtb = x(i).M(i) =0,0024.44,01+0,0134.28,02+0,9732.16,05+0,011.30,08 = 16,43g.mol-1. ntng = 45,04mol. n(CH4) = 43,83mol. n(C2H6) = 0,495mol. b)Ethiu nhit (H2O(k)) = n(i).cHo(i) = -35881kJ. EPUC(H2O(k)) = 9,981.1.3600 = 35932kJ chnh lch = -0,14%. 3) * Vnc = 22,5m3. nnc = V. .M-1 = 1,249.106mol. Enc = nnc.Cp.T = 1316MJ. * nkk = PV(RT)-1 = 2,065.104 mol Cp(kk) = 29,16JK-1mol-1. Ekk = nCpT = 12,05MJ J = E(A.t)-1 = w.T.d-1. E = 1556MJ. 5) Etng = Enc + Ekk + Ephng = 2884MJ Th tch kh tng ng = 2884.106(3600.9981.0,9)-1 = 89,13m3. My t nng kh tn = 0,4.89,18 + 150 = 185,67euro. L si in tn = 2884.106.0,137.(3600.1)-1 + 100 = 209,75euro. III. BI TP CHUN B CHO CC K THI OLYMPIC HA HC QUC T: BI TP CHUN B CHO IChO LN TH 30 : Diliti, mt cht thit yu cho h thng y ca tu khng gian Enterprise, l mt tiu phn thc s bit (d n khng th hin mi tnh cht m Roddenberry v cng s gn cho n!). Diliti c to thnh do s kt ni hai nguyn t liti pha kh. Li(k) + Li(k) Li2(k) (1) a) Entanpi to thnh ca diliti kh c th o c trc tip. Tuy nhin bit c cc tham s nhit ho hc sau y: Hof(Li(k)) = 159,4kJ.mol-1. IE(Li(k)) = 5,392eV (1eV = 96,486kJ.mol-1) Do(Li2+(k)) = 129,8kJ.mol-1 (Do(Li2+(k)) l mnh lin kt ca Li2+(k)) IE(Li2+(k)) = 5,113eV. Dng cc ga tr trn, xc nh Hof(Li2(k)) v Do(Li2(k)) b) Nh ha hc ch huy l plasma xon trn tu Enterprise ang th nghim hot ng ca h thng. ng ta np 122,045g liti nguyn cht vo bung phn ng trng. Bung phn ng c th tch 5,9474.105m3 v c duy tr ti nhit hot ng l 610,25K. Mt thit b o p sut rt nhy cho thy p sut trong bung n nh ti 9,462.10-4Torr (1Torr = 0,133322kPa); phn tch bng phng php quang ph ca mt trong bung phn ng cho thy ton b liti ha hi

(Bung phn ng lm bng hp kim durani c p sut hi bng khng ti 610,2