hv 2014-hdc hoa hoc 10
DESCRIPTION
Hv 2014-Hdc Hoa Hoc 10TRANSCRIPT
S GD&T HA BNH
S GD&T QUNG NINH
TRNG THPT CHUYN H LONG
THI OLYMPIC TRI H HNG VNG LN TH X
MN: HA HC - KHI: 10 Ngy thi: 01 thng 08 nm 2014
Thi gian: 180 pht thi gm: 03 trang.
Bi 1 (2, 5 im)
1. 134Cs v 137Cs l sn phm phn hch ca nhin liu urani trong l phn ng ht nhn. C hai ng v ny u phn r -. Vit phng trnh phn ng ht nhn biu din cc phn r phng x ca 134Cs v 137Cs, tnh nng lng (ra eV) c gii phng trong phn r ca 134Cs. Cho bit: Nguyn t khi (u) ca 55134Cs l 133,906700; 56134Ba l 133,904490.
S Avogaro NA = 6,02.1023; 1eV = 1,602.10-19J; c = 2,998.108 ms-1
2. Tnh nng lng ion ho I4 v I5 ca nguyn t 5X3. ng v phng x 13N c chu k bn r l 10 pht, thng c dng chp cc b phn trong c th. Nu tim mt mu 13N c hot phng x l 40 (Ci vo c th, hot phng x ca n trong c th sau 25 pht s cn li bao nhiu?Bi 2 (2,5 im) 1. Cho cc phn t XeF2, XeF4, XeOF4. a. Vit cng thc cu to Li - uyt (Lewis) cho tng phn t.
b. D on cu trc hnh hc ca cc phn t .
c. Hy cho bit kiu lai ho ca nguyn t trung tm trong mi phn t trn.
2. Cho kim loi A tn ti c 2 dng lp phng tm khi v lp phng tm din. Khi A tn ti dng lp phng tm khi th khi lng ring ca A l 15g/cm3. Hy tnh khi lng ring ca A dng lp phng tm din. Cho rng bn knh ca A nh nhau trong c 2 loi tinh th. (Ch : Hc sinh khng bt buc phi v mng c s)3. Bit rng mono clobenzen c momen lng cc (1 = 1,53 D. Hy tnh momen lng cc (m ; (p ca meta, para diclobenzen.
Bi 3 (2,5 im)
1. nhit cao, c cn bng : I2 (k) 2 I (k) (1)
Bng sau y tm tt p sut ban u ca I2 (k) v p sut tng cn bng t c nhng nhit nht nh.T (K)10731173
P(I2) (atm)0.06310.0684
P tng (atm)0.07500.0918
Tnh (H, (G v (S 1100 K. (Gi s (H v (S khng ph thuc nhit trong khong nhit nht nh). Cho: hng s kh R = 8,314(J.mol-1.K-1)2. t0C v 1 atm, cn bng (1) c Kp = 4, 9.10-3. Tnh phn ly ca I2 iu kin ny. Cho bit gi tr Kp s thay i th no khi cn bng cho c vit di dng: 1/2I2(k)( I(k).
Bi 4 (2,5 im)
1. Cc qu trnh trao i cht din ra trong c th ng vt c th sn sinh ra cc cht c hi, th d O2(. Nh tc dng xc tc ca mt s enzim (E) m cc cht ny b ph hu.
Th d 2 O2( + 2 H+
O2 + H2O2 (() Ngi ta nghin cu phn ng (() 25o C vi xc tc E l supeoxieimutaz (SOD). Cc th nghim c tin hnh trong dung dch m c pH = 9,1. Nng u ca SOD mi th nghim u bng 0,400.10(6 mol.L(1. Tc u Vo ca phn ng nhng nng u khc nhau ca O2( c ghi bng di y:
Co (O2() (mol.L(1)7,69.10(63,33.10(52,00.10(4
Vo (mol.L(1.s(1)3,85.10(31,67.10(2 0,100
a. Thit lp phng trnh ng hc ca phn ng (() iu kin th nghim cho.
b. Tnh hng s tc phn ng. (Hc sinh ch cn tnh 1 gi tr ca k).
2. C phn ng bc mt : CCl3COOH (k) ( CHCl3 (k) + CO2 (k) tin hnh 30oC, nng cht phn ng gim i mt na sau 1 gi 23 pht 20 giy. 70oC, nng cht phn ng gim i mt na sau 16 pht 40 giy.
a. Tnh thi gian cn nng gim xung cn 25% lng cht ban u 70oC.
b.Tnh nng lng hot ha ca phn ng.Bi 5 (2,5 im)
1. Dung dch X gm Cu(NO3)2 0,06M v Pb(NO3)2 0,04M. Tnh pH ca dd X. Bit:
2. Ngi ta iu ch mt dung dch Y bng cch ho tan 0,05 mol axit axetic v 0,05 mol Natri axetat trong nc ri thm nc n th tch 1 lt .Tnh pH ca dung dch Y. Cho Ka(CH3COOH) = 1,8.10-5.Bi 6 (2,5 im)
1. Lp s pin in theo quy c da trn hai bn phn ng di y:
Au3+ + 2e Au+ Fe3+ + 1e Fe2+ Ch r anot, catot v vit cc qu trnh oxi ha kh cc in cc, phn ng tng xy ra trong pin. Tnh sc in ng chun ca pin v hng s cn bng ca phn ng xy ra tron g pin. Gi s tt c cc nng l 1 M v cc p sut ring phn l 1, 0 atm. Cho ; ; .2. Cn bng cc phn ng oxi ho kh sau y bng phng php thng bng ion - electron:
a. Fe3P + NO3- + ....... NO +H2PO4- +...
b. Cr3+ + ClO3- + OH- CrO42- + Cl- +....
3. Cho gin Latimer ca photpho trong mi trng kim:
Tnh th kh chun ca cp HPO32- / H2PO2-.Bi 7 (2,5 im)
1. Cho 3 nguyn t A, B v C. Bit n cht A tc dng vi n cht B nhit cao sinh ra hp cht D. Cht D b thu phn mnh trong nc to ra kh chy c v c mi trng thi. n cht B v n cht C tc dng vi nhau cho kh E, kh ny tan c trong nc to dung dch lm qu tm ho . Hp cht ca A vi C (hp cht F) c trong t nhin v thuc mt trong nhng cht c cng rt cao. Xc nh A, B, C, D, E, F v vit phng trnh cc phn ng nu trn.2. Cc kh X, Y khc nhau c ng trong hai bnh. C hai kh u c mi kh chu, khng mu v c tng khi lng l 6,8 gam. Khi t chy hon ton trong O2 d, ton b lng kh X to ra 5,4 gam H2O v kh Z rt t tan trong nc. Khi t chy trong oxi d, ton b lng kh Y to ra nc v kh T c th lm mt mu dung dch nc Brom. Nu cho ton b cng lng kh Y nh trn i qua dung dch Pb(NO3)2 d thu c 23,9 gam kt ta en. Hn hp kh Z v T cn nng 9,2 gam v chim th tch 4,48 lt (ktc). Xc nh kh X, Y, Z, T v vit cc phng trnh phn ng xy ra. Gi s cc phn ng xy ra hon ton.Bi 8 (2,5 im) 1.Vit cc phng trnh phn ng xy ra (dng ion thu gn):
a. Ion I- trong KI b oxi ho thnh I2 bi FeCl3, O3.
b. Ion Br- b oxi ho bi H2SO4 c, BrO3- trong mi trng axit.
2. Mt dung dch A cha 2 mui Na2SO3 v Na2S2O3. Cho Cl2 d i qua 100 ml dung dch A ri thm vo hn hp sn phm mt lng d dung dch BaCl2 thy tch ra 6,524 gam kt ta. Mt khc, nu thm vo 100ml dung dch A mt t h tinh bt, sau chun dung dch A n khi mu xanh bt u xut hin th dng ht 29 ml dung dch it 0,5 M. Vit phng trnh ho hc dng ion thu gn v tnh nng mol mi cht trong dung dch A. Gi s cc phn ng xy ra hon ton.Cho nguyn t khi ca cc nguyn t :
S = 32; Ba = 137; O = 16; H = 1; Pb =207; S = 32; N = 14 ....................................................Ht....................................................
(Gim th coi thi khng gii thch g thm)H tn th sinh:............................................................SBD:..................................Gim th 1:...................................................................
Gim th 2:.................................................................
S S GD&T QUNG NINH
TRNG THPT CHUYN H LONG
P N THI HC SINH GII TRI H HNG VNG LN TH X
MN: HA HC
LP: 10
Ngy thi: 01 thng 08 nm 2014(Hng dn chm c 11 trang)
CuNi dungim
Bi 12,5(2,5 im)
1. 134Cs v 137Cs l sn phm phn hch ca nhin liu urani trong l phn ng ht nhn. C hai ng v ny u phn r -. Vit phng trnh phn ng ht nhn biu din cc phn r phng x ca 134Cs v 137Cs, tnh nng lng (ra eV) c gii phng trong phn r ca 134Cs. Cho bit: Nguyn t khi (u) ca 55134Cs l 133,906700; 56134Ba l 133,904490.
S Avogaro NA = 6,02.1023; 1eV = 1,602.10-19J; c = 2,998.108 ms-1
2. Tnh nng lng ion ho I4 v I5 ca nguyn t 5X3. ng v phng x 13N c chu k bn r l 10 pht, thng c dng chp cc b phn trong c th. Nu tim mt mu 13N c hot phng x l 40 (Ci vo c th, hot phng x ca n trong c th sau 25 pht s cn li bao nhiu?
Hng dn chm:
111. 55134Cs 56134Ba + e (1)
55137Cs 56137Ba + e (2)
Nng lng thot ra trong phn r phng x ca 55134Cs:
E = m.c2 = (133,906700 - 133,904490) (10-3/6,02.1023)( 2,998.108)2(J)
= 3,3.10-13 J = 3,3.10-13/1,602.10-19 = 2,06.106 eV
0,25
0,250,50,25
25X4+ 5X5+ + 1e- I5 1s1 1s0E(1s1) = -13,6.52 /12 = -340 (eV) E(1s0) = 0 (eV)I5 = 0 (-340) = 340 (eV)5X3+ 5X4+ + 1e- I4 1s2 1s1E(1s2) = -2.13,6.(5 0,3)2 /12 = - 600,848 (eV) E(1s1) = - 340 (eV)
I4 = (-340) (-600,848)= 260,848 (eV) Ch : hc sinh c nhiu cch khc nhau ng u c im tuyt i0,250,250,25
3A = = (. N0. e((t = (. N A0 = (. N0( A = A0. e((t = A0.= 40. e( 2,5.ln2 = 7,071 (Ci.0,250,5
Bi 22,5 (2,5 im) 1. Cho cc phn t XeF2, XeF4, XeOF4. a. Vit cng thc cu to Li - uyt (Lewis) cho tng phn t.
b. D on cu trc hnh hc ca cc phn t .
c. Hy cho bit kiu lai ho ca nguyn t trung tm trong mi phn t trn.
2. Cho kim loi A tn ti c 2 dng lp phng tm khi v lp phng tm din. Khi A tn ti dng lp phng tm khi th khi lng ring ca A l 15g/cm3. Hy tnh khi lng ring ca A dng lp phng tm din. Cho rng bn knh ca A nh nhau trong c 2 loi tinh th. (Ch : Hc sinh khng bt buc phi v mng c s)3. Bit rng mono clobenzen c momen lng cc (1 = 1,53 D. Hy tnh momen lng cc (m ; (p ca meta, para diclobenzen.
Hng dn chm:
211. a. Cng thc cu to Li-uyt (Lewis)
b. Cu trc hnh hc
XeF2 : thng XeOF4 : thp vungc. Kiu lai ho ca nguyn t trung tm Xe: XeF2:sp3d XeF4: sp3d2XeF4 : vung phngXeOF4: sp3d2
0,25
0,25
0,25
2 a Mt mng lp phng tm khi:
Cnh a1 = 4r/ 3
Khi lng ring d1 = 15g/ cm3 S n v nguyn t: n1 = 8.1/8 + 1 = 2
Mt mng lp phng tm din:
Cnh a2 = 2 r2
Khi lng ring d2 (g/ cm3)
S n v nguyn t: n2 = 8.1/8 + 6.1/2 = 4
d = nM/ ( NA. V); V = a3Do :
d1: d2 = (n1 .a23) : (n2 .a13) =[ 2. (2 r2)3 ] : [ 4. (4 r/3)3 ] = 0,919
Suy ra: d2 = 16,32 g/cm30,250,250,250,25
3clo c m in ln, (1 hng t nhn ra ngoi
( = ( ( = 0
Dn xut meta: (m = (1 (tam gic u)Dn xut para: (p = (1 ( (1 = 0
0,250,25
0,25
Bi 3
2,5 1. nhit cao, c cn bng : I2 (k) 2 I (k) (1)
Bng sau y tm tt p sut ban u ca I2 (k) v p sut tng cn bng t c nhng nhit nht nh.T (K)
1073
1173P(I2) (atm)
0.0631
0.0684
P tng (atm)
0.0750
0.0918
Tnh (H, (G v (S 1100 K. (Gi s (H v (S khng ph thuc nhit trong khong nhit nht nh). Cho: hng s kh R = 8,314(J.mol-1.K-1)
2. t0C v 1 atm, cn bng (1) c Kp = 4,9.10-3. Tnh phn ly ca I2 iu kin ny. Cho bit gi tr Kp s thay i th no khi cn bng cho c vit di dng: 1/2I2(k)( I(k).
Hng dn chm:
31 I2 (k) 2I(k) Ban u P(I2) 0 Cn bng P(I2) x 2xPtng = P(I2)b x + 2x = P(I2)b + x ( x = Ptng P(I2) b
* 1073 K, x = 0.0750 0.0631 = 0.0119 atmP(I)cb = 2x = 0.0238 barP(I2)cb = 0.0631 0.0119 = 0.0512 atmK = = (khng quan trng n v ca K) * 1173 K, x = 0.0918 0.0684 = 0.0234 bar
P(I)cb = 2x = 0.0468 barP(I2)cb= 0.0684 0.0234 = 0.0450 barK = = 0,04867 = 0,0487
ln , ln = 1,4817
= 7,945(10(5 K(1 ( (Ho = = 155052 J = 155 kJ
* 1100K ; ln ( K1100 = 0,0169 = 0,017
(Go = (RTlnK = ( 8,314 (1100 ( ln(0,0169) = 37248,8 J = 37,2488 kJ
(Go = (Ho ( T(So ( (So = = 107,1 J.K(1
0,250,25
0,25
0,25
0,25
2 I2 (k) 2I(k) Ban u P 0 Cn bng P x 2xPtng = P x + 2x = P + x = 1 P = 1 x Vy : hay Gii phng trnh ta c : x = 0,0338 suy ra
hay = 3,5%Gi tr
khi cn bng cho c vit di dng: 1/2I2(k)( I(k).
0,25
0,25
0,50,25
Bi 4
2,5(2,5 im)
1. Cc qu trnh trao i cht din ra trong c th ng vt c th sn sinh ra cc cht c hi, th d O2(. Nh tc dng xc tc ca mt s enzim (E) m cc cht ny b ph hu.
Th d 2 O2( + 2 H+
O2 + H2O2 (() Ngi ta nghin cu phn ng (() 25o C vi xc tc E l supeoxieimutaz (SOD). Cc th nghim c tin hnh trong dung dch m c pH bng 9,1. Nng u ca SOD mi th nghim u bng 0,400.10(6 mol.L(1. Tc u Vo ca phn ng nhng nng u khc nhau ca O2( c ghi bng di y:
Co (O2() mol.L(17,69.10(63,33.10(52,00.10(4Vo mol.L(1.s(13,85.10(31,67.10(2 0,100
a. Thit lp phng trnh ng hc ca phn ng (() iu kin th nghim cho.
b. Tnh hng s tc phn ng.
2. C phn ng bc mt : CCl3COOH (k) ( CHCl3 (k) + CO2 (k) tin hnh 30oC, nng cht phn ng gim i mt na sau 1 gi 23 pht 20 giy. 70oC, nng cht phn ng gim i mt na sau 16 pht 40 giy.
a. Tnh thi gian cn nng gim xung cn 25% lng cht ban u 70oC.
b.Tnh nng lng hot ha ca phn ng.
Hng dn chm:
41a.
v = = k [H+]( [O2(]( ;
V [H+] l mt hng s nn v = k [O2(](
3,85.10-3 = k (7,69.10-6)(1,67.10-2 = k (3,33.10-5)( 0,100 = k (2,00.10-4)( =
0,231 = (0,231)( ( = 1
=
0,167 = (0,167)( ( = 1
=
0,0385 = (0,0385)( ( = 1
v = k [O2(]b.
k = = 501 s(1
k = = 502 s(1
k = = 500 s(1
ktb = 501 s(1 Hc sinh ch cn tnh 1 trong cc gi tr k, hoc l cho im0,250,250,250,25
2 = 6,93(10(4 s(1.
Theo t = ( t= = 2000 s
(tc l 2 ( t)
* k= = 1, 39(10(4 s(1. b) Theo
(
Ea = 34,704 (KJ/mol)0,250,250,25
0,25
0,25
Bi 5
2,5(2,5 im)
1. Dung dch X gm Cu(NO3)2 0,06M v Pb(NO3)2 0,04M. Tnh pH ca dd X. Bit:
2. Ngi ta iu ch mt dung dch Y bng cch ho tan 0,05 mol axit axetic v 0,05 mol Natri axetat trong nc ri thm nc n th tch 1 lt .Tnh pH ca dung dch Y. Cho Ka(CH3COOH) = 1,8.10-5.
51. Ta c cc cn bng:
Cu2+ + H2O Cu(OH)+ + H+ (1)K1 = 10-8
Pb2+ + H2O Pb(OH)+ + H+
(2)K2 = 10-7,8
H2O H+ + OH-
(3)Kw = 10-14V >> Kw nn ta c th tnh pH theo cn bng (1) v (2), b qua cn bng (3).
Theo iu kin proton, ta c:
Theo cn bng (1), (2), ta c:
Gi s nng cn bng ca Cu2+, Pb2+ l nng ban u, ta tnh c:
h = 3,513.10-5(M)
Tnh li nng cn bng ca Cu2+, Pb2+ theo gi tr H+ trn
Gi x, y ln lt l nng cn bng ca Cu(OH)+, Pb(OH)+Theo cn bng (1), (2) ta c
.Gi tr ca x, y rt nh so vi nng ban u nn nng cn bng ca Cu2+, Pb2+ coi nh bng nng ban u( kt qu lp)
Vy [H+] = 3,513.10-5; pH = 4,454
0,25
0,25
0,250,25
0,25
0,25
2a. Dung dch X l dung dch m axit.
pH = pK + lg = 4,745 + 0 = 4,745
1,00,25
Bi 6(2,5 im)
1. Lp s pin in theo quy c da trn hai bn phn ng di y:
Au3+ + 2e Au+ Fe3+ + 1e Fe2+ Ch r anot, catot v vit cc qu trnh oxi ha kh cc in cc, phn ng tng xy ra trong pin. Tnh sc in ng chun ca pin v hng s cn bng ca phn ng xy ra tron g pin. Gi s tt c cc nng l 1 M v cc p sut ring phn l 1, 0 atm. Cho ; ; .2. Cn bng cc phn ng oxi ho kh sau y bng phng php thng bng ion - electron:
a. Fe3P + NO3- + ....... NO +H2PO4- +...
b. Cr3+ + ClO3- + OH- CrO42- + Cl- +....
3. Cho gin Latimer ca photpho trong mi trng kim:
Tnh th kh chun ca cp HPO32- / H2PO2-
Hng dn chm:
61Fe2+(aq) +2e Fe
Fe3+(aq) +3e Fe
Fe3+(aq) + e Fe2+(aq)
= - ( = 3 - 2= 0,77VVy suy ra catot (+) l cp Au3+/Au+ v anot l cp Fe3+/Fe2+ Anot (-) Pt | Fe2+(aq), Fe3+(aq) || Au3+(aq), Au+(aq) | Pt (+) Catot
(-) Anot 2* | Fe2+(aq) Fe3+(aq) + e
(+) Catot Au3+(aq) + 2e Au+(aq)
Phn ng trong pin: Au3+(aq) + 2Fe2+(aq) 2Fe3+(aq) + Au+(aq) K
K = ()2. = (1)
thay vo (1), ta c: K = 1016,61 iu kin chun, sc in ng chun ca pin:
= = 0,49V0,250,25
0,25
2Cn bng phn ng bng phng php ion electron
a) Fe3P + NO3- + ....... NO +H2PO4- +...
3x Fe3P + 4H2O 3Fe3+ + H2PO4-+ 6H+ + 14e
14x NO3- + 4H+ + 3e NO + 2H2O
3Fe3P + 14NO3- +38H+ 9Fe3+ +3H2PO4-+14NO + 16H2O
b) Cr3+ + ClO3- + OH- CrO42- + Cl- +....
2 x Cr3+ +8OH- CrO4- + 4H2O + 3e
1 x ClO3- + 3H2O + 6e Cl- +6OH-
2 Cr3+ + ClO3- + 10 OH- 2CrO42- + Cl- +5H2O0,250,25
0,25
0,25
3(1) PO43- + 2H2O + 2e HPO32- + 3OH-. (G01 = -2FEo1.
(2) HPO32- + 2H2O + 2 e H2PO2- + 3OH- .
(G01 = -2FEo2.
(3) PO43- + 4 H2O + 4 e H2PO2- + 6OH-.
(G03 = -4FEo3.
2. T hp cc phng trnh ta c:
* (3) = (1) + (2)
( 4E3 = 2(E1+ E2)
(E (HPO32- / H2PO2-)= E2= (4E3 2E1)/2 = [4 . (-1,345) 2. (-1,12) ]/2 = -1,57 V0,250,25
0,25
Bi 72,5 (2,5 im)
1. Cho 3 nguyn t A, B v C. Bit n cht A tc dng vi n cht B nhit cao sinh ra hp cht D. Cht D b thu phn mnh trong nc to ra kh chy c v c mi trng thi. n cht B v n cht C tc dng vi nhau cho kh E, kh ny tan c trong nc to dung dch lm qu tm ho . Hp cht ca A vi C (hp cht F) c trong t nhin v thuc mt trong nhng cht c cng rt cao. Xc nh A, B, C, D, E, F v phng trnh cc phn ng nu trn.2. Cc kh X, Y khc nhau c ng trong hai bnh. C hai kh u c mi kh chu, khng mu v c tng khi lng l 6,8 gam. Khi t chy hon ton trong O2 d, ton b lng kh X to ra 5,4 gam H2O v kh Z rt t tan trong nc. Khi t chy trong oxi d, ton b lng kh Y to ra nc v kh T c th lm mt mu dung dch nc Brom. Nu cho ton b cng lng kh Y nh trn i qua dung dch Pb(NO3)2 d thu c 23,9 gam kt ta en. Hn hp kh Z v T cn nng 9,2 gam v chim th tch 4,48 lt (ktc). Xc nh kh X, Y, Z, T v vit cc phng trnh phn ng xy ra. Gi s cc phn ng xy ra hon ton.
Hng dn chm:
71 Hp cht AxBy l mt mui. Khi b thu phn cho thot ra H2S.
Hp cht F AnCm l Al2O3Vy A l Al, B l S, C l O
2 Al + 3 S Al2S3 (1)
(A) (B) (D)Al2S3 + 6 H2O 2 Al(OH)3 + 3 H2S (2)(D) 4 Al + 3 O2 2 Al2O3 (3) (A) (C) (F)S + O2 SO2 (4)
(B) (C) (E)
0,250,25
0,25
0,25
2A: NH3; B: H2S; C: N2; D: SO2H2S + Pb(NO3)2 ( PbS + 2HNO30,1 0,1 0,1
n = n = 0, 1 V = 2,24 lt, m = 3,4 gam.
2H2S + 3O2 ( 2SO2 + 2H2O
0,1 0,1
SO2 + Br2 + H2O ( H2SO4 + 2HBr
mA = 6,8 3,4 = 3,4 gam
mC = 9,2 0,1 x 64 = 2,8 gam; nC = (4,48 0,1 x 22,4) : 22,4 = 0,1 mol.
Vy A cha N v H: mH = 5,4 x 2 : 18 = 0,6 mol.
A:NxHy vi x : y = A: NH30,250,25
0,25
0,250,5
Bi 8Bi 8 (2,5 im) 1.Vit cc phng trnh phn ng xy ra (dng ion thu gn):
a. Ion I- trong KI b oxi ho thnh I2 bi FeCl3, O3.
b. Ion Br- b oxi ho bi H2SO4 c, BrO3- trong mi trng axit.
2. Mt dung dch A cha 2 mui Na2SO3 v Na2S2O3. Cho Cl2 d i qua 100 ml dung dch A ri thm vo hn hp sn phm mt lng d dung dch BaCl2 thy tch ra 6,524 gam kt ta. Mt khc, nu thm vo 100ml dung dch A mt t h tinh bt, sau chun dung dch A n khi mu xanh bt u xut hin th dng ht 29 ml dung dch it 0,5 M. Vit phng trnh ho hc dng ion thu gn v tnh nng mol mi cht trong dung dch A. Gi s cc phn ng xy ra hon ton.
Hng dn chm:
81a
b
Vit PTP
2I- + 2Fe3+ 2Fe2+ + I22I- + O3 + H2O
2OH- + O2 + I22Br- + 4H+ + SO42-(c) Br2 + SO2 + 2H2O
5Br- + BrO3- + 6H+
3Br2 + 3H2O0,50,75
2Phng trnh ha hc:
5H2O + S2O+ 4Cl2 ( 2SO+ 8Cl ( + 10H+ (1) x 2x
H2O + SO+ Cl2 ( SO+ 2Cl ( + 2H+ (2) y y
Ba2+ + SO ( BaSO4 ( (3) 2S2O + I2 ( S4O + 2I ( (4) x x/2
H2O + SO + I2 ( SO + 2I ( + 2H+ (5) y y
Ta c h phng trnh: 2x + y = = 0,028 (I)
+ y = 0,0145 (II)
Gii h phng trnh trn: x = 0,009 v y = 0,01.
Nng mol ca Na2S2O3 = 0,09 M v Na2SO3 = 0,1 MHc sinh vit ng mi phng trnh c 0,2 im.
Tnh ton ng kt qu c 0,25 im
1,25
CHNH THC
CHNH THC
d [O2]
dt
3,85.10-3 1,67.10-2
7,69.10-6 3,33.10-5
(
1,67.10-2
0,100
3,33.10-5
2,00.10-4
(
3,85.10-3
0,100
7,69.10-6
2,00.10-4
(
3,85.10-3 mol.L(1.s(1
7,69.10-6 mol.L(1
1,67.10-2 mol.L(1.s(1
3,33.10-5 mol.L(1
0,100 mol.L(1.s(1
2,00.10-4 mol.L(1
1413
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