hv 2014-hdc hoa hoc 11
DESCRIPTION
Hv 2014-Hdc Hoa Hoc 11TRANSCRIPT
S GD&T QUNG NINH
TRNG THPT CHUYN H LONG
P N OLYMPIC TRI H HNG VNG LN TH X
MN: HA HC - KHI: 11
Ngy thi: 01 thng 08 nm 2014
Thi gian: 180 pht
Bi 1: (2,5 im)
1. Xt phn ng: IO3- + 5I- + 6H+ ( 3I2 + 3H2O (1)
Vn tc ca phn ng (1) o 25oC c gi tr theo bng sau:
Th nghim[I-][IO3-][H+]Vn tc
(mol.l-1.s-1)
10,010,10,010,6
20,040,10,012,4
30,010,30,015,4
40,010,10,022,4
a. Lp biu thc tnh tc phn ng.
b. Tnh hng s tc phn ng v xc nh n v ca hng s tc . (Hc sinh ch cn tnh mt gi tr k da trn cc s liu cho)
c. Nng lng hot ha ca phn ng E = 20kJ/mol 25oC. Vn tc ca phn ng
thay i th no nu nng lng hot ha gim i mt na?
2. un nng hn hp kh gm O2 v SO2 c cht xc tc, xy ra phn ng:
O2 + SO2 SO3
(2)
Tnh hng s cn bng Kp ca phn ng 60oC (chp nhn hiu ng nhit ca phn ng khng ph thuc nhit ). Nhit nh hng nh th no ti trng thi cn bng ca phn ng (2)? Cho cc s liu nhit ng nh sau:
Kh(kJ.mol1)(J.K1.mol1)
SO3-395,18256,22
SO2-296,06248,52
O2 0,0205,03
p n:
1
a.v = k [I- ]x. [IO3- ]y. [H+ ]z
Thay cc gi tr nng thch hp vo mi th nghim
0,6 = k [0,01 ]x. [0,1 ]y. [0,01 ]z
2,4= k [0,04 ]x. [0,1 ]y. [0,01 ]z
5,4 = k [0,01 ]x. [0,3 ]y. [0,01 ]z
2,4 = k [0,01 ]x. [0,1 ]y. [0,02 ]z
Gii cc phng trnh ta tm c x = 1, y = 2, z = 2.
Biu thc tnh tc l: v = k [I- ]. [IO3- ]2. [H+ ]20.250.25
0.25
b. Thay x,y,z vo mt trong cc phng trnh ta c k = 6.10-7 (mol-4.l4.s-1)0.25
c. ta c: k1 = A e -E1/RT , k2 = A e -E2/RT , ln k2/k1 =
Th vo biu thc ta c: ln = ( k2 = 56,6k1
Vy tc phn ng tng 56,6 ln0.5
2Ta c: = - RTlnKp
25 oC: . T phn ng: O2 + SO2 SO3, suy ra:
= (- 395,18 + 296,06) 298.10-3 . (256,22 248,52 - . 205,03)
= - 99,12 - 298.10-3.(- 94,815) - 70,87 (kJ.mol-1 )
= 2,65.1012.
Khi = const, ta c:
3,95.1010 (atm- ).
(Hc sinh tnh Kp c th ghi n v hoc khng ghi u tnh im)
Khi tng nhit t 25 oC n 60 oC, hng s cn bng Kp gim t 2,65.1012 xung 3,95.1010 (atm- ), iu ny hon ton ph hp vi nguyn l Le Chatelier (L Satli), do phn ng (1) ta nhit.0.25
0.25
0.25
0.25
Bi 2: (2,5 im)
1. Mt dung dich X gm hn hp HNO3 0,001M va H3PO4 0,001M.
a. Tinh pH cua dung dich X.
b. Thm 50 ml dung dich NaOH 0,002M vao 50 ml dung dich X thu c dung dich Y. Tinh pH cua dung dich Y. Bit cac hng s axit cua H3PO4 ln lt la: pKa1 = 2, 15; pKa2 = 7, 21; pKa3 = 12, 32.
2. Thit lp s pin khi pin hot ng th xy ra cc phn ng sau:a. MnO + Cr3+ Cr2O +...b. Cu2+ + 4NH3 Cu(NH3)
p n:
1a. Tinh pH cua dung dich X
HNO3 ( H+ + NO3-
0,001M 0,001M
H3PO4 H+ + H2PO4- (1) Ka1= 10-2,15 H2PO4- H++ HPO42- (2) Ka2 = 10-7,21 HPO42- H+ + PO43- (3) Ka3 = 10- 12,32 H2O H+ + OH- (4) Kw = 10-14Do Ka1 >> Ka2 >> Ka3 va Kw nn cn bng (1) la chu yu
H3PO4 H+ + H2PO4- (1) Ka1= 10-2,15Co : 0,001M 0,001M
[ ] : 0,001-x 0,001 + x x
Theo LTDKL ta co :
EMBED Equation.DSMT4 = 10-2,15 x = 7,975.10-4 [H+] = 0,001 + 7,975.10-4 = 1,7975.10-3 M
Vy pH cua dd X = 2,750.5
b. Tinh pH cua dung dich Y
Tinh lai nng : CNaOH = 0,001M ; CHNO3 = 5.10-4M; CH3PO4 = 5.10-4M
P: NaOH + HNO3 ( NaNO3 + H2O
5.10-4 5.10-4
NaOH + H3PO4 ( NaH2PO4 + H2O
5.10-4 5.10-4 5.10-4
DD Y : NaH2PO4 : 5.10-4 M
NaH2PO4 ( Na+ + H2PO4-
H2PO4- H++ HPO42- Ka2 = 10-7,21 H2PO4- + H+ H3PO4 (Ka1)-1 = 102,15 H2O H+ + OH- Kw = 10-14iu kin proton mc khng : H2PO4- , H2O
[H+] = [OH-] + [ HPO42-] -[H3PO4]
h = Kw/h + Ka2. - (Ka1)-1.h.[H2PO4-]
h = Coi gn ung: [H2PO4-] = CH2PO4-= 5.10-4 M
Ka2.[H2PO4-] >>Kw
h= = 5,367.10-6 pH cua dd Y = 5,27
0.5
0.5
2a) Cp oxi-ha kh: ,
Catot (+): qu trnh kh:
Anot (-): qu trnh oxi ha:
S pin (-) Pt Cr2O72-, Cr3+, H+ MnO4-, Mn2+, H+ Pt (+)
0.25
0.25
b) Chn in cc lm vic thun nghch vi ion Cu2+: Cu2+/Cu, [Cu(NH3)4]2+/Cu
; [Cu2+] ln E ln cc (+)(+) Cu2+/Cu: Cu2+ + 2e Cu(-) [Cu(NH3)4]2+/Cu: Cu + 4NH3 [Cu(NH3)4]2+ + 2e
S pin (-) Cu [Cu(NH3)4]2+, NH3 Cu2+ Cu (+)
0.25
0.25
Bi 3: (2,5 im)
1. Nguyn t X (c nhiu dng th hnh) c mt anion cha oxi ng vai tr quan trng trong nhim nc. m in ca n nh hn ca oxi. N ch to hp cht phn t vi halogen. Ngoi hai oxit n phn t cn c nhng oxit cao phn t. X cn c vai tr rt quan trng trong sinh ha. Cc orbital p ca nguyn t X c 9 electron.
a. l nguyn t no? Vit cu hnh ca n.
b. Nguyn t X to c nhng axit cha oxi (oxoaxit) c cng thc chung H3XOn vi n = 2, 3, 4. Vit cng thc cu to ca 3 axit ny. nh du cc nguyn t H axit v ghi s oxi ha ca X trong cc hp cht ny.
2. A la mt hp cht cua nit va hiro vi tng s in tich hat nhn bng 10. B la mt oxit cua nit, cha 36,36% oxi v khi lng.
a. Xac inh cac cht A, B, D, E, G, X va hoan thanh cac phng trinh phan ng:
A + NaClO X + NaCl + H2O A + Na G + H2
G + B E + H2O E + H2SO4 D + NaHSO4
b. Vit cng thc cu tao cua D.
p n:
1a. Vit cu hnh ca X v suy ra X l nguyn t photpho 0.25
b.
0.75(mi ct ng 0.25)
2a. Gi s hp cht ca N v H c cng thc NxHy. V tng in tch ht nhn ca phn t bng 10, m N c Z = 7 v H c Z = 1 nn hp cht A ch c th l NH3.
- Oxit ca N cha 36,36% khi lng l O do , nu gi thit rng trong phn t B c 1 nguyn t O (M = 16) th s nguyn t N trong phn t l: N = 16(100-36,36) : 36,36x14 = 2. Nh vy B l N2O.Cc phn ng ho hc ph hp l:
2NH3 + NaClO N2H4 + NaCl + H2O
2NH3 + 2Na 2NaNH2 + H2
NaNH2 + N2O NaN3 + H2O
NaN3 + H2SO4 HN3 + NaHSO4Nh vy: A = NH3; B = N2O; D = HN3; E = NaN3; G = NaNH2; X = N2H40.250.250.25
0.25
0.25
b. Cng thc cu to ca cht D (HN3 - axit hirazoic) l:
H N(-3) = N(+5) N(-3).0.25
Bi 4: (2,5 im)
1. Nitro ha phenol (vi axit nitric long) thu c ch yu p-nitrophenol v o-nitrophenol, cng vi mt lng nh m-nitrophenol. Theo c ch phn ng th electrophin, nhm NO2+ c th tn cng vo cc v tr trn vng benzen to ra phc ( . bn ca cc phc ( s quyt nh sn phm no c to thnh nhiu hn.
a. Vit 1 cng thc cng hng bn nht ca phc ( ng vi v tr tn cng octo/ para hoc
meta ca phenol gii thch kt qu ny.
b. So snh v gii thch (i)- nhit si, (ii)- ha tan trong nc ca p-nitrophenol v o-
nitrophenol.
2. Nguyn t -H cnh nhm carbonyl (C=O) kh linh ng. Chn nguyn t H linh ng nht mi hp cht sau v so snh linh ng ca chng. Gii thch s la chn ny.
3. Phn ng Diels-Alder ca hai hp cht sau cho cc ng phn-dn xut xiclohexen.
a. V mt l thuyt c bao nhiu ng phn (cu to, lp th) c to thnh, gii thch.
b. Vit 2 ng phn bt k l ng phn cu to (khc nhau v v tr).
c. Vit 2 ng phn bt k l cp ng phn i quangd. Vit 2 ng phn bt k l ng phn dia (khng phi i quang).p n:
1a. (0,5 ) Nitro ha phenol. Phc ( (ng vi cc v tr octo, para v meta) bn th sn phm to ra nhiu hn.
Lu : HS khng cn phi vit ht cc cu trc cng hng, nhng phi vit cu trc bc 3o (1), (2) trng hp octo, para v 1 cu trc bc 2o trng hp meta.
Trong cc cng thc cng hng ca phc (, to thnh khi nhm NO2+ tn cng vo v tr octo hay para c cation bc 3; cation ny c bn ha nh gii ta in tch dng bi i e ca nhm HO, nn cc sn phm to ra t chng l chnh.
Cn khi NO2+ tn cng vo v tr meta ch hnh thnh cation bc 2 v khng c gii ta in tch dng nh 2 trng hp trn, nn sn phm to ra rt t.
b. (0,5 ) So snh v gii thch (i)- nhit si; (ii)- ha tan. ng phn p-nitrophenol to c lin kt-H lin phn t, cn o-nitrophenol ch c lin kt-H ni phn t, nn
Nhit si ca p-nitrophenol > ts ca o-nitrophenol.
Cng vi l do trn, p-nitrophenol d to c lin kt-H vi nc, cn o-nitrophenol kh to lin kt-H vi nc, nn
tan trong nc ca p-nitrophenol > tan ca o-nitrophenol.
2(0,25 ) Xp cc cht theo chiu tng linh ng-H.
(0,25 ) Gii thch: cc nhm ht e c nh ln n linh ng ca -H.
(IV) c nhm NO2 l nhm ht e mnh (hiu ng -I)
(II) c nhm CHO l nhm ht e yu hn NO2
(I) c nhm CH3CO l nhm ht e yu hn CHO
(V) c nhm C2H5CO l nhm ht e yu hn CH3CO
(III) c nhm CH2Cl l nhm ht e yu nht.
3a) (0,25 ) C 8 ng phn.
C th to ra 2 sn phm l ng phn cu to (khc nhau v v tr). Mi ng phn ny c th tn ti 4 ng phn lp th (2 C*).
b) (0,25 ) Cu trc ca hai sn phm bt k l ng phn v tr (HS khng cn v lp th).
0,125
0,125
c) (0,25 ) Cu trc ca hai sn phm bt k l ng phn i quang (HS ch cn vit 1dng i quang).
0,125
0,125
d) (0,25 ) Cu trc ca hai sn phm bt k l ng phn dia (HS ch cn vit 1loi ng phn lp th). 0,125
0,125
Bi 5: (2,5 im)
1. Phn ng ca brm vi buten-2 xy ra theo hng cng trans vo ni i. Vit c ch phn ng ca:
a. Brm trong CH2Cl2 vi cis-buten-2; Sn phm l dng erythro hay threo ? (v 1 dng minh ha). K hiu cu hnh tuyt i (R hay S) mi tm bt i trong phn t sn phm (ch cn vi 1 i quang).
b. Brm/CH2Cl2 vi trans-buten-2; Sn phm l dng erythro hay threo ? (v 1 dng minh ha). K hiu cu hnh tuyt i (R hay S) mi tm bt i trong phn t sn phm (ch cn vi 1 i quang).
2. Phn ng Robinson xy ra qua cc bc: cng Michael, ngng t ng vng andol v loi nc.Vit c ch gii thch s to thnh sn phm ca phn ng sau. 3. Vit c ch gii thch s to thnh sn phm ca phn ng sau.
p n:
1a. (0,5 ) C ch phn ng ca brm/CH2Cl2 vi cis-buten-2
Sn phm l dng threo.
Cu hnh tuyt i: 2R v 3R hoc 2S v 3S (HS vit 1 trong 2 dng threo v 1trong 2 cp cu hnh trn u c)
(0,5 ) C ch phn ng ca brm/CH2Cl2 vi trans-buten-2.
Sn phm l dng erythro.
Cu hnh tuyt i: 2R v 3S hoc 2S v 3R. (HS vit 1 trong 2 dng erythro v 1trong 2 cp cu hnh trn u c)
2(1,0 ) Vit c ch phn ng Robinson.
3(0,5 ) C ch chuyn v ca ancol khi c mt ca axit.
Bi 6: (2,5 im)
1. Anetol (cng thc phn t C10H12O) c trong hoa hi. Cu trc ca anetol c xc nh da vo cc thng tin sau:
(i)- lm mt mu nc brm; (ii)- oxi ha thu c axit metoxi benzoic A (C8H8O3) v axit B (C2H4O2); (iii)- Nitro ha A ch thu c mt dn xut mononitro duy nht C.
a) Lp lun v vit cng thc cu trc ca anetol, A, B v C.
b) V cu trc 2 ng phn hnh hc ca anetol.
2. Cu trc ca hp cht X (cng thc phn t C9H8) c xc nh da vo cc thng tin sau:
(i)- hidro ha trong iu kin m du cho D (C9H10);
(ii)- hidro ha hon ton cho E (C9H16);
(iii)- oxi ha mnh thu c axit phtalic [o-C6H4(COOH)2].
Lp lun v vit cng thc cu trc ca E v X.
3. Hp cht Y (C13H18O) c hot tnh quang hc, khng phn ng vi initrophenylhydrazin, nhng tham gia phn ng iodoform. Phn ng zon phn (kh ha) A cho metyl xeton F v cht G. Nu ly sn phm axit to ra t G vi thuc th Tolenx em un nng th thu c H (C6H8O4). V tr ca cc nhm trong F c xc nh nh sau. Khi nitro ha F, cc nhm ny u nh hng ng thun nhm NO2 gn vo cng 1 v tr v cho 1dn xut mononitro.
Lp lun v vit cng thc cu trc ca Y, F, G v H.
p n:
1(1,0 ) Anetol l dn xut ca benzene, c cng thc phn t C10H12O ( bt bo ha: = 5).
Anetol lm mt mu nc brm nn mch nhnh c mt lin kt i.
Oxi ha thu c axit metoxi benzoic A (C8H8O3) v axit axetic B (CH3COOH).
Nitro ha A ch thu c mt dn xut mononitro duy nht C: vy 2 nhm COOH v OCH3 trong A nm v tr 1,4 (para).
0.25 0.25 0.25 2 ng phn hnh hc ca anetol. (0.125 x 2 = 0.25)
2(0,5 ) Hp cht X (C9H8) l mt dn xut ca benzen c bt bo ha, = 6.
X c mt lin kt i (hidro ha thnh D, C9H10).
E (C9H16) khng cn lin kt i na nhng vn c = 2, nn E phi l hp cht bixiclic (2 vng gip v sn phm oxi ha l axit phtalic): 1 vng 6 v 1 vng 5.
Vy, X v E (cho im ch vi 2 cng thc ny):
0.25 0.25
3(1,0 ) Hp cht Y (C13H18O) c bt bo ha, = 5), d on y l dn xut ca benzene vi 1 lin kt i mnh nhnh.
G l andehit, khi oxi ha cho axit lactic (C3H6O3), un nng cht ny to ra dimer H (loi 2H2O).
Cng thc phn t ca metylxeton F c th xc nh c thng qua CTPT ca Y v CTPT ca G. Suy ra, F l dn xut 2 ln th ca benzene vi cc nhm C2H5 v CH3CO. Khi nitro ha F, c 2 nhm ny cng nh hng ng thun nhm NO2 gn vo 1 v tr (trn vng benzene) v cho 1dn xut mononitro. iu ny ch ph hp khi 2 nhm ny nm v tr 1,4 (para).
Do Y quang hot nn cu trc phi l:
0.25 0.25 0.25 0.25
Bi 7: (2,5 im)
1. Hp cht F c tng hp theo s sau.
a) Vit cng thc cu trc ca A, B, C v D.
b) Vit cng thc ca tc nhn E.
Gi . * Phn ng t A n B: EtAlCl2 ng vai tr l axit Lewis;
* C c cng thc phn t C13H22O.
2. Cho axetilen, mt hp cht carbonyl (ty chn) v cc tc nhn cn thit khc, hy vit cc bc tng hp cis-2-metylhex-4-en-3-ol.
3. T benzen v cc ha cht cn thit khc hy vit cc phn ng tng hp K. p n:
1(0,75 )
Cng thc cu trc ca A, B, C, D v E. Mi cng thc ng: 0.15
2(0,75 ) Tng hp cis-2-metylhex-4-en-3-ol.
Phn t gm ba mnh cu trc:
C th ni 2 phn no trc cng c!
Nhn c CH3CCH: 0,25
Nhn c ancol : 0,25 Kh vi xt Lindlar: 0,25 Kh lin kt ba
3(1,0 ) T benzen v cc ha cht cn thit hy vit cc phn ng tng hp K.
Tng hp n B, F, H, J v phn ng to ra K: mi giai on 0,15
Bi 8: (2,5 im)
Cht X dng tinh th mu trng c cc tnh cht sau:
- t X nhit cao cho ngn la mu vng.
- Ha tan X vo nc c dung dch A, cho kh SO2 i t t qua dung dch A thy xut hin mu nu. Nu tip tc cho SO2 qua th mu nu bin mt thu c dung dch B; thm mt t HNO3 vo dung dch B, sau thm d dung dch AgNO3 thy to thnh kt ta mu vng.
xc nh cng thc phn t ca X ngi ta ha tan hon ton 0,1 g X vo nc thm d KI v vi ml H2SO4 long, lc c mu nu, chun bng Na2S2O3 0,1 M ti mt mu tn ht 37,4 ml dung dch Na2S2O3. Tm cng thc phn t ca X. Gi s cc phn ng xy ra hon ton.
p n:
X chy cho ngn la mu vng ( thnh phn nguyn t ca X c natri.
Dung dch X tc dng vi SO2 n d thu c dung dch B to kt ta vng vi AgNO3( thnh phn nguyn t ca X c iot.
Phn ng ca X vi SO2 chng minh X c tnh oxi ha.
T lp lun trn X c cation Na+ v anion IO
t cng thc ca X l NaIOx.
IO + (2x-1) I + 2x H( x I2 + x H2O (1)
I2 + 2Na2S2O3( 2NaI + Na2S4O6 (2)
1,87.10-3( 3,74.10-3S mol Na2S2O3 = 0,1.0,0374 = 3,74.10-3Theo (2) ( S mol I2 = (S mol Na2S2O3) = 1,87.10-3Theo (1) ( S mol IO= (s mol I2) = .1,87.10-3( = .1,87.10-3( = 1,87.10-3 0,1x = 0,2805 + 0,02992x
( x = 4
Cng thc phn t ca X: NaIO40.5
0.5
0.25
0.25
0.25
P N CHNH THC
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
(2)
(1)
t
(0.25)
(0.25 )
(0.25)
(0.25)
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0.25
0.25
0.25
0,25
0,25
0,25
0,15
0,15
0,15
0,15
0,15
1
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