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Transcript of Bộ Giáo Trình Hóa Học Đại Cương - Huỳnh Thiên Lương, Từ Anh Phong, Phan an,...
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8/11/2019 B Gio Trnh Ha Hc i Cng - Hunh Thin Lng, T Anh Phong, Phan an, Phm Th Thanh, Nguyn Ph Huyn Chu, Nguyn Th Minh Minh,
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TRNG I HC TR VINH
BI GING
HA HC I CNG
Bin son: Th.s Hunh Thin Lng
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Bi ging Ha hci cng 1
MU
Ha hc l mt trong nhng lnh vc khoa hc tnhin nghin cu vthgii vt chtv svn ng ca n, nhm tm ra cc quy lut vn ng vn dng vo cuc sng.
Svn ng ha hc ca vt cht l qu trnh bin i cht ny thnh cht khc. V dnhsoxi ha kim loi bi oxi ca khng kh, sphn hy cc cht hu cbi cc vi khun, squang hp bin kh cacbonic v hi nc thnh cc hp cht gluxit, st chy nhin liu to ra
nng lng dng trong i sng v sn xut.Nhng schuyn ha cc cht nhtrn gi l hin tng ha hc hay phn ng ha hc.
Cc phn ng ha hc xy ra thng km theo sbin i nng lng di cc dng khc nhau(nhit, in, quang, c,...) c gi l nhng hin tng km theo phn ng ha hc.
Khnng phn ng ha hc ca cc cht phthuc vo thnh phn, cu to phn tvtrng thi tn ti ca chng, iu kin thc hin phn ng, l tnh cht ha hc ca cc cht.
Bi vy i tng ca ha hc c tm tt nhsau:Ha hc l khoa hc vcc cht,n nghin cu thnh phn, cu to, tnh cht ca cc cht, schuyn ha gia chng, cc hintng km theo schuyn ha v cc quy lut chi phi chng.
Cc qu trnh ha hc khng ngng xy ra trn v tri t, trong lng t, trong khngkh, trong nc, trong cc cthng vt, thc vt,...
Nhiu ngnh khoa hc, kinh t lin quan cht ch vi ha hc: cng nghip ha hc,luyn kim, a cht, sinh vt hc, nng nghip, y hc, dc hc, xy dng, giao thng vn ti,ch to vt liu, cng nghip nh, cng nghip thc phm,... S dnh vy l v cc ngnh usdng cc cht l i tng; do cn phi bit bn cht ca chng.
Slin quan cht chgia ha hc v cc ngnh khoa hc khc lm ny sinh cc mnha hc phc vcho tng ngnh: ha nng, ha hc t, ha hc trong xy dng, ha hc nc,sinh ha, ha hc bo v thc vt, ha hc bo vmi trng, ha dc, ha thc phm, haluyn kim...
Chng 1MT SKHI NIM V NH LUT CBN CA HA HC
I. CC KHI NIM MU
1. Nguyn tNguyn tl ht nhnht cu to nn cc cht khng thchia nh hn na bng phng
php ha hc.2. Nguyn tha hc
Nguyn tha hc l khi nim ch mt loi nguyn t. Mt nguyn tha hc cbiu thbng k hiu ha hc. V d: nguyn toxi O, canxi Ca, lu hunh S...3. Phn t
Phn tc to thnh tcc nguyn t, l ht nhnht ca mt cht nhng vn mangy tnh cht ca cht .
V d: Phn tnc H2O gm 2 nguyn thidro v 1 nguyn toxi, phn tClo Cl2gm 2 nguyn tclo, phn tmetan CH4gm 1 nguyn tcacbon v 4 nguyn thidro...4. Cht ha hc
Cht ha hc l khi nim chmt loi phn t. Mt cht ha hc c biu thbngcng thc ha hc. V d: mui n NaCl, nc H2O, nitN2, st Fe...5. Khi lng nguyn t
l khi lng ca mt nguyn t ca nguyn t. Khi lng nguyn tc tnhbng n vcacbon (vC). Mt vC bng 1/12 khi lng nguyn tcacbon (12C).
V d: khi lng nguyn toxi = 16 vC, Na = 23 vC...6. Khi lng phn t
l khi lng ca mt phn t ca cht. Khi lng phn t cng c tnh bngvC. V d: khi lng phn tca N2 = 28 vC, HCl = 36,5 vC...
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Bi ging Ha hci cng 2
7. Mol l lng cht cha N = 6,02 .10
23phn tvi m (phn tnguyn t, ion electron...).
N c gi l sAvogaro v n bng snguyn tC c trong 12 gam12
C.8. Khi lng mol nguyn t, phn t, ion
l khi lng tnh bng gam ca 1 mol nguyn t(phn thay ion...). Vs trnng bng trskhi lng nguyn t(phn thay ion). V d: khi lng mol nguyn tcahidro bng 1 gam, ca phn tnitbng 28 gam, ca H2SO4 bng 98 gam...
9. Ha trHa trca mt nguyn tl slin kt ha hc m mt nguyn tca nguyn t to
ra vi cc nguyn t khc trong phn t. Mi lin kt c biu th bng mt gch ni hainguyn t. Ha trc biu thbng chsLa M.
Nu qui c ha trca hidro trong cc hp cht bng (I) th ha trca oxi trong H2Obng (II), ca nittrong NH3 bng (III)... Da vo ha tr(I) ca hidro v ha tr(II) ca oxi cthbit c ha trca nhiu nguyn tkhc.
V d: Ag, cc kim loi kim (ha trI); Zn, cc kim loi kim th(II) Al (III), cc khtr(ha tr0) ;Fe (II, III); Cu (I, II); S (II, IV, VI)10. Soxi ha
Soxi-ha c qui c l in tch ca nguyn ttrong phn tkhi ginh rng cpelectron dng lin kt vi nguyn tkhc trong phn tchuyn hn vnguyn tc inm ln hn.
tnh soxi-ha ca mt nguyn t, cn lu : Soxi-ha c thl sdng, m, bng 0 hay l sl; Soxi-ha ca nguyn ttrong n cht bng 0; Mt snguyn tc soxi-ha khng i v bng in tch ion ca n- H, cc kim loi kim c soxi ha +1 (trong NaH, H c soxi-ha -1)- Mg v cc kim loi kim thc soxi-ha +2- Al c soxi - ha +3; Fe c hai soxi-ha +2 v +3- O c soxi - ha -2 (trong H2O2O c soxi-ha -1)
Tng i ssoxi - ha ca cc nguyn ttrong phn tbng 0.
V d:2,50 0 1 1 1 6 2 4 7 1
22 4 2 22 3 2 6 4 2, , , , , , ,Zn Cl Na Cl K S O Na S O Na S O K MnO H O
4 1 1 0 3
2 2 2 22 5 4 3 4 2 3 2 4, , ( ), ( ),C O C H OH C H O CH CHO C H O CH COOH H C O
II. CCNH LUT CN BN1.nh lut bo ton khi lng
Nh bc hc Lomonossov Mikhail Vasilyevich cp n nh lut bo ton khi lngnm 1948 ti Nga, nh lut c Antonine Laurent Lavoisier pht minh ti Php trong khongcc nm 1772 1777.
- nh lut: Khi lng ca cc cht tham gia phn ng bng khi lng ca cc cht tothnh sau phn ng.
- ng dng: - Dng cn bng cc phng trnh phn ng- Tnh khi lng cc cht tham gia v to thnh sau phn ng.
2.nh lut thnh phn khng i (Joseph Louis Proust 1801 n 1807)nh lut: l mt hp cht ho hc d c iu chbng cch no th bao gicng c
thnh phn khng i.V d: Nc d iu chbng nhiu cch khc nhau nht hidr v xi trong khng kh,
thc hin phn ng gia mt axit v mt baz, t cht hydrcacbon cng lun lun u chahydr v xi theo tlkhi lng hydr v xi 1:8.
Ngy nay ngi ta thy nh lut thnh phn khng i p dng ng cho cht kh v chtlng khi lng phn tthp. Vi mt scht rn, do nhng khuyt tt ca mng tinh th, thnh
phn ca chng c thkhng tng ng chnh xc vi cng thc ha hc.V d: c thiu chcc tinh thtitan oxit TiO ng vi tlnguyn t1:1; tuy nhin nu
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Bi ging Ha hci cng 3
thay i iu kin iu ch, ta c th thu c cc tinh thng vi thnh phn tTi0,75O tiTi1,44O.
3.nh lut tlbi (John Dalton 1803)nh lut: Nu hai nguyn tkt hp vi nhau cho mt shp cht th ng vi cng
mt khi lng nguyn tny, cc khi lng ca nguyn tkia tlvi nhau nhnhng snguyn n gin
V d: Nit to vi oxi nm oxit, nu ng vi mt n vkhi lng nitth khi lng
ca oxi trong cc oxit ln lt l 0,57; 1,14; 1,71; 2,28; 2,85; cc khi lng ny t lvinhau nhl: 0,57 : 1,14 : 1,71 : 2,28 : 2,85 = 1 : 2 : 3 : 4 : 5iu trn y c thhiu c ddng nu cht nhn cc oxit c CTPT ln lt l N2O,
NO, N2O3, NO2, N2O54. Phng trnh trng thi kh l tng (Clapeyron Mendeleev)
Nhng nghin cu vtnh cht ca cc cht kh cho thy rng nhit khng qu thp vp sut khng qu cao (so vi nhit v p sut thng), phn ln cc kh u tun theo mt hthc gi l phng trnh trng thi kh l tng.
PV = nRTTrong : P: p sut cht kh
V: thtch cht khn: smol kh
T: K (T = to+ 273)R: hng skh. Gi trca R phthuc vo cc n vo:P V R n v
atm lt 22,4: 273 0,08205 l atm mol-1K-1
mmHg mililit 62400 ml mmHg mol-1K-1
N/m2 m3 8,3144 J mol-1K-1
Ta bit:m m mRT
n PV RT M M PV
V vy nh lut ny c ng dng xc nh phn tgam ca cc cht kh bng thcnghim.
Vi hai khi kh, tphng trinh cn bn ta c:
2 2 2 2
1 1 1 1
P V n TP V n T
Cc trng hp ring:
Trng hp 1: V1= V2, T1= T2rt ra: 1 1
2 2
P n
P n
Trng hp 2: P1= P2, T1= T2rt ra: 1 1
2 2
V n
V n
Nhcc nh lut cht kh, ta c thxc nh c khi lng phn tca cc cht kha. Nu bit khi lng m ca mt khi kh iu kin xc nh (V, T, P) p dng phng
trnh Clapeyron Mendeleev xc nhphn tgam ca cc cht kh bng thc nghim.
Ta bit:m m mRT
n PV RT M M PV b. Nu bit khi lng ring D0ca mt cht kh iu kin chun (khi lng 1 lt kh
ktc), ta c M = 22,4.D0c. Nu bit tkhi hi cht kh A (m ta cn nh gi trkhi lng phn t) so vi mt
cht B r, ta cng suy ra c:
/ /A
B A B A BA
Md M M d
M
5.nh lut Avgadr- nh lut: cng mt iu kin nht nh v nhit v p sut, nhng th tch
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Bi ging Ha hci cng 4
bng nhau ca cc cht kh u cha cng mt sphn t.T iu kin chun (i vi phn ng xy ra gia cc cht kh) ta c "iu kin
chun (OoC, 1 atm), mt mol ca bt kmt cht kh no cng u chim mt thtch bng nhauv bng 22,4lt".
6.nh lut ng lng6.1. ng lng ca mt nguyn tTrong cc phn ng ho hc, cc nguyn tkt hp vi nhau theo nhng tlxc nh
gi l tlkt hp hay ng lng ca chng.Vy "ng lng ca mt nguyn tl sphn khi lng ca nguyn tt c thtcdng hoc thay thva vi mt phn khi lng hydr hoc 8 phn khi lng xi.
ng lng k hiu l .V d: HCl c Cl = 35,5 v H = 1ng lng ca mt nguyn tthc cht l sphn khi lng ca nguyn t ng
vi mt n vho trm n tham gia phn ng.
A
n =
Trong :A: khi lng mol nguyn t: ng lng ca nguyn tn : ho trca nguyn t
* Ch :V ho trca mt nguyn tc ththay i nn ng lng ca n cng thay i.V d: ng lng ca C trong CO l C= 12/2 = 6
ng lng ca C trong CO2l C= 12/4 = 3i vi cc nguyn tc ho trkhng i th ng lng cng khng i.- ng lng gam ca mt nguyn tl khi lng ca nguyn t tnh ra gam ng
lng ca nguyn t.6.2. ng lng ca mt hp chtng lng ca mt hp cht l sphn khi lng ca cht tc dng va vi
mt ng lng ca mt nguyn thay mt hp cht khc bt k.ng lng ca mt hp cht thng c tnh theo cng thc:
n =
Trong : M: khi lng mol phn tca hp cht- Trong phn ng trao in: + sion H+m mt phn taxit tham gia trao i
+ sion OH-m mt phn tbaztham gia trao i+ Tng sin tch ion m hoc dng m mt phn t mui
tham gia trao i.- Trong phn ng xi ho khn: secletron m mt phn tcht xi ho thu vo hay mt phn tcht khmt i.V d: ng lng gam ca KMnO4trong cc mi trng nhsau
- Mi trng axit:4
24 25 8 4 5KMnO
MnO e H Mn H O
- Mi trng trung tnh:4
4 2 23 2 43
KMnOMnO e H O MnO OH
- Mi trng baz:4
24 41 1KMnO
MnO e MnO
- ng lng gam ca mt hp cht l gi trng lng ca cht tnh ra gam.V d: ng lng gam ca HCl bng 36,5gam
ng lng gam ca H2bng 2gam6.3. Nng ng lng (N)Nng ng lng gam ca mt dung dch l sng lng gam ca cht tan c
trong mt lt dung dch
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Bi ging Ha hci cng 5
V d: dd HCl 1N c 36,5gam HCl nguyn cht trong mt lt.dd H2SO40,1N c 4,9 gam H2SO4trong 1 lt
6.4. nh lut ng lng"Cc cht phn ng vi nhau theo nhng khi lng tlvi ng lng ca chng"
hay "cc cht tham gia phn ng vi nhau theo nhng slng ng lng gam nhnhau".
Ta c: hayA A A B
B B A B
m m m
m
Trong mA, mBl khi lng hai cht A, B phn ng va vi nhau. A, Bl nglng ca hai cht A, B.
p dng nh lut ng lng cho phn ng xy ra trong dung dch:Gisc 2 cht A v B phn ng vi nhau theo phng trnh:A + B CGi NA, NBln lt l nng ng lng ca dd A v B. VA, VBl thtch ca dung
dch A v dung dch B phn ng va vi nhau.Theo nh lut ng lng ta c: cc cht A v B phn ng va vi nhau theo s
ng lng nhnhau nn: VA.NA= VB.NBTy ta c thxc nh nng ng lng ca mt cht khi bit nng ng
lng ca cht kia bng thc nghim.
7.nh lut tlthtch cht kh (Gay Lussac Nh khoa hc Php)cng iu kin nhit v p sut, thtch cc cht kh tham gia phn ng (chttc dng ban u cng nhsn phm) tlvi nhau nhnhng snguyn n gin
V d: N2 + 3H2 2NH3Quan hgia cc cht kh tham gia phn ng l nhsau:
2 2 3: : 1: 3 : 2N H NHV V V
8. p sut ring cht kh v nh lut Daltonp sut ring ca mt cht kh trong mt hn hp l p sut do cht kh to nn khi n
chim thtch ca ton bhn hp kh trong cng nhng iu kin vt l.Trn 2 l kh oxi vi 3 l kh nitc cng p sut 1 atm c 5 l hn hp, p sut ring ca
oxi v ca nittrong hn hp l:
2
21 0,4a
5Op tm 2
31 0,6a
5Np tm
nh lut: p sut chung ca hn hp cc cht kh khng tham gia tng tc ha hcvi nhau bng tng p sut ring ca cc cht kh to nn hn hp
Trn 3 l CO2(960 mmHg) vi 4 l O2 (1080 mmHg) v 6 l N2(906 mmHg) c 10 l hnhp mi. Tm p sut chung ca hn hp
2
3960 288
10COp mmHg
2
41080 432
10Op mmHg
2
6906 544
10Np mmHg
p sut chung ca hn hp: p = 288 + 432 + 544 = 1264 mmHg9. Qui tc Dulong Petit
nh ngha 1: T nhit (tc nhit dung ring) mt cht l nhit lng cn thit anhit ca mt gam cht ln 1o.
nh ngha 2: Nhit dung nguyn tl tch sgia khi lng mol nguyn t(A) vi tnhit ca n cht do nguyn tto nn.
Qui tc Dulong Petit: Nhit dung nguyn t ca a s n cht rn c gi tr vokhong 6,3 (cal -1mol-1)
A tnhit 6,3(khi lng mol nguyn tc n v: g mol-1, tnhit: cal -1g-1)Nu bit tnhit ca n cht rn ta c thsuy ra gi trkhi lng nguyn tgn ng
ca nguyn tnhqui tc trn.
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Bi ging Ha hci cng 6
Cu hi v bi tp1. Vit cng thc cu to ca cc cht c cng thc phn tdi y:
a. SO2, SO3, P2O5; b. HBr, H2S, PH3, CH4c. KOH, Ba(OH)2, Al(OH)3 d. CaCl2, AlBr3, Al2S3;e. HNO3, H2SO4, H3PO4 f. KNO3, Ca(NO3)2, Na2SO4, AlPO4
2. Xc nh soxi ha ca cc nguyn tdi y:a. MnO, MnO2, Mn(OH)3, Mn
2+, MnO, 24MnO
b. Na2S, S, FeS2, SO2, SO3, Na2S2O3, Na2S4O6, 4HSO , 23SO
, HS
c. NH3, NO, N2O3, N2O, NO2, 4NH , NH4Cl, HNO3, NH4NO3, 3NO
, N2H4.
3. Qui ra smol cc lng nguyn tsau: 5,4 g nhm, 16 g lu hunh; 6,4 g oxi; 71 g clo.4. C bao nhiu mol nguyn tv bao nhiu nguyn ttrong 4 g oxi, 20 g hidro, 12 g magie, 112g st, 7 g silic?5. Khi xc nh thnh phn axit axetic, ngi ta thy c 2,1 phn khi lng cacbon; 0,35 phnkhi lng H; 2,8 phn khi lng O. Khi lng phn taxit axetic bng 60. nh CTPT axitaxetic.6. t chy hon ton 13,8 g mt hp cht hu cc 26,4 g CO2; 16,2 g H2O. Tkhi hi cahp cht so vi hidro bng 23. nh CTPT hp cht.7. Nung m gam hn hp rn gm Fe2O3v Fe, ri cho mt lung kh CO (thiu) qua hn hptrn. Sau phn ng thu c 30,2 g cht rn vi 22 g CO2. Tnh m8. 8 gam hn hp kim loi ha tan hon ton trong dung dch HCl cho 4,48 l H2(ktc). Khi ccn fung dch thu c bao nhiu gam mui khan?9. Canxi clorua cha 36% Ca v 64% Cl. nh ng lng canxi bit ng lng clo bng35,5.10. Sunfua mt kim loi cha 52% kim loi. nh ng lng kim loi bit ng lng luhunh l 16.11. nh ng lng cc nguyn tdi y trong cc phn ng:
a. S + O2 SO2 S= ?b. Fe + Cl2 FeCl3 Fe= ?c. C + O2 CO2 C= ?
d. C + O2 CO C= ?12. nh ng lng tng axit, tng baztrong cc phn ng:a. H3PO4 + NaOH NaH2PO4 + H2Ob. H3PO4 + 3NaOH Na3PO4 + 3H2Oc. 2HCl + Cu(OH)2 CuCl2 + H2Od. HCl + Cu(OH)2 Cu(OH)Cl + H2O
13. nh ng lng cc cht gch di:a. FeSO4 + BaCl2 BaSO4 + FeCl2b. Al2O3 + 6HCl 2AlCl3 + 3H2Oc. CO2 + NaOH NaHCO3d. CO2 + 2NaOH Na2CO3 + H2O
e. Al2O3 + 2NaOH 2NaAlO2 + H2Of. KCr(SO4)2. 12H2O + 3KOH Cr(OH)3+ 2K2SO4+ 12H2O14. nh ng lng cc cht gch di:
a. 2FeCl3 + SnCl2 2FeCl2 + SnCl4b. 2KMnO4+ 5HNO2+ 3H2SO4 2MnSO4+ 2K2SO4+ 5HNO3+ 3H2Oc. K2Cr2O7+ 3H2S + 4H2SO4 Cr2(SO4)3+ 3S + K2SO4+ 7H2O
15. nh ng lng gam cc cht gch di, qui ra sgam cc cht tng ng thnh snglng gam ca chng:
Ca(OH)2 + H3PO4 CaHPO4 + 37 g 24,5 g
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Bi ging Ha hci cng 7
K2Cr2O7 + FeSO4 + Fe2(SO4)3+ Cr2(SO4)4+ 29,4 g 15,2 g
16. nh khi lng axit oxalic (ng lng 45) va lm mt mu 0,79 g thuc tm(ng lng 31,6).17. Mt kim loi tvi oxi hai oxit. Khi un nng 3 g mi oxit trong mt lung kh hidro c d,ngi ta thu c ln lt 0,679 g v 0,377 g H2O
a. Tnh ng lng ca kim loi trong tng oxit;
b. nh tn kim loi.18. 1,355 g mui st clorua tc dng va 1,000 g NaOH. Tnh ng lng ca mui stclorua, nh cng thc phn tca n.19. nh thtch 1 g kh hidro iu kin chun.20.
a. Tm thtch cc lng kh sau y ktc : 80 g O2, 3g NO, 128 g SO2b. Tm th tch ktc ca 1 mol cc cht sau: O2, CO2, glixerin C3H8O3 (cht lng c
khi lng ring D= 12,6 g/ml)21. Tm khi lng ca 0,56 l H2; 1,12 l CO2; 4,48 l N2ktc.22. Tm khi lng phn t cc cht kh c khi lng ring ktc l: 1,339 g/l; 1,07 g/l;0,7142g/l.23. Tm khi lng kh butan C4H8cha trong mt bnh kn dung tch 25 l (p = 1,64 atm, t
o=
87oC)24. Tnh khi lng phn tca benzen bit 600 ml hi benzen 87oC, 624 mmHg nng 1,3 g.25. Mt bnh kn dung tch khng i cha hn hp cng thtch H2v O2150
oC, 1 atm. tchy hn hp a bnh v150oC, p sut bnh l bao nhiu?26. Bnh kn cha 2 thtch nitv 3 thtch kh hidro, sau mt thi gian tham gia phn ng,mt na lng N2bin thnh NH3. Hi p sut bnh thay i thno nu nhit v th tchbnh khng i?27. Mt bnh kn dung dch khng i cha 3 th tch oxi v 1 thtch metan 120oC, p sut0,5 atm. t chy hn hp, phn ng hon ton, a nhit bnh v120oC. Hi p sut mica bnh l bao nhiu?28. Mt bnh kn dung tch cnh cha hn hp cng thtch oxi v axetilen 150oC. t chyhn hp, phn ng hon ton. Hi p sut mi ca bnh tng hay gim nhthno so vi banu v150oC?29. Mt bnh kn dung tch khng i cha hn hp cng thtch nitvi hidro 0oC, 10 atm.Sau khi tin hnh tng hp amoniac, a nhit bnh v0oC.
a. Tm p sut bnh sau phn ng, gisc 60% lng hidro phn ng.b. Nu p sut trong bnh sau phn ng l 9 atm th c bao nhiu phn trm mi kh
tc dng?30. Trn 30 l CH4vi 40 l H2v 10 l Co cng nhit . p sut ban u ca CH4, H2v COtng ng l 720, 630 v 826 mmHg. Th tch ca hn hp bng 80 l. Tnh p sut ring catng kh v p sut chung ca hn hp.31. Mt bnh 5 l cha oxi p sut 9 atm ni vi mt bnh khc 10 l cha oxi p sut 6 atm. Mkha cho thng kh hai bnh, hi p sut cui cng l bao nhiu?
32. Cho 2,69 g PCl5ha hi ti 250
o
C trong mt bnh kn dung tch 1 l. Mt phn PCl5bphnhy theo phng trnh phn ng:PCl5(k) PCl3(k) + Cl2(k)p sut cui cng l 1 atm. p sut ring tng kh trong hn hp.
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Bi ging Ha hci cng 8
Chng 2: CU TO NGUYN T
Khi nim nguyn t"atom" (khng thphn chia) c cc nh trit hc cHy Lpa ra cch y hn hai nghn nm. Tuy nhin mi n thk19 mi xut hin nhng githuytvnguyn tv phn t.
Nm 1861 thuyt nguyn t, phn tchnh thc c tha nhn ti Hi nghha hcthgii hp Thy S.
Chn cui thk19 v u thk20 vi nhng thnh tu ca vt l, cc thnh phncu to nn nguyn tln lt c pht hin.1. Thnh phn cu to ca nguyn t
Vmt vt l, nguyn tkhng phi l ht nhnht m c cu to phc tp, gm t nhtl ht nhn v cc electron. Trong ht nhn nguyn tc hai ht c bn: proton v ntron.
Ht Khi lng (g) in tch (culong)electron (e) 9,1.10-28 -1,6. 10-19
proton (p) 1,67.10-24 +1,6. 10-19ntron (n) 1,675.10-24 0
- Khi lng ca e 1/1840 khi lng p.- in tch ca e l in tch nhnht v c ly lm n vin tch, ta ni electron
mang in tch -1, cn proton mang in tch dng +1.- Nu trong ht nhn nguyn tca mt nguyn tno c Z proton th in tch ht
nhn l +Z v nguyn t phi c Z electron, v nguyn ttrung ha in.- Trong bng tun hon, sthtca cc nguyn tchnh l sin tch ht nhn hay s
proton trong ht nhn nguyn tca nguyn t.2. Nhng mu nguyn tcin2.1. Mu Rzfo (Anh) 1911
Tthc nghim Rzfo a ra mu nguyn thnh tinh nhsau:- Nguyn tgm mt ht nhn gia v cc electron quay xung quanh ging nhcc
hnh tinh quay xung quanh mt tri (hnh 1).- Ht nhn mang in tch dng, c kch thc rt nhso vi kch thc ca nguyn t
nhng li chim hu nhton bkhi lng ca nguyn t.Mu Rzfo cho php hnh dung mt cch n gin cu to nguyn t. Tuy nhin khnggii thch c s tn ti ca nguyn t cng nhhin tng quang ph vch ca nguyn t.
2.2. Mu Bo (an Mch), 1913Da theo thuyt lng tca Plng v nhng nh lut ca vt l cin, Bo a rahai nh :
- Trong nguyn t, electron quay trn nhng quo trn xc nh (hnh 2). Bn knh ccquo c tnh theo cng thc:
rn= n2. 0,53 . 10-8cm = n2. 0,53 A
o
(1)n l cc stnhin 1, 2, 3,..., n
Nhvy cc quo thnht, thhai... ln lt c cc bn knh nhsau:r1= 1
2. 0,53 A = 0,53 A
Hnh 2Hnh 1
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Bi ging Ha hci cng 9
r2= 22. 0,53 A = 4. 0,53 A = 4r1
- Trn mi quo, electron c mt nng lng xc nh, c tnh theo cng thc:
2
113,6nE eVn
(2)
Khi quay trn quo, nng lng ca electron c bo ton. N chpht hay thu nnglng khi bchuyn tmt quo ny sang mt quo khc. iu gii thch ti sao li thuc quang phvch khi kch thch nguyn t.
Thuyt Bo nh lng c cc quo v nng lng ca electron trong nguyn tng thi gii thch c hin tng quang phvch ca nguyn thidro l nguyn tn ginnht (chc mt electron), tuy nhin vn khng gii thch c quang ph ca cc nguyn t
phc tp.iu cho thy rng i vi nhng ht hay hht vi m nhelectron, nguyn t th
khng thp dng nhng nh lut ca chc cin. Cc hny c nhng c tnh khc vi hvm v phi c nghin cu bng phng php mi, c gi l chc lng t.3.c tnh ca ht vi m hay nhng tin ca chc lng t3.1. Bn cht sng ca ht vi m (electron, nguyn t, phn t...)
Nm 1924, Bri (Php) trn csthuyt sng - ht ca nh sng ra thuyt sng- ht ca vt cht:
Mi ht vt cht chuynngu lin kt vi mt sng gi l sng vt cht hay sng linkt, c bc sngtnh theo hthc:
h
mv
h: hng sPlanckm: khi lng ca htv: tc chuyn ng ca ht
Nm 1924, ngi ta xc nh c khi lng ca electron, ngha l tha nhnelectron c bn cht ht.
Nm 1927, Davison v Gecme thc nghim cho thy hin tng nhiu x chmelectron. iu chng tbn cht sng ca electron.
Nhvy:Electron va c bn cht sng va c bn cht ht.
3.2. Nguyn l btnh (Haixenbec -c), 1927i vi ht vi m khng thxcnh chnh xcng thi ctcv vtr.
.2
hx v
m
x: bt nh vvtrv: bt nh vtc m: khi lng htTheo h thc ny th vic xc nh v tr cng chnh xc bao nhiu th xc nh tc
cng km chnh xc by nhiu.4. Khi nim cbn vchc lng t4.1. Hm sng
Trng thi ca mt h v m s hon ton c xc nh nu bit qu o v tc chuyn ng ca n. Trong khi i vi nhng h vi m nh electron, do bn cht sng - ht vnguyn l bt nh, khng th vc cc quo chuyn ng ca chng trong nguyn t.
Thay cho cc quo, c hc lng tm t th mi trng thi ca electron trong nguyn tbng mt hm sgi l hm sng, k hiu l (pxi).
Bnh phng ca hm sng 2c ngha vt l rt quan trng:2biu thxc sut c mt ca electron ti mtim nhtnh trong vng khng gian
quanh ht nhn nguyn t.Hm sng nhn c khi gii phng trnh sng i vi nguyn t.
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Bi ging Ha hci cng 12
Thng thng chcn vit cu hnh electron i vi cc phn mc lp ngoi cng vphn mc d hoc f lp st ngoi cng m cha bo ha.
Cn lu rng cu hnh ni trn l i vi cc nguyn ttrng thi cbn. Khi bkchthch electron c thnhy ln nhng phn mc cao hn trong cng mt mc.
V d
Nhvy trng thi cbn C c hai electron c thn, cn trng thi kch thch n cbn electron c thn. Chnh cc electron c thn ny l cc electron ha tr.6. Hthng tun hon cc nguyn tha hc6.1.nh lut tun hon
"Tnh cht ca cc n cht, thnh phn v tnh cht ca cc hp cht ca cc nguyn tho hc bin thin tun hon theo chiu tng ca in tch ht nhn nguyn t".
Vy sbin i tun hon tnh cht ca cc nguyn sl do sbin i tun hon cu
trc e- ca nguyn tca cc nguyn t.
6.2. Bng hthng tun hon cc nguyn t(Xt bng hthng tun hon dng di)6.2.1. Bng hthng tun hon dng di c xy dng trn cssau:- Cc nguyn tc xp theo chiu tng dn in tch ht nhn.- Cc nguyn tc cng slp electron c xp thnh mt hng ngang gi l chu k.
- Cc nguyn tc cng cc electron ho trc xp thnh ct dc gi l nhm.6.2.2. Sthtcc nguyn tMi nguyn tc xp vo 1 c sthtng bng in tch ht nhn Z ca nguyn
t.V d: Nguyn tS c Z = 16 c 16p nn c 16e- trng thi cbn.
6.2.3. Cu to bng hthng tun honCc nguyn tho hc trong bng hthng tun hon c xp theo thtin tch ht
nhn Z tng dn v c phn thnh cc chu kv nhm.Chu k:
L dy nguyn txp theo hng ngang gm cc nguyn tc cng s(trchu k1 bt
u bng Hydro) v kt thc v mt kh him.Sthtca chu kbng slp electron m mi nguyn ttrong chu kc. Hin nay
bng hthng tun hon gm 7 chu k, nh sthtt1 n 7.Trong :
Chu k1 gm 2 nguyn tH (Z = 1) v He (Z = 2)Chu k2 gm 8 nguyn ttLi (Z = 3) n Ne (Z = 10)Chu k3 gm 8 nguyn ttNa (Z = 11) n Ar (Z = 18)Chu k4 gm 18 nguyn ttK (Z = 10) n Kr (Z =36)Chu k5 gm 18 nguyn ttRb (Z = 37) n Xe (Z = 54)
Chu k6 gm 32 nguyn ttCr (Z = 55) n Rn (Z = 86)
(c bit trong chu k6 c 14 nguyn txp sau La (Z = 57) c xp tch ring thnh
1 hng ngang di bng gi l hLantan).
- Chu k7 gm cc nguyn tang xy dng tFr (Z = 87) tri.
Trng thi kch thch
Trng thi cbn C (Z = 6)
C*
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Bi ging Ha hci cng 13
(Trong chu k7 c 14 nguyn txp sau Ac (Z = 89) c tch ring thnh hActini)* Nhm:
Gm cc nguyn tc cng selectron ho trsthtca nhm bng selectron hotrm cc nguyn tc. Mi nhm c chia thnh 2 phn nhm: chnh v ph.
- Phn nhm chnh (nhm A): gm cc nguyn ts hoc p- Phn nhm ph(nhm B) gm cc nguyn td hoc f
Cc xc nh selectron ho trca mt nguyn t
Loi nguyn t - Selectron ho tr
s hoc p - Tng selectron lp ngoi cng
d hoc f - Selectron trn ns (nu (n-1) d bo ho
- Tng selectron trn ns v (n-1)d [nu (n-1)d cha bo ho]V d:
- Cu (z = 29) c cu hnh [Ar]3d10
4s1: thuc nhm IB
- Zn (z = 30) c cu hnh [Ar]3d10
4s2: thuc nhm IIB
- Fe (z = 26) c cu hnh [Ar]3d64s2: thuc nhm VIIIB
- Li, Na, K c cu hnh ns1: thuc nhm IA* Khi:
Ngy nay ngi ta thng xp cc nguyn tthnh cc khi da vo cu hnh electron
ca chng, c 4 khi l:
- Khi s: gm cc electron c cu hnh electron ngoi cng l ns1, ns2 l cc nhm IA,
IIA.- Khi p: gm cc nguyn tc cu hnh cc electron ngoi cng tns2np1 n ns2np6 tc
l cc nguyn tthuc nhm IIIA n VIIIA.- Khi d: gm cc nguyn tc cu hnh (n-1)d1-10ns hoc 2 thuc cc nhm B
- Khi f: gm 2 dy lantanic va actinic xp cui bng.6. 3. Sbin i tun hon tnh cht cc nguyn t
6.3.1. Bin i tun hon cu hnh electron ca cc nguyn t
- i vi cc nguyn tnhm A, cc electron lp ngoi cng tng dn t1e- nhm IA
n 8 electron lp VIIIA, iu ny c lp li tt ccc chu k.
- i vi cc nguyn tnhm B: Cc electron lp ngoi cng ging nhau l ns2 (trmt strng hp ngoi l).
+ Cc nguyn td th phn lp electron d tng t1 electron nhm IIIB n 9 electron nhm IB v 10 electron nhm IIB.
+ Cc nguyn tf: cu hnh electron ngoi cng v hngoi ging nhau chkhc nhau phn lp (n-2)f.
6.3.2. Bn knh nguyn tTrong mt chu kkhi i ttri sang phi, in tch ht nhn tng dn nn lc ht gia ht
nhn v in tngoi cng tng dn, do bn knh nguyn tgim dn.Trong mt nhm khi i ttrn xung bn knh nguyn ttng dn do slp in ttng
dn.6.3.3. Nng lng ion ho (I)L nng lng ti thiu cn cung cp bc mt electron ra khi nguyn ttrng thi
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Bi ging Ha hci cng 14
cbn.- Trong mt chu kkhi i ttri sang phi, in tch ht nhn, lc ht gia cc ht nhn
v electron ngoi cng tng dn, electron bht cht hn nn nng lng ion ho tng dn.
- Trong mt nhm khi i ttrn xung do bn knh nguyn ttng nn khong cch gia
ht nhn v electron ngoi cng tng dn, lc ht gia ht nhn v in tngoi cng gim dn
do nng lng ion ho gim dn.6.3.4. i lc vi electron (E)L nng lng c gii phng khi nguyn ttrng thi kh nhn thm mt electron
trthnh anion kh.A + e- = A-
Cho n nay ngi ta chmi xc nh c i lc i vi electron ca mt st nguynt.
Trong cc nguyn tnhm VIIA c i lc vi electron ln nht. Cc nguyn t s2,
s2p
6, s
2p
3c i lc vi electron rt km.
6.3.5. Soxi ho
- Soxi ho dng cao nht ca cc nguyn tbng sthtca nhm cha chng (trcc nhm VIIIB, IB, Lantanic, Actinic, O, F v kh him).- Soxi ho m thp nht hu nhchc cc nguyn tphi kim v soxi ho m thp
nht ca nguyn tbng sthtca nhm cha nguyn t tri 8.
V d:Nitc soxi ho m thp nht l: 5 - 8 = -36.3.6. Hp cht vi hydro v oxi- Hp cht vi hydro:
Cc kim loi nhm IA, IIAto cc ion H- (NaH, CaH2)
Cc nguyn tphi kim thng to cc hp cht cng ho tr, trong hydr c xi ho
+1, phi kim c soxi ho m (CH4, NH3, H2O, HF, HCl )- Hp cht vi oxi:
xi to c oxit vi hu ht cc nguyn t(trPt, Au).Trong mt chu kkhi i ttri sang phi soxi ho ca cc nguyn tto oxit tng dn
t1 n 7.V d:
Chu k3: Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7Cc bazv axit tng ng cng nhcc oxit c tnh axit tng dn ttri sang phi.
NaOH Mg(OH)2 Al2(OH)3 H2SiO3 H3PO4 H2SO4 HClO4tnh axit tng dn.
Trong mt nhm khi i ttrn xung tnh kim loi tng nn tnh bazca mt oxit vhidroxit cng tng dn.
BeO MgO CaO SrO BaO
Be(OH)2 Mg(OH)2 Ca(OH)2 Sr(OH)2 Ba(OH)2
lng tnh baz bazmnh
6.4. Quan hgia cu hnh electron v vtr ca nguyn ttrong bng hthng tun hon
6.4.1. Bit sin tch ht nhn ca nguyn t(Z) ta c thxc nh c nguyn tthuc chu kmy, nhm my v cu hnh electron ca nguyn t.
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Bi ging Ha hci cng 15
V d: Cho X c Z = 16
Ta c thvit cu hnh electron ca X: 1s22s22p63s22p4
Vy X thuc chu k3, nhm VIA l nguyn tlu hunh (S).6.4.2. Bit cu hnh electron ca nguyn t, c thxc nh in tch ht nhn Z, sth
tca nguyn t, chu k, nhm, hp cht vi hydro, hp cht vi oxi V d1: Nguyn tX c cu hnh 1s22
s22p
63s
23p
5(Z = 17)
V n = 3 nn nguyn tny thuc chu k3. Selectron ho tr (2 + 5 = 7) thuc nhm
VIIA l nguyn tClo.
- Hp cht vi hydro l HCl
xit cao nht vi oxi l Cl2O7
V d2:
Nguyn tY c cu hnh 1s2 2s22p63s
23p
63d
54s
2 (Z = 25)
V lp d cha bo ho nn Y thuc nhm B c eho trl (2 + 5 = 7) nn Y thuc nhmVIIB c lp electron ngoi cng l 4 nn thuc chu k 4 l nguyn tMn.
- Khng c hp cht vi hydro
- xit cao nht ca Mn vi xi l Mn2O76.4.3. Bit vtr ca nguyn tc thsuy ra mt stnh cht ho hc cbn ca mt s
cc hp cht ca nguyn t.V d: Bit nguyn tM thuc chu k4, nhm IIA vy l kim loi kim th, chng c
thnhng 2 electron trthnh cation: M = M2++ 2e to c cc hp cht c lin kt ion lchnh, hp cht vi hydr c cng thc l MH2, hp cht vi xi l MO v hidroxit tng ng lM(OH)2.
Cu hi v bi tp:1. Ni dung nguyn l bt nh v thuyt sng vt cht.
2. Hy cho bit khi nim vhm sng v ngha vt l ca 2
.3. Obitan nguyn tl g? Thno l my electron?4. Hy cho bit hnh dng ca m my electron 2s; 2pxv c im ca cc m my . Skhc nhau gia cc m my 1s v 2s; 2px v 2py, 2pz.5. Hy cho bit ni dung ca nguyn l vng bn v ngha ca nguyn l ny. Vit dy thtnng lng ca cc obitan trong nguyn t.6. Pht biu qui tc Hund v nu ngha ca qui tc ny.7. Vit cu hnh electron ca cc nguyn tc sthtz = 28; 36; 37; 42; 47; 53; 56; 80. Hycho bit vtr ca nguyn ttrong HTTH v tnh cht ha hc c trng.8. Gii thch v saoO (z = 8) c ha tr2, cn S (z = 16) li c cc ha tr2, 4, 6
N (z = 7) c ha tr3, cn P (z = 15) li c cc ha tr3, 5
F (z = 9) c ha tr1, cn Cl (z = 17) li c cc ha tr1, 3, 5, 7.9. Vit cu hnh electron ca cc ion: Cu+, Cu2+.10. Vit cu hnh electron ca Ar. Cation, anion no c cu hnh e ging Ar?11. Trn cscu trc nguyn t, c thphn cc nguyn tha hc thnh my loi? Hy nuc im cu to electron ca mi loi.12. Nu c im cu hnh electron ca cc nguyn tphn nhm chnh nhm I v tnh cht hahc c trng ca chng.13. Nu c im cu hnh electron ca cc nguyn tphn nhm chnh nhm VII v tnh chtha hc c trng ca chng.
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Bi ging Ha hci cng
IA
1H
1,008IIA IIIA IV
3Li
6,94
4Be
9,01
HTHNG TUN HON
CC NGUYN THA HC 5 B10,81
6
12,011
Na22,99
12Mg
24,31IIIB IVB VB VIB VIIB VIIIB IB IIB
13Al
26,98
14
28,019
K39,10
20Ca
40,08
21Sc
44,96
22Ti
47,90
23V
50,94
24Cr
51,99
25Mn
54,94
26Fe
55,85
27Co
58,93
28Ni
58,71
29Cu
63,54
30Zn
65,38
31Ga
69,72
32
72,537
Rb85,47
38Sr
87,62
39Y
88,91
40Zr
91,22
41Nb
92,91
42Mo
95,94
43Tc
(99)
44Ru
101,07
45Rh
102,91
46Pd
106,40
47Ag
107,87
48Cd
112,41
49In
114,82
50
118
55 Cs132,91
56 Ba137,31
57* La138,49
72 Hf178,49
73 Ta180,95
74 W183,85
75 Re186,20
76 Os190,20
77 Ir192,20
78 Pt195,09
79 Au196,97
80 Hg200,59
81 Tl204,37
82
20787
Fr(223)
88Ra
226,03
89 **Ac
(227)
104Ku
(261)
105Rf
(261)
106Db
(261)
* HLantan
58Ce
140,12
59Pr
140,91
60Nd
144,24
61Pm
(147)
62Sm
150,35
63Eu
151,96
64Gd
157,25
65Tb
158,93
66Dy
162,50
67Ho
164,93
68
167** Hactini
90Th
232,4
91Pa
(231)
92U
238,03
93Np
(237)
94Pu
(242)
95Am
(243)
96Cm
(247)
97Bk
(247)
98Cf
(251)
99Es
(254)
100
(25
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Bi ging Ha hci cng 17
Chng 3: LIN KT HA HC V CU TO PHN TTrmt skh tr, cc nguyn tkhng tn ti c lp m chng thng lin kt vi nhau
to nn cc phn t. Vy cc phn tc hnh thnh nhthno? Bn cht ca cc lin kt l g?1. Mt si lng c lin quan n lin kt1.1.in m ca nguyn t
in m l i lng cho bit khnng nguyn tca mt nguyn tht electron linkt vpha n. cng ln th nguyn tcng dthu electron.
Trong lin kt gia 2 nguyn tA v B to ra phn tAB.Nu A> Bth electron lin kt slch hoc di chuyn vpha nguyn tB.Ngi ta qui c ly in m ca Li l 1 th cc nguyn tkhc sc in m tng
i nhsau:Bng 1. m in ca mt snguyn t:
IA IIA IIIA IVA VA VIA VIIA VIIIAH
2,20He
-Li
0,98Be
1,57B C
2,55N
3,04O
3,44F
3,98Ne
-Na
0,93Mg
1,31Al
1,61Si
1,90P
2,19S
2,58Cl
3,16Ar
K 0,82 Ca 1,00 Ga 1,81 Ge 2,01 As 2,18 Se 2,55 Br 2,96 Kr 2,90Rb
0,82Sr
0,95In
1,78Sn
1,96Sb
2,05Te
2,10I
2,66Xe
2,6Cs
0,79Ba
0,89Tl
2,04Pb
2,33Bi
2,02Po
2,00At
2,20Rn
Fr0,7
Ra0,89
Nhn xt:- Trong mt chu k, ttri sang phi in m ca cc nguyn ttng dn.- Trong mt phn nhm chnh, ttrn xung di in m gim dn.- Cc nguyn tkim loi kim c < 1, Fr c nhnht.- Cc nguyn tphi kim c > 2, F c ln nht.
1.2. Nng lng lin kt l nng lng cn thit ph vmi lin kt v to ra cc nguyn tthkh.
Nng lng lin kt thng k hiu E v tnh bng Kcalo cho mt mol lin kt.V d: EH-H= 104 Kcal/mol, EO-Htrong H2O = 110 Kcal/mol
Nng lng lin kt cng ln th lin kt cng bn.1.3.di lin kt
Bng2.di lin kt v nng lng lin kt ca mt slin ktLin kt Phn t r0 (A) E (kcal/mol)
C - HC - Cl
C - FC - CC - CC = CC CH - HO = OO - HS - H
N - H
CH4CHCl3
CH3FC6H6
CnH2n+2CnH2n
CnH2n-2H2O2
H2OH2S
NH3
1,091,77
1,38
1,541,341,200,741,210,961,351,01
98,775,8
116,3
79,3140,5196,7104,0118,2109,496,892,0
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Bi ging Ha hci cng 18
l khong cch gia hai nhn nguyn tkhi hnh thnh lin kt. di lin ktthng k hiu r0v tnh bng A (1A = 10
-8cm).di lin kt cng nhth lin kt cng bn vng.
1.4.bi ca lin ktSlin kt c hnh thnh gia hai nguyn tcho trc c gi l bi ca lin kt v
c k hiu l . V dbi ca lin kt gia cc nguyn tC trong etan, etilen, axetilen ln ltl 1, 2, 3.
bi ca lin kt cng ln th lin kt cng bn, nng lng lin kt cng ln v di lin ktcng nh(bng 2).
1.5. Gc lin kt (gc ha tr) l gc to bi hai mi lin kt gia mt
nguyn tvi hai nguyn tkhc.V dgc lin kt trong cc phn tH2O, CO2,
C2H4nhsau:
1.6.phn cc ca lin kt. M men lng ccTrong nhng lin kt gia hai nguyn tkhc nhau, do c schnh lch vin m,
electron lin kt blch vpha nguyn tc in m ln hn, to ra y mt in tch m no (thng k hiu -), cn nguyn tkia mang mt in tch +. Khi ngi ta ni lin kt b
phn cc.V d: H Cl
2
O C O
phn cc ca lin kt c nh gi qua m men lng cc (muy). thng c tnh
bng n vgi l bai (D).phn cc ca lin kt phthuc vo in tch trn cc v di lin kt.
Bng 3. Gi trm men lng cc ca mt slin ktLin kt H-F H-Cl H-Br H-I N=O C=O
(D) 1,91 1,07 0,79 0,38 0,16 0,11Nhn xt: Nguyn tca hai nguyn tc chnh lch in m cng ln th lin kt
gia chng cng phn cc.2. Nhng thuyt cin vlin kt
2.1. Qui tc bt tNhng thuyt kinh in vlin kt da trn qui tc bt t(octet). Xut pht tnhn xt sau
y:- Tt ccc kh tr(trHeli) u c 8 electron lp ngoi cng.- Chng rt t hot ng ha hc: khng lin kt vi nhau v hu nhkhng lin kt vi
nhng nguyn tkhc to thnh phn t, tn ti trong tnhin di dng nguyn ttdo.V vy cu trc 8 electron lp ngoi cng l mt cu trc c bit bn vng. Do cc
nguyn tc xu hng lin kt vi nhau t c cu trc electron bn vng ca cc kh trvi 8 (hoc2 i vi heli) electron lp ngoi cng.
Da trn qui tc ny ngi ta a ra mt sthuyt vlin kt nhsau:2.1. Lin kt ion (Kotxen -c), 1916
Lin kt ion c hnh thnh gia nhng nguyn t ca hai nguyn t c s chnh lchnhiu vdin m (thng > 2).
Khi hnh thnh lin kt, nguyn t ca nguyn tc nhnhng hn 1, 2 hay 3 electron chonguyn tca nguyn tc ln hn, khi n trthnh cc ion dng v nguyn tnhn electrontrthnh cc ion m c cu trc electron ging kh tr. Cc ion dng v m ht nhau to ra phn t.
V d:
Na + Cl Na+ + Cl-
2s22p63s1 3s23p5 2s22p6 3s23p6Nhvy bn cht ca lin kt ion l lc ht tnh in gia cc ion tri du.Trong lin kt ion, ha trca nguyn tbng sin tch ca ion vi du tng ng. Trong
180o
104,5o
120o o
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Bi ging Ha hci cng 20
Lin kt hidro c thhnh thnh gia cc phn t.... ...H F H F
| |...H O H O
H H
| |
...H O H O
H R
hoc trong cng mt phn tgi l lin kt hidro ni phn t.
Lin kt hidro l lin kt yu, nng lng lin kt nhv di lin kt ln. Tuy nhin nc nh hng nhiu n tnh cht vt l v ha hc ca phn t. V d:
- Do c lin kt hidro, H2O c nhit si cao hn H2S c cu to tng tvi n.- Cc phn thu cmang nhm O - H c nhit si cao hn cc ng phn ca chng khng
cha lin kt ny: ancol so vi ete; axit so vi este...- Ancol tan v hn trong nc l do to c lin kt hidro vi nc.- Lin kt hidro to ra gia cc nhm -C = O v -NH ca axit amin trong cc chuipolypeptit
duy tr c cu trc khng gian ca phn tprotein.Tm li, cc thuyt cin vlin kt cho php m tv phn loi mt cch n gin lin kt ha hc, tgii thch c mt stnh cht ca phn t. Tuy nhin cc thuyt ny c mt shn chsau y:
- Nhiu hp cht hay ion khng tha mn qui tc bt tnhng vn tn ti mt cch bnvng, v d: NO, NO2, Fe
2+...- Cha ni c bn cht ca lc lin kt gia cc nguyn ttrong phn tl g.- Khng cho bit cu trc khng gian ca cc phn t.Phn tl nhng hht vi m, v vy l thuyt vlin kt v cu to phn tphi c xy
dng trn csca chc lng t(CHLT).Nm 1927 ra i hai thuyt CHLT vlin kt bsung cho nhau, l thuyt lin kt ha tr
(vit tt l VB - valence bond) v thuyt obitan phn t(vit tt l MO molecular obitan).Lun im chyu ca cc thuyt ny l lin kt ha hc c hnh thnh do sthp cc AO ca
cc nguyn tlin kt to ra mt hmi c nng lng nhhn hban u m chnh l phnt.3. Thuyt lin kt ha tr
Thuyt lin kt ha tr(cn gi l thuyt cp electron lin kt) do Haile, Lnn (c) xut nm 1927, sau c Poling v Slyt(M) pht trin.3.1. Shnh thnh lin kt trong phn tH2
Thuyt VB c ra trn csnghin cu shnh thnh lin kt trong phn tH2.Mi nguyn tH c mt electron trng thi cbn 1s. Khi hai nguyn tH tin li gn
nhau sc hai khnng xy ra.- Nu hai electron c spin cng du, khi khong cch r gim, nng lng ca htng lin
tc, l trng thi khng bn,khng to ra lin kt ha hc.
- Nu hai electron cspin khc du nhau, nng lngca hgim dn, v ti khongcch r0= 0,74A c gi trcctiu tng ng vi nng lngES< 2E0, khi htrng thi
bn vng, trng thi hnh thnhlin kt (hnh 1). Hnh 1
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Bi ging Ha hci cng 21
Nu lu rng mi obitan s (m my s) c bn knh 0,53A th khi tip xc nhau khong
cch gia hai ht nhn phi l 1,06A. Trong khi khong chc khi hnh thnh lin kt ch cn0,74A. iu chng tkhi hnh thnh lin kt, hai obitan s c xen phvo nhau lm tng xcsut c mt electron vng gia hai ht nhn, mt in tch m tng ln gy ra sht hai htnhn v lin kt chng vi nhau.
Nhvy lc lin kt ha hc cng c bn cht tnh in.3.2. Nhng lunim cbn ca thuyt VB
Tnghin cu ca Haile v Lnn vphn t H2, Poling v Slyt pht trin thnhthuyt lin kt ha tr.- Lin kt cng ha trc hnh thnh do sghpi hai electronc thn c spin ngc du cahai nguyn tlin kt, khi c sxen phhai AO.- Mcxen phca cc AO cng ln th lin kt cng bn, lin ktc thc hin theo phngti sxen phl ln nht.
Nh vy, theo VB, khi hnh thnh phn t, cc nguyn t vn ginguyn cu trc electron, linkt c hnh thnh chdo s t hp (xen ph) ca cc electron ha tr(electron c thn).Trong thuyt VB, ha trca nguyn tbng se c thn ca nguyn ttrng thi cbn haytrng thi kch thch.
V d:
3.3. Snh hng lin kt. Lin kt(xch ma) v lin kt(pi)Ty theo cch thc xen phca cc m my electron, ngi ta phn bit lin kt , lin kt
...- Lin kt ha hc to ra do sxen phcc m my electron trn trc ni hai nhn ca nguyn t
c gi l lin kt xch ma. Lin kt c thhnh thnh do sxen phcc m my s - s, s - p hayp - p (hnh 2).
Nhvy, khi to ra lin kt th t c sxen phln nht, v vy lin kt xch ma llin kt bn. Nu gia hai nguyn tchc mt lin kt th lin kt lun lun l lin kt .
- Lin kt ha hc to ra do sxen phcc m my electron hai bn ca trc ni hainhn nguyn t, c gi l lin kt pi. Lin kt c thhnh thnh do sxen phcc m my p -
p (hnh 2), p - d...So vi lin kt th lin kt bn hn v mc xen phln hn v vng xen phnm trn
trc ni hai nhn nguyn t.Khi gia hai nguyn tc thai lin kt trln th chc mt lin kt cn li l cc lin
kt .V d: Trong phn tH2 c 1 lin kt do sxen ph2 m my s.Phn tCl2c mt lin kt do sxen ph2 m my p.Phn tHCl c mt lin kt do sxen phm my s ca H v m my px ca Cl.Phn tO2c mt lin kt do sxen phma m my px-pxv mt lin kt do sxen
ph2 m my py-pyca 2 nguyn toxi.
C Ha tr2 C Ha tr2 C* Ha tr4
Hnh 2
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Bi ging Ha hci cng 23
ca mt tdin u. Trc ca cc AO ny to ra gc 109o28'. V dslai ha ca m my
s vi 3 m my p trong nguyn tC khi hnh thnh phn tCH4.
3.5. Hnh hc phn tca mt shp chtThuyt VB cho php hnh dung c cu trc khng gian ca phn t.V d: CH4 Metan
C c lai ha sp3. 4 obitan lai ha xen phvi 4AO s ca H to 4 lin kt . Hnh hc phn tc dng tdin u. Gc lin kt 109
o28'.
NH3Amoniac
N c lai ha sp3. 3 obitan lai ha xen phvi 3AO s ca H to 3 lin kt . Hnh hc phntc dng chp. Gc lin kt l 107
o18'.
H2ONc
O c lai ha sp3. 2 obitan lai ha xen phvi 2AO s ca H to 2 lin kt . Hnh hc phn tc dng gc. Gc lin kt l 104
o30'.
3.6. Lin kt khngnh cPhn tbenzen c cu trc nhthy hnh 9. C6 nguyn tC u c lai ha sp2.Mi C to 2 lin kt vi 2 C bn cnh v 1 lin kt vi H. Cc obitan p thun cn li (c
trc vung gc vi mt phng ca cc lin kt ) xen phvi nhau ti ra cc lin kt . Nhvy ccelectron c gii ta trn c6 nguyn tC. Ngi ta gi cc lin kt l cc lin kt khngnh c. Mt cch tng tcng thy cc phn tbutadien (hnh 10). Cc lin kt khng nhcc m tbng nhng du chm thay cho cc gch.
Hnh 5
Hnh 6
Hnh 7
Hnh 8
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Bi ging Ha hci cng 24
C6H6 Benzen
C4H6
Butadien
4. Phng php obitan phn t(Phng php MO)
4.1. Csphng phpBa nh khoa hc Hund Mulliken Lennard Jones l nhng ngi t nn tng cho phng
php orbitan phn tbng cch mrng khi nim orbitan nguyn tchophn t.1/ Nu trong nguyn t, mi electron c miu tbng mt hm sng (obitan nguyn t
AO) th trong phn tmi electron c miu tbng mt hm sng (orbitan phn t, vit ttMO).
2/ Nu trong mi orbitan nguyn t, electron c c trng bng mt bcc slng tvta c cc orbitan nguyn tvi cc tn s, p, d, f th trong mi obitan phn t, electron cng cc trng bng mt bcc slng t, ta c cc orbitan phn tvi cc tn , , ,
3/ Nu mi obitan nguyn tng vi mt mc nng lng xc nh th mi orbitan phn tcng tng ng vi mt mc nng lng xc nh.
4/ Vic in electron vo AO tun theo nguyn l vng bn, nguyn l Pauli, qui tc Hund,th vic in electron vo MO cng tun theo nhng qui tc, nguyn l .
5/ Tcc AO ca cc nguyn ttham gia vo vic to thnh lin kt ngi ta c thsuy raMO ca phn tmi to thnh bng cch thp tuyn tnh cc AOban u: n AO thp li cho nMO
V dsthp tuyn tnh hai AO sav sbca hai nguyn thidro a v b cho hai MO( )a bc s s ( )a bc s s
a/ Ta c thxc nh c+v ctiu kin chun ha hm sng.2 2 2 2 2 21 ( ) ( 2s )a b a b a bdv c s s dv c s s s dv
2 2 21 [ 2s ]a b a bc s dv s dv s dv Cc orbitan sa, sbl nhng obitan chun h ngha l:
2 2 1a bs dv s dv Tch phn a bs s dv c k hiu l S v c gi l tch phn ph. Ta c:
2 2 1(1 1 2 ) [2(1 )] 12(1 )
c S c S cS
V S
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Bi ging Ha hci cng 25
1( )
2a bs s
1( )
2a bs s
b/ Ta cn c thxc nh biu thc tnh nng lng E+, Etng ng vi hai MO trn.T phng trnh H = E, nhn hai v vi , ri ly tch phn ton khng gian
2H dv E dv E (v cc hm chun ha)Xt trng hp = +
1 (21 1 1 1
= s s s s2 2 2 2
a b
a a b b a b b a
E H dv s s dv
s H dv s H dv s H dv s H dv
V sav sbl nhng AO tng ng nn ta c tht:
s sa a b bs H dv s H dv (a c gi l tch phn Coulomb)s sa b b as H dv s H dv (b c gi l tch phn trao ihoc tch phn cng hng)
T ta c: E+= + Trng hp , chng minh tng tta c E= Tch phn Coulomb l nng lng electron orbitan sa(hay sb); n bng nng lng ca
nguyn thiro trng thi cbn.Tch phn trao i l nng lng tng tc gia orbitan savi sb; n c gi tr m nnE E . Khi electron tnguyn tchuyn vMO +hthng trnn bn hn; khi electron t
nguyn tchuyn vMO htrnn km bn hnMt khc, mt electron c xc nh bgi tr2.i vi +ta c:
2 2 21 [ 2s ]2 a b a b
s s s
Do c thm shng + 2sasbnn khong gia hai ht nhn nguyn tc s tng mt electron c tc dng lin kt hai ht nhn thnh phn t.
2 2 21 [ 2s ]2 a b a b
s s s
Do c shng 2sasbnn khong gia hai ht nhn nguyn tc sgim mt electronc tc dng lm yu lin kt gia hai ht nhn nguyn t.
V nhng l do trn +c gi lMO lin kt, c gi lMO phn lin kt.4.2. Mt sv d
- Hai AO 1sa, 1sbca hai nguyn thiro a v b c ththp tuyn tnh to thnh hai MO*,lks s nhsau:
+
sbsa
lk
s
lk
s
*
= c+(sa + sb) *
s = c+(sa + sb)
s
slk
Nnglngtng
1s 1s
s*
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Bi ging Ha hci cng 26
- Hai AO pz(a), pz(b) c ththp dc theo trc lin kt z ni lin tm hai nguyn ta v b
cho hai MO *,lkz z nhsau:
- Hai AO px(a), px(b) c ththp dc theo trc lin kt z ni lin tm hai nguyn ta v b
cho hai MO *,lkx x nhsau:
Mt cch tng t, hai MO *;lky y cng hnh thnh nhthp hai AO py(a) vi py(b)
vung gc vi trc lin kt z.*
5 5[ ( ) ( )]; [ ( ) ( )]lky y y y y yc p a p b c p a p b
Nhng trng thi c cng mc nng lng l nhng trng thi suy bin. V dhai trng thilkx v
lky l hai trng thi suy bin.
Gin mc nng lng ca cc MO cho cc phn thai nguyn tng hch A2:
Skhc bit nng lng gia s v p ca cc AO l khng ln, trnh tnng lng ca cc
MO tng theo trt t:* * * *lk lk lk lk
s s x y z x y z
Tri li, khi skhc bit nng lng gia s v p ca cc AO l ln nhtrng hp F2, O2th
trnh tnng lng c hi khc:* * * *lk lk lk lk
s s z x y x y z
lk
z = c3[px(a)+px(b)]
*
z
lk
z
*
z
px px
= c4[px(a)-px(b)]
z
2p
2s
2p
2s
Nnglngtng
*z
*s
lkx
lky
s
lkz
*y
*x
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Bi ging Ha hci cng 27
Cu hi v bi tp:1. Theo qui tc bt t, hy biu din lin kt trong cc phn tv ion sau y: H2O; NH3; NH4
+;CO2; SO2; SO3; HNO2; HNO3; H2SO4.2. Sphn loi cc lin kt da vo in m. Cho cc v dv nu cc iu kin, qu trnh hnhthnh cc lin kt ion, lin kt cng, lin kt cho nhn.3. iu kin hnh thnh lin kt hidro, so snh nng lng lin kt ion v lin kt cng. Nu mt viv dcho thy nh hng ca lin kt hidro n tnh cht vt l ca cc cht.4. Hy nu nhng lun im cbn ca thuyt lin kt ha tr(VB).5. Cho v d, c im ca lin kt v lin kt . So snh v gii thch vbn ca hai lin ktny.6. Lai ha l g? c im cc m my lai ha sp; sp
2; sp
3. Cho cc v d vnhng nguyn tc
slai ha ny.7. Hnh hc phn tv sxen phcc m my electron trong cc phn t: H2; O2; N2; HCl;CO2 ; CH4; NH3; H2OTrong cc phn ttrn C; N; O; S c kiu lai ha g?8. Hnh hc phn tca cc phn tbutadien, benzen. Cho bit cch biu din chng bng cng
thc: CH2= CH - CH = CH2; c chnh xc khng? Ti sao?9. Vit cu hnh electron ca cc tiu phn: 2 2 2 2; ; ;H H He He
10. Vit cu hnh electron ca cc phn t: O2, F2, Ne2, NO, CO, BNC thgii thch tnh thun tca c phn tO2, NO, BN, nhthno?11. Vit cu hnh electron ca cc phn t: LiF, BeF, BF. So snh bn v tnh cht tca ccphn tny.12. M tphn tBeH2, CH4theo phng php MO.13. Cho ba nguyn tA (Z = 1), B (Z = 17) v C (Z = 8)
a. Vit cu hnh electron ca ba nguyn ttrnb. Vit CTPT v gii thch shnh thnh cc phn tA2C, B2C7c. Vit cng thc phn tv gii thch shnh thnh cc phn t ABC, ABC2, ABC3 v
ABC4.14. Cho bit trng thi lai ha ca nguyn ttrung tm trong phn tCBr4, BeF2, BBr3, CS2. Nucu trc khng gian ca cc phn t.15. Hy gii thch l do:
a. V sao trong phn tH2O, NH3cc gc ha trHOH (104o29) v HNH (107o) li nhhn
gc tdin (109o28)b. V sao gc ha trHSH (92o15) ca phn tH2S li nhhn gc ha trHOH ca phn
tH2O?c. V sao gc ha trFOF (103o15) ca phn tF2O li nhhn gc ha trHOH ca phn
tH2O?16. i vi mi hp cht sau y: F2O, NH3, BH3, hy cho bit:
a. Scp electron lin kt, scp electron khng lin kt ca nguyn ttrung tm;b. Hnh hc ca phn t;c. nh gi cc gc ha tr
17. M tcu to phn tv ion di y theo phng php VB.a. H2SO3v23SO b. H2SO4v
24SO
c. HNO3v 3NO d. 23CO
18. M tcu to phn tNH3, CO2, SO2theo phng php VB.
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Bi ging Ha hci cng 28
Chng 4: NHIT NG HA HC
Nhit ng ha hc l mn hc nghin cu vnng lng v chuyn ha nng lng mtrc ht l nhit v mi tng quan chuyn ha gia nhit vi cng v cc dng nng lng khc.
Nhit ng hc da trn hai nguyn l cbn rt ra tthc tin ca loi ngi.Nguyn l thnht ca nhit ng hc vbn cht l nh lut bo ton nng lng trong
qu trnh chuyn nhit thnh cng v cc dng nng lng khc.Nguyn l thhai ca nhit ng hc cp n mt tnh cht khc ca nhit l trong
khi cc dng nng lng khc c thchuyn hon ton thnh nhit th nhit khng thchuyn thnhcc dng nng lng khc m khng c mt mt.
V cc phn ng ha hc lun lun km theo sbin i vnng lng (chyu di dngnhit) cho nn vic nghin cu nhit ng hc sc mt ngha nht nh i vi ha hc.1. Nguyn l thnht ca nhit ng hc - nhit ha hc1.1. Ni dung ca nguyn l
T lu con ngi bit s dng nhng ngun nng lng t nhin, bin chng thnhnhng dng thch hp phc vcho cuc sng ca mnh. Tnhng my th snhci xay chybng sc gi, ci gi go dng sc nc, h i n nhng pht minh vi nhng chinc, nh my thy in... Nhng trc nhiu ngi mc chto ra nhng my c thsnsinh ra cng mt cch lin tc m chcn cung cp cho n mt lng nng lng ban u. Mi c
gng to ra nhng chic my nhvy u i n tht bi. T con ngi rt ra c mtkt lun:Khng thno chto c ng clin tc sinh cng m khng cn cung cp mt lng
nng lng tng ng. ng cnhvy sau ny c gi l ng cvnh cu loi 1. Kt lunny l mt trong nhng cch pht biu nguyn l thnht ca nhit ng hc.
nh lut bo ton v bin ha nng lng do Lmanxp pht biu nm 1787 cng l mtcch pht biu khc ca nguyn l ny: "Nng lng khng tsinh ra v khng tmt i, n chcthchuyn tdng ny sang dng khc theo nhng tltng ng nghim ngt".
Nhng th nghim chnh xc ca Jun (Joule) (1848 - 1873) chra rng khi bin mt lngc nng hay mt lng in nng tng ng thnh nhit th lun nhn c cng mt nhitlng nhnhau.1.2. Nhit ha hc
Mt phn ca nhit ng ha hc nghin cu qu trnh nhit trong cc phn ng ha hcc gi l nhit ha hc.
1.2.1. Khi nim vdtrnhit hay entanpi:Vbn cht, tt cnhng bin i ha hc u xy ra km theo vi sta ra hay hp th
nng lng m trc ht l di dng nhit. Stng hay mt nhit c thxem nhkt quca sbin i ca mt i lng gi l dtrnhit (hay entanpi) ca cc cht tham gia qu trnh .
Dtrnhit c k hiu bng H. Sbin i dtrnhit (sthay i entanpi) H c thvit di dng:
H = H (sn phm cui) - H (cht u)Trong trng hp nu tt csn phm cui cng v cc cht u ly trng thi tiu chun
(p = 1at, T = 298oK) th bin thin entanpi c k hiu l H
ov c gi l bin thin entanpi
tiu chun.V d: i vi phn ng to ra H2O tH2v oxiH2+ 1/2 O2= H2OC H
o= - 285,7 KJ/mol mol: phn tgam
Hoc gi trm, dtrnhit ca sn phm phn ng nhhn ca cc cht u. iu c
ngha rng trong qu trnh ny c thot ra nhit.Trong trng hp chung, ta c: Nhit thot ra H < 0 Nhit hp thvo H > 0Nhng qu trnh trong nhit c ta ra (H < 0) gi l exotecmic v ngc li qu trnh
trong nhit c hp thvo (H > 0) gi l endotecmic.1.2.2. Nhngnh lut ca nhit ha hc:nh lut Lavoazi - Laplax (Lavoisie - Laplas) (1780): Lng nhit cn thit phn hy
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Bi ging Ha hci cng 29
mt hp cht ha hc bng lng nhit thot ra khi to thnh cht .nh lut ny cho php vit cc phng trnh nhit ha hc ca phn ng theo chiu thun
hay nghch ty , chcn thay i du ca nhit phn ng (H).V dc thvit: 1/2 H2+ 1/2 I2= HI H = + 6,2 Kcalohay: HI = 1/2 H2+ 1/2 I2 H = - 6,2 Kcalonh lut Getxow (Hess) (1840):Nhit ca phn ng chph thuc vo bn cht v trng
thi ca cc cht u v cui m khng phthuc vo cch thc din bin ca phn ng.V d: Thc hin phn ng t cacbon bng 2 con ng.
Kt quo nhit ta ra trong cc qu trnh trn cho thy:H1= - 94,05 KcaloH2= - 26,42 KcaloH3= - 67,63 KcaloNhcc sliu chr: H1= H2+ H3Nhvy th nghim chng ti t trng thi u nhnhau (cacbon v oxi n trng thi
cui nhnhau - cacbon dioxit) th d bng con ng no nhit ca qu trnh vn khng i. nhlut Getxl nh lut cbn ca nhit ha hc, n cho php chng ta tnh c nhit ca nhiuphn ng khng thdo trc tip c. Chng hn trong v dtrn ta c thtnh c mt trong bai lng H1, H2, H3khi bit hai i lng cn li.
iu c ngha quan trng khi nghin cu chuyn ha nng lng ca thc n trong cth. Nhchng ta bit thc n khi c a vo cthchu sbin i qua hng lot cc phnng phc tp khc nhau. Qua cc phn ng nng lng c gii phng cung cp cho cth.Nhit ca cc phn ng ny khng tho trc tip c. Tuy nhin, da vo nh lut Getx, ta cthtnh c gi trnng lng ca tng loi thc n.
V dSacaroza khi vo c thqua rt nhiu phn ng nhng sn phm cui cng l khicacbon dioxit v nc. V vy theo nh lut Getxlng nhit do sacaroza boxi ha trong cthcng phi bng lng nhit do cht ny ta ra khi t n vi oxi bn ngoi cth, m lng nyc thxc nh c bng php o nhit lng.
Da vo nh lut Getxow cng c th tnh c nhit ca mt phn ng bt knu bitnhit sinh v nhit chy ca cc cht tham gia v to thnh ca phn ng.Tnh nhit ca mt phn ng da vo nhit sinh ca cc cht.
Nhit sinh ca mt cht l nhit ca phn ng to ra mt mol cht tcc nguyn ttrng thi bn vng nht.
V d: H2(k) + 1/2 O2(k) = H2O (1) Ho= - 68,3 Kcalo
Nhit ca phn ng ny Ho
= - 68,3 Kcalo, chnh l nhit sinh ca nc.Tnh ngha trn ta cng thy rng nhit sinh ca tt ccc nguyn t trng thi bn
vng nht u bng 0.Nhit sinh cng nhnhit ca mt phn ng bt kphthuc vo iu kin phn ng.V vy cho thng nht v tin so snh, chng c quy viu kin tiu chun: p sut 1
at v 298oK. Nhit sinh tiu chun c k hiu l H 0s .
Di y l nhit sinh tiu chun ca mt scht.
H3
CO
C CO2
H2
H1
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Bi ging Ha hci cng 31
Bng 2. Nhit sinh tiu chun ca mt shp cht
Hp cht Cng thc HSo Kcalo/mol
Metan
Axetylen
Benzen
Ancol etylic
Phenol
Axeton
Axit axetic
Cacbon tetraclorua
Clorofom
Anilin
CH4
C2H2
C6H6
C2H5OH
C6H5OH
(CH3)2CO
CH3COOH
CCl4
CHCl3
C6H5NH2
-212,8
-810,62
-780,98
-326,7
-372,0
-430,9
-208,3
-37,3
-89,2
-811,9
V d: Tnh nhit ca phn ng( ) 2( ) 2 ( ) 2 5 ( )2 4k k l l CO H H O C H OH
0CH -267,63 - 468,32 0 -326,66
Theo nh lut Getxta c: H1+ H2= H3H1chnh l nhit ca phn ng cn xc nh 2CO2+ 4H2OH2 l tng nhit chy ca H2O v C2H5OH tc l cc cht sn phm phn ngH3l tng nhit chy ca CO v H2tc l cc cht tham gia phn ngT ta c:H1= H3- H2H1= -135,26 - 237,28 - (-326,7) = -81,88 KcalVy ta c thrt ra quy tc sau:Nhit ca phn ng bng tng nhit chy ca cht tham gia phn ng tri tng nhit chy
ca cc cht sn phm phn ng trong nhit chy ca tng cht c nhn ln vi hs t
lng. H = A Hctg- A Hcsp2. Nguyn l thhai ca nhit ng hc - nng lng tdo
2.1. Ni dung ca nguyn lNguyn l thnht ca nhit ng hc mi chcho thy tnh cht bo ton ca nng lng
trong cc qu trnh bin i m cha cho bit khnng bin i ca nng lng tmt dng nysang mt dng khc cng nhgii hn ca sbin i .
Nguyn l thhai ca nhit ng hc cp n mt tnh cht khc ca nng lng: trongkhi cc dng nng lng khc c thbin hon ton thnh nhit th nhit li khng thbin honton thnh cc dng nng lng khc.
Nguyn l thhai cng c rt ra tkinh nghim thc tin ca con ngi v c pht
H1
H2O + C2H5OH
2CO + 4H2 CO2+ H2O
H2
H3
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Bi ging Ha hci cng 34
V d: nh nng lng lin kt trung bnh ca cc lin kt O H trong phn tnc, bitrng nng lng lin kt H H v O = O tng ng 435,9 v 498 kJ, khi t chy 2 mol ta ra483,68 kJ.
Gii: 2H2(k) + O2(k)2H2O H = -483,68 kJ- 483,68 = 2(+435,9 + 498,7 - 4Hlk (O-H)2. Bin thin entropiS = S (sn phm) - S (tc cht)
Nu iu kin tiu chun thS 0298p= S0298 (sn phm) - S
0298 (tc cht)
3. Thng p to thnh tiu chun (nng lng tdo Gibbs)G 0298p= G
0298 (sn phm) - G
0298 (tc cht)
Cu hi v bi tp:1. Thno l ni nng ca h? Ni nng phthuc vo nhng yu tno?2. Ni dung v biu thc ca nguyn l thnht ca nhit ng hc.3. Hiu ng nhit ca phn ng. Pht biu nh lut Hess vhiu ng nhit ca phn ng. Ti saoni nh lut Hess l hquca nguyn l thnht ca nhit ng hc.4. nh ngha nhit sinh, nhit chy ca mt cht. Cng thc tnh hiu ng nhit ca phn ng da
vo nhit sinh, nhit chy.5. nh ngha nhit nguyn tha ca mt cht. Nng lng lin kt l g? Chng c quan hvinhau nhthno?6. Tnh HS
0ca Ca(OH)2bit hiu ng nhit ca phn ng:
CaO + H2O = Ca(OH)2H = -15,26 Kcal/molHS
0ca CaO v H2O tng ng l: -151,8 v -68,3 Kcal/mol
7. Khi t chy 1 mol glucoza thy thot ra 673 Kcal. Tnh HS0ca glucoza bit HS
0ca CO2v
H2O tng ng l: -94,1 v -68,3 Kcal/mol.8. Ni dung v cch pht biu nguyn l thhai ca nhit ng hc. Entropi l g? ngha ca hmsny.9. Nng lng tdo l g? Da vo n khi no mt phn ng c thtdin bin?10. Tc ng ca cc yu tentanpi v entropi ln chiu hng din bin ca mt phn ng.11. Cc phn ng sau y c xy ra khng iu kin chun? (khng cn tnh).a. 3O22O3H > 0b. C6H6(h) + 15/2 O2(k) 6CO2+ 3H2O H = -718,6 Kcal/molc. CaCO3(r) CaO(r) + CO2 H = 42,6 Kcal/mold. SO2(k) + 1/2O2SO3(k) H = -23,7 Kcal/mol12. Phn ng sau y xy ra theo chiu no iu kin tiu chun:
H2S(k) + 0,5 O2 H2O(h)+ S(r)Bit HS0 -4,8 -57,8 Kcal/mol
S0298 49,1 49,0 45,1 7,6 cal/mol
13. Cho phn ng:
C2H4 + H2O(h) C2H5OH(h)Bit GS0 16,3 -54,6 -40,3 Kcal/molS
0298 52,5 45,1 54,5 cal/mol
Hi: 25oC phn ng din ra theo chiu no? ta nhit hay thu nhit.
14. 3,27 g km khi t chy ta ra 4,15 kcal. Tm nhit to thnh km oxit.15. 9 g nhm kt hp vi oxi cho 65,5 kcal. nh nhit to thnh ca nhm oxit.
p: Htt(Al2O3) = -393 kcalmol-1
16. t chy 0,532 g C6H6lng trong mt lng kh oxi dtrong iu kin ng tch 25oC cho
kh cacbonic v nc lng; ng thi ta ra 5,33, kcal. Tm nhit t chy ca benzen.p: -782,354 kcal.
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Bi ging Ha hci cng 35
16. Tnh hiu ng nhit phn ng tcc dkin:Al2O3(r) + 3SO3(k) Al2(SO4)3(r)
Htt -399,09 -94,45 -820,98 kcal p: -138,54 kcal17. Tm hiu ng nhit ca phn ng:
Fe2O3(r) + 2Al(r) 2Fe (r) + Al2O3(r). Bit nhit to thnh (Htt) ca Fe2O3l -196,30kcal, ca Al2O3= -399,09 kcal.18. Tnh hiu ng nhit ca phn ng: 2 4 2 2 6( ) ( ) ( )C H k H k C H k tcc dkin sau:
a) Nhit to thnh tiu chun ca 2 4 2( ), ( )C H k H k ln lt l: + 12,5; -20,24 kcal;b) Nhit t chy tiu chun ca 2 4 2 2 6( ), ( ), ( )C H k H k C H k ln lt l: -337,3kcal; -68,3
kcal; -372,86 kcalc) Nng lng lin kt H H; C = C; C C; C H ln lt l: 104; 147; 83; 99 kcal.p: a) - 32,74 b) - 32,74; c) -30
19. nh nhit to thnh ca kh NO bit nng lng lin kt N N, O = O, N = O tng ng l+945,6; + 498,7 v + 631 kJ. p: + 91,15 kJ.20. nh hiu ng nhit ca phn ng: 2 2 2 2 2( ) 2 ( ) ( )C H k Cl k CHCl CHCl k
Bit nng lng lin kt C C, C C, Cl Cl; C Cl tng ng l + 347,3; + 823,1; +242,3; 345,2 kJ p: -420,5 kJ21. Tm nng lng lin kt trung bnh ca lin kt N H trong phn tNH3, bit rng:
2 2 31 3( ) ( ) -46,192 2N k H NH k H kJ v nng lng lin kt ca N2, H2 tng ng l
945,6 v 435,9 kJ p: + 390,947 kJ22. Tnh nhit to thnh tiu chun ca Fe2O3(r) tcc dkin sau:
02 3 2 298( ) 3 ( ) 2 ( ) 3 ( ) 6,74Fe O r CO k Fe r CO k H kcal
0r 2 2 298( ) ( ) 94,10g aphitC O k CO k H kcal
0r 2 298
1( ) ( ) 26,42
2g aphitC O k CO k H kcal p: -196,3 kcal
23. Tnh nhit to thnh tiu chun ca ru etylic lng tcc dkin sau:a) 2 5 2 2 2 1( ) 3 ( ) 2 ( ) 3 ( ) 327C H OH l O k CO k H O l H kcal
b) r 2 2 2( ) ( ) 94,10g aphitC O k CO k H kcal
c) 2 2 2 31
( ) ( ) ( ) 68,32
H k O k H O l H kcal p: -65,9 kcal
24. Tm nhit to thnh tiu chun ca Ca3(PO4)2tinh thbit:a) 12 g Ca chy ta ra 45,57 kcalb) 6,2 g P chy ta ra 37,00kcalc) 168 g CaO tc dng vi 142 g P2O5ta ra 160,5 kcalHiu ng nhit o trong iu kin ng p. p: - 986,2 kcal
25. Tnh nhit to thnh tiu chun ca FeCl2(r) bit:
2 2( ) (dd) ( ) (dd) 21,00Fe r HCl H k FeCl H kcal
2 2( ) (dd) 19,50FeCl r aq FeCl H kcal
( ) (dd) 17,5HCl k aq HCl H kcal
2 2( ) ( ) 2 ( ) 44,48H k Cl k HCl k H kcal
K hiu aq chmt lng nc ln p: - 80,98 kcal26. Tm nhit tiu chun ca AsCl3rn bit cc dkin sau:
02 3 2 3 3 298As ( ) 3 3 As (dd) 7,55O r H O H O H kcal
03 2 3 3 298As ( ) 3 ( ) As (dd) 3 (dd) 17,58Cl r H O l H O HCl H kcal
02 3 298As( ) 3 / 2 ( ) As ( ) 71,39r Cl k Cl r H kcal
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Bi ging Ha hci cng 36
0298( ) (dd) 17,31HCl k aq HCl H kcal
02 2 2981/ 2 ( ) 1/ 2 ( ) 22,31H k Cl HCl k H kcal
02 2 2 298( ) 1/ 2 ( ) 68,30H k O H O l H kcal
p: -82,77 kcal27. Tm bin thin entropi tiu chun ca cc phn ng sau:
CaCO3(r) CaO (r) + CO2(k)0
298S 22,16 9,5 51,06 (cal.mol
-1
K
-1
Cgraphit + CO2(k) 2CO (k)0298S 5,74 313,63 197,565 (Jmol
-1K-1)
H2(k) + 1/2O2(k) H2O (k)0298S 130,575 205,029 188,716 (Jmol
-1K-1)
28. Tnh bin thin nng lng tdo Gibbs ca qu trnh hnh thnh 1 mol nc tcc n cht 25oC, 1 atm. Bit:
a) nhit to thnh tiu chun ca nc lng l -68,317 kcalb) H2(k) + 1/2O2(k) H2O (k)
0298S 31,20 49,01 16,78 (calmol
-1K-1)
29. Lu hunh thoi v lu hunh n t l hai dng th hnh ca lu hunh. Hi:
a)
25oC dng no bn hn? Nhit ti hai dng cn bng nhau?Bit rng 0298S Sthoi=31,80,
0298S S n t= 32,77
30. Mg c thchy trong kh cacbonic khng? Cho bit2Mg(r) + CO2(k) 2MgO (r) + C(than ch)
0,298ttH 0 -94,1 -143,83 0 (kcal.mol
-1)0298S 7,77 51,1 6,43 1,36 (cal.K
-1.mol-1)
31. Cho phn ng: 2 4 2( ) 2 ( )N O k NO k 0298H (kcal.mol
-1) 2,31 8,090
298S (cal.K-1.mol-1) 72,73 57,46
a) Phn ng sdin ra theo chiu no: 0oC, v 100oCb) Phn ng t trng thi cn bng nhit no?32. Bit entanpi to thnh v entropi cc cht di y iu kin chun:
Fe O2 FeO Fe2O3 Fe3O40298H (kcal.mol
-1) 0 0 -64,3 -269,5 -266,90298S (cal.K
-1.mol-1) 6,5 49 14,0 20,9 36,2
a) Tnh bin thin thng p to thnh ca ba loi oxit ca st kcb) kc. oxit no ca st l bn hn?
33.1. Tnh entanpi sinh tiu chun ca ozon, kim cng tcc dkin sau:
Cthan ch+ O2(k) CO2(k)0
298 393,14H kJ Ckim cng+ O2 CO2(k)
0298 395,03H kJ
3As2O3+ 3O2(k) 3S2O5(r)0298 811,34H kJ
3As2O3+ 2O3(k) 3S2O5(r)0298 1090,98H kJ
2. Tkt quthu c trn v cc dkn sau: H(O-O)(tnh tO2) = -493,24 kJmol-1v
H(O-O)(tnh tH2O) = -137,94 kJmol-1, hy chng minh rng khng thgn cho phn tO3cu
trc vng kn.p 1. 0 1,298 1,89sH kJmol
(kim cng) ; 0 1,298 141,70sH kJmol (ozon)
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Bi ging Ha hci cng 38
V d: Cc phn ng (a), (b), (c) tng ng l cc phn ng bc 1, 2, 3.Bc phn ng c thl snguyn nhng cng c thl phn s(thng i vi phn ng
phc tp).* Phn tsca phn ng:Sphn loi phn ng theo phn ts lin quan trc tip vi cch thc ca phn ng.
Phn tsl stiu phn (phn t, nguyn thay ion) ng thi tng tc vi nhau trong mt phnng n gin. V vy phn tschc thl snguyn.
V d: Trong phn ng:CH3-N=N-CH3 CH3-CH3+ N2tham gia vo tng tc chc mt phn t. V vy phn ng c phn tsl mt hay phn
ng n phn t.Trong phn ng: H2+ I22HI to thnh sn phm hai phn tH2v I2phi ng thi
tham gia vo mt tng tc, v vy phn ng c phn tsbng hai hay phn ng lng phn t.Nhng phn ng c phn tsbng 3 hay cao hn thng t gp v xc sut ng thi 3 phn tphn ng vi nhau rt nh.
Lu : Trong nhng phn ng n gin th bc phn ng thng trng vi phn ts.3.nh hng ca nhit n tc phn ng
Nghin cu nh hng ca nhit n tc phn ng cho php tm hiu bn cht canhng tng tc ha hc ng thi tm c chnhit ti u cho phn ng ha hc.
Nhit nh hng n tc phn ng theo nhng cch khc nhau.
Dng ng cong (1a) l phbin i vi phn ng ha hc.Dng ng cong (1b) thng gp nhng phn ng c lin quan n cc hp cht sinh
hc nhcc protein enzym. Vi cc protein, trng thi tnhin, tc tng theo nhit . Nhngkhi t n mt nhit no chng bbin tnh, mt hiu quxc tc v do tc phn nggim.
Nhng ni chung tc ca a sphn ng ha hc tng ln khi tng nhit . nh hngny tun theo mt squi tc sau y.3.1. Qui tc Van Hp
"Khi nhitca phnng tng ln 10oth hng stcphnng (cng l tc
phnng) tng ln t2n 4 ln".
10 2 4T
T
k
k (2)
kT+10: hng stc nhit T + 10o
kT: hng stc nhit T: c gi l hsnhit ca phn ngTrong trng hp tng qut, biu thc ca nh lut Van Hp c dng:
.10n T n
T
k
k (3)
V d: Mt phn ng c hsnhit = 3. Hi tng nhit ln 40oth tc phn ngtng ln bao nhiu ln.
Gii: Theo qui tc Van Hp, ta c:
Hnh 1 a, b
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Bi ging Ha hci cng 39
44.10 3 81n T
T
k
k
3.2. Biu thc Arniuxnh hng ca nhit n tc phn ng c biu thmt cch chnh xc hn v p
dng c trong mt khong nhit rng hn qua biu thc Arniux:
lnE
k BRT
(4)
R : hng skh, c gi trbng 1,98 cal/mol.KB : hng sE : hng si vi mt phn ng xc nh, ngha l n phthuc vo bn cht ca cc cht
phn ng. E c gi l nng lng hot ha ca phn ngTbiu thc Arniux, ta thy khi nhit tng, tc phn ng tng. Mt khc, phn ng
c nng lng hot ha cng ln sdin ra vi tc cng nh.3.3. Thuyt va chm hotng v nng lng hot ha
* Thuyt va chmcc nguyn thay phn tc thphn ng c vi nhau chng phi va chm vo nhau. Do
tc phn ng stng ln khi sva chm (hay tn sva chm) tng.Thuyt va chm khng gii thch c skhc bit rt ln gia kt qutnh ton l thuyt v cc kt quthc nghim.
V d: Theo tnh ton khi nhit tng 10o, sva chm chtng ~ 2% ngha l tc chcthtng ~ 2% nhng trong thc ttheo qui tc Van Hp tc phn ng li tng t nht l 200%.iu ny khng thgii thch c nu chda vo sva chm n thun.
* Thuyt va chm hot ng (hay thuyt hot ha) v nng lng hot ha Thuyt ny chorng khng phi mi va chm m chnhng va chm ca cc nguyn thay phn thot ng (gil cc va chm hot ng) mi dn n phn ng.
Cc nguyn thay phn thot ng l cc nguyn thay phn tc mt nng lng dln so vi nng lng trung bnh ca chng.
Nng lng ti thiu m mt mol cht phnng cn phi cchuyn cc phn tcachng ttrng thi khng hotng trthnh hotng gi l nng lng hot ha ca phnng.
Nhvy, theo thuyt hot ha phn ng: A + Bk
P c thhnh dung nhsau:'* * *kA B A B AB P Cc phn tA v B cn phi c hot ha thnh A* v B*, khi
to thnh hp cht trung gian hot ng AB* v cui cng phn hy to ra sn phm P.
Nhvy c thphn ng c vi nhau, phn tcc cht phnng dng nhphi vt qua mt hng ro nng lng. chnh l nnglng hot ha ca phn ng (hnh 2).
Nu nng lng hot ha cng nhth tc phn ng scngln. V vy khi xt khnng phn ng, ngi ta thng dng i lngny so snh.
* Xc nh nng lng hot ha ca phn ng:
Bng thc nghim xc nh hng stc ca phn ng t nhthai nhit khc nhau T1, T2, khi ta c:
1
1
ln TE
k BRT
v2
2
ln TE
k BRT
T 1
1
1 2
2 1
ln T
T
kRTTE
T T k
(5)
V d: Xc nh nng lng hot ha ca mt phn ng bit rng trong khong nhit t17 n 27
oC phn ng c hsnhit = 2,8.
Gii:
Hnh 2
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T1= 17 + 273 = 290oK
T2= 27 + 273 = 300oK
1,98.290.3002,303lg2,8 17850 /
10E cal mol
4.nh hng ca xc tc n tc phn ng4.1. Mt skhi nim vxc tc
Xc tc l hin tng lm tng tc ca phn ng khi c mt nhng cht c bit, gi l
nhng cht xc tc, cc cht ny sau khi tham gia vo phn ng c hon trli vlng v cht.Thng xc tc c chia thnh 3 loi:* Xc tcng th: cc cht phn ng v cht xc tc to thnh mt pha ng nht kh hoc
lng. V d:
2 2 32 2NOSO O SO ng pha kh
3 2 5 2 3 2 5HCH COOC H H O CH COOH C H OH
ng pha lng
Trong xc tc ng th, phn ng xy ra trong ton b th tch ca h phn ng (trongkhng gian ba chiu), tc phn ng tlthun vi nng cht xc tc.
* Xc tc d th: Cc cht phn ng v cht xc tc to thnh mt hd th(khng ngnht)
2 2 22 ( ) 2PtH O l H O O dthlng - kh
2 4 2 2 6NiC H H C H dthkh rn
Trong xc tc dth, phn ng din ra trn bmt cht xc tc (trong khng gian hai chiu).Tc phn ng tlthun vi bmt cht xc tc.
* Xc tc enzym (xc tc men)V d: 6 12 6 2 5 22 2
zimazaC H O C H OH CO
Enzym l cc cht xc tc sinh hc - cht xc tc ca cthsng, n c bn cht l protein.4.2. Cchv vai tr ca xc tc
Phnng c xc tc thng din ra qua nhiu giaion trung gian (to ra cc hp chttrung gian).
V dphn ng gia hai cht A v B khi c mt cht xc tc K.A + B K C + D
sdin ra nhsau: Trc ht mt trong nhng cht phn ngsphn ng vi cht xc tc to ra mt hp cht trung gian [AK]*,sau hp cht ny li phn ng tip vi cht phn ng thhai to rahp cht trung gian [ABK]*. Cui cng [ABK]* phn hy to ra snphm v hon trli cht xc tc.
A + K [AK]*[AK]* + B [ABK]*[ABK]* C + B + K
Cc hp cht trung gian (c du sao) thng c nng lngcao, khng bn, ch tn ti trong mt khong thi gian rt ngn catin trnh phn ng xc tc.
Nhvy, sc mt ca cht xc tc lm cho phn ng din raqua mt sphn ng trung gian c nng lng hot ha thp hn sovi phn ng khng c xc tc (hnh 3) v do lm tng tc phn ng.
Thng thng nng lng hot ha ca nhiu phn ng nm trong nhng gii hn sau:Phn ng E (Kcal/mol)
Khng xc tcXc tc v cXc tc enzym
30 4515 308 - 12
Cc phn ng xc tc bng enzym c nng lng hot ha thp hn nhiu so vi phn ngkhng c xc tc hoc c xc tc v c. Do n c thlm cho cc phn ng trong cthdin ra
Hnh 3
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v cng nhanh chng.V dphn ng phn hy hidro peoxit 2H2O2 2H2O + O2nu khng c xc tc th i
hi mt nng lng hot ha l 35,96 Kcal/mol. Khi c xc tc platin E = 24,02 Kcal/mol, cn khic xc tc bng enzym catalaza chcn mt nng lng hot ha 14 Kcal/mol.4.3. Mt scim ca xc tc
- Cht xc tc chc thlm tng tc ca phn ng m khng thgy ra c phn ng.iu ny c ngha l i vi nhng phn ng khng c khnng xy ra khi xt vtiu chun nhitng hc th khng thtm c cht xc tc cho n.
- Cht xc tc lm tng tc phn ng thun bao nhiu ln th cng lm tng tc phnng nghch by nhiu ln.
- Cht xc tc c tnh chn lc. Mt cht xc tc thng chc thxc tc cho mt phnng hoc mt loi phn ng nht nh. Tnh chn lc thhin c bit r i vi cc enzym, v vyngi ta thng ni cc enzym c tnh c hiu cao.
- Mt lng nhcht xc tc c thxc tc cho mt lng ln cht phn ng. Sdnhvyl v, vnguyn tc, xc tc khng bthay i sau phn ng.5. Cn bng ha hc
5.1. Phnng thun nghch - hng scn bngPhn ng thun nghch l phn ng din ra theo hai chiu:
1kA B C D 2k
3 2 5 2 3 2 5CH COOC H + H O CH COOH + C H OH Tc phn ng thun: vt= k1[A] [B]Tc phn ng nghch: vn= k2[C] [D]Trong qu trnh phn ng, tc phn ng thun gim dn, cn tc phn ng nghch
tng dn. Khi vt= vnth ngi ta ni phn ng t ti trng thi cn bng:k1[A]' [B]' = k2[C]' [D]'T rt ra:
' '1
' '2
[ ] [D]
[A][B]ck C
Kk
(6)
V d1:
3 33 ( ) 3FeCl KSCN Fe SCN KCl
vang nhat o am
Khi cn bng c thit lp nu ta thm vo hmt t tinh thKCl snhn thy mu nhti. iu chng tcn bng chuyn dch theo chiu nghch tc l chiu lm gim bt nng KCl. Ngc li, mu sm ln nu ta thm KSCN hay FeCl3, chng tcn bng chuyndch theo chiu thun lm gim nng cc cht thm vo.
V d2: 2 2 33 2N H NH Q
Phn ng theo chiu thun ta nhit lm cho hnng ln ng thi li to ra t sphn thn do lm gim p sut trong h. V vy nu gim nhit (lm lnh hphn ng) th cn bngschuyn dch theo chiu thun l chiu ta nhit. Nu tng p sut (v dbng cch nn h) th cnbng sphi chuyn dch theo to ra t sphn t hn, tc l chiu thun.Trong tng hp amoniac, tng hiu sut ca phn ng,ngi ta thng thc hin p sut rtcao v nhit tng i thp.6. Cc phn ng phc tp
Cc phn ng phc tp l cc phn ng din ra qua nhiu giai on hay gm nhiu phn ngthnh phn (hay phn ng cs).6.1. Phnng thun nghch
Gm hai phn ng thnh phn: phn ng thun v phn ng nghchB C D
2 2 2H I HI
Khi vt= vnphn ng thun nghch t ti trng thi cn bng.trng thi cn bng, nng cc cht khng thay i.
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Phn ng (1) c gi l lin hp vi phn ng (2) v khi tin hnh n cung cp nnglng lm cho phn ng (2) cng xy ra c.
V d: Stng hp glucoza-6-photphat (G6P) trong cthc thc hin do lin hp giahai phn ng:
Acginin photphat + H2OAcginin + H3PO4 sinh nng lngGlucoza + H3PO4G6P + H2O cn nng lng
Khi lin hp, phn ng tng cng sl:Acginin photphat + Glucoza G6P + Acginin
TM TTXt phn ng gm 4 cht kh A, B, C, D
aA( ) ( ) dD( ) eE( )k bB k k k Phn ng chuyn sang chiu thun khi: 0G , chiu nghch khi: 0G Phn ng t trng thi cn bng khi: 0G , lc ta c:
0 ( ) ( )ln( ) ( )
d eD E
a bA B
P PG RT
P P Nu t:
( ) ( )
( ) ( )
d eD E
P a bA B
P PK
P P ta c:
0
ln PG
KRT
Thay cho p sut hi, lng cht c biu thdng nng C.Lin hgia Kpv KC: ( )
nP CK K RT
vi ( ) ( )n d e a b
Cu hi v bi tp:5.1. Nng cc cht phn ng nh hng n tc phn ng nhthno? Hy pht biu v vitbiu thc ca nh lut tc dng khi lng. Bc phn ng l g?2. Phn bit bc phn ng v phn tsca phn ng.3. Phn ng 2NO + O22NO2l mt phn ng n gin. Tc phn ng thay i nhthnokhi: Tng nng O2ln 4 ln.4. Nhit nh hng nhthno n tc phn ng, thhin qua biu thc v qui tc no?5. Mt phn ng c hsnhit = 3,1. Hi khi tng nhit thm 40
o, tc phn ng tng ln
bao nhiu ln?6. Hsnhit ca mt phn ng bng 3. Hi phi tng nhit ln bao nhiu tc phnng tng ln 243 ln?
7. Trnh by ni dung cbn ca thuyt hot ha. Nng lng hot ha ca mt phn ng l g?8. Ti sao sc mt ca cht xc tc li lm tng tc phn ng. Vv gii thch gin nnglng ca phn ng khi c v khng c mt cht xc tc.9. Hng scn bng ca mt phn ng l g? Hy pht biu v minh ha nguyn l chuyn dchcn bng qua cc v d.10. Cc cn bng sau y chuyn dch thno khi tng nhit , tng p sut:
a. N2+ O22NO Qb. 2CO + 2H2CH4+ CO2+ Qc. CaO + CO2CaCO3+ Qd. N2O42NO2- Q
11. Thno l phn ng thun nghch, phn ng ni tip? Cho v d.12. Cho v dvphn ng dy chuyn. Nhng giai on cbn ca mt phn ng dy chuyn.
13. Trn 1 mol kh CO vi 3 mol H2O 850oC trong mt bnh phn ng dung tch 1 lt. Khi cnbng sau c thit lp: 2 2 2( ) ( ) ( ) ( )CO k H O k CO k H O k smol CO2 thu c l 0,75 mol.
Tnh gi trhng scn bng Kpv KC.14. C cn bng: 2 2 24 ( ) 2 ( ) 2 ( )HCl k O H O k Cl k . Trn 1 mol HCl vi 0,48 mol O2 c
0,402 mol Cl2, hc p sut p = 1 atm, t = 386oC. Tnh KP. p: 79,28
15. Hng scn bng KCca phn ng sau:
2 2 2( ) ( ) ( ) ( )CO k H k CO k H O k 850oC bng 1. Nng u ca kh CO2v H2l 0,2 v 0,8
M. Tm nng 4 cht thi im cn bng.
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16. 986oC cn bng sau y c KC= 0,63: 2 2 2( ) ( ) ( ) ( )CO k H O k CO k H k
Cho 1 mol nc, 3 mol cacbon oxit vo mt bnh kn dung tch khng i vi p sut hnhp kh ban u l 2 atm 986oC. Tnh:
a) p sut hn hp lc cn bng.b) Smol kh hiro lc cn bng.c) p sut ring tng kh lc cn bng.
17. Cho phn ng: H2+ I2 2HI
a)
Vit biu thc tc ca phn ng, bit rng:Nu tng nng hiro gp i, ginguyn nng iot th tc cng tng gp iNu ginguyn nng hiro v tng nng iot ln gp 3 th tc cng tng gp 3.b) Cho bit bc phn ng v phn tsca phn ng.
18. Phn ng A + 2B 3C c hng stc k = 0,6, nng u CA= 2,0M; CB= 2,5M. Dophn ng nng B cn bng 0,5M. Tm nng ca A v tc ca phn ng ti thi im ny.19. Phn ng phn hy N2O nhit 1173
oCc hng stc k = 5.10-4:2N2O (k) 2N2(k) + O2(k) Nng u ca N2O l 3,2M. nh tc phn ng
lc u v khi c 25% N2O bphn hy.20. Phn ng 2NO + O2NO2ti mt thi im no nng cc cht CNO= 0,8; CO2= 0,6M.Nu tng nng NO thnh 1,2 v O2thnh 0,9M th tc phn ng sbin i ra sao?21. Phn ng: 2NO (k) + O2(k) 2NO2(k).
Tc phn ng sthay i nhthno khi:a) tng nng oxi ln 4 ln;b) gim bt nng nitoxit 1/3 so vi ban u;c) nng NO v O2u tng 3 ln;d) gim nng NO2hai ln?
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Chng 6: I CNG VDUNG DCH* Cc hphn tn v dung dch:
Hphn tn l nhng htrong c t nht mt cht phn b(gi l cht phn tn) vo mtcht khc (gi l mi trng phn tn) di dng nhng ht c kch thc nhb.
Da vo kch thc ht, ngi ta chia thnh:- Hphn tn phn t- ion hay cn gi l dung dch thc. V ddung dch mui axit, baz...
Kch thc ht y < 1 nm.- Hphn tn keo hay cn gi l dung dch keo. V dgelatin, htinh bt, keo axit silixic...
c kch thc ht t1 - 100 nm.- Hphn tn th c hai dng l huyn ph v nhtng. V dnc sng cha nhng ht
ph sa: sa... Kch thc ht ca nhng hny > 100 nm.Trong chng ny chng ta cp n dung dch phn tv nhng tnh cht chung ca
chng.1.nh ngha v phn loi dung dch
Dung dch l mt hng nht ca hai hay nhiu cht c tlkhc nhau thayi trongmt phm vi rng.
Tnh ngha c thc:- Dung dch rn, v dcc hp kim.- Dung dch kh, v dkhng kh.
- Dung dch lng, v ddung dch ca cc cht rn (ng, NaCl...), kh (O2, NH3...), lng(C2H5OH, benzen...) trong nc.Cc nh ha hc v sinh hc thng tip xc vi cc dung dch lng m cht lng y
thng l nc. Trong cc dung dch ny, nc l mi trng phn tn c gi l dung mi, cccht phn tn gi l cc cht tan.
Theo bn cht ca cht tan, ngi ta phn chia thnh:- Dung dch khng in li: Cht tan c mt trong dung dch di dng phn t. V ddung
dch ng, C2H5OH, O2trong nc.- Dung dch in li: Trong dung dch c mt cphn tv ion. V ddung dch ca cc
mui, axit, baz... trong nc.2. Nng dung dch
Nng dung dch l i lng biu thcht tan trong dung dch.
C mt scch biu thnng ty thuc vo mc ch sdng.2.1. Nngphn trm.
K hiu %Nng phn trm biu thbng sgam cht tan trong 100 gam dung dch.V d: Dung dch huyt thanh ngt l dung dch glucoza 5% (5 gam glucoza ha tan trong
95 gam nc).2.2. Nngmol hay mol/lt.
K hiu MNng mol c biu thbng smol cht tan trong 1 lt dung dch.V d: Dung dch NaOH 0,1 M l dung dch c 4 gam NaOH trong 1 lt dung dch.Mun c dung dch ny, ngi ta phi cn chnh xc 4 gam NaOH v thm nc n th
tch cui cng l 1 lt.
2.3. Nngng lng.K hiu NNng ng lng c biu th bng sng lng gam cht tan trong 1 lt dung
dch.ng lng gam ca mt cht l lng cht tnh bng gam khi phnng tng
ng (kt hp hay thay th) 1 nguyn tgam hidro (1,008 gam).ng lng gam ca mt cht phthuc vo phn ng m n tham gia vo.
*ng lng gam can chtV d1: Trong phn ngH2+ 1/2 O2H2O
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16 gam oxi kt hp vi 2 nguyn tgam hidro. Vy ng lng gam ca oxi (2O
) l 16/2 = 8
gam.V d2: Trong phn ng:
Mg + 2HCl H2+ MgCl224 gam Mg thay th2 nguyn tgam hidro. Vy Mg= 24/2 = 12 gamNhvy: ng lng gam ca mt n cht bng nguyn tgam ca n chia cho ha tr.Lu : i vi nhng nguyn tc nhiu ha trth ng lng gam ca n c thkhc
nhau.V d: Trong cc phn ng sau y:Fe + 1/2 O2= FeO Fe= 56 g/22Fe + 3/2O2= Fe2O3 Fe= 56 g/3
*ng lng gam ca hp cht tham gia phnng traoi:V d1: Trong phn ng:NaOH + HCl NaCl + H2O40 g NaOH phn ng tng ng vi 1 phn tHCl (36,5 g) tc l tng ng vi 1
nguyn tgam hidro. V vy ENaOH= 40g/l v EHCl= 36,5 g/l.V d2: Trong phn ng
3NaOH + H3PO4Na3PO4+ 3H2O1 phn tgam H3PO4 khi phn ng tng ng vi 2 nguyn tgam hidro:
3 4 3 4 / 2H PO H POM Nhvy: ng lng gam ca mt cht trong phnng traoi bng phn tgam
chia cho sin tch dng hay m m mt phn tcht traoi.*ng lng gam ca hp cht tham gia phnng oxi - ha kh
V d: Trong phn ng:KMnO4+ FeSO4+ H2SO4MnSO4+ Fe2(SO4)3+ K2SO4+ H2OMt phn tgam KMnO4nhn 5 mol electron (tng ng vi 5 nguyn tgam hidro).
V vy4 4
/ 5KMnO KMnOM Tng t: 4 4 /1FeSO FeSOM
Nhvy:ng lng gam ca mt cht trong phnng oxi - ha khbng phntgam ca cht chia cho selectron m mt phn tcht cho hoc nhn.
V d: Tnh ng lng gam ca axit oxalic trong cc phn ng sau y v cho bit mun
pha c dung dch 0,1N ca axit ny cn phi tin hnh nhthno?H2C2O4+ 2NaOH = Na2C2O4+ 2H2O (1)H2C2O4+ KMnO4+ H2SO4CO2+ MnSO4+ K2SO4+ H2O (2)
Gii: Trong phn ng (1) H2C2O4trao i 2 in tch dng (2H+) hay 2 in tch m ( 22 4C O
). V
vy:2 2 4 2 2 4
/ 2H C O H C OM
Trong phn ng (2) mt phn tH2C2O4 cho i 2e (2C+3C
+4). V vy ng lng gam
ca H2C2O4trong phn ng ny cng bng M/2.Mun pha dung dch H2C2O4 0,1 N ta phi cn chnh xc 4,5 gam H2C2O4 v thm nc
n thtch 1 lt.Nng ng lng gam c s dng rng ri trong ha hc, c bit trong ha hc
phn tch.
Tnh ngha vng lng gam c thsuy ra rng: Khi hai cht phn ng va vinhau th sng lng gam phn ng ca cht ny ng bng sng lng gam phnng ca cht kia.
V d: Nu VAlt dung dch cht A nng NAphn ng va vi VBlt dung dch cht Bc nng NB. Khi ta c:VA. NA= VB. NB
l biu thc ca nh lut ng lng: "Cc cht ha hc (n cht hay hp chtphnng vi nhau theo cng sng lng gam"c sdng trong ha hc phn tch xcnh nng ca cht tan trong dung dch.
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2.4. Nngmolan.K hiu mNng molan biu thsmol cht tan trong 1000 gam dung mi.V d: Dung dch glucoza 0,5 m l dung dch gm 90 gam glucoza trong 1000 gam
2.5. Nngphn mol hay nngmol ring phnNng phn mol ca mt cht i no c tnh bng tssmol ca cht v tng s
smol ca tt ccc cht to nn dung dch:
ii
i
nN n Ni: nng phn mol ca cht i
ni: smol cht ini: tng smol ca cc cht to nn dung dch
3. p sut thm thu ca dung dch3.1. Hin tng thm thu
Hai nhnh A v B camt ng hnh ch U cngn cch bng mt mngthm thu tc l mng ckch thc cc lchcho ccphn tdung mi i qua cn
cc tiu phn cht tan b gili (hnh 1).
Bn nhnh A chamt dung dch ng bnnhnh B cha nc nguyncht (hay mt dung dch ng c nng nhhn nng dung dch nhnh A). Sau mt thigian nht nh, nhn thy mc cht lng trong nhnh A nng ln mt cao h no , cn mc chtlng trong nhnh B bhthp xung. iu chng t c nhng phn tdung mitnhnh B chuyn sang nhnh A.
Hin tng cc phn tdung mi khuch tn mt chiu qua mng thm thu tdung misang dung dch (hoc tdung dch c nng thp sang dung dch c nng cao hn) c gi lhin tng thm thu.3.2. p sut thm thu -nh lut Van Hp, 1887 (Vant' Hoff - H Lan)
p sut thm thu l p sut gy nn bi hin tng thm thu. Vln n c gi trbngp sut gy nn bi ct nc c chiu cao h trong th nghim trn hoc bng p sut cn t lndung dch lm ngng hin tng thm thu.
p sut thm thu (thng k hiu ) phthuc vo nng v nhit ca dung dch theonh lut Van Hp:
p sut thm thu ca mt dung dch t lthun vi nng v nhit ca dung dch.= R . C . TR: hng skh l tng, bng 0,082 lt.at/mol/KC: nng mol/lt ca dung dchT: nhit tuyt i ca dung dch
Hin tng thm thu c ngha sinh hc rt quan trng v mng ca cc tbo l cc mngthm thu.- Nhc hin tng thm thu nc c vn chuyn trcy ln ngn.
4. Nhit si v nhit ng ca dung dch4.1. p sut hi ca dung dch
- p sut hi ca mt cht lng l p sut gy nn bi nhng phn tca n trn mt thongca cht lng.
- p sut hi bo ha l p sut to ra trn mt thong khi qu trnh bay hi t ti trng thicn bng.
- p sut hi tng khi tng nhit ca cht lng.
Hnh 1
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- cng mt nhit , p sut hi hay p sut hi bo ha ca dung dch lun lun nh hnp sut ca dung mi nguyn cht do trn mt thong ca dung dch c cc tiu phn cht tan nng(hnh 2)
4.2. Nhitsi ca dung dch
Mt cht lng ssi khi p sut hi bo ha ca n bng p sut kh quyn. V dnc si100
oC v nhit ny p sut hi ca n bng p sut 1 at. Trong khi t c p sut 1
at, cn phi tng nhit ca dung dch hn 100oC.
Tm li: Mt dung dch s sinhitcao hn nhit si ca dung mi. Nng dung dch cng ln th nhit si ca n cng cao. Hiu nhit si ca dung dch v dung mic gi l tngim sica dung dch, k hiu TS.4.3. Nhitngc ca dung dch
Mt cht lng sng c nhit ti p sut hi bo ha trn pha lng bng p suthi bo ha ha trn pha rn. Trn hnh 2 ng biu din bin i p sut hi bo ha trn pha rn(on OA) ct ng p sut hi trn dung dch im tng ng vi nhit thp hn 0
oC.
Tm li:Mt dung dch sngcnhitthp hn nhitngc ca dung
mi. Nng dung dch cng ln th nhit ng ca n cng thp. Hiu nhit ng ca dungmi v dung dch c gi l himngca dung dch, k hiu Td.4.4.nh lut Raun, 1886 (Raoult - Php)
tngim si hayhimng ca dung dch tlthun vi nngmolan cadung dch.
TS= kS. CmTd= kd. Cmksv kdtng ng c gi l hng snghim si v hng snghim ng ca dung mi.
N l nhng i lng c trng i vi mt dung mi nht nh.Bng 1. Nhit si v nhit ng ca mt sdung mi
Dung mi Ts oC ks Td
oC kd
H2OC6H6C2H5OHC6H5OHC6H12
1008079
81
0,522,571,193,042,79
05,5
406,5
1,865,12
7,272,02
Da vo nh lut Raun v bng thc nghim xc nh h im ng (phng phpnghim ng) hay tng im si (phng php nghim si) ca dung dch, ngi ta c thtmc phn tgam ca mt cht tan nht nh.
V d: Ha tan 10 gam cht A trong 100 gam nc. Dung dch nhn c ng c nhit-2,12
oC. Tnh phn tgam ca cht A.
Hnh 2
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8/11/2019 B Gio Trnh Ha Hc i Cng - Hunh Thin Lng, T Anh Phong, Phan an, Phm Th Thanh, Nguyn Ph Huyn Chu, Nguyn Th Minh Minh,
50/644
Bi ging Ha hci cng 49
0 ( 2,12) 2,12odT C
10 1000 100
100A Am
M
1002,12 1,86
A
T: