Bai Tap Hoa Hoc Dai Cuong

Bi tp nng cao chuyn i cng Trang: 1 *****************************************************************************************************************************************

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Phn 1: CU TO NGUYN T BI TP PHN R PHNG X - PHN GN HT NHN

Cau 1: Cht phng x 210Po c chu k bn r T = 138 ngy. Tnh khi lng Po c phng x l 1 Ci (S: 0,222 mg) Cu 2: Tnh tui ca mt pho tng c bng g bit rng phng x ca n bng 0,77 ln phng x ca mt khc g cng khi lng va mi cht. Bit 14CT 5600= nm. (S: 2100 nm) Cu 3: Xt phn ng ht nhn xy ra khi bn cc ht vo bia Al: 27 3013 15Al P n+ + . Cho bit: mAl = 26,974u ; mP = 29,970u ; m = 4,0015u ; mn = 1,0087u ; mp = 1,0073u.Hy tnh nng lng ti thiu ca ht cn thit phn gn xy ra. (S: 3MeV) Cu 4: Mt mu poloni nguyn cht c khi lng 2 (g), cc ht nhn Poloni ( )21084Po phng x pht ra ht v chuyn thnh mt ht nhn AZ X bn. a. Vit phng trnh phn ng v gi tn AZ X . b. Xc nh chu k bn r ca poloni phng x bit trong 365 ngy n to ra th tch V = 179 cm3 kh He

(ktc) c. Tm tui ca mu cht trn bit rng ti thiim kho st t s gia khi lng AZ X v khi lng cht

l 2:1. (S: a. 82Pb207 Ch b. 138 ngy ) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

BI TP HO LNG T - MOMEN LNG CC NNG LNG LIN KT Cu 1: Thc nghm xc nh c momen lng cc ca phn t H2O l 1,85D, gc lin kt HOH l 104,5o, di lin kt OH l 0,0957 nm. Tnh ion ca lin kt OH trong phn t oxy (b qua momen to ra do cc cp electron ha tr khng tham gia lin kt ca oxy). 1D = 3,33.10-30 C.m. in tch ca electron l -1,6.10-19C ; 1nm = 10-9m.

Hng dn gii: Gi thit ion ca lin kt O H l 100%

ta c: -9 -19

-300,0957.10 .1,6.10

= =4,600D3,33.10

. => ion ca lin kt O H l 32,8%

Cu 2: nh sng nhn thy c phn hy c Br2(k) thnh cc nguyn t khng. Bit rng nng lng ph v lin kt gia hai nguyn t l 190kJ.mol-1. Ti sao hi Br2 c mu? Bit h = 6,63.10-34 J.s ; c = 3.108 m.s-1 ; NA = 6,022.1023 mol-1.

Hng dn gii -7

AcE = h .N = 6,3.10 m

.

Do nm trong vng cc tia sng nhn thy nn phn hy c v c mu.

Cu 3: Bit 2

n 2ZE = -13,6 (eV)n

(n: s lng t chnh, Z: s n v in tch ht nhn). a. Tnh nng lng 1e trong trng lc mt ht nhn ca mi h N6+, C5+, O7+. b. Qui lut lin h gia En vi Z tnh c trn phn nh mi lin h no gia ht nhn vi electron trong

cc h ? Hng dn gii

a. Theo u bi, n phi bng 1 nn ta tnh E1. Do cng thc l E1 = 13,6 Z2 (ev) (2) Th t theo tr s Z: Z = 6 C5+ : (E1) C5+ = 13,6 x 62 = 489,6 eV Z = 7 N6+ : (E1) N6+ = 13,6 x 72 = 666,4 eV Z = 8 O7+ : (E1) O7+ = 13,6 x 82 = 870,4 eV



b. Quy lut lin h E1 vi Z : Z cng tng E1 cng m (cng thp). Qui lut ny phn nh tc dng lc ht ht nhn ti e c xt: Z cng ln lc ht cng mnh nng lng cng thp h cng bn, bn nht l O7+. Cu 4: Vic gii phng trnh Schrodinger cho h nguyn t 1electron ph hp tt vi l thuyt c in ca Bohr v s lng t ha nng lng.

2

n 2ZE = -13,6 (eV)n

. cho tin s dng th cc gi tr s ca cc

hng s xut hin trong cng thc trn c chuyn ht v n v eV. iu th v l khi ta s dng cng thc trn cho phn t heli trung ha. Trong nguyn t heli lc ht nhn tc dng ln electron b gim bt do electron khc chn mt. iu ny c ngha l in tch ca ht nhn tc dng ln electron khng phi l Z = 2 na m s nh hn gi l in tch hiu dng (Zeff). Nng lng ion ha ca nguyn t heli trng thi c bn l 24,46eV. Tnh Zeff.

Hng dn gii Mi electron lp n = 1 ca nguyn t heli c nng lng Z2eff = 13,6eV Mc nng lng thp nht ca heli Z2eff = 27,2eV trng thi c bn ion He+ c nng lng = -4.13,6 = -54,4eV Nng lng ion ho = (-54,4 + Z2eff. 27,2) = 24,46 => Zeff = 1,70 Cu 5: Bng phng php quang ph vi sng ngi ta xc nh phn t SO2 trng thi hi c: 2SO 1,6D =

o

o

S Od 1,432 A ; OSO 109 5 = = . a. Tnh in tch hiu dng ca nguyn t O v nguyn t S trong phn t SO2 b. Tnh ion ca lin kt S-O

Hng dn gii a. i vi phn t SO2 c th xem trung tm in tch dng trng vi ht nhn nguyn t S cn trung tm in tch m s nm im gia on thng ni hai ht nhn nguyn t O. Nh vy momen lng cc ca phn t SO2: 2SO 2 = l . Trong l l khong cch gia hai trong tm in tch v c tnh nh sau:

oo1,432 cos59 45' 0,722A= =l . Theo d kin cho:

2SO1,6D = nn t y rt ra:

18

8 101,6 10 0,23

2 0,722 10 4,8 10

= =

Vy in tch hiu dng ca nguyn t O l -0,23 cn in tch hiu dng ca nguyn t S l +0,46 in tch tuyt i ca electron b. Mt khc nu xem lin kt S-O hon ton l lin kt ion th momen lng cc ca phn t l:

2

8 10SO 0,722 10 2 4,8 10 6,93D

= =

Vy ion x ca lin kt S-O bng: 1,6x 100% 23%6,93

= =

Cu 6: Tnh nng lng lin kt ion ENa-F ca hp cht ion NaF. Bit cc tr s (kJ/mol): INa = 498,5 ; FF = -328 ; khong cch ro = 1,84

o

A , nNaF = 7 l h s y Born, 12o 8,854.10 = l hng s in mi trong

chn khng. ENa-F c tnh theo cng thc: 2

AA B A B

o o

N .e 1E 1 I F4 . .r n

=

pi . (S: ENa-F = 497,2)

Phn 2: S BIN THIN TUN HON CA MT S TNH CHT THEO CHIU TNG DN IN TCH HT NHN Cau 1: Tnh nng lng mng li ca LiF da vo cc s liu cho bi bng sau: Nng lng (kJ/mol) Nng lng (kJ/mol) i lc electron ca F(k) : AF = 333,000 Lin kt FF: Elk = 151,000 Ion ho th nht ca Li(k): I1 = 521,000 Sinh nhit ca LiF(tinh th) = 612,300 Entanpi nguyn t ho Li(tinh th) = 155,200 Umng li= ? S: Uml = 1031 kJ.mol-1



Cu 2: Nng lng ion ha th nht ca cc nguyn t chu k 2 nh sau

a. Hy cho bit v sao khi i t Li n Ne, nng lng ion ha th nht ca cc nguyn t nhn chung tng dn nhng t: Be sang B ; t N sang O th nng lng ion ho th nht li gim dn

b. Tnh in tch ht nhn hiu dng Z i vi mt electron ha tr c nng lng ln nht trong cc nguyn t trn v gii thch chiu bin thin gi tr Z trong chu k. Bit rng: 13,6eV = 1312kJ/mol ;

2

1 2Z'I 13,6 (eV)n

=

S: 1,26 ; 1,66 ; 1,56 ; 1,82 ; 2,07 ; 2,00 ; 2,26 ; 2,52 Cu 3: Nng lng lin kt n gin gia hai nguyn t A v B l EAB lun ln hn gi tr trung bnh cng

cc nng lng lin kt n EAA ; EBB l AB : ( )AB AA BB AB1E E E2= + + . Gi tr AB (kJ/mol) c trng cho phn c tnh ion ca lin kt AB lin quan n s khc nhau v m in gia A v B, tc l hiu s

A B . Theo Pauling: A B AB0,1 = . thu c gi tr m in ca nguyn t cc nguyn t khc nhau, Pauling gn gi tr m in ca hiro l 2,2

a. Tnh m in ca Flo v Clo da vo cc s liu nng lng lin kt: HF HCl F2 Cl2 H2 565 431 151 239 432

b. Tnh nng lng lin kt ECl-F Hng dn gii

a. F F12, 2 0,1 565 (151 432) 3,852

= + => =

Cch tnh tng t: Cl 3,18 =

b. 1Cl F13,85 3,18 0,1 x (151 239) x E 240kJ.mol2

= + => = =

Cu 4: Da vo phng php gn ng Slater, tnh nng lng ion ha th nht I1 cho He (Z = 2). Hng dn gii:

He c cu hnh 1s2, ( )2* 2

*

He *2 2

13,6 2 0,313,6(Z )E 2 2 78,6eVn 1

= = =

He+ c cu hnh 1s1, 2 2

*

2 2He

13,6Z 13,6 2E 54, 4eVn 1+

= = =

Qu trnh ion ho: * *1 HeHeHe He 1e I E E ( 54,4) ( 78,6) 24,2eV++ + = = = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Phn 3: CU TRC MNG TINH TH Cau 1: Tinh th NaCl c cu trc lp phng tm din. Tnh bn knh ca ion Na+ v khi lng ring ca

tinh th NaCl bit cnh ca mng c s a = 5,58 o

A ; bn knh ion o

Clr 1,810 A = ; khi lng mol ca Na

v Cl ln lt l: 22,99 g.mol-1 v 35,45 g.mol-1 (S: r+ = 0,98o

A ; d = 2,23 g/cm3) Cau 2: Tinh th Fe c cu trc tinh th lp phng tm khi v cng a ca mng c s l

o

a 2,860A= cn Fe kt tinh dng lp phng tm din vi o

a 3,560 A= . Tnh bn knh kim loi v khi lng ring ca st thuc hai loi cu trc trn bit Fe = 55,800 g/mol

S: Fe : r = 1,24o

A ; d = 7,92 g/cm3 ; Fe : r = 1,26o

A ; d= 8,21 g/cm3

Nguyn t Li Be B C N O F Ne I1 (kJ/mol) 521 899 801 1087 1402 1313 1681 2081

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Cu 3: Tinh th MgO c cu trc kiu NaCl vi cnh ca mng c s: o

d 4,100A= . Tnh nng lng mng li ca MgO theo phng php Born-Landr v phng php Kapustinxki bit rng s Madelung ca mng li MgO: a = 1,7475 ; e = 1,602.10-19C ; 12o 8,85.10

= ; NA = 6,023.1023 ; nB = 7

Theo Born-Landr: 2

A

o B

Z Z e aN 1U (1 )4 R n

+ =

pi vi R = r+ + r-

Theo Kapustinxki: 7Z Z n

U 1,08.10R

+ =

Hng dn gii Thay s vo hai phng trnh trn ta suy ra: Theo Born-Landr: U = 4062 kJ/mol ; theo Kapustinxki: U = 4215 kJ/mol Cu 4: St kim loi nng chy 1811K. Gia nhit phng v im nng chy ca n, st kim loi c th tn ti cc dng th hnh v cc dng tinh th khc nhau. T nhit phng n 1185K, st c cu to tinh th dng lp phng tm khi (bcc) quen gi l st- . T 1185K n 1667K st kim loi c cu to mng lp phng tm din (fcc) v c gi l st- . Trn 1167K v cho ti im nng chy st chuyn v dng cu to lp phng tm khi (bcc) tng t st- . Cu trc sau cng (pha cui) cn c gi l st- 1. Cho bit khi lng ring ca st kim loi nguyn cht l 7,874g.cm-3 293K,

a. Tnh bn knh nguyn t ca st (cm). b. c lng khi lng ring ca st (tnh theo g.cm-3) 1250K

Ch : B qua cc nh hng khng ng k do s gin n nhit ca kim loi. Thp l hp kim ca st v cacbon, trong mt s khong trng gia nguyn t st (cc hc) trong mng tinh th b chim bi cc nguyn t nh l cacbon. Hm lng cacbon trong hp kim ny thng trong khong 0,1% n 4%. Trong l cao, s nng chy ca st cng d dng khi thp cha 4,3% theo khi lng. Nu hn hp ny c lm lnh qu nhanh (t ngt) th cc nguyn t cacbon c phn tn trong mng st- . Cht rn mi ny c gi l martensite - rt cng v gin. D hi b bin dng, cu to tinh th ca cht rn ny l ging nh cu to tinh th ca st- (bcc). 2. Gi thit rng cc nguyn t cacbon c phn b u trong cu trc ca st.

a. c tnh hm lng nguyn t cacbon trong mt t bo n v ( mng c s) ca st- trong martensite cha 4,3%C theo khi lng.

b. c tnh khi lng ring (g.cm-3) ca vt liu ny. Khi lng mol nguyn t v cc hng s: MFe = 55,847g.mol-1 ; MC = 12,011g.mol-1 ; NA = 6,02214.1023mol-1.

Hng dn gii 1. Cc bc tnh ton:

1. nh ngha cc tham s ca chiu di (a, b, c, d1, d2 v r) v th tch (V1 v V2) cho c hai cu to bcc v fcc ca st.

2. Tnh th tch V1 ca mng n v ca st - nh khi lng ring ca n (bcc) 293K, khi lng mol nguyn t ca st (MFe), v s Avogadro NA.

3. Tnh chiu di d1 cnh ca mng n v bcc t th tch ca n. 4. Tnh bn knh nguyn t r ca st t chiu di d1. 5. Tnh chiu di d2 ca cnh mng n v fcc ( 1250K) t bn knh nguyn t r ca st. 6. Tnh th tch V2 ca mng n v fcc ca st - t chiu di d2 ca cnh. 7. Tnh khi lng m ca s nguyn t st trong mt mng n v ca st - t khi lng mol

nguyn t MFe ca st v s Avogadro NA. 8. Tnh khi lng ring (fcc) ca st - t cc ga tr ca m v V2. Mt hng khc tm khi lng

ring fcc ca st - l tnh ti l phn trm khong khng gian chim ch trong c hai loi mng n v bcc v fcc,

c th thay th cc bc t 5 n 8 bng cc bc t 5 n 8 sau y: 5. Tnh t l phn tm khong khng gian chim ch ca mng n v bcc.



6. Tnh t l phn tm khong khng gian chim ch ca mng n v fcc. 7. T t l fcc/bcc ta suy ra c t l: bcc/fcc. 8. T ga tr cho trc bc 7 ta tnh c fcc.

2. Cc chi tit: 1. 293K st - c cu trc tinh th bcc. Mi mng n v thc s cha hai nguyn t, trong mt

nguyn t tm ca mng. 1250K, st - c cu to tinh th fcc. Mi mng n v thc s cha 4 nguyn t v tm ca mi mt c mt na nguyn t.

- r: bn knh nguyn t ca st - a: chiu di ng cho mt mt ca mng n v bcc. - b: chiu di ng cho qua tm ca mng n v bcc. - c: chiu di ng cho mt mt ca mng n v fcc. - d1: chiu di cnh ca mng n v bcc ca st - . - d2: chiu di cnh ca mng n v bcc ca st - . - V1: Th tch ca mng n v bcc ca st - . - V2: Th tch ca mng n v bcc ca st - . - Va: th tch chim bi mt nguyn t. - Va1: Th tch chim bi hai nguyn t trong mt mng n v bcc. - Va2: Th tch chim bi bn nguyn t trong mt mng n v fcc. - R1: T l phn trm khong khng gian chim ch trong mt mng n v bcc. - R2: T l phn trm khong khng gian chim ch trong mt mng n v fcc.

3 2a a1 a2 a2 a 1

32 2

2 2 2 2 31 1 1 1 1

32 2

2 2 32 2 2 2

4V = r ; V = 2V ; V = 4V ; b = 4r ; a = 2d ; 3

16r 16rb = d a = 3d d = V = d = 3 3

16r 16rc = 4r ; c = 2d d = V = d =

2 2

pi

+

2. 1,000cm3 st c khi lng 7,874g 293K (bcc). 1 mol st c khi lng 55,847g (MFe). Vy 0,1410mol ca st chim trong th tch 1,000cm3 hoc 1mol st s chim th tch 7,093cm3. 1 mol tng ng chim 6,02214.1023 nguyn t.

-23 31

7,093.2V = = 2,356.10 cm 6,02214.1023

mi n v mng.

1. d1 = V11/3 = 2,867.10-8 cm. 2. Vi cu to bcc, ga tr ca d1 c th c biu th l: d1 = (16r2/3)1/2. Vy ga tr ca r s l: r = (3d12/16)1/2 = 1,241.10-8cm. 3. 1250K, trong cu to fcc, d2 = (16r2/2)1/2 = 3,511.10-8cm. 4. V2 = d23 = 4,327.10-23cm3. 5. Khi lng m ca 4 nguyn t st trong mng n v fcc s l:

m = 55,847.4/(6,02214.1023) = 3,709.10-22g 6. fcc = m/V2 = 8,572g/cm3.

Cch gii khc tm khi lng ring fcc ca st - : 5. R1 = [(Va1)/V1].100% = 68,02% 6. R2 = [(Va2)/V2].100% = 74,05% 7. bcc/fcc = 74,05/68,02 = 1,089 8. fcc = 8,572g/cm3.

3. Cc bc tnh ton: 1. T phn trm cu thnh ca martensite (theo khi lng), tnh s mol tng ng ca cacbon v st.



2. a t l mol C/Fe v mt mng n v (Ghi ch: Hai nguyn t Fe trong mi mng n v). 3. Tm s nguyn be nht cc nguyn t C trong s nguyn b nht ca mng n v (khng bt buc). 4. Tnh khi lng st trong mt mng n v 5. Tnh khi lng cacbon trong mt mng n v 6. Tnh tng khi lng st v cacbon trong mt mng n v 7. Tnh khi lng ring ca martensite [(martensite c 4,3%C)] t tng khi lng ca C v Fe v th tch V1

ca mng n v st - cu to bcc. 4. Chi tit:

1. Trong 100,0g martensite c 4,3%C nC = 0,36mol v nFe = 1,71mol. Vy c 1 nguyn t cacbon c 4,8 nguyn t st hay 0,21 nguyn t cacbon cho mi nguyn t st.

2. Martensite c cu to tinh th bcc (2 nguyn t st cho mi mng n v). Nh vy s nguyn t cacbon trong mi mng n v l: 2.(1/4,8) = 0,42 nguyn t.

3. 5 nguyn t C [(0,42 nguyn t C/0,42).5] trong 12 mng n v [1 mng n v/0,42).5] 4. S gam Fe trong mi mng n v l: 55,847.2/(6,02214.1023)= 1,8547.10-22 g 5. S gam C trong mi mng n v l: 12,011/(6,02214.1023) = 1,9945.10-23 g 6. Tng khi lng C v Fe = 1,8457.10-22 + 0,42.1,9945.10-23 = 1,938.10-22 g. 7. Mi mng n v ca st - chim th tch V1 = 2,356.10-23 cm3. 8. (martensite c 4,3%C) = 1,938.10-22/(2,356.10-23) = 8,228 g.cm-3.

Cu 5: Cho cc d kin sau:

Nhit hnh thnh ca NaF(rn) l -573,60 KJ.mol-1 ; nhit hnh thnh ca NaCl(rn) l -401,28 KJ.mol-1

Tnh i lc electron ca F v Cl. So snh kt qu v gii thch. Hng dn gii:

p dng nh lut Hess vo chu trnh M(r) MX(r)X2(k)

M(k)

M+(k) X-(k)

HTH

HML

HHT

+

+ AE

X(k)I1

+

HLK 12

12

+

AE (F) > AE (Cl) d cho F c m in ln hn Cl nhiu. C th gii thch iu ny nh sau: Phn t F2 t bn hn phn t Cl2, do HLK (F2) < Hpl (Cl2) v dn n AE (F) > AE (Cl). Cng c th gii thch: F v Cl l hai nguyn t lin nhau trong nhm VIIA. F u nhm. Nguyn t F c

bn knh nh bt thng v cn tr s xm nhp ca electron. Phn 4: NHIT NG HA HC

BI TP NHIT HA HC Cu 1: Tnh nng lng lin kt trung bnh ca lin kt OH v OO trong phn t H2O2 da vo cc s liu (kJ/mol) sau:

2 2 2

o o o o

(H O,k ) (H,k) (O,k) (H O ,k )H 241,8 ; H 218 ; H 249, 2 ; H 136,3 = = = =

Cu 2: Tnh oH ca phn ng sau 423K: 2(k) 2(k) 2 (h)1H O H O2

+

Nng lng KJ.mol-1 Nng lng KJ.mol-1 Thng hoa Na 108,68 Lin kt ca Cl2 242,60 Ion ha th nht ca Na 495,80 Mng li ca NaF 922,88 Lin kt ca F2 155,00 Mng li ca NaCl 767,00

Ta c: AE = HHT - HTH - I1 - HLK + HML (*) Thay s vo (*), AE (F) = -332,70 kJ.mol-1 v AE (Cl) = -360 kJ.mol-1.



Bit rng: 2

o 1H O( )H 285, 200(kJ.mol ) = long ; nhit ha hi ca nc lng: o 1373H 37,5(kJ.mol ) = v nhit

dung mol oPC (J.K-1.mol-1) ca cc cht nh sau: H2 (k) O2 (k) H2O (h) H2O (l)

27,3 + 3,3.10-3T 29,9 + 4,2.10-3T 30 + 1,07.10-2T 75,5 Cu 4: Cho cc phng trnh nhit ha hc sau y: (1) 2 ClO2 (k) + O3 (k) Cl2O7 (k) H0 = - 75,7 kJ (2) O3 (k) O 2 (k) + O (k) H0 = 106,7 kJ (3) 2 ClO3 (k) + O (k) Cl2O7 (k) H0 = -278 kJ (4) O2 (k) 2 O (k) H0 = 498,3 kJ. k: k hiu cht kh. Hy xc nh nhit ca phn ng sau: (5) ClO2 (k) + O (k) ClO3 (k).

Hng dn gii Kt hp 2 pt (1) v (3) ta c

ClO2 (k) + 1/2 O3 (k) 1/2 Cl2O7 (k) H0 = - 37,9 kJ 1/2 Cl2O7 (k) ClO3 (k) + 1/2 O (k) H0 = 139 kJ

(6) ClO2 (k) + 1/2 O3 (k) ClO3 (k) + 1/2 O (k) H0 = 101,1 kJ Kt hp 2 pt (6) v (2) ta c

ClO2 (k) + 1/2 O3 (k) ClO3 (k) + 1/2 O (k) H0 = 101,1 kJ 1/2 O2 (k) + 1/2 O (k) 1/2 O3 (k) H0 = -53,3 kJ

(7) ClO2 (k) + 1/2 O2 (k) ClO3 (k) H0 = 47,8 kJ

Kt hp 2 pt (7) v (4) ta c ClO2 (k) + 1/2 O3 (k) ClO3 (k) + 1/2 O (k) H0 = 101,1 kJ O (k) 1/2 O2 (k) H0 = - 249,1 kJ

(5) ClO2 (k) + O (k) ClO3 (k) H0 = - 201,3 kJ. l pt nhit ha (5) ta cn tm. Cu 5: Cho hai phn ng gia graphit v oxi:

(gr) 2(k) (k)

(gr) 2 (k) 2(k)

1(a) C + O CO2

(b) C + O CO

Cc i lng Ho

, So

(ph thuc nhit ) ca mi phn ng nh sau:

o

TH (J/mol) oTS (J/K.mol) (a) - 112298,8 + 5,94T - 393740,1 + 0,77T (b) 54,0 + 6,21lnT 1,54 - 0,77 lnT

Hy lp cc hm nng lng t do Gibbs theo nhit G0

T(a) = f(T), G

0

T(b) = f(T) v cho bit khi tng

nhit th chng bin i nh th no? Cu 6: Trong mt th nghim ngi ta cho bt NiO v kh CO vo mt bnh kn, un nng bnh ln n 1400oC. Sau khi t ti cn bng, trong bnh c bn cht l NiO (r), Ni (r), CO (k) v CO2

(k) trong CO chim 1%, CO2

chim 99% th tch; p sut kh bng 1bar (105 Pa). Da vo kt qu th nghim v cc d kin nhit ng cho trn, hy tnh p sut kh O2

tn ti cn bng vi hn hp NiO v Ni 14000C.

Cu 7: Cn bng gia Cgr vi Ckc c c trng bi nhng s liu sau: gr kcC C 0 0298K 298KH 1,9kJ / mol ; G 2,9kJ / mol = =



a. Ti 298K, loi th hnh no bn hn b. Khi lng ring ca Cgr v Ckc ln lt l: 2,265 v 3,514 g/cm3. Tnh hiu s H U ca qu trnh

chuyn ha trn ti p sut P = 5.1010 Pa (S: a. Cgr ; b. -94155 J/mol) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

BI TP NG HA HC CN BNG HA HC Cu 1: i vi phn ng : A k1

k2 B. Cc hng s tc k1 = 300 giy -1 ; k2 = 100 giy -1 . thi im

t = 0 ch c cht A v khng c cht B. Hi trong bao lu th mt na lng ban u cht A bin thnh cht B. (S: 2,7.10-3 s) Cau 2: Ngay nhit thng gia NO2 v N2O4 tn ti cn bng sau: 2(k) 2 4(k )2NO N O . 24

oC, hng s cn bng ca phn gn trn l KP = 9,200. Ti nhit ny, cn bng s dch theo chiu no nu p sut ring phn ca cc cht kh nh sau

a. 2 4 2N O NO

P 0,900atm;P 0,100atm= = b.

2 4 2N O NOP 0,72021atm;P 0, 27979atm= =

c. 2 4 2N O NO

P 0,100atm;P 0,900atm= =

Cu 3: Xt phn ng: I ClO IO Cl + + . Thc nghim xc nh vn tc ca phn ng ny xc nh bi

biu thc: [I ][ClO ]v k [OH ]

= . Chng minh c ch sau gii thch c thc nghim

(1) 1K2H O ClO OH HClO + + (nhanh) (2) 2KHClO I HIO Cl + + (chm) (3) 2K 2HIO OH H O IO + + (nhanh) Cu 4: i vi phn ng thun nghch pha kh 2 SO2 + O2 2 SO3

a. Ngi ta cho vo bnh kn th tch khng i 3,0 lt mt hn hp gm 0,20 mol SO3 v 0,15 mol SO2. Cn bng ha hc (cbhh) c thit lp ti 25oC v p sut chung ca h l 3,20 atm. Hy tnh t l oxi trong hn hp cn bng.

b. Cng 25oC, ngi ta cho vo bnh trn y mol kh SO3. trng thi cbhh thy c 0,105 mol O2. Tnh t l SO3 b phn hy, thnh phn hn hp kh v p sut chung ca h

Hng dn gii a. Xt 2 SO2 + O2 2 SO3 (1)

ban u 0,15 0,20 lc cbhh ( 0,15 + 2z) z (0,20 2z) Tng s mol kh lc cbhh l n1 = 0,15 + 2z + z + 0,20 2z = 0,35 + z T pt trng thi: P1V = n1RT n1 = P1V / (RT) = 3,2.3/(0,082.298) = 0,393 => z = 0,043. Vy x

O 2 = z/n1 = 0,043/0,393 = 0,1094 hay trong hhcb oxi chim 10,94% b. 2 SO2 + O2 2 SO3 (2)

ban u 0 0 y lc cbhh 2. 0,105 0,105 (y 2. 0,105). Trng thi cbhh c xt i vi (1) v (2) nh nhau v T (v cng V) nn ta c: K = const ; vy: n

3

2SO / (n 2

2SO .n 2O ) = const.

Theo (1) ta c n3

2SO / (n 2

2SO .n 2O ) = ( 0,20 2. 0,043)

2 / (0,15 + 0,086)2. 0,043 = 5,43. Theo (2) ta c n

3

2SO / (n 2

2SO .n 2O ) = (y 0,21)

2/ (0,21)2.0,105 = 5,43. T c phng trnh: y2 0,42 y + 0,019 = 0. Gii pt ny ta c y1 = 0,369 ; y2 = 0,0515 < 0,105 (loi b nghim y2 ny). Do ban u c y = 0,369 mol SO3 ; phn li 0,21 mol nn t l SO3 phn li l 56,91%

Ti cbhh: tng s mol kh l 0,369 + 0, 105 = 0,474 nn:



SO3 chim ( 0,159 / 0,474).100% = 33,54% SO2 chim ( 0,21 / 0,474).100% = 44,30%; O2 chim 100% - 33,54% - 44,30% = 22,16%.

T pt trng thi: P2V = n2RT P2 = n2 RT/ V = 0,474.0,082.298/3 P2 = 3,86 atm. Cu 5: NOCl b phn hy theo phn ng: (k) (k) 2(k)2NOCl 2NO Cl + . Lc u ch c NOCl. Khi cn bng 500K c 27% NOCl b phn hy v p sut tng cng ca h l 1atm. Hy tnh 500K b. Kp v oG ca phn ng. c. Nu h p sut xung di 1atm th s phn hy NOCl tng hay gim? V sao? Cu 6: i vi phn ng: A + B C + D (phn ng l n gin) 1. Trn 2 th tch bng nhau ca dung dch cht A v dung dch cht B c cng nng 1M:

a. Nu thc hin phn ng nhit 333,2K th sau 2 gi nng ca C bng 0,215M. Tnh hng s tc ca phn ng.

b. Nu thc hin phn ng 343,2K th sau 1,33 gi nng ca A gim i 2 ln. Tnh nng lng hot ha ca phn ng (theo kJ.mol-1).

2. Trn 1 th tch dung dch cht A vi 2 th tch dung dch cht B, u cng nng 1M, nhit 333,2K th sau bao lu A phn ng ht 90%?

Cu 7: N2O5 d b phn hy theo phn ng sau: 2 5(k ) 2(k) 2(k)N O 4NO O + . Phn ng l bc nht vi hng s tc phn ng l: k = 4,8.10-4 s-1

a. Tnh thi gian m mt na lng N2O5 phn hy b. p sut ban u ca N2O5 l 500 mmHg. Tnh p sut ca h sau 10 pht

(S: a. 1444s ; b. 687,5 mmHg) Cu 8: nhit T(K), hp cht C3H6O b phn hy theo phng trnh: 3 6 (k) 2 4(k) (k) 2(k)C H O C H CO H + + o p sut P ca hn hp phn ng theo thi gian ta thu c kt qu cho bi bng sau:

a. Chng minh phn ng l bc nht theo thi gian b. thi im no p sut ca hn hp bng 0,822 atm

Cu 9: Vi phn ng pha kh: 2 2A B 2AB (1)+ , c ch phn ng c xc nh: (nhanh) (nhanh) (chm) Vit biu thc tc phn ng (1) v gii thch.

Cu 10: Xc nh cc hng s tc k1 v k2 ca phn ng song song (S trn). Bit rng hn hp sn phm cha 35% cht B v nng cht A gim i mt na sau 410 s. (k1 = 0,591.10-3 ; k2 = 1,099.10-3 s-1) Cu 11: Thc nghim cho bit s nhit phn pha kh N2O5

0t NO

2 + O2 (*) l phn ng mt chiu bc nht. C ch c tha nhn rng ri ca phn ng ny l N2O5 1k NO 2 + NO3 (1) NO

2 + NO3 1k N2O5 (2) NO

2 + NO3 2k NO + NO 2 + O2 (3) N2O5 + NO 3k 3 NO 2 (4). a. p dng s gn ng trng thi dng cho NO, NO3 c ch trn, hy thit lp biu thc tc ca (*).

Kt qu c ph hp vi thc nghim khng? b. Gi thit rng nng lng hot ha ca (2) bng khng, ca (3) bng 41,570 kJ.mol-1. Da vo c im

cu to phn t khi xt c ch trn, phn tch c th a ra biu thc tnh k-1/ k2 v hy cho bit tr

s ti 350 K.

t (pht) 0 5 10 15 ? P (atm) 0,411 0,537 0,645 0,741 0,822

2

2 2

2

(a) A 2A(b) A B AB(c) A AB 2AB

+

+

k1 A k2

B

C



c. T s phn tch gi thit im b) khi cho rng cc phn ng (1) v (2) dn ti cn bng ha hc c hng s K, hy vit li biu thc tc ca (*) trong c hng s cbhh K.

Hng dn gii: a. Xt d[NO3]/dt = k1[N2O5] k -1[NO2][NO3] k2[NO2][NO3] 0 (a) [NO3] = k1[N2O5] / {(k -1 + k2)[NO2]} (b). Xt d[NO]/dt = k2[NO2][NO3] - k3[NO][N2O5] 0 (c) [NO] = k2[NO2][NO3] / k3[N2O5] / {(k -1 + k2)[NO2]} (d). Th (b) vo (d) ta c [NO] = k1k2 / k3(k -1 + k2) (d). Xt d[N2O5]/dt = - k1[N2O5] + k -1[NO2][NO3] - k3[NO][N2O5] (e). Th (b), (d) vo (e) v bin i thch hp, ta c d[N2O5]/dt = { - k1 + (k -1 k2)/ (k -1 + k2)}[N2O5] = k`[N2O5] (f) b. Trong (2) do s va chm gia NO2 vi NO3 nn N2O5 O2NONO2 c ti to, tc l c s va chm ca

1 N vi 1 O. Ta gi y l trng hp 1. Trong (3) NO c to ra do 1 O b tch khi NO2 ; NO2 c to ra t s tch 1O khi NO3. Sau 2 O kt hp to ra O2. Ta gi y l trng hp 2. Nh vy y s va chm gia cc phn t ng chng gp 2 so vi trng hp 1 trn.

Phng trnh Archniux c vit c th cho mi phn ng xt: P. (2): k

-1 = A2e 2 /E RT (*); P. (3): k2 = A3e 3 /E RT (**) Theo lp lun trn v ngha ca i lng A trong pt Archniux c trng cho s va chm dn ti phn ng, ta thy A3 = 2A2. Ta qui c A 2 = 1 th A3 = 2. Theo bi: E2 = 0; E3 = 41,570 kJ.mol -1; T = 350. Thay s thch hp, ta c:

k -1/ k2 = e 3 /E RT = e

341,578/8,314.10 .350 8.105(ln).

c. Kt hp (1) vi (2) ta c cbhh: N2O5 NO2 + NO3 (I) K = k1 / k -1 = [NO2][NO3] / [N2O5] (I.1) a (I.1) vo b/ thc (c): [NO] = k2[NO2][NO3] / k3[N2O5] = k2K/k3 (I.2). Th b/ thc (I.2) ny v (b) trn vo (e), ta c d[N2O5]/dt = - k1[N2O5] + k -1[NO2]{ k -1[NO2](k1[N2O5]/ (k -1 + k2)[NO]}- k3(k2K/k3). Thu gn b/ t ny, ta c d[N2O5]/dt = {- k1+ (k-1k1/(k -1 + k2)) - k2K}[N2O5] (I.3) Gi thit k

-1>> k2 ph hp vi iu kin Ea2 0. Cbhh (I) nhanh chng c thit lp. Vy t (I.3) ta c d[N2O5]/dt = {- k1+ (k -1k1/ k -1) - k2K}[N2O5] (I.4). Ch K = k1 / k -1, ta c: d[N2O5]/dt = {- k1+ (k -1- k2)K}[N2O5] (I.5). Cu 12: Trong mt h c cn bng 3 H2 + N2 2 NH3 (*) c thit lp 400 K ngi ta xc nh c cc p sut ring phn sau y: p(H2) = 0,376.105 Pa , p(N2) = 0,125.105 Pa , p(NH3) = 0,499.105 Pa 1. Tnh hng s cn bng Kp v G0 ca phn ng (*) 400 K. 2. Tnh lng N2 v NH3, bit h c 500 mol H2. (S: 1. 38,45 ; -12,136 kJ.mol-1 ; 2. n

(N2) = 166 mol ; n (NH3) = 644 mol) Cu 13: Cho phn ng A + B C + D

(*) din ra trong dung dch 25 OC. o nng A trong hai dung dch cc thi im t khc nhau, thu c kt qu: Dung dch 1 [A]0 = 1,27.10-2 mol.L-1 ; [B]0 = 0,26 mol.L-1

t(s) 1000 3000 10000 20000 40000 100000 [A] (mol.L-1) 0,0122 0,0113 0,0089 0,0069 0,0047 0,0024

Dung dch 2 [A]0 = 2,71.10-2 mol.L-1 ; [B]0 = 0,495 mol.L-1



t(s) 2.000 10000 20000 30000 50000 100000 [A] (mol.L-1) 0,0230 0,0143 0,0097 0,0074 0,0050 0,0027

1. Tnh tc ca phn ng (*) khi [A] = 3,62.10-2 mol.L-1 v [B] = 0,495 mol.L-1. 2. Sau thi gian bao lu th nng A gim i mt na? (S: 1. v = 4,32.10 6 mol.L-1. s-1 ; 2 T = 8371 s) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Phn 5: IN HA HC Cu 1: xc nh hng s to phc (hay hng s bn) ca ion phc [Zn(CN)4]2-, ngi ta lm nh sau: - Thm 99,9 ml dung dch KCN 1M vo 0,1 ml dung dch ZnCl2 0,1 M thu c 100ml dung dch ion

phc [Zn(CN)4]2- (dung dch A). - Nhng vo A hai in cc: in cc km tinh khit v in cc so snh l in cc calomen bo ho c

th khng i l 0,247 V (in cc calomen trong trng hp ny l cc dng). - Ni hai in cc vi mt in th k, o hiu in th gia chng c gi tr 1,6883 V. Hy xc nh hng s to phc ca ion phc [Zn(CN)4]2-. Bit th oxi ho - kh tiu chun ca cp Zn2+/Zn bng -0,7628 V. (S: 1,4 = 1018,92 ) Cu 2: Dung dch A gm CrCl3 0,010 M v FeCl2 0,100 M.

a. Tnh pH ca dung dch A. b. Tnh pH bt u kt ta v kt ta hon ton Cr(OH)3 t dung dch CrCl3 0,010 M (coi mt ion

c kt ta hon ton nu nng cn li ca ion trong dung dch nh hn hoc bng 1,0.10-6 M). c. Tnh 2

4 2

o

CrO / CrOE

. Thit lp s pin v vit phng trnh phn ng xy ra trong pin c ghp bi cp 2- -4 2CrO /CrO v -3NO /NO iu kin tiu chun.

Cho: Cr3+ + H2O CrOH2+ + H+ 1= 10-3,8 Fe2+ + H2O FeOH+ + H+ 2 = 10-5,92 Cr(OH)3 Cr3+ + 3 OH KS = 10-29,8

Cr(OH)3 H+ + CrO2- + H2O K = 10-14 H2O H+ + OH- Kw =10-14

24 3 3

o o o

CrO / Cr(OH) ,OH NO ,H / NORTE 0,13V;E 0,96V;2,303 0,0592(25 C)F +

= = =

p s: a. pH = 2,9 b. e kt ta hon ton Cr(OH)3 t dung dch Cr3+ 0,010 M th: pH 7,2 c. Eo = -0,13 V ; s o pin: (-) Pt | CrO42- 1M ; CrO2- 1M ; OH- 1M || NO3- 1M ; H+ 1M | (Pt) NO, pNO = 1atm (+) Cu 3: Trong khng kh dung dch natri sunfua b oxi ho mt phn gii phng ra lu hunh. Vit phng trnh phn ng v tnh hng s cn bng.

Cho: E0(O2/H2O) = 1,23V ; E0(S/S2-) = - 0,48V; 2,3 RT/F ln = 0,0592lg Cu 4: sn xut 1 tn nhm ngi ta in phn boxit cha 50% Al2O3. Hi cn lng Boxit v nng lng kWh l bao nhiu, bit rng in p lm vic l 4,2V. Tnh thi gian tin hnh in phn vi cng dng in 30000A (S: 12509 kWh ; t = 99h) Cu 5: Thit lp mt pin ti 25oC: Ag | [Ag(CN)n(n-1)-] = C mol.l-1, [CN-] d || [Ag+] = C mol.l-1 | Ag 1. Thit lp phng trnh sc in ng E (n,[CN ], p )= f , l hng s in li ca ion phc 2. Tnh n v p , bit Epin =1,200 V khi [CN-] = 1M v Epin = 1,32 V khi [CN-] = 10M Cu 6: Da vo cc s liu th kh chun sau xy dng gin th kh chun ca Urani (gin Latime) v cho bit ion no khng bn trong dung dch.

22 2UO / UO

+ +

42UO / U+ +

4U / U+ 3U / U+ Eo, V 0,062 0,612 -1,5 -1,798



Cu 7: 25oC xy ra phn ng sau: 2 4 3 3Fe Ce Fe Ce+ + + ++ + .Cho cc s liu v th kh chun ca cc cp: 3 2 4 3o oFe / Fe Ce / CeE 0,77V;E 1,74V+ + + += =

1. Tnh hng s cn bng K ca phn ng 2. Tnh th phn ng ti thi im tng ng, bit ban u s mol ca Fe2+ v Ce4+ l bng nhau.

(S: K = 2,76.1016 ; E = 1,255V) Cu 8: Thit lp gin Latime ca Vanai da vo cc d kin sau:

2

2

/SO Zn / Zn

V / V V / V

0,170V 0,760V

1,180V +

= =

= =

2- 2+4

2+ 3+

+ 2+ 2- o4 2 4 2+ + 2+ 2+ o4 2

o o

SOo o

(1) 2V(OH) + SO 2VO + SO + 4H O ; E = 0,83V(2) 2V(OH) + 3Zn + 8H 2V + 3Zn + 8H O ; E =1,129VE ; E

E ; E 0, 255V

Cu 9: Chun 10 cm3 dung dch FeCl2 0,1 N bng dung dch K2S2O3 0,1N 25oC. Phn ng c theo di bng cch o th in cc platin. Tnh th im tng ng bit rng gi tr th in cc chun:

3 2 2 22 8 4

o o

Fe / Fe S O /SOE 0,77V;E 2,01V+ + = = (S: 1,62V) Cu 10: Cho bit cc s liu sau ti 25oC:

2 2

o o

O / H OAu / AuE 1,7V;E 1, 23V+ = = . Hng s in li tng ca ion phc [Au(CN)2]- l 7,04.10-40. Chng minh rng khi c mt ion CN- trong dung dch kim th

2

o

[Au(CN) ] / AuE nh

hn 2

o

O / OHE

ngha l oxi c th oxi ha c vng. (S: -0,61V < 0,404V => pcm) Cu 11: xc nh s tn ti ca ion thy ngn s oxi ha +I trong dung dch, ngi ta thit lp mt pin sau ti 25oC: Hg | Hgn(ClO4)n 2,5.10-3M || Hgn(ClO4)n 10-2M| Hg. Sut in ng o c l 0,018V. Tnh gi tr ca n t suy ra s tn ti ca nnHg

+ trong dung dch. (S: n = 2) Cu 12: 25oC ta c: 2 2

2

o o

Hg / Hg Hg / HgE 0,85V;E 0,79V+ += = ; 2

2 2 4

-28 30t (Hg I ) 4[HgI ]T =10 ; 10

= ; 244[HgI ]

l hng s in li tng ca [HgI4]2-. 1. Tnh 2 2

2

o

Hg / HgE + +

2. Tnh hng s cn bng ca phn ng sau trong dung dch: 2 22Hg Hg Hg+ + + . Ion 22Hg

+ bn hay

khng bn trong dung dch. 3. Trong dung dch 22Hg

+ 10-2M cha I- s to ra kt ta. Tnh nng I- khi bt u kt ta Hg2I2

4. Tnh 2 2

o

Hg I / HgE . Thit lp phng trnh 2 2o

Hg I / HgE ([I ])= f 5. 2 24Hg 4I [HgI ]+ + , nng no ca I- th 2 24[Hg ] [HgI ]+ = 6. Tnh 2

4 2 2

o

HgI / Hg IE

. Thit lp phng trnh 24 2 2

o 24HgI / Hg I

E ([HgI ],[I ])

= f - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Phn 6: NNG DUNG DCH S IN LI Cu 1: Tnh pH ca dung dch KHSO3 1M bit cc hng s in li ca axit H2SO3 ln lt l:

1 2

-2 -7a aK =1,3 10 ; K =1,23 10

Cu 2: Tnh tan ca FeS pH = 5 cho bitt: Fe2+ + H2O [Fe(OH)]+ + H+ c lg = -5,92 TFeS = 10-17,2 ; H2S c Ka1 = 10-7,02 ; Ka2 = 10-12,9 (S: S = 2,43.10-4 M) Cu 3: Cho 0,01 mol NH3, 0,1 mol CH3NH2 v 0,11 mol HCl vo H2O c 1 lt dung dch. Tnh pH ca dung dch thu c ? Cho +

4NHpK = 9,24 , +

3 3CH NHpK = 10,6 ,

2H OpK = 14

Hng dn gii Xt cc cn bng sau CH3NH2 + HCl CH3NH3Cl



0,1 0,1 0,1 (mol) NH3 + HCl NH4Cl 0,01 0,01 0,01 (mol) Do V= 1 (l) nn CM = n. Dung dch cha CH3NH3Cl 0,1M v NH4Cl 0,01M CH3NH3Cl CH3NH3+ + Cl- NH4Cl NH4+ + Cl- CH3NH3+ CH3NH2 + H+ K1 = 10-10.6 (1) NH4+ NH3 + H+ K2 = 10-9.24 (2) Tnh gn ng v do (1) v (2) l s in li ca 2 axit yu nn ta c:

10,6 9.24 61 1 2 2H C .K C .K 0,1.10 0,01.10 2,875.10

pH lg H 5,54

+

+

= + = + =

= =

Cu 4: nh gi kh nng ha tan ca HgS trong: a. Axit Nitric HNO3 b. Nc cng toan

Bit: - 223 4

0 0 -51,8 7 12,92 14,92S/H S HgS 2 a1 a2NO /NO HgCl

E = 0,96V ; E = 0,17V ; T = 10 ; H S: K = 10 ; K 10 ; = 10

= Cu 5: Tnh pH ca dd NH4HCO3 0,1M. Cho bit: +

4

6,35 10,332 3 a1 a2 aNHH CO : K = 10 ; K 10 ; pK = 9,24 =

Cu 6: Ion 22 7Cr O b thy phn theo phng trnh sau: 2 22 7 2 4Cr O H O 2CrO 2H ++ + ; K = 10-14,4

1. Thm KOH vo dung dch K2Cr2O7 nng ban u ca hai cht u bng 0,1M. Tnh pH ca dung dch thu c

2. Trn 20 ml dung dch K2Cr2O7 0,1M vi 20 ml dung dch Ba(NO3)2 1M s to kt ta BaCrO4 (Tt = 10-9,7) Tnh pH ca dung dch thu c sau khi trn.

(S: 1. pH = 6,85 ; 2. pH = 1) Cu 7: Dung dch MgCl2 0,01M 25oC bt u kt ta Mg(OH)2 ti pH = 9,5. 1. Tnh tch s tan ca Mg(OH)2 2. Tnh th kh ca cp Mg2+/Mg khi pH = 11, bit rng th kh chun ca n l 2,36 3. Gii thch ti sao khi ghp Mg vo cc thit b bng thp th c th bo v c thp khi b n mn

S: Tt = 10-11 ; 22 2 Mg(OH)o 2Mg / Mg Mg / MgT0,0592E E lg 2,51V

n [OH ]+ + = + = ; 2 2o o

Mg / Mg Fe / FeE E+ +< => n mn in ha

Cu 8: Tnh nng ti thiu ca NH3 c th ha tan hon ton 0,1 mol AgCl bit rng TAgCl = 10-10, hng s in li tng ca phc [Ag(NH3)2]+ bng 10-7,2 (S: 2,7M) Cu 9: Tnh ha tan (mol.l-1) ca AgCl trong dung dch NH3 1M bit rng TAgCl = 10-10, hng s bn tng ca phc [Ag(NH3)2]+ bng 1,6.107 (S: 0,037M) Cu 10: Hg2+ to vi I- kt ta mu HgI2 (Tt = 10-28). Nu d I- th HgI2 tan to thnh [HgI4]2- ( 304 10 = ) Thm dung dch KI 1M vo 10 ml dung dch Hg2+ 0,01M. Tnh th tch V1 dd KI cn thm vo bt u kt ta HgI2 v th tch V2 dung dch KI cn thm vo HgI2 bt u tan ht. Tnh nng cc ion trong dung dch khi cn bng trong c hai trng hp S: Khi bt u kt ta V1 = 10-12 cm3 ; [Hg2+] = 0,01M ; [I-] = 10-13 M ; [HgI4]2- = 10-24 M Khi kt ta bt u ha tan ht: V2 = 0,5 cm3 ; [HgI4]2- = 0,01M ; [I-] = 0,1M ; [Hg2+] = 10-24 M Cu 11: Dung dch cha ion Fe(SCN)2+ c mu bt u t nng 10-5M. Hng s bn ca ion Fe(SCN)2+ l 2b 2 10 = 1. Trong 500 cm3 dung dch cha 10-3 mol FeCl3 v 5.10-3 mol KSCN. Tnh nng ion Fe(SCN)2+ ti

trng thi cn bng. Hi dung dch c mu khng 2. Ha tan tinh th NaF vo dung dch trn (th tch dung dch khng bin i) to thnh ion FeF2+ vi hng

s bn l 5b 1,6 10 = . Hi bt u t lng no th mu bin mt. S: 1. 1,27.10-3M > 10-5M nn c mu ; 2. 0,0938 gam



Cu 12: Mt sunfua kim loi MS c tch s tan Tt. Tnh pH ca dung dch M2+ 0,01M bt u kt ta MS bng dung dch H2S bo ha 0,1M v pH ca dung dch khi kt thc s kt ta ca sunfua ny, nu chp nhn nng ca M2+ cn li trong dung dch l 10-6M

S: Bt u kt ta : t1pH lg T 122

= + , kt thc kt ta: t1pH lg T 142

= +

Cu 13: Thm 1 ml dung dch 4NH SCN 0,10 M vo 1ml dung dch 3Fe + 0,01 M v F 1M. C mu ca phc 2+FeSCN hay khng? Bit rng mu ch xut hin khi 2+ 6eSCN 7.10

>F

C M v dung dch c axit ha

s to phc hidroxo ca Fe (III) xy ra khng ng k. Cho 1 13,103 eF3 10F

= ; 12 3,03

eSCN 10F + = ( l hng s bn). Cu 14: nh gi thnh phn cn bng trong hn hp gm Ag+ 1,0.10-3 M; 3NH 1,0 M v Cu bt. Cho

3 2

7,242Ag(NH ) 10 + = ; 23 4 12,034 ( ) 10+ =Cu NH ; 20 0Ag / Ag Cu / CuE 0,799V;E 0,337V+ += = ( 250C)

Cu 15: Cho: H2SO4 : pKa2 = 2 ; H3PO4 : pKa1 = 2,23 , pKa2 = 7,21 , pKa3 = 12,32 1. Vit phng trnh phn ng v xc nh thnh phn gii hn ca hn hp khi trn H2SO4 C1M vi

Na3PO4 C2M trong trng hp sau: 2C1 > C2 > C1 2. Tnh pH ca dung dch H3PO4 0,1M 3. Cn cho vo 100ml dung dch H3PO4 0,1M bao nhiu gam NaOH thu c dung dch c pH= 4,72.

B N TP HA HC 12 K THI HC SINH GII QUC GIA LP 12 THPT KIM TRA S 1

MN THI: HA HC HA HC I CNG



Thi gian lm bi: 180 pht (Khng k thi gian pht ) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Cu 1: (2.0 im) 1. Ti sao ion phc spin thp [Co(NH3)6]3+ li c mu. Gii thch da vo 1o 22900(cm ) = . Cho bit:

1 11 cm 11,962 J.mol = . 2. Da trn m hnh VSEPR, gii thch dng hnh hc ca NH3, ClF3, XeF4. 3. Qu trnh: O O 1e+ + c I1 = 13,614 (eV). Da vo phng php Slater xc nh hng s chn ca cc

electron trong nguyn t i vi electron b tch. So snh bn tng i ca hai cu hnh electron ca O v O+, gii thch.

Cu 2: (2.0 im) Thit lp biu thc ph thuc gia th oxi ha kh vi pH ca mi trng trong 2 trng hp sau: 1. 2 3

2 7

2 32 7 2 Cr O / 2Cr

Cr O 14H 6e 2Cr 7H O 1,33V +

+ ++ + + = + o ; E . pH = 7, 22 7Cr O c oxi ha c I

khng? Bit rng: 2

o

I / 2IE 0,6197V

=

2. 3 23 2 Co(OH) / Co(OH)Co(OH) 1e Co(OH) OH 0,17V

+ + = + o ; E

Bit tch s ion ca nc KW = 10-14 v RT2,303 ln x 0,0592lg xF

= ( ti 25oC, 1atm) Cu 3: (1.5 im) Cho phn ng: A B C D+ + (1) l phn ng n gin. Ti 27oC v 68oC, phng trnh (1) c hng s tc tng ng ln lt l k1 = 1,44.107 mol-1.l.s-1 v k2 = 3,03.107 mol-1.l.s-1, R = 1,987 cal/mol.K 1. Tnh nng lng hot ha EA (theo cal/mol) v gi tr ca A trong biu thc ( )

ERTk A e

= mol-1.l.s-1.

2. Ti 119oC, tnh gi tr ca hng s tc phn ng k3. 3. Nu CoA = CoB = 0,1M th 1/2 nhit 119oC l bao nhiu. Cu 4: (2.5 im) in phn dung dch NaCl dng in cc Katode l hn hng Hg dng chy u v dng cc titan bc ruteni v roi l Anode. Khong cch gia Anode v Katode ch vi mm 1. Vit phng trnh phn ng xy ra ti in cc khi mi bt u in phn pH = 7. Tnh cc gi tr th

in cc v th phn gii 2. Sau mt thi gian, pH tng ln n gi tr pH = 11. Gii thch ti sao. Vit cc phng trnh xy ra ti pH

. Tnh th in cc v th phn gii Cho bit:

2 23 2

o o o

O / H ONa / Na 2H O / HE 2,71V ; E 0,00V ; E 1, 23V+ += = = . Vi dung dch NaCl 25% v 0,2% Na

trong hn hng Na/Hg: oNa / Na(Hg)E 1,78V+ = . 2

o

Cl / ClE 1,34V

= cho dung dch NaCl 25% theo khi lng

2H1,3V = trn Hg ;

2O0,8V = trn Ru/Rd

Cu 5: (1.5 im) Trong cc tinh th (cu trc lp phng tm khi) cc nguyn t cacbon c th chim cc mt ca mng c s

1. Bn knh kim loi ca st l 1,24 o

A . Tnh di cnh a ca mng c s

2. Bn knh cng ha tr ca cacbon l 0,77 o

A . Hi di cnh a s tng ln bao nhiu khi st c cha cacbon so vi cnh a khi st nguyn cht

3. Tnh di cnh mng c s cho st (cu trc lp phng tm din) v tnh tng chiu di cnh mng bit rng cc nguyn t cacbon c th chim tm ca mng c s v bn knh kim loi st l

1,26 o

A . C th kt lun g v kh nng xm nhp ca cacbon vo 2 loi tinh th st trn Cu 6: (1.5 im) Kt qu phn tch mt phc cht A ca Platin (II) cho bit c: 64,78 % khi lng l Pt, 23,59 % l Cl, 5,65 % l NH3 v 5,98 % cn li l H2O 1. Tm cng thc phn t ca phc cht bit rng A l phc cht 1 nhn v Pt c s phi tr l 4. Vit cng

thc cu to 2 ng phn cis v trans ca n



2. Entanpi t do chun to thnh 25oC ca cc ng phn cis, trans ln lt l: -396 v -402 kJ.mol-1. Tnh hng s cn bng K ca phn ng sau: cis(A) trans(A)

3. Tnh nng mol/lit mi ng phn trong dung dch, bit rng lc u ch c ng phn cis nng 0,01M. Cho Pt = 195 ; Cl = 35,5 ; N = 14 ; O = 16 ; H = 1

Cu 7: (2.0 im) Nitramit c th b phn hy trong dd H2O theo phn ng: NO2NH2 N2O(k) + H2O Cc kt qu thc nghim cho thy vn tc phn ng tnh bi biu thc: 2 2

3

[NO NH ]v k [H O ]+=

1. Trong mi trng m bc ca phn ng l bao nhiu 2. Trong cc c ch sau c ch no chp nhn c:

a. C ch 1: 1k2 2 2 (k) 2NO NH N O + H O b. C ch 2:

2

3

k +2 2 3 2 3 2

k+ +2 3 2 3

NO NH H O NO NH + H ONO NH N O + H O

++

c. C ch 3:

4

5

6

k +2 2 2 2 3

k2 2

k+3 2

NO NH H O NO NH + H ONO NH N O + OHH O OH 2H O

+

+

Cu 8: (3.0 im) 1. C 3 nguyn t A, B v C. A tc dng vi B nhit cao sinh ra D. Cht D b thu phn mnh trong nc

to ra kh chy c v c mi trng thi. B v 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 c trong t nhin v thuc loi cht cng nht. Hp cht ca 3 nguyn t A, B, C l mt mui khng mu, tan trong nc v b thu phn. Vit tn ca A, B, C v phng trnh cc phn ng nu trn.

2. kho st s ph thuc thnh phn hi ca B theo nhit , ngi ta tin hnh th nghim sau y: Ly 3,2 gam n cht B cho vo mt bnh kn khng c khng kh, dung tch 1 lt. un nng bnh B ho hi hon ton. Kt qu o nhit v p sut bnh c ghi li trong bng sau:

Nhit (oC) p sut (atm) 444,6 0,73554 450 0,88929 500 1,26772 900 4,80930

1500 14,53860 Xc nh thnh phn nh tnh hi n cht B ti cc nhit trn v gii thch.

Cu 9: (1.5 im) C th vit cu hnh electron ca Ni2+l: Cch 1: Ni2+ [1s22s22p63s23p63d8] Cch 2: Ni2+ [1s22s22p63s23p63d64s2].p dng phng php gn ng Slater, tnh nng lng electron ca Ni2+ vi mi cch vit trn (theo n v eV). Cch vit no ph hp vi thc t. Ti sao. Cu 10: (2.5 im) 1. Phng th nghim c mu phng x Au198 vi cng 4,0 mCi/1g Au. Sau 48 gi ngi ta cn mt dung

dch c phng x 0,5 mCi/1g Au. Hy tnh s gam dung mi khng phng x pha vi 1g Au c dung dch ni trn. Bit rng Au198

c t1/2 = 2,7 ngy m. 2. Hy chng minh rng phn th tch b chim bi cc n v cu trc (cc nguyn t) trong mng tinh th kim

loi thuc cc h lp phng n gin, lp phng tm khi, lp phng tm din tng theo t l 1 : 1,31 : 1,42.

Nhanh Chm

Nhanh Chm Nhanh


Cu Ni dung im 1 Tnh c: 437nm = . S hp th nh sng nm trong ph nhn thy nn c mu. 0,5

2

Cu to ca NH3 cho thy quanh nguyn t N trung tm c 4 vng khng gian khu tr electron, trong c 1 cp electron t do (AB3E) nn phn t NH3 c dng thp y tam gic vi gc lin kt nh hn o109 28' (cp electron t do i hi mt khong khng gian khu tr ln hn)

Cu trc thp y tam gic tm l nguyn t N Phn t ClF3 c 5 khong khng gian khu tr electron, trong c 2 cp electron t do (AB3E2) nn phn t c dng ch T (Cc electron t do chim v tr xch o)

Phn t XeF4 c 6 vng khng gian khu tr electron, trong c hai cp electron t do (AB4E2) nn c dng vung phng (trong cu trc ny cc cp electron t do phn b xa nhau nht)

0,25 3 = 0,75

1

3

- Cu hnh electron: O 1s22s22p4 km bn hn O+ 1s22s22p3 do lc y ln nhau ca 2 trong mt orbital ca phn lp 2p v do O+ t cu hnh bn bo ha phn lp 2p nn bn

- t b l hng s chn ca cc electron trong nguyn t i vi electronb tch.

Ta c: *2

*2 2 21 2

ZI 13,6 13,614 Z n 4 (8 b) 4 b 6n

= = = = = =

0,25

0,5

2.0

2 1

2 3 2 32 7 2 7

2 32 7

2 32 7

1422 7

2Cr O / 2Cr Cr O / 2Cr 3

214 2 7

2Cr O / 2Cr 3

22 7

2Cr O / 2Cr 3

Cr O H0,0592 lg6 Cr

Cr O0,0592 0,0592lg H lg6 6 Cr

Cr O0,05920,138pH lg6 Cr

+ +

+

+

+

+

+

+

+

+

= + +

= +

o

o

o

E = E

E

E

t: 2 3 2 32 7 2 7Cr O / 2Cr Cr O / 2Cr

0,138pH + +=

' oE E

2 32 7

'

Cr O / 2CrE

+ l th iu kin v ph thuc vo pH. pH cng gim th dung dch

cng c mi trng axit th E cng tng, tnh oxi ha ca 22 7Cr O

cng mnh. - Ti pH = 0, [H+] = 1M th E = Eo = 1,33V - Ti pH = 7 th E = 0,364 <

2

o

I / 2IE 0,6197V

= nn khng oxi ha c I-

0,5

0,25

0,25

2 3 2 3 2Co(OH) / Co(OH) Co(OH) / Co(OH)

10,0592lg [OH ]+oE = E . Thay WK[OH ] [H ]

+= ta c:

0,25


3 2 3 2

3 2

3 2

Co(OH) / Co(OH) Co(OH) / Co(OH)W

Co(OH) / Co(OH) W

Co(OH) / Co(OH) W

[H ]0,0592lgK

0,0592lg[H ] 0,0592lg K0,0592pH 0,0592lg K

+

+

+

= +

=

o

o

o

E = E

E

E

. Thay 3 2o

Co(OH) / Co(OH)14

W

0,17

K 10+

=

E =

3 2Co(OH) / Co(OH) 0,996 0,0592pH E = pH cng tng th E cng gim ngha l tnh oxi ha ca Co(OH)3 gim, tnh kh ca Co(OH)2 tng

0,5

0,25

2.0

1

- Phn ng ng hc bc hai, p dng phng trnh Archnius ta c: A

11

Eln k ln ART

= + ; A22

Eln k ln ART

= +

A A2 1

2 1

2 A

1 1 2

2 1 2A

2 1 1

E Eln k ln k ln A ln ART RT

k E 1 1lnk R T T

T T kE R ln 3688, 2(cal / mol)T T k

= + +

=

=

-

( )( ) ( )

ERT

E ERT RT1

9 1 11kkk A e A 7 10 (mol .l.s )e e

= = =

0,5

0,25

0,25

2 ( )ERT3 7 1 13k A e 6,15 10 (mol .l.s )

= = 0,25

3

3 71/ 23 oA

1 1,63 10 (s)k .C

= = 0,25

1.5

1

Trong dung dch NaCl c: NaCl Na++Cl- ; 2H2O H3O + + OH- Khi in phn c th c cc qu trnh sau xy ra: Catode: Na+ + Hg + e Na(Hg) 1 (1) 2H2O H3O+ + OH- 2 2 H3O+ + 2e H2 + 2H2O 1 2H2O + 2e H2 + OH- (2) Anode: 6 H2O O2 + 4H3O+ + 4e (3) 2 Cl- Cl2 + 2e (4)

Na / Na(Hg)E 1,78V+ = , 3 27

2H O / HE 0,00V 0,0592lg10 0, 413V+ = + =

23 2 3 2

' o

H2H O / H 2H O / HE E 1,713+ += + = . Do

3 2

' o

2H O / H Na / Na(Hg)E E+ +> nn khi mi bt u in phn, Katode qu trnh (2) s xy ra, c H2 thot Anode Anode: T (3) ta c:

2 2 2 2 2 2 2 2 2

o '

O / H O O / H O 3 O / H O O / H O OE E 0,0592lg[H O ] 0,817V ; E E 1,617V+= + = = + = Bi v:

2 22

'

O / H OCl / 2ClE E

< nn Anode xy ra qu trnh (4) v c Cl2 bay ra Phng trnh in phn: 2Cl- + 2H2O H2 + Cl2 2OH- Th phn gii: V = ' 'A KE E = 3,053V

4.0,125 = 0,5

0,25

0,25 0,25 0,25

4

2 Sau mt thi gian, do [OH-] tng nn pH cng tng. Khi pH = 11, phn ng in

phn xy ra nh sau:


Ti Catode: [H+] =10-11. 3 2 3 2

' o

2H O / H 2H O / H Na / NaE 0,649V ; E 1,949V E+ + += = > nn

Anode c qu trnh (1) xy ra Ti Anode:

2 2 2 2 2 2

o '

O / H O O / H O 3 O / H OE E 0,0592lg[H O ] 0,581V ; E 1,381V+= + = = Do

2 22

'

O / H OCl / 2ClE E

< nn Anode vn c Cl2 bay ra Phng trnh in phn: 2Na+ + 2Cl- + 2Hg Cl2 + 2Na(Hg) Th phn gii: V = ' 'A KE E = 3,12V

0,25

0,25

0,25 0,25

2.5

1 di cnh a ca mng c s ca st l: o4r 4 1, 24

a 2,86A3 3

= = = 0,25

2. Khi st c cha cacbon, tng chiu di cnh a ca mng c s l:

o

Fe C2 (r r ) a 2(1,24 0,77) 2,86 1,16 A = + = + = 0,25

5

3

di cnh a ca mng c s ca st l: o4r 4 1,26

a 3,56A2 2

= = =

Khi st c cha cacbon, tng chiu di cnh a ca mng c s l: o

Fe C2 (r r ) a 2(1, 26 0,77) 3,56 0,5A = + = + = Kt lun: Kh nng xm nhp ca cacbon vo st kh hn vo st , do c ha tan ca C trong st nh hn trong st

0,25

0,25

0,5

1.5

1

t CTPT ca A l: PtxCly(NH3)z(H2O)t. V phc cht A l phc 1 nhn nn phn t khi ca A: PtA

M 100% 195 100M 301(g / mol)%Pt 64,78

= = = . T % ca cc thnh phn

c trong A x = 1, y = 2, z = 1, t = 1 CTPT l: PtCl2(NH3)(H2O)

CTCT 2 ng phn cis, trans:

Cis Trans

0,25

0,125.2 = 0,25

2

Xt phn ng chuyn ha: Cis Trans Cn bng: 10-2 x x

o

298KG 402 396 6kJ = + = = -6000J ; 6000

8,314 298K e 11, 27= =

0,5

6

3

Xt phn ng chuyn ha: Cis Trans K = 11,27 Cn bng: 10-2 x x

[trans]K [cis]= =3 4

2x 11, 27 x [trans] 9,2 10 [cis] 8 10

10 x

= = = =

0,25.2 = 0,5

1.5

7 1 Do trong mi trng m [H3O]+ = const nn biu thc tc phn ng l:

2 2v k[NO NH ]= l phn ng bc nht theo thi gian 0,5


2

- C ch 1: 1 2 2v k [NO NH ]= loi - C ch 2: 3 2 3v k [NO NH ]+=

M: 2 2 32 3 22

[NO NH ][H O ][NO NH ] k [H O]+

+=

Vy: 2 2 33 22

[NO NH ][H O ]v k k [H O]

+

= loi

- C ch 3: 5 2v k [NO NH ]=

M: 2 2 22 43

[NO NH ][H O][NO NH ] k [H O ]

+=

Vy: 2 2 25 43

[NO NH ][H O]v k k [H O ]+=

Trong mi trng dung dch nc [H2O] = const. Chn c ch 3

0,25 0,25

0,25

0,25

0,5

2.0

1

- Hp cht AxBy l mt mui. Khi b thu phn cho thot ra H2S. - Hp cht AnCm l Al2O3 nhm oxi - Vy A l Al nhm, B l S lu hunh, C l O oxi - Hp cht AoBpCq l Al2(SO4)3 nhm sunfat 2 Al + 3 S Al2S3 Al2S3 + 6 H2O 2 Al(OH)3 + 3 H2S 4 Al + 3 O2 2 Al2O3 S + O2 SO2 Al3+ + 2 H2O Al(OH)2+ + H3O+

5 cht 0,125.5 = 0,625

5 ptrnh 0,125.5 = 0,625

8

2

S mol nguyn t S trong 3,2 gam lu hunh: S3,2

n 0,1mol32

= =

Dng cng thc: PVnRT

= tnh c s mol cc phn t lu hunh trng thi hi ti

cc nhit : * 444,6oC: n1 = 0, 0125 mol gm cc phn t S8 v 0, 0125 8 = 0,1 mol

* 450oC: n2 = 0,015 mol, s nguyn t S trung bnh trong 1 phn t: 0,1 6,67

0,015

Thnh phn hi lu hunh nhit ny c th gm cc phn t lu hunh c t 1 n 8 nguyn t.

* 500oC: n3 = 0,02 mol, s nguyn t S trung bnh trong 1 phn t: 0,1 5

0,02=

Thnh phn hi lu hunh nhit ny c th gm cc phn t lu hunh c t 1 n 8 nguyn t hoc ch gm cc phn t S5.

* 900oC: n4 = 0,05 mol, s nguyn t S trung bnh trong 1 phn t:0,1 2

0,05=

Thnh phn hi lu hunh nhit ny c th gm cc phn t lu hunh c t 1 n 8 nguyn t hoc ch gm cc phn t S2. * 1500oC : n5 = 0,1 mol : Hi lu hunh ch gm cc nguyn t S.

0,25

0,25

0,25 0,25

0,25

0,25

0,25

3.0

9 Nng lng ca mt electron phn lp l c s lng t chnh hiu dng n* c tnh theo biu thc Slater: 1 = -13,6 x (Z b)2 /n* (theo eV)

0,25


Hng s chn b v s lng t n* c tnh theo quy tc Slater. p dng cho Ni2+ (Z=28, c 26e) ta c: Vi cch vit 1 [Ar]3d8: 1s = -13,6 x (28 0,3)2/12 = -10435,1 eV 2s,2p = -13,6 x (28 0,85x2 0,35x7)2/ 22 = - 1934,0 eV 3s,3p = -13,6 x (28 1x2 0,85x8 0,35x7)2/32 = - 424,0 eV 3d = - 13,6 x (28 1x18 0,35x 0,35x7)2/32 = - 86,1 eV E1 = 2 1s + 8 2s,2p + 8 3s,3p + 8 3d = - 40423,2 eV Vi cch vit 2 [Ar]sd64s2: 1s, 2s,2p, 3s,3p c kt qu nh trn . Ngoi ra: 3d = -13,6 x (28 1x18 0,35x5)2/32 = - 102,9 eV 4s = - 13,6 x (28 1x10 0,85x14 0,35)2/3,72 = - 32,8 eV Do E2 = - 40417,2 eV. E1 thp (m) hn E2, do cch vit 1 ng vi trng thi bn hn. Kt qu thu c ph hp vi thc t l trng thi c bn ion Ni2+ c cu hnh electron [Ar]3d8.

0.125.5 = 0,625

0,125.3 = 0,375

0,25

1.5

1

- t = 48 h = 2 ngy m. - p dng biu thc tc ca phn ng mt chiu bc mt cho phn ng phng x, ta c: = 0,693/t1/2; Vi t1/2 = 2,7 ngy m, = 0,257 (ngy m)-1. T pt ng hc p. mt chiu bc nht, ta c: =(1/t) ln N0/N. Vy: N/N0 = e- t = e-0,257 x 2 = 0,598. Nh vy, sau 48 gi phng x ca mu ban u cn: 0,598 x 4 = 2,392 (mCi). Do s gam dung mi tr cn dng l: (2,392 : 0,5) 1,0 = 3,784 (g)

0,25

0,25 0,25 0,25

10

2

Phn th tch b chim bi cc nguyn t trong mng tinh th cng chnh l phn th tch m cc nguyn t chim trong mt t bo n v ( mng c s).

- i vi mng n gin: + S nguyn t trong 1 t bo: n = 8 x 1/8 = 1 + Gi r l bn knh ca nguyn t kim loi, th tch V1 ca 1 nguyn t kim loi:

V1 = 4/3 xpi r3 (1) + Gi a l cnh ca t bo, th tch ca t bo l:

V2 = a3 (2) Trong t bo mng n gin, tng quan gia r v a c th hin trn hnh sau:

hay a = 2r (3). Thay (3) vo (2) ta c: V2 = a3 = 8r3 (4) Phn th tch b chim bi cc nguyn t trong t bo l: V1/V2 = 4/3 pi r3 : 8r3 = pi /6 = 0,5236

i vi mng tm khi: + S nguyn t trong 1 t bo: n = 8 x 1/8 + 1 = 2. Do V1 = 2x(4/3)pi r3 . + Trong t bo mng tm khi quan h gia r v a c th hin trn hnh sau:

0,5

r

a


HA HC: NGH THUT, KHOA HC V NHNG BT NG TH V

Do : d = a 3 = 4r. ra a = 4r/ 3 Th tch ca t bo: V2 = a3 = 64r3/ 3 3 Do phn th tch b chim bi cc nguyn t trong t bo l: V1 : V2 = 8/3 pi r3 : 64r3/3 3 = 0,68 i vi mng tm din: + S nguyn t trong 1 t bo: n = 8 x 1/8 + 6 x = 4. Do th tch ca

cc nguyn t trong t bo l: V1 = 4 x 4/3pi r3 + Trong t bo mng tm din quan h gia bn knh nguyn t r v cnh a ca t bo c biu din trn hnh sau:

T d ta c: d = a 2 = 4r, do a = 4r/ 2 Th tch ca t bo: V2 = a3 = 64r3/2 2 Phn th tch b cc nguyn t chim trong t bo l: V1/V2 = 16/3 pi r3: 64r3/ 2 2 = 0,74 Nh vy t l phn th tch b chim bi cc nguyn t trong 1 t bo ca cc mng n gin, tm khi v tm din t l vi nhau nh 0,52 : 0,68 : 0,74 = 1 : 1,31 : 1,42.

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0,5

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