Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 1 - CN BNG HA HC Cu 1: Phn ng di y t n cn bng 109K vi hng s cn bng Kp = 10: C (r) + CO2 (k) 2CO (k) (a)Tm hm lng kh CO trong hn hp cn bng, bit p sut chung ca h l 1,5atm. (b) c hm lng CO bng 50% v th tch th p sut chung l bao nhiu? Gii: (a) C +CO2 2COn [ ] (1 - x) 2x1 + x (mol) Ta c:5 , 1x 1x 1x 1x 2PPK2CO2COP2+ ((

+= == 10x = 0,79 Vy hn hp cn bng cha 2.0,79 = 1,58 molCO (88,27%) v 1 0,79 = 0,21 mol CO2(11,73%) (b)T10 P5 , 0) 5 , 0 (K2P= =P = 20 atm. Cu 2:Cho phn ng: 2SO2 (k)+O2 (k)2SO3 (k)H =- 198 kJ Cho10,51molkhSO2v37,17molkhngkh(20%vthtchlO2cnlilN2)cxctcl V2O5. Thc hin phn ng 4270C, 1 atm th phn ng t hiu sut 98%. Tnh hng s cn bng KC, KP ca phn ng 4270C. Gii: nO2 b = 7,434 (mol), nN2 b = 29,736 (mol) 2SO2(k) + O2 2SO3 (k)H =- 198 kJ Ban u:10,51 (mol)7,434 (mol) 0 Lc phn ng: 10,3 (mol)5,15 (mol) 10,3 (mol) Lc CB: 0,21 (mol)2,284 (mol)10,3 (mol) s mol hn hp TTCB = 0,21 + 2,284 + 10,3 + 29,736 = 42,53 (mol) Pi = xi.P = xi.1 = xi

23P 22 2(Pso )K =(Pso ) .Po v- nC PK =K (RT) A (R = 0,082, T = 427 + 273 = 7000K, n = -1) 24P 2(10,3) .42,53K =>> 4,48.10(0,21) .2,284v4 -(-1) 4CK =4,48.10 .(0,082.700)257.10 ~Cu 3: Cho cn bng ha hc sau: N2O4 (k) 2NO2 (k)(1) Thc nghim cho bit khi lng mol phn t trung bnh ca hai kh trn 35oC bng 72,45 g/mol v 45oC bng 66,80 g/mol. (a)Tnh phn li ca N2O4 mi nhit trn? (b)Tnh hng s cn bng KP ca (1) mi nhit trn? Bit P = 1 atm (c)Cho bit theo chiu nghch, phn ng thu nhit hay ta nhit? Gii:Xt cn bng: N2O4 (k) 2NO2 (k)(1) (a)Gi a l s mol ca N2O4 c trong 1 mol hn hp s mol NO2 trong 1 mol hn hp l (1 - a) mol * 350CcM= 72,45 g/mol = 92a + 46(1 - a) a = 0,575 mol = nN2O4 v nNO2= 0,425 mol N2O4 (k) 2NO2 (k) Ban ux0 Phn ng0,2125 0,425 Cn bngx - 0,2125 0,425 x- 0,2125=0,575 x= 0,7875 mol , vy= = o % 1007875 , 02125 , 026,98% * 450C cM= 66,80 g/mol = 92a + 46(1 - a)a = 0,4521mol = nN2O4 v nNO2= 0,5479 mol N2O4(k) 2NO2(k) Ban ux 0 Phn ng0,273950,5479 Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 2 - Cn bngx- 0,273950,5479 x- 0,27395=0,4521 x= 0,72605 mol , vy= = o % 10072605 , 027395 , 037,73% (b)PnnPhhNONO22 = ,PnnPhhO NO N4 24 2=v P = 1 atm 350C= = =575 , 0) 425 , 0 (P) P (K2O N2NOP4 220,314 450C = = =4521 , 0) 5479 , 0 (P) P (K2O N2NOP4 220,664 c) T kt qu thc nghim ta thy, khi nhit tng t 350C ln 450C th o tng. C ngha khi nhit tng cn bng dch chuyn theo chiu thun. Vy theo chiu thun phn ng thu nhit, nn theo chiu nghch phn ng ta nhit. Cu 4:Tnh xem c bao nhiu % cht ban u (N2 + 3H2 ) chuyn thnh amoniac, nu phn ng cthc hin 500 atm ,1000atm vnhn xt kt qu vi nguyn l chuyn dch cn bng? Bit hng s cn bng ca phn ngiu ch amoniac 5000C l 1,5.105 atm2 . : Gii:N2 + 3H2 2NH3

53210 . 5 , 1.2 23 =H NNHP PP => 2 23H NP P = => p P P PNH H N= + +3 2 2 => ) ( 4 / 13 2NH NP P P =) ( 4 / 33 2NH HP P P ==> 42) ( 2725633NHNHP PP= 1,5.10-5 => 2) .( 271633NHNHP PP =510 . 5 , 1 =>P =500atm =>3NHP= 152atm : P = 1000atm =>3NHP= 424 atm Tnh % chuyn ho:N2+ 3H2 2NH3

Ban u: 1 3 0mol Phn nga3a2amol Sau phn ng(1-a) (3-3a)2a mol =>4662 , 05001522 42= => =aaa=> % hn hp ban u: 4a/4 = a => 46,62% =>5955 , 05004242 42= => =aaa=> % hn hp ban u: 4a/4 = a => 59,55% => p sut tng cn bng chuyn dch theo chiu lm gim p sut ca h. Cu 5: N2O4 phn li 20,0% thnh NO2 27oC v 1,00 atm. Hy xc nh (a) gi tr Kp;(b) phn li ca N2O4 ti 27oC v 0,10 atm;(c) phn li ca 69g N2O4 trong bnh 20 L 27oC.Gii: Xt phn ng phn li: N2O4 2NO2

n0 no2no n-no2no Phn mol: o +o 11 o +o12,P14PPPK22O N2NOO N2NOP4 224 22o o= = =(a)17 , 0 1) 2 , 0 ( 1) 2 , 0 ( 4P14K2222P= = o o=Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 3 - (b)%) 6 , 54 ( 546 , 0 17 , 0 10 , 01422= o = o o (c)mol 75 , 09269n = = . S dng cng thc ) 1 ( 9225 , 020300 082 , 0 ) 1 ( 75 , 0P4 2O No = o =o = o= 845 , 120300 082 , 0 . 75 , 0 . 2P2NO 17 , 0) 1 ( 9225 , 0) 845 , 1 (K2P=o o=%) 27 , 19 ( 1927 , 0 = oBi 6: Xt phn ng tng hp amoniac : N2 (k) + 3H2 (k) 2NH3 (k) 450oC hng s cn bng ca phn ng ny l KP = 1,5.10-5.(a)Ban u trn N2 vH2 theo t l 1:3 v th tch. Tnh hiu sut phn ng tng hp NH3 khi p sut h bng 500 atm v bng 1000 atm. (b) Cc kt qu tnh c c ph hp vi nguyn l chuyn di cn bng ha hc hay khng? 1.Gii: (a) Gi x v h ln lt l s mol ban u ca N2 v hiu sut phn ng. N2 (k) + 3H2 (k) 2NH3 (k) nox3x0 nhx3hx2hx x(1-h)3x(1-h)2hx En = x(4-2h) 323H N2NHPP) h 2 4 ( x) h 1 ( x 3P) h 2 4 ( x) h 1 ( xP) h 2 4 ( xxh 2P . PPK2 23||.|

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\|= = K P) h 1 ( 2 , 5) h 2 4 ( h 22 = (*) -Ti 500 atm, (*)0 1 , 10 h 2 , 28 h 1 , 142= + vi1 h s467 , 0 h = , vy hiu sut phn ng bng 46,7% -Ti 1000 atm, (*)0 1 , 10 h 2 , 28 h 1 , 142= + vi1 h s593 , 0 h = , vy hiu sut phn ng bng 59,3% (b)Khipsuttng,hiusutphnngtnghpNH3tng.iunyphhpvinguynl chuyndicnbng.Khitngpsut,cnbngchuynditheochiulmgims phntkh (vi phn ng tng hp NH3 l chiu thun). Cu 7: Cho phn ng: H2 (K)+I2 (K) 2HI (K)

Thc hin phn ng trong bnh kn 0,5lt toC vi 0,2mol H2 v 0,2mol I2 .Khi phn ng t trng thi cn bng nng ca HI l 0,3mol/lt. 1.1.Tnh hng s cn bng ca phn ng toC. 1.2..Thm vo cn bng trn 0,1mol H2 th cn bng dch chuyn theo chiu no tnh nng cc cht trng thi cn bng mi. 1.3..Tnh hng s cn bng ca phn ng sau toC. HI (K) 21H2 (K)+21I2 (K) Gii:1.1.Tnh hng s cn bng ca phn ng toC. Theo gi thit:[H2] = [I2] = 0,4mol/lt

H2 (K)+I2 (K) 2HI (K) Trc phn ng:0,4 (mol/lt) 0,4(mol/lt) Lc cn bng:0,25(mol/lt)0,25 (mol/lt ) 0,3 (mol/lt )Kcb= 44 , 125 , 0 . 25 , 03 , 02=1.2.Nng cc cht trng thi cn bng mi: Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 4 - -Nu thm vo cn bng 0,1mol H2nng H2 tng cn bng dch chuyn theo chiu thun. -Khithm 0,1 mol H2 nng H2 = 0,25 + 0,2 = 0,45mol/lt Gi x l nng H2 tham gia phn ng t cn bng mi. H2 (K)+ I2 (K) 2HI (K) Trc phn ng: 0,45 (mol/lt)0,25(mol/lt) 0,3(mol/lt) Lc cn bng:0,45-x (mol/lt ) 0,25-x (mol/lt)0,3+2x (mol/lt) Kcb= 44 , 1) 45 , 0 ).( 25 , 0 () 2 3 , 0 (2= +x xx 2,56x2+ 2,208x 0,072 = 0K: 0< x < 0,25 Gii phng trnhta c :x = 0,03146nhn ;x/ = -0,89loi. Vy : [H2] =0,41854 (mol/lt) [I2] =0,21854 (mol/lt) [HI] =0,36292 (mol/lt) 1.3..Hng s cn bng ca phn ng sau toC.HI (K) 21H2 (K)+21I2 (K) Gi K/cb l hng s cn bng ca phn ng: Ta c K/cb = | | | || | HII H 212212=cbK1 =44 , 11= 65 Cu 8: Cho 2 mol N2 v 8 mol H2 vo bnh kn c th tch l 2 lt, sau khi phn ng:N2(K)+3H2(K)2NH3(K) t trng thi cn bng , a nhit v nhit ban u , th p sut trong bnh bng 0,9 ln p sut u. Tnh K cn bng. Gii:Tng s mol ban u trong bnh kn : bn = 2+ 8 = 10 mol Trong cng iu kin t0 v V : T l mol = T l p sut. Ta c: sPP= snn 9 , 01 = sn10 ns = 0,9 x 10 = 9 mol Gi x l s mol N2 tham gia phn ng: N2(K)+3H2(K) 2NH3(K) Trc p: 2 mol8 mol Phn ng:x mol3x mol2x mol Sau p:(2 x)(8-3x)2x mol Tng s mol cc cht kh sau phn ng: sn = 10 2x = 9 mol x = 0,5mol trng thi cn bng :nN2= 2 0,5 = 1,5 mol [N2] = 25 , 1 = 0,75 mol/l nH2=8 3 x 0,5 = 6,5 mol/lt[H2] = 25 , 6 = 3,25 mol/lt ; nNH3 = 2 x 0,5 = 1 mol [NH3] = 21 = 0,5 mol/lt Kcb = 22332. HNNHC CC = 32) 25 , 3 ).( 75 , 0 () 5 , 0 (= 9,71. 10-3

Cu 9 : Cho cn bng : N2O4

2NO2 Ly 18,4 gam N2O4 vo bnh chn khng c dung tch 5,9 lt 27OC. Khi t ti cn bng, p sut l 1 atm. Cng vi khi lng ca N2O4 nhng nhit 110OC th trng thi cn bng, nu p sut vn l 1 atm th th tch hn hp kh t 12,14 lt. a/. Tnh thnh phn % N2O4 b phn li 27OC v 110OC. b/. Tnh hng s cn bng 2 nhit trn, t rt ra kt lun phn ng ta nhit hay thu nhit. Gii:a)S mol N2O4 ban u: 18,4/92 = 0,2 mol N2O4 2NO2 b (mol):0,20 Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 5 - p: x2x cn : 0,2 x2x * to = 27oC th: (0,2 + x1) = 5,9.273 / 22,4(273 + 27) = 0,23969 x1 = 0,03969 % N2O4 b phn hy: [0,03969/ 0,2] 100% = 19,845% * to = 110oC th: (0,2 + x2) = 12,14.273 / 22,4(273 + 110) = 0,3863 x2 = 0,1863 % N2O4 b phn hy: [0,1863/ 0,2] 100% = 93,15% b)K = [NO2]2 / [N2O4] = {2x/V)2 / (0,2 x)/V = 4x2 / V(0,2 x) * to = 27oC: V1 = 5,9 l; x1 = 0,03969 mol K1 = 6,66.10-3

* to = 110oC: V2 = 12,14 l; x2 = 0,1863 mol K2 = 0,8347 Nhn xt: Ta thy khi tng s mol N2O4 phn hy tng v hng s cn bng cng tng phn ng thu nhit Cu 10:C cn bng: N2O4 (K) 2NO2 (K). Cho 18,4gam N2O4 vo bnh dung tch l 5,904 lt 270C. Lc cn bng p sut ca hn hp kh trongbnh l 1atm. Tnh p sut ring phn ca NO2 v N2O4 lc cn bng. Nu gimp sutca hlc cn bngxung bng0,5 atmth p sutring phn ca NO2, N2O4 lc ny l bao nhiu? Kt qu c ph hp nguyn l ca Le Chatelier khng ? Gii: ) mol ( 2 , 0924 , 18n4 2O N= =N2O4 2NO2 Ban u0,2 0 Cn bng0,2 - x2x Tng s mol c trong h lc cn bng: 0,2 x + 2x = 0,2 + x 24 , 0) 27 273 ( 082 , 0904 , 5 1RTPVx 2 , 0 =+= = +=> x = 0,04 2NOn (lc cn bng) = 0,08 mol 4 2O Nn (lc cn bng) = 0,2 0,04 = 0,16 mol V s mol N2O4 gp i s mol NO2 nn p sut N2O4 cng gp i ca NO2

Vy:(atm)32P), atm (31P4 2 2O N NO= =;( )61PPK32221O NNOP4 22= = =t 2NOP khi cn bng l P th p sut ca N2O4 khi cn bng l: 0,5 P. T : 0 5 , 0 P P 6P 5 , 0P61K22P= + = =(atm) 283 , 0 P(atm); 217 , 0 P5 2 2O N NO= =Kt qu:77 , 0283 , 0217 , 0PP4 22O NNO~ =So snh vi trng hp trn:5 , 02331PP4 22O NNO= =Vy:Khipsutcahxungthcnbngdchchuynsangphalmtngpsutcahln, ngha l sang pha c nhiu phn t kh hn (ph hp nguyn l). Cu 11:Cn bng: N2O4(k) 2NO2(k)

nhn c xut pht t a mol N2O4. Gi ol phn li ca N2O4. a/ Tnh s mol NO2, N2O4 v tng s mol ca h khi cn bng theo a vo ? b/Tnh p sut ring phn ca NO2, N2O4 khi cn bng theoov p sut tng P ca h?Tnh KP theoov P? c/ Nu ban u c 1,588 gam N2O4 trong bnh 0,5 lt 250C v P = 760 mmHg th o , p sut ring phn ca NO2, N2O4 lc cn bng l bao nhiu? Gii:NO2 =2ao , N2O4=a(1-o ), PNO2=2o P/(1+o ), PN2 O4=(1-o )P/(1+o ). Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 6 - P =4o2P/(1-o )(1+o );o = 0,1587,KP=0,103, PNO2=0,274 atm, PN2 O4= 0,726 atm. Cu 12:Cn bng trong hH2(k) +I2(k)

2HI(k)c thit lp vi cc nng sau : [H2] = 0,025 M ; [I2] = 0,005 M ; [HI] = 0,09M . a) Khi h t trng thi cn bng p sut ca h bin i nh th no ? b) Tnh hng s cn bng Kcb v nng ban u ca I2 v H2 ? Gii:H2 + I2 2HI Trong qu trnh phn ng s mol kh khng i . Nu th tch v nhit khng i th p sut ca h khng iKcb = 22 2[HI][H ].[I ]= 64,8.T [HI] = 0,09 M=> [H2]p = [I2]p = 0,045M => Do nng ban u: [H2] = 0,07M; [I2 ]= 0,05M Cu 13: Nung FeS2 trong khng kh, kt thc phn ng thu c mt hn hp kh c thnh phn: 7 SO2; 10 O2; 83 N2 theo s mol.un hn hp kh trong bnh kn (c xc tc) 800K, xyra phn ng: 2SO2 + O2

2SO3Kp = 1,21.105. a)Tnh chuyn ho ( s mol) SO2 thnh SO3 800K, bit p sut trong bnh lc ny l 1 atm, s mol hn hp kh ban u (khi cha un nng) l 100 mol. b)Nu tng p sut ln 2 ln, tnh chuyn ho SO2 thnh SO3, nhn xt v s chuyn dch cn bng. a) Cn bng: 2SO2 + O2 2SO3 Ban u: 7 10 0 (mol) lc cn bng:(7-x) (10 - 0,5x)x(x: s mol SO2 phn ng). Tng s mol cc kh lc cn bng: 100 0,5x = n. p sut ring ca cc kh:2SOP = (7-x). np ; 2PO= (10 0,5x). np; 3PSO=x . np Kp = 2 23P . ) (P) (P22O SOSO= ) 5 , 0 10 .( ) 7 () 5 , 0 100 (22x xx x = 1,21. 105

do K>> x ~ 7 Ta c : 5 , 6 . ) 7 (5 , 96 . 492x = 1,21. 105 Gii c x = 6,9225. Vy chuyn ha SO2 SO3: 7% 100 . 9225 , 6= 98,89. b) Nu p sut tng 2 ln tng t c:7- x'= -210 . 5 . 0,300 = 0,0548x' = 6,9452. chuyn ho SO2 SO3: (6,9452 . 100)/7 = 99,21 Ktqu ph hp nguyn l Lsatlie: tngpsut phn ng chuyn theo chiuv pha cs phn t kh t hn. Bi 14: Xt qu trnh cn bng sau ti 686oC : CO2 (k) + H2 (k) CO (k) + H2O (k) Nngccchtticnbnglnltbng[CO]=0,050 M,[H2]=0,045M,[CO2]= 0,086 M v [H2O] = 0,040 M. Nu tng nng CO2 ln n gi tr 0,500 M (nhit khng i) th nng mi cht cn bng mi c thit lp li bng bao nhiu ? Hng s cn bng nng :52 , 0045 , 0 086 , 0050 , 0 040 , 0] H ][ CO [] CO ][ O H [K2 22C== = Thm CO2, cn bng chuyn di theo chiu thun : CO2 (k) + H2 (k) CO (k)+ H2O (k) 0,5000,0450,0500,040 -x -x+x +x 0,500-x0,045-x0,050 +x0,040 + x Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 7 - T52 , 0) x 045 , 0 ( ) x 500 , 0 () x 050 , 0 ( ) x 040 , 0 (] H ][ CO [] CO ][ O H [K2 22C= + += = 0,48x2 + 0,373x 9,7.10-3 = 0 x = 0,025M Vy [CO2] = 0,48M, [H2] = 0,020M, [CO] = 0,075M v [H2O] = 0,065M. Cu 15:Khi nung nng n nhit cao PCl5 b phn li theo phng trnh PCl5 (k) PCl3 (k) + Cl2 (k) 1. Cho m gam PCl5 vo mt bnh dung tch V, un nng bnh n nhit T (K) xy ra phn ng phn li PCl5. Sau khi t ti cn bng p sut kh trong bnh bng p. Hy thit lp biu thc ca Kp theo phn li ov p sut p. Thit lp biu thc ca kc theo o, m, V. 2. Trong th nghim 1 thc hin nhit T1 ngi ta cho 83,300 gam PCl5 vo bnh dung tch V1. Sau khi t ti cn bng o c p bng 2,700 atm. Hn hp kh trong bnh c t khi so vi hiro bng 68,862. Tnh o v Kp. 3. Trong th nghim 2 gi nguyn lng PCl5 v nhit nh th nghim 1 nhng thay dung tch l V2 th o c p sut cn bng l 0,500 atm. Tnh t s 21VV. 4. Trong th nghim 3 gi nguyn lng PCl5 v dung tch bnh V1 nh th nghim 1nhng h nhit ca bnh n T3 = 0,9 T1 th o c p sut cn bng l 1,944 atm. Tnh Kp v o. T cho bit phn ng phn li PCl5 thu nhit hay pht nhit. Cho: Cl = 35,453 ; P : 30,974 ; H = 1,008 ; Cc kh u l kh l tng. Li gii: 1. Thit lp biu thc cho Kp, Kc PCl5 (k) PCl3 (k) + Cl2 (k) ban ua mol cn bnga x x x (mol) Tng s mol kh lc cn bng : a + x = n o = xa; Khi lng mol: 5PClM= 30,974 + 5 x 35,453 = 208,239 (g/mol)

3PClM= 30,974 + 3 x 35,453 = 137,333 (g/mol)

2ClM = 70,906 (g/mol) gam208,239 gam/molm = a mol PCl5 ban u *p sut ring phn lc cn bng ca mi kh:

5PClP = pa xa x+trong 3PClP= 2ClP=+xPa x

Kp = 2 35Cl PClPClP PP = | | |+\ .| | |+\ .2 - xpa xa xpa x = ( ) +222xpa x + | | |\ .a xa x 1p

Kp =+ 2( ) ( )x pa x a x = 22 2xa x p ;Kp = 222 22 2xapa xa a = oo 221p* Kc = [PCl5] = o (1 ) aV trong [PCl3] = [Cl2] = o aV Kc = | | | |3 25Cl[ ]PClPCl = ( ) o22aV ( ) o 1Va = oo 2(1 )aV = oo 2 208, 239 (1 )mV Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 8 - Hoc: Kp = Kc (RT)V Vkh = 1 Kp = Kc (RT)pV = nRT = (a + x) RT RT = +pVa x = o + (1 )pVa

Kp = Kc +pVa x oo 21p= Kc+pVa x

Thay x = ao oo 221p= Kco + (1 )pVa Kc = o oo+22(1 )1aV Kc = ( )o oo o++2(1 )1(1- )aV= oo 2(1 )aV =oo 2 208, 239 V (1 )m * Quan h Kp v Kc. T cch 1 : Kc = Kp1RT Thay RT = pVa(1 ) o + Kc = Kp a(1 )pVo + =2a(1 )1 pVpo oo+ = 2aV(1 )oo 2. Th nghim 1 : 5PClnban u = a = 83, 30 g208,239 g/mol =0,400 mol M ca hn hp cn bng: 68,826 2,016 = 138,753 g/mol Tng s mol kh lc cn bng: n1 = a (l + o1) = 83, 30 g138, 753 g/mol= 0,600 mol n1 = a (1 + o1) = 0,400 (1 + o1) = 0,600 o1 = 0,500 * Tm Kp ti nhit T1 :Kp = 22 1oo p = 22(0, 5)1 (0, 5) 2,70 = 0,900 3. Th nghim 2: - Gi nguyn nhit Kp khng i. - Gi nguyn s mol PCl5 ban u: a = 0,400mol. - p sut cn bng P2 = 0,500 atm. Ta c22221oo p2 = Kp = 22221oo 0,500 = 0,900 o22 = 0,64286o2 = 0,802 Tng s mol kh lc cn bng: n2 = 0,400 + (1+ o2) ~ 0,721 (mol). * Th tch bnh trong TN 2: V2 = 2 12n RTpso viV1 = 1 11n RTp 21VV = 2 11 2n pn p= 0, 721 2, 7000, 600 0, 500= 6,486 (ln) 4. Th nghim 3: - Thay i nhit Kp thay i. - Gi nguyn s mol PCl5 ban u a = 0,400 mol v V1 - p sut cn bng P3 thay i do: nhit gim (T3 = 0,9 T1),tng smol kh thay i (n3 = n1). P3 = 1,944 atm ; Tnh o3 : n3 = a (1+ o3) = 0,400 (1+ o3) ; p3V1 = n3RT3 = 0,9 n3RT1 ; P1V1 = n1RT1.

3 31 1P 0, 9nP n=30, 400 (1 ) 0, 9 1, 9442, 700 0, 600o + =o3 =0,200 n3 =0,48 mol * KP(T3 ) = 233 23p1oo =22(0, 200)1 (0, 200) 1,944 = 0,081 * Khi h nhit , Kp gim cn bng chuyn dch theo chiu nghch. Chiu nghch l chiu pht nhit Chiu thun l chiu thu nhit. Bi 16:Cho cn bng ho hc: 2NO2 N2O4kJ H 04 , 58 = ATrng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung - 9 - Cn bng s chuyn dch nh th no , gii thch, khi: 1/ Tng nhit . 2/ Tng p sut. 3/ Thm kh tr Ar trong 2 trng hp: a) Gi p sut khng i. b) Gi th tch khng i. 4/ Thm xc tc. Gii: 2NO2 N2O458, 04 H kJ A = (phn ng ta nhit) a) Tng nhit cn bng chuyn sang tri theo chiu phn ng thu nhit. b) Tng p sut cn bng chuyn sang phi theo chiu lm gim s mol. c) Thm kh tr : * p sut khng i Th tch tng gim p sut ring phn ca cc kh . Kp =2 422N ONOPP = 22 42N ONOnn. VRT khi thm kh trQ = 22 42N ONOnn. /VRT v V/ >VQ>Kp vy Q Kp : s mol N2O4 phi gim cn bng chuyn theo chiu t phi sang tri ( to NO2) *Th tch khng i p sut ring phn ca cc kh khng i cn bng khng chuyn dch.d) Xc tc tm tng hoc gim tc c phn ng thun v nghch khng lm chuyn dch cn bng. Cu 17: 1)Cn bng: N2 (k) + 3H2 (k) 2NH3 (k)s chuyn dch chiu no khi(a) thm Ar vo hn hp cn bng nhng gi cho th tch khng i? (b) thm Ar vo hn hp cn bng nhng gi cho p sut khng i? Gii:(a) Ta c: 2n23H N2NH3H N2NHPRTVKRTVn nnP PPK2 232 23|.|

\|= |.|

\|= =. V V, KP v T = const nnconstn nnK3H N2NHn2 23= = . Nh vy c s tng p sut ca h nhng khng c s chuyn dch cn bng. (b) 2n PRTVK K |.|

\|=Vi T, P v Kp = const, khi thm Ar lm V tng nn Kn phi gim. S thm agon lm cn bng chuyn dch theo chiu nghch (chiu lm phn li NH3). Ch thch: Trong trng hp ny, ta khng th da vo nguyn l Le Chatelier d on chiu din bin ca phn ng. 2) Phn ng nhit phn CaCO3 c tin hnh trong 1bnh kn. Khi p sut ca CO2 trong bnh ln n 0,236 atm th khng thay i na mc d trong bnh vn cn CaCO3 v c CaO.1)Tnh Kp, Kc ca phn ng 800oC 2)Trong bnh dung tch 10 lt, nu ta b vo 5 gam CaCO3 v 2 gam CaO, nung nng bnh n 800oCtcnbngthsaukhicnbng,khilngmichtrntrongbnhlbaonhiu gam? P N1) CaCO3 (r) = CaO(r) + CO2 (k) (*) Ti 800oC khi 2COP= 0,236 th khng thay i na t cn bng Nn Kp = 2COP= 0,236 (atm) ;n = n kh sau - n kh trc 3n10 68 , 2) 273 800 (2734 , 22236 , 0(RT)KpKcA =+= =(mol/lt) Kc = [CO2] = 2,6810-3 (mol/lt) 2)V nung nng bnh n 800oC: do nhit khng i nn Kc khng i Kc = 2,6810-3 (mol/lt)2COn= 2,6810-3 10 = 2,6810-2 (mol)t phng trnh (*)3CaCOn nhit phn = 2COn v to ra 2,6810-2 mol CaO 3CaCOm nhit phn = 2,6810-2 100 = 2,68 (gam) Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 10 - Vy trong bnh cn li lng cht rn l: 3CaCOm = 5 2,68 = 2,32 (gam) CaOm= 2,6810-2 56 + 2 = 3,5008 (gam) Bi 18:Ngi ta trn CO v hi H2O ti nhit 1000K vi t l 1 : 1. Tnh thnh phn ca h lc cn bng, bit rng: 2 2 22H O 2H O + c pkp1 = 20,113 2 22CO 2CO O + c pkp2 = 20,400 T cc d kin bi ta c: 2 2 P3P,22 2 2 P4 P,11 1CO O CO K2 K1H O H O K K2+ =+ = 20,1130,1435 P12 2 2 P P3 P4 20,400P2K 10CO H O CO H K K .K 10 1, 392K 10+ + = = = = ~ Gi s ban u ly 1 mol CO v 1 molH2O 2 2 2 PCO H O CO H K 1, 392 + + = Ban u1mol 1mol Lc cu bng1-a(mol) (1-a)mola mola mol 2 222CO HP2CO H Oa aP. PP .Pa2 2K1 a 1 aP .P (1 a)P. P2 2= = = vi P l p sut chung PaK 1,17981 a = = =>a 0, 54mol ~2 2a x 100%%H %CO 27%2= = =;2(1 a)x100%%CO %H O 23%2= = =Cu 17:H2(kh)+I2(kh) 2HI(kh) Thc hin phn ng tng hp hiro iouatrong mt bnh kn, dung tch 2 lit nhit T, c hng s cn bng K = 36. a, Nu nng ban u ca H2 v I2 bng nhau v bng 0,02M th nng ca cc cht ti thi im cn bng l bao nhiu? b, cn bng trn, ngi ta thm vo bnh 0,06gam hiro th cn bng cng b ph v v hnh thnh cn bng mi. Tnh khi lng hiro ioua cn bng cui? a)H2(kh)+I2(rn) 2HI(kh) Trc phn ng: 0,02M0,02M 0 Phn ng: x x2x Cn li:0,02 x 0,02 x 2x Vy : ( )( ) ( )( )2236 2 6 0, 02 0, 0150, 02 . 0, 02xx x xx x = = = Kt lun: cn bng: [HI]= 0,03M, [H2] = [I2] = 0,005M b)S mol H2 thm:0,06 : 2 = 0,03 (mol) nng tng thm: 0,03: 2 = 0,015M H2(kh)+I2(kh) 2HI(kh) Ban u:0,02M 0,005M0,03M Phn ng:a a2a Cn bng: 0,02 a 0,005 a 0,03 + 2a ( )( )( )20, 03 2360, 02 0, 005aKa a+= = a = 2,91.10-3 v2,89.10-2 M. V a < 0,005 nn ch nhn a = 2,91.10-3 Khi lng HI cn bng cui:(0,03 + 2. 0,0029). 2. 128 = 9,165(gam) Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 11 - Cu 18: Trong bnh kn dung tch khng i cha 35,2x (g) oxi v 160x (g) SO2. Kh SO2 136,5oC c xc tc V2O5. un nng bnh mt thi gian, a v nhit ban u, p sut bnh l P. Bit p sut bnh ban u l 4,5 atm v hiu sut phn ng l H%. a, Lp biu thc tnh p sut sau phn ng P v t khi hi d ca hn hp kh sau phn ng so vi khng kh, theo H. b, Tm khong xc nh P, d? c, Tnh dung tch bnh trong trng hp x = 0,25? Hng dn gii: 235, 21,1 ( )32O bdauxx mol n = = ;21602,5 ( )64SO bdauxx mol n = = 2SO2 +O2 2SO3 Ban u:2,5x1,1x0 Phn ng:2,2xH 1,1xH 2,2xH Sau phn ng:(2,5x 2,2xH)(1,1x 1,1xH)2,2xH n2 =2,5x-2,2xH+ 1,1x- 1,1xH+2,2xH= x(3,6-1,1H) (mol)Trng hp bi ton ng V, ng T. ( )( )1 22 13, 6 1,1 4,5' 1, 25 3, 6 1,1' 3, 6x Hn n P PP HP n n x = = = = b, Khi H = 0 P = 4,5 (atm) ;H = 1 P = 3,125 (atm) Vy trong thi gian phn ng th3,125 < P < 4,5 .T khi hi so vi khng kh: 160 35, 2 195, 2(3, 6 1,1 ) 3, 6 1,1sau truocsausau truocm m x xMn n x H H+= = = = ;( )/195, 2 6, 73129 3, 6 1,1 29 3, 6 1,1sauhhsau kkMH Hd = = = Khi H = 0 d = 1,87; H = 1 d = 2,69; Vy 1,87 < d < 2,69 c, p dng cng thc: PV = nRT; Pu = 4,5atm;nu = 3,6x = 3,6.0,25 = 0,9(mol) ( )22, 40,9. 273 136,52736,72( )4,5nRTV lP+= = = Cu 19: Trong mt bnh kn A dung tch 1 lt 500 0C, hng s cn bng ca phn ng tng hp HI t H2 v I2 bng 46.a)Tnh nng mol cc cht trng thi cn bng? Bit ban u trong bnh A c 1mol H2 v 1mol I2 b) Nu ban u cho 2 mol HI vo bnh A nhit 500 0C th nng cc cht lc cn bng l bao nhiu? c) Nu h ang trng thi cn bng cu a, ta thmvo h 1,5 mol H2 v 2,0 mol HI th cn bng dch chuyn theo chiu no? a) Cn bng : H2 + I22HI Ban u1M1M 0 Phn ngxx2x Cn bng 1-xx2x Ta c biu thc cn bng : Kc = | || || | ( )46x 1x 4I HHI222 22== (iu kin x x > 0 32 s PP' s 1Cau 33: 8170C hang so ca n bang Kp cua phan ng gia CO2 va C(r) d e tao thanh CO bang 10. Xac nh : a/ Phan mol cua cac kh trong hon hp luc can bang, khi ap sua t chung bang 4 Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 19 - b/ Ap suat rieng cu a kh CO2 luc can ba ng c/ A p suat chung cua hon hp sao cho luc can bang CO2 chiem 60% ve the tch 2. a/C(r) + CO2 (K)2CO(K) Kp= 10 B1mol 0 | | 1-o 2oKp=22COCOPP =10 ( viPi= xi Pva xi =iinn )ta co 22111COCOP PP Poooo=+=+ thay va o bieu thc tnh Kp 21 20, 62 0, 234, 0, 7661 1CO COx xo ooo o= = = = =+ +

b/ 2 2. 0, 234.4 0, 936CO COP x P = = = atmc/ Luc ca n bang CO2 chiem 6% the tch , nen( )220, 06;0, 940, 9410 0, 679 .0, 06COCOPxxK P P atm=== = =

Cu 34:X l nguyn t thuc nhm A, hp cht vi hidro c dng XH3. Electron cui cng trn nguyn t X c tng 4 s lng t bng 4,5. a) iu kin thng XH3 l mt cht kh. Vit cng thc cu to, d on trng thi lai ho ca nguyn t trung tm trong phn t XH3. b)Cho phn ng: 2XOCl2XO + Cl2, 5000C c Kp= 1,63.10-2. trng thi cn bng p sut ring phn ca PXOCl =0,643 atm, PXO = 0,238 atm. a)Tnh PCl2 trng thi cn bng. b)Nu thmvo bnh mt lng Cl2 trng thi cn bng mi p sut ring phn ca XOCl bng 0,683 atm th p sut ring phn ca XO v Cl2 l bao nhiu? a> V X thuc nhm A, hp cht vi hidro c dng XH3 nn l nhm VA (ns2np3). Vy: ms = +1/2; l = 1 ; m = +1suy ra: n = 4,5 2,5 = 2. Vy X l Nit ( 1s22s22p3) Cng thc cu to cc hp cht v d on trng thi lai ha ca nguyn t trung tm: NH3 : N c trng thi lai ho sp3. NHHH b> Xt phn ng 5000C: 2NOCl 2NO+Cl2Kp = 1,63.10-2 trng thi cn bng c: PNOCl = 0,643 atm, PNO = 0,238 atm. Vy da v biu thc Kp ta c:PCl2 = Kp. 2||.|

\|NONOClPP = 0,119 atm. Nu thm vo bnh mt lng Cl2 PNOCl = 0,683 atmth PNO = 0,238 0,04= 0,198 atm=> v PCl2 = 0,194 atm. 2. 8200C hng s cn bng Kp ca cc phn ng nh sau: CaCO3 (tt) CaO (tt)+ CO2 (k)K1 = 0,2 C gr + CO2(k) 2CO (k) K2 = 2 Cho 1 mol CaCO3 v 1 mol C vo bnh chn khng dung tch 22,4 lt duy tr 8200C. a. Tnh s mol cc cht khi cn bng. b. th tch no ca bnh th s phn hy CaCO3 l hon ton. Hng dn:a. K1=PCO2~ 0,2 atmK2 = 22PCOCO P =>Do PCO = 2 , 0 . 2 ~0, 632 atm Gi x,y l s mol CaCO3 v CO2 phn ng. T suy ra s mol cc cht Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 20 - CaCO3 (tt) CaO (tt)+CO2 (k) K1 = 0,2 (1) Ban u1 mol0 0 Phn ngx x x [ ] 1- x x x C gr+CO2(k) 2CO (k) K2 = 2(2) Ban u 1x0 mol Phn ng yy2y [ ]1-y x-y 2y trng thi cn bng l:CaCO3CaOCO2CCO 1 - xxx - y1 - y2y x - y =2CO2P V0,05 mol CORT=;2y =COP V0,158 mol CORT = nCaO = 0,129 mol ; 3CaCOn= 0,871mol ; nC = 0,921 molb. S phn hy hon ton th x = 1 2COn =(1- y) molvnCO=2y (mol).p sut CO2 v CO khng i. ( )0, 632V 2yRT0, 2V1y RT= = =>gii hai phng trnh ta c V ~ 173,69 lt Vy khi th tch tng, p sut chung gim th cn bng chuyn dch theo chiulm tng p sut, tc l theo chiu to ra kh CO2.(kt qu ph hp nguyn l Le Chaterlier)Cu 35: . 6000K i vi phn ng: H2 +CO2

H2O(k) + CO c nng cn bng ca H2, CO2, H2O v CO ln lt bng 0,600; 0,459; 0,500 v 0,425 mol./1. a) Tm Kc, Kp ca phn ng. b) Nu lng ban u ca H2 v CO2 bng nhau v bng 1 mol c t vo bnh 5 lt th nng cn bng cc cht l bao nhiu? Gii:a) Kc = | | | || | | |22 2..H O COH CO = 0, 5 0, 4250, 6 0, 459 = 0,7716 ; Kp = Kc(RT)n = 0,7716(do n = 0) b) Ti CBHH: [H2O] = a ;[CO] = a ; [H2] = [CO2] = 0,2 aTa c : 22(0, 2 )aa = 0,7716 a = 0,094 v 0,2 a = 0,106p s: Kc = Kp = 0,772; [H2] = [CO2] = 0,106 M v [H2O] = [CO] = 0,094 M. Cu 36:. 1000K hng s cn bng Kp ca phn ng 2SO2 + O2 2SO3bng3,50 atm1. Tnh p sut ring lc cn bng ca SO2 v SO3 nu p sut chung ca h bng 1atm v p sut cn bng ca O2 bng 0,1atm.Gii:Gi x l p sut ring ca SO2 thp sut ring ca SO3 = 1 0,1 x = 0,9 x Kp = 22(0, 9 )0,1xx= 3,50 x = 0,57 atmv 3PSO= 0,33 atm Cu 37: . a) Tnh hng s cn bng Kp i vi phn ng: N2 + 3H2 2NH3. 250C Bit G0htca NH3 = 16,64 kJ/mol b) Kp s thay i th no khi phn ng cho c vit di dng: 12N2 + 32H2 NH3. + G = 2. 16,64 = 33,28 kJ/molG0 = RTlnKp lnKp = 33, 288, 314 298 | | |\ .= 13,43 Vy Kp = 6,8. 105. Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 21 - Kp = 32 223.NHN HPP P nn i vi phn ng 12N2 + 32H2 NH3. c Kp = 32 21 32 2.NHN HPP P = PK= 825 Cu 38. Cn bng ca phn ng: NH4HS (r) NH3 (k) + H2S (k) c thit lp 2000C trong mt th tch V. Phn ng cho l thu nhit. Cho bit p sut ring ca NH3 s thay i th no khi cn bng c ti lp sau khi:a) Thm NH3 ; b) Thm H2S ;c) Thm NH4HS ;d) Tng nhit ;e) psut ton phn s tng do thm Ar vo h ;f) Th tch bnh tng ti 2V. Gii:a) tng ; b) gim ; c) khng i ;d) tng ; e) khng i ;f) tng. Cu 39. Cn bng ca phn ng kh CO2 bng C :C + CO2 2CO xy ra 1090K vi hng s cn bng Kp = 10. a) Tm hm lng kh CO trong hn hp cn bng, bit p sut chung ca h l 1,5atm. b) c hm lng CO bng 50% v th tch th p sut chung l bao nhiu? Gii:a)C+ CO2 2CO n [ ](1 - x) 2x 1 + x (mol) Phn mol11xx+ 21xx + Ta c : Kp = 22COCOPP = 22111xxxx ( (+ + . 1,5 = 10x = 0,79 mol Vy hn hp lc cn bng cha 2. 0,79 = 1,58 mol CO (88%)v 1 0,79 = 0,21 mol CO2(12%) b) Suy ra Kp = ( )20, 50, 5. P = 10P = 20 atm. Cu 40. 500C v di p sut 0,344 atm phn ly o ca N2O4 (k) thnh NO2(k) bng 63%. Xc nh Kp; Kc; Kx. Gii: N2O4 (k)NO2(k) n []1 - o2o1 + o(o l phn ly) Phn mol 11 oo+ 21 oo + Kp = 22COCOPP = 22111oooo ( (+ + . 0,344 thayo = 0,63 tnh c Kp = 0,9 p dng Kc = Kp.(RT)n vin = 1 vKx = Kp. P ntnh cKc = 0,034 v Kx = 2,63 Cu 41. 8200C c cc phn ng sau vi hng s cn bng tng ng: CaCO3 (r) CaO (r) + CO2 (k) K1 = 0,2 C (r) + CO2 (k) 2CO (k)K2 = 2,0 Ly hn hp gm 1 mol CaCO3 v 1 mol C cho vo bnh chn khng c th tch 22,4 lt gi 8200C. 1- Tnh s mol cc cht lng c trong bnh khi phn ng t ti trng thi cn bng. 2-S phn hu CaCO3s hon ton khi th tch bnh bng nhiu (psutring ca cc kh khng i). Kt qu ny c ph hp vi nguyn l L-Sa-T - Lie khng? Lin h thc t sn xut vi sng. Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 22 - Gii:a) K1 = P2CO= 0,2 (atm) K2 = 22COCOPP = 2 PCO=2 0, 2 = 0,63 atm T PV = nRTvi R = 0,082S mol CO2 =0,05 v s mol CO = 0,16 Gi s mol CaCO3 v C phn ng l x v y CaCO3 CaO+CO2 v C+ CO2 2CO [](1 x) x (x y)(1 y)y2y Ta c: x y = 0,05 v 2y = 0,16 x = 0,13 v y = 0,08 S mol: CaCO3 = 0,87;CaO = 0,13;C = 0,92;CO2= 0,05;CO = 0,16 b) Khi CaCO3 phn hu hon ton th x = 1 s mol CO2 = 1 x ta c: 0,2V = (1 y)RTv0,63V = 2yRT. Gii 2 phng trnh cho V = 174 lt Vy khi th tch tng, p sut chung gim th cn bng chuyn dch theo chiulm tng p sut, tc l theo chiu to ra kh CO2.(kt qu ph hp nguyn l Le Chaterlier)Cu 42 1. 137Csl nguyn t phng x dng trong l phn ng ht nhn, c chu k bn hul 30,2 nm. Sau bao lu th lng cht ny cn li 1,0%. 2. Ti 250CG A to thnh ca cc cht nh sau (theo kJ.mol-1): H2O (k)CO2 (k)CO (k)H2O (l) -228,374-394,007-137,133-236,964 a) Tnh Kp ca phn ng: CO(k) + H2O(l) H2(k) + CO2(k) ti 250C b) Tnh p sut hi nc ti 250C. c)Hn hp cc kh CO, CO2, H2 m mi kh u cp sut ring phn l 1,0atm c trn vi H2O (l), d. Tnh p sut ring phn ca mi kh c trong hn hp cn bng ti 250C, bit qu trnh xy ra khi th tch coi nh khng i. Hng s tc ca qu trnh phn r ht nhn l: 1120, 693 0, 693k 0, 023(nam )t 30, 2= = =Ta c : 1371372, 303 ( Cs)dau 2, 303 100t log log t 200, 2608696(nam)k ( C)sau 0, 023 1= = => =2.a)CO(k) + H2O(l) H2(k) + CO2(k)2 2 20 0 0 0 0298(pu) H (k) CO (k) CO(k) H O(l)0 1298(pu)G G G G GG 0 ( 394, 007) 137,133 236,964 19, 91 kJ.molA = A + A A A A = + + + = p dng phng trnh ng nhit Van Hof, ta c: 0 3T0TG 19,91.103,49 2,303.RT 2,303.8,314.298G RTln Kp RT.2, 303.lg KpKp 10 10 10 3086, 045027A A = = = = = = b) xt 2H O(h)P 250C ta xt cn bng ti 250C: H2O(l)H2O(h)2 20 0 0298(pu) H O(h) H O(l)0 1298(pu)G G GG 228, 374) 236,964 8, 59 kJ.molA = A A A = + = Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 23 - 0T38,59 G2,303.RT 2,303.8,314.298.10Kp 10 10 0, 03122677 A = = = V:2H O(l)P =const =1,0atm=> 2H O(h)P 0, 03122677 = atm ( 250C) c)V iu kin T, V u l const => psut ring phn t l vi s mol mi khnn c th tnh p sut ring phn theo phn ng: CO(k) + H2O(l) H2(k) + CO2(k)Ban u1 11 (atm) Cn bng 1-x 1+x 1+x 2 22H O COCO[P ] [P ](1 x)Kp 3, 086 x 0, 9987[P ] (1 x)= = = =+ Vy ti thi im cn bng 250C: 2 23COH CO[P ] 1 x 1, 3.10 atm[P ] [P ] 1 x 1, 9987 atm= == = + = Cu 43: Nn 2 mol Nit v 8 mol Hidr vo mt bnh kn c th tch 2 lt (ch cha sn cht xc tc vi th tch khng ng k), c gi nhit khng i. Khi phn ng trong bnh t cn bng,p sut cc kh trong bnh bng 0,8 p sut lc u (lc mi cho xong cc kh vo, cha phn ng) .Tm hng s cn bng caphn ng. Gii:Tm hng s cn bng: Phng trnh phn ng: N2+3H2 2NH3

Ban u 2 mol 8 mol0molPhn ng xmol 3x mol 2x molCn bng (2- x mol)( 8- 3x mol)2x molV phn ng xy ra nhit khng i v trong bnh kn nn gia p sut v s molta c t lP1; p sut trc phn ng P2; p sut sau phn ng n1; s mol trc phn ng n2; s mol sau phn ng 2 11108 , 0 n PP=mol n 82 =. Tng s mol cc cht sau phn ng: (2 - x)+(8 - 3x) + 2x = 8 x = 1 mol 2Nn (sau) = 2 1 = 1 mol [N2] = l molV/21 1= 2Hn (sau) = 8 3 = 5 mol [H2] = l molV/25 5= 3NHn = 2 x 1 = 2 mol [NH3] = l molV/ 122 2= = Hng s cn bng ca phn ng: 128 , 01251625211] ][ [] [3232 223= =|.|

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\|= =H NNHKcb Mt s bi tp t luyn 2121nnPP= Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 24 - 1. Mt bnh 5 lt cha 1 mol HI (k) c un nng ti 8000C. Xc nh phn trm phn li ca HI 8000C theo phn ng 2HIH2 + I2 (k) Bit Kc = 6,34. 10 4

p s: 4,8% 2. 250c hng s cn bng Kp i vi phn ng N2 + 3H2 2NH3 bng 6,8.105. a) TnhAG0 ca phn ng. b) Nu cng nhit trn, p sut u ca N2, H2, NH3 l 0,250; 0,550 v 0,950 atm. Tm AG ca phn ng. p s:a) -33,28 kJ;b) -25,7kJ 3. Ngi ta tin hnh phn ng: PCl5

PCl3 + Cl2 vi 0,3 mol PCl5; p sut u l 1 atm. Khi cn bng c thit lp, p sut o c bng 1,25 atm (V,T = const) a) Tnh phn li v p sut ring ca tng cu t. b) Thit lp biu thc lin h gia phn li v p sut chung ca h. p s: a) 0,25;P = P0( 1 + o ) 4. Xc nh nhit i vi phn ng CaCO3 CaO + CO2 bit rng 8000C p sut phn li bng 201,8mm Hg v 9000C bng 992 mm Hg. p s: -166,82 kJ/mol 5. Trong mt th nghim ngi ta t mt mpun cha N2O4 lng c m = 4,6 g vo mt bnh phn ng ui ht khng kh c dung tch 5,7lt. p v mpun ri a nhit ca bnh phn ng ln 500C; Kt qu l N2O4 bay hi v b phn li, p sut trong bnh o c l 0,4586atm. Tnh phn li ca N2O4 v hng s cn bng Kc i vi phn ng N2O4 2NO2 . p s: 97,4% ;Kc = 8,58 6. Mt hn hp u gm 7% SO2, 11% O2 v 82% N2 di p sut 1 atm, c un nng ti 10000K vi s c mt ca mt cht xc tc. Sau khi cn bng c thit lp, trong hn hp cn bng SO2 chim 4,7%. Tm mc oxi ha SO2 thnh SO3 v hng s cn bng Kp v Kc ca phn ng: 2SO2 + O2

2SO3 (ghi ch: mc oxi ha c o bng t s gia p sut cn bng v p sut u) . p s: 32,9% , Kp = 2,44 ; Kc = 200 7. Trong s tng hp NH3 4000C theo phn ng N2 + 3H2 2NH3 hn hp u gm N2 v H2 c ly theo ng t l hp thc ri a vo bnh phn ng dung tch 1lt. Trong hn hp cn bng, ngi ta thy c 0,0385 mol NH3. Tnh Kc, Kp. p s: Kc = 5,12. 107; Kp = 1,68.104 8. 250C hng s cn bng Kp ca phn ng thu nhit 2NO + Br2 (k)2NOBr (k) bng 116,6 atm 1. a) Nu em trn NOBr c P = 0,108 atm vi NO c P = 0,1atm v Br2 c P = 0,01 atm to ra mt hn hp kh 00C th v tr cn bng s nh th no (cu tr li phi nh lng). b) a NOBr c P = 5 atm vo bnh phn ng 500C th thy trong hn hp cn bng cNOBr P = 4.30 atm. Tnh Kp 500C . So snh gi tr Kp ny vi Kp 250C. Gii thch? p s: Kp (500C) = 179 atm -1 9. Ngi ta cho 1 mol axit axetic tc dng vi 1 mol ru izopropylic. nhit t, cn bng : CH3COOH + CH3CHOHCH3CH3COOCH(CH3)2+H2O s t c khi 0,6 mol este c to ra. a/ Tnh thnh phn ca hn hp cn bng. b/ Nu thm vo hn hp cn bng ny 1 mol CH3COOH hoc 1 mol CH3CHOHCH3 hoc 1 mol este th thnh phn ca hn hp cn bng mi s l bao nhiu ? S :a. axit : 0,4 mol ; ru : 0,4 mol ; este : 0,6 mol ; nc : 0,6 molb. Thm 1 mol axit :1,22 mol axit ; 0,22 mol ru ; 0,78 mol este ; 0,78 mol nc10. 270C v di p sut 1 atm phn ly ca N2O4 l 20%. Theo phn ng : N2O4 (k) 2NO2 (k) a/ Tnh : -KP v cho bit phn ng thu nhit hay ta nhit, bit 630C phn ng c KP = 1,27 ? - phn ly p sut 0,05 atm. b/ Ngi ta cho 46 gam N2O4 vo bnh kn c dung tch 20 lt 270C. Tnh thnh phn ca hn hp kh lc cn bng. Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 25 - S : a. KP = 0,167phn ng thun thu nhit b. 0,385 mol N2O4 ; 0,230 mol NO2 11. Mt bnh phn ng c dung tch 10 lt cha 0,1 mol H2 v 0,1 mol I2 698 K, bit hng s cn bng KC = 54,4. Tnh nng cn bng ca H2, I2 v HI S :[H2] = [I2] = 0,00213 mol/l [HI] = 0,0157 mol/l 12. Cn bng ca phn ng kh CO2 bng C :C (r) +CO2 (k) 2CO (k)xy ra 1090 K vi hng s cn bng KP = 10 a/ Tm hm lng kh CO trong hn hp kh cn bng, bit p sut chung ca h l 1,5 atm b/ c hm lng CO bng 50% v th tch th p sut chung l bao nhiu ? S : a. 88%b. 20 atm 13. 500C v di p sut 0,344 atm, phn lyoca N2O4 (k) thnh NO2 (k) bng 63%. Xc nh KP, KC S :KP = 0,867 atm; KC = 0,034 14. 00C v di p sut 1 atm, phn lyoca N2O4 (k) thnh NO2 (k) bng 11%. a/ Xc nh KP. b/ Cng ti 00C khi p sut gim t 1 atm xung 0,8 atm th phn lyothay i nh th no ? c/ Cn phi nn ng nhit hn hp kh ti p sut no phn lyobng 8% ? S : a. 0,049 atmb. o = 12,3% (tng) c. P = 1,9 atm. 15. 630C hng s cn bng Kp ca phn ng : N2O4 (k)2NO2 (k) Kp = 1,27 Tnh thnh phn ca hn hp theo s mol khi p sut ca h ln lt bng : 1 atm; 10 atm. T rt ra kt lun g v nh hng ca p sut n s chuyn dch cn bng ? S : P = 1 atm : NO2 = 66% N2O4 = 34%P = 10 atm : NO2 = 29,8%N2O4 = 70,2% 16.Phn ng di y s din ra theo chiu no khi 00C v 1000C ? N2O4 (k) 2NO2 (k) Bit0298AH (kcal/mol)2,31 8,09 0298S A(cal/K.mol) 72,73 57,46 17. 8500C phn ng : CO(k) + H2O(k) CO2 (k)+H2 (k)ti trng thi cn bng c hng s cn bng KC = 1. Cho bit nng cc cht thi im u : CCO = 2 mol/l ; CH2O = 0,1 mol/l v CCO2 = 0, CH2 = 0. Tnh nng cc cht trng thi cn bng.S :[CO] = 1,133M ; [H2O] = 0,033M ; [CO2] = [H2] = 0,067M18.Cho phn ng : N2 + 3H22NH3tin hnh nhit v th tch khng i. Bit rng ti trng thi cn bng cc cht c nng nh sau : [N2] = 0,3M ; [H2] = 0,9M ; [NH3] = 0,4M a/ Tnh hng s cn bng ca phn ng b/ Tnh nng ban u ca N2 v H2

S : a. KC = 0,73 b. N2 = 0,5M ; H2 = 1,5M 19. Bng cch tnh ton hy cho bit cn bng sau y : 2SO2 (k)+O2 (k)2SO3 (k) chuyn dch theo chiu no khi tng p sut ca h ln 3 ln (bng cch nn) ? Cho bit nhit khng i. 20. Tin hnh phn ng thun nghch trong bnh kn c dung tch 1 lt. CO(k)+ Cl2 (k) COCl2 (k) nhit khng i nng cn bng ca cc cht l : [CO] = 0,02M ; [Cl2] = 0,01M ; [COCl2] = 0,02M. Bm thm vo bnh 0,03 mol Cl2. Tnh nng cc cht trng thi cn bng mi. S :[CO] = 0,01M ; [Cl2] = 0,03M ; [COCl2] = 0,03M 22 Cho cn bng ha hc :N2 +3H2 2NH3vi AH= - 92 KJ.mol-1 Nu xut pht t hn hp cha N2 v H2 theo t l s mol ng bng h s t lng, tc t l 1:3 th khi t ti trng thi cn bng ( 4500C v 300 atm) NH3 chim 36% v th tch. a. Tnh hng s cn bng Kp

b. Gi nhit khng i (4500C) cn tin hnh di p sut bao nhiu khi t ti trng thi cn bng NH3 chim 50% v th tch ? c. Gi p sut khng i (300 atm) cn bng cn tin hnh nhit no khi t ti trng thi cn bng NH3 chim 50% v th tch ? Trng THPT K Thut L Thy Ti liu luyn thi HSG Gio vin: Nguyn Cao Chung- 26 - S :a. Kp = 8,14.10-5b. P = 680 atm c. 653K 23 Cho phn ng :CO2 (k)+H2 (k)CO (k)+ H2O (k) (1) a. TnhAG0 ca phn ng (1) 1000K, bit 01000KAH= 35040 J.mol-1. 01000KS A= 32,11 J.mol-1.K-1 b. Tnh hng s cn bng KC, KP ca phn ng (1) 1000K. c. Mt hn hp kh cha 35% th tch H2, 45% th tch CO v 20% th tch hi nc c nung nng ti 1000K. Tnh thnh phn hn hp trng thi cn bng. S : a. 2930 J b. KC = KP = 0,703c. CO = 34,6% ; CO2 = 10,4% ; H2O = 9,6% ; H2 = 45,4%