Bài giảng Hoá học đại cương - Lê Minh Thành

7/25/2019 Bi ging Ho hc i cng - L Minh Thnh

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1

GV: L Minh ThnhGV: L Minh ThnhGV: L Minh ThnhGV: L Minh Thnh

1/19/2011

Chng 6

1/19/2011

3

Nng lng l kh nng lm thay itrng thi hoc thc hin cng nng lnmt h vt cht.

Nng lng c chia thnh 2 dng .

Cu hi: Nhit nng, ha nng, c nng,in nng, nng lng tnh in, mthanhthuc loi nlng no?

1/19/2011

4

ng nngng nngng nngng nng ThThThTh nngnngnngnng

C nngC nngC nngC nngNng lng ca vt v

m chuyn ng

NhiNhiNhiNhit nngt nngt nngt nngChuyn ng vi mca ngt, pht, ion

iiiin nngn nngn nngn nngChuyn ng ca

electron trong vt dn

BBBBc xc xc xc xBc x in t truyn

trong khng gian

ThThThTh nng hnng hnng hnng hp dp dp dp dnnnnDo tng tc hp dn

Nng lNng lNng lNng lng tnh ing tnh ing tnh ing tnh innnnDo tng tc tnh in

Ha nngHa nngHa nngHa nngLc ht ca electron v ht

nhn trong nguyn t

1/19/2011

5

Ni dung:.

V d v s bin i + bo ton n.lng:

1/19/2011 6

Nhit lng (nhit), l mt dng nnglng d tr trong vt cht nh vo

chuyn ng nhit hn lon ca cc ht

cu to nn vt cht.

Nhit : l tnh cht vt

l ca vt cht.

1/19/2011


2/63

7

Khi nim h:

Khi nim mi trng xung quanh: .

Nhit lun c truyn t vt nng hn

sang vt lnh hn. Qu trnh thu nhit Q > 0,

Qu trnh ta nhit Q < 0

1/19/2011

VD: Hy xc nh du ca nhit lngvi mi qu trnh sau:

Thu nhit Q > 0 Ta nhit Q < 0

81/19/2011

1 J = 1 kg.m2/s2

1 J = 0,23901 cal

1 cal = 4,184 J

1 Cal = 1000 cal

1 BTU = 1054,35 J9

J, cal, Cal, eV, J/mol, cal/molJ, cal, Cal, eV, J/mol, cal/molJ, cal, Cal, eV, J/mol, cal/molJ, cal, Cal, eV, J/mol, cal/mol

Bng chuyn i n v nng lng:

1/19/2011 10

Khi nim: l lng nhit cn cung cp lm tng nhit ca 1 gam cht ln 1K

Biu thc

n v o: J/g.K, i khi J/K....

Ch : + T v Q l cng du

+ Nhit dung mol

+ Dng khi nung nng, lm lnh

T.m

QC

= (6.1)

1/19/2011

11

Khi vt c t nng

T > 0 Q > 0; qt thu nhit

Tng nhit trao i trong h bng khng.

Q1 + Q2 + . = 0 (6.3)

Khi vt c lm lnh

T < 0 Q < 0; qt ta nhit

1/19/2011 12

Qu trnh bin i trng thi:

rn lng hi ; T = const

(ng vi cc tn gi l qu trnh bay hi; ngngng c; nng chy)

Nhit bin i trng thi: nhit bay hi;nhit nng chy,

V d: Hbhi [H2O] = 2256 J/g

Hnchy[H2O] = 333 J/g1/19/2011


3/63

13

VD:VD:VD:VD: Tnh nhit lng cn thit a500 g nc t -50oC n 200oC ? ChoHnc v Hbh (J/g), C (J/g.K).

HHHH2222O (r)O (r)O (r)O (r)----50505050ooooCCCC

HHHH2222O (r)O (r)O (r)O (r)0000ooooCCCC

(1)(1)(1)(1) HHHH2222O (l)O (l)O (l)O (l)0000ooooCCCC

(2)(2)(2)(2) HHHH2222O (l)O (l)O (l)O (l)100100100100ooooCCCC

(3)(3)(3)(3) HHHH2222O (h)O (h)O (h)O (h)100100100100ooooCCCC

(4)(4)(4)(4) HHHH2222O (h)O (h)O (h)O (h)200200200200ooooCCCC

(5)(5)(5)(5)

Q1 = C.m. T

Qtng= Q1 + Q2 + Q3 + Q4 + Q5

Q2 = m. HncQ3 = C.m. T Q

4= m. H

bhQ5 = C.m. T

1/19/2011 14

Khi nim nhit ng hc: l ngnh vt l

nghin cu dng nhit ca chuyn ng

vt cht v nhng qui lut ca chuyn

ng .

Nhit ha hc l mt phn ca nhit nhc nhm mc ch kho st s trao i

nng lng i km theo nhng bin i

vt l, ha hc ca vt cht.1/19/2011

Trong trng hp gin n ng p,cng W c tnh theo:

W = -P.V (6.5)

15

Biu thc :

Ni dung nguyn l 1, chnh l nh lutbo ton nng lng, khng nh rngnng lng lun c bo ton.

U = Q + W (6.4)

1/19/2011

Quy cdu

nh hng ti h

H thu nhit Q > 0 (+) U tng

H ta nhit Q < 0 (-) U gim

H nhn cng W > 0 (+) U tng

H sinh cng W < 0 (-) U gim

Bng quy c du ca Q v W:

161/19/2011

a. Entanpi (H) v bia. Entanpi (H) v bin thin entanpi (n thin entanpi (H)H)

Biu thc bin thin entanpi:

Khi nim entanpi (H): nng lng camt h nhit ng m trao i nhit v

cng vi mi trng, H = U + pV

H = Hcui Hu

H = U - W

n v o:171/19/2011

S khc nhau gia entanpi (H) v ni nn(U)

Lin h gia bin thin entanpi (H), bithin ni nng (U) v nhit lng (Q)trong tng qu trnh:

P = const th Qp = H

V = const th Qv = U

Quy c v du ca U v H

181/19/2011


4/63

b. Hm trb. Hm trb. Hm trb. Hm trb. Hm trb. Hm trb. Hm trb. Hm trng thing thing thing thing thing thing thing thi

Khi nim hm trng thi: l mt c tnh m s

bin thin gi tr ca n trong bt c qu trnh

no cng chph thuc vo gi tr u v gi tr

cui m khng ph thuc vo con ng chuyn

bin.

Mt s hm trng thi: Ni nng (U), nng lng

t do (F), th nhit ng (Z hay G), entanpi (H),

entropi (S)...... l nhng hm trhm trhm trhm trng thing thing thing thi191/19/2011

Mt s ch khi tnh H:tr.279

20

o Bin thin entanpi ca phn ng ph thuc vo s

cht tham gia, cht to thnh v trng thi tn ti (r, l,

ca cc cht.

o Phn ng to nhit th gi tr H mang du m, phng thu nhit gi tr H mang du dng.

o i lng H c gi tr tng ng nhng ngc

du trong phn ng phn hy v to thnh.1/19/2011

VD: chra lin h ca H trong t/h:

H2(k) + O2(k) H2O (k) ; H1

2H2(k) + O2(k) 2H2O (k) ; H2

H2(k) + O2(k) H2O (k) ; H1

H2O (k) H2 (k) + O2(k) ; H3

H2(k) + O2(k) H2O (k) ; H4

Vi H4 to ra 9g nc1/19/2011 21

Dng c: nhit k cc caf,

22

a. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, oa. Khi P = constant, o H

Quy trnh:

Cch tnh:

Qp + Qdd = 0 Qdd = C.m.T

1/19/2011

23

H = Q p : nMg

VDVDVDVD:::: Cho 0,5g Mg vo 1 nhit lng k cc caf,

thm 100ml dd HCl 1M. Nhit dd tng t

22,2oC n 44,8oC. Tnh H ca phn ng trn 1

mol Mg? (Cdd = 4,2J/g.K v DddHCl= 1 g/ml)

m (g) dd

22,2oC44,8oC

(1) Qdd

(J)

(2) Qp

(J)

(3) Hp

(J/mol Mg)

Qdd = C.m.T

Qp + Qdd = 0

1/19/2011

Dng c: bomnhit k,

24

b. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, ob. Khi V = constant, o U

Quy trnh:

Cch tnh:

Qnc = C.m.T

Qp + Qnc + Qbom = 0

1/19/2011


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VDVDVDVD:::: t chy 1 g octan trong nhit lng k (V

khng i). Dng c c t vo 1 bnh cha 1,2

kg nc. Nhit ca nc v bom tng t 25oC

ti 33,2oC. Cbom = 837 J/K. Tnh nhit t chy

trn 1 gam octan?

Qnc = C.m.T

Qbom = Cbom .T

Qp = -(Qnc + Qbom)

1/19/2011

Ni dung: Hiu ng nhit ca

phn ng ha hc ph thuc vo

trng thi ca cc cht u v

ca cc sn phm cui, ch

khng ph thuc vo cc giai

on trung gian ca phn ng.

26

H qu: Nu mt phn ng ha hc l tng chai hay nhiu phn ng khc, th H ca phnng tng c tnh bng tng cc gi tr H ctt c cc phn ng cng li.

G.I.Hess nh bNga (1802-18

1/19/2011

Cch p dng:+ pp i s:

A + B C ; H1

2B + D E ; H2

2A + E D + 2C H3 = ?

V d:

CH4(k) + 2O2(k) CO2(k)+ 2H2O(k) H1 = - 802 kJ

H2O(l) H2O(k) H2 = + 44 kJ.

CH4(k) + 2O2(k) CO2(k) + 2H2O(l) H3 = ?

1/19/2011 27

+ pp th (biu mc nng lng)

H3=?

E

CO2 (k)

C(r) + O2 (k)

CO(k) + O2(k)

H2

H1

Hop = Hos[CaO (r)]

+Hos[CO2(k)]- Hos[CaCO3 (r)]

= +179,0 kJ

E

CaCO3 (r)

Ca(r) + C(r) + 3/2 O2 (k

CaO(r) + CO2

H1

H

H2

281/19/2011

29

Entanpi sinh:

Entanpi sinh tiu chun: Entanpi sinh tiu

chun ca mt cht l bin thin entanpi ca

phn ng hnh thnh 1 mol hp cht t cc

nguyn t ban u trng thi chun.

Trng thi chun (k chun): l trng thi ti

dng tn ti ca vt cht bn vng nht,

p sut 1 atm v nhit 25oC (298K)

1/19/2011

VD: chra H no trong cc p saul Hs :

CaO + CO2 CaCO3 H1

2Ca + O2 2CaO H2

Na + Cl2 NaCl H3

HCl + NaOH NaCl + H2O H4

p n ng l:

1/19/2011 30


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K hiu: Hs, Ho, hay Ho298,s

Ch : (tr.288)

vi n cht: Ho298,s =0.

Gi tr ca Hos c th s dng so snhkh nng bn vng nhit gia cc hp chtcng nhm.

Hu ht gi tr entanpi sinh tiu chun umang du m, iu ny cho bit qu trnh tothnh cc hp cht t cc nguyn t ban uthng l to nhit.

1/19/2011 32

Hop = Hos,(sn phm) - H

os,(tham gia)

P/: aA + bB cC + dD

Hop= [c.Hos(C)+d.H

os(D)] - [a.H

os(A)+b.H

os

1/19/2011

33

Hp < 0 phn ng c xu hng din ratheo chiu to sn phm

Hp > 0 phn ng c xu hng din ratheo chiu to cht tham gia

1/19/2011

Bi sau:

Chng 7: Cu to nguyn t

Bi tBi tBi tBi tp chng 6:p chng 6:p chng 6:p chng 6:

11, 15, 25, 29, 33, 39, 45, 53, 79, 93.11, 15, 25, 29, 33, 39, 45, 53, 79, 93.11, 15, 25, 29, 33, 39, 45, 53, 79, 93.11, 15, 25, 29, 33, 39, 45, 53, 79, 93.

1/19/2011

1/19/2011 35

1. Nng lng v phn ng ha hc c li

quan nh th no vi nhau?2. Nhng hnh thc

tnh ton c linquan n s linh l g?

1/19/2011


7/63

HA HCI CNG

GV: L Minh ThnhGV: L Minh ThnhGV: L Minh ThnhGV: L Minh Thnh1

CU TO

NGUYN T

Chng 7

2

7.1. BC X IN To Cc khi nim n tp

bc sng () l khong cch gia hai nh sng (cao nht

hoc thp nht) lin tip ; n v o.

tn s sng (, f) l s dao ng ca sng ti mt im cho

trc trong mt n v thi gian ; n v o.

bin sng l lch cc i ca dao ng sng so vi v

tr cn bng.

tc sng l khong cch lan truyn sng trong mt n v

thi gian ; n v o.Tc sng (m.s-1) = (m) f (s-1)

c = f (7.1)

3

7.1.a. Sng dng sng dng l sng c cc bng v cc nt c nh trong

khng gian.

c im: c 2 nt; khong cch gia hai im nt lin

lun lun l /2; c th c nhng bc sng xc nh.

4

5

7.1.b. Ph in t v ph kh kin

Ph in t l mt khong no ca sng in t.

6

Cu hi: Sng in thoi di ng c bc sn

nm trong khong no


8/63

7.2. PLANCK, EINSTEIN, NNG LNG VPHOTON

Thuyt Planck: Bc x in t c hp th hoc pht x

di dng nhng lng gin on gi l lng t nng lng

Phng trnh Planck:

ngha ca thuyt Planck: gii quyt c vn khng

hong t ngoi m thuyt Maxwell cha gii thch c.

Khng hong t ngoi (ultraviolet catastrophe) l mu thun gia l

thuyt v thc nghim khi nghin cu thc nghim v s bc x nhit,

ngi ta thu c nhng kt qu khng th gii thch ni bng l

thuyt pht x c in. 7Max Planck(1858-1947)

Albert Einstein1879 - 1955

James Clerk Maxwell

(1831 1879)

8

Khi nim hiu ng quang in: l hin tng ccht electron bn ra khi b mt kim loi khi c nh

sng p vo.

iu kin xy ra hiu ng quang in: nng

lng in t chiu vo phi ln hn cng thot ca

eletron lin kt vi kim loi: f fo

Cng thc ca Einstein

E = m.c2 = h.fc.m

h

p

h== (tnh cho mtht photon )

ngha ca lun im Einstein v nh sng.. 9

Cu hi: Hy so snh nng lng ca 1 mol photo

nh sng c = 625 nm vi nng lng ca 1 mol

photon vi sng c f = 2,45GHz?

E1 = NA.h.f1 = NA.h.(c /1) E2 = NA.h.f2

Trong hai cng thc trn, ch n v o ca l

cn n v ca f l ..

Gi tr ca h = . cn gi tr ca c = .. 1

HD:

Phn bit ph lin tc, ph vch, ph pht x v ph hp th

11

7.3. PH VCH CA NGUYN T

.

.

2 2

1 1 1= R -

2 n

vi n > 2 (7.3)

hng s Rydberg: R = 1,097

107

m-1

1


9/63

13

7.3.a. M hnh nguyn t H ca Bohr

Cng thc ca Bohr

n 2

R hcE = -

n

(7.4)

n: S lng t chnh, n = 1,2,3

E: n v o (J/nguyn t)

ngha m hnh ca Bohr

R.h.c = 1312 kJ/mol th E (kJ/mol)1

M hnh Bohr: Electron chuyn ng

trn nhng qu o nht nh, trn

nng lng ca e khng i

15

7.3.b. Thuyt Bohr v quang ph vch

Cu hi: quan st video sau v gii thch quang ph vch

c?

Cc cng thc:

E = Esau Etrc

E1 photon

= (E.103):NA

n 2

R hcE = -

n

E1 photon = (h.c):1

(J/photon)

17

7.4. TNH CHT SNG CA ELECTRON

Cu hi: Cc ht vt cht (khc nh sng) c tnhcht lng tnh sng-ht khng?

Quan im ca Louis Victor de Broglie.

h =

m v

(7.6)

Mi ht vt cht khi lng m chuyn ng vi tc vs c bc sng

ngha ca thuyt de Broglie. 1


10/63

7.5. C HC LNG T V NGUYN Ta. Nguyn l bt nh Heisenberg

Ni dung: Khng th xc nh c chnh xc ng thi cta v nng lng ca electron trong nguyn t

Biu thc: x . p > h

b. M hnh nguyn t theo Schrdinger

ngha ca nguyn l

Coi e chuyn ng sng, m t bng hm sng

ngha ca hm sng v hm mt xc sut 2

Bn cht m hnh nguyn t theo Schrdinger

19

i = n v o (r,t) = hm sng,. = hng s Planck rt gn = ton t Hamilton.

Phng trnh Schrdinger:

Cc kt qu gii phng trnh Schrdinger gm c hm

sng , nng lng E v b cc s lng t 2

Kt qu hm sng (v d).

0-r/a01s

1/a = e

Kt qu nng lng (v d).

2

2

2)( ax

Aex

=

2

22

2

222

82 ma

hn

manE

n == h

nEnE =+= )

2

1(h 21

Louis Victor de Broglie1892 - 1987

Werner Karl Heisenberg1901 - 1976

Erwin Rudolf JoseAlexander Schrdin

1887 - 1961

B cc s lng t:

S lng t chnh n: (1,2,3n)

S lng t ph ( s lng t xung lng):

(0, 1, 2, .(n-1))

S lng t t m: (-, -(-1),0(-1), )

ngha ca b s lng t.

H qu: khi nim obitan.

23

S lng t chnh n; n = 1, 2, 3, ,

L tha s u tin trong qu trnh xc nh nnglng e:

En = - R.h.c/n2

n tng En tng v cc mc nng lng cng xnhau

Cc e c cng gi tr n thuc cng 1 lp:

2

Gi tr n: 1 2 3 4.

K hiu lp: K L M N.


11/63

Gi tr : 0 1 2 3

Phn lp: s p d f

S lng t xung lng , = 0,1, 2,, n-1

Gi tr ca s tng ng vi k hiu phn lp:

Gi tr ca s tun theo gi tr ca n:

Lp n = 1 = 0 obitan: 1s

Lp n = 2 = 0, 1 obitan: 2s, 2p

25

Phn lp: s p d f

S obitan: 1 3 5 7

S lng t t m, m = 0,1, 2,,

m cho bit cch nh hng ca obitan

Mi phn lp () c (2+1) obitan c nh hngkhc nhau, nhng c nng lng bng nhau

2

ngha b s lng t (n, , m): mi b s lngt ny i din cho mt obitan duy nht trong nt

27 2

Cch khai trin mt b s lng t xc nobitan tng ng v ngc li

4pyn = 4

= 1

m = 0

Cu hi: T mt s lng t (n=2) ta c th c

nhng b s lng t (n, , m) no?

HD: t n, suy ra cc gi tr , sau t mi gi

tr , suy ra cc gi tr m tng ng. Kt hpng thi c 3 gi tr n, , m ta s c cc b slng t theo yu cu.

7.6. HNH DNG CC OBITAN

Obitan s:

Obitan p:

Obitan d:

29 Obitan f: ..

CU HNH ELECTRON NGUYN T VTNH TUN HON HA HC

Bi sau: Chng 8

BT chng 7

7, 11, 17, 25, 27, 33, 37, 55, 61, 65


12/63

February 11, 2011

HA HHA HHA HHA HCCCC

I CNGI CNGI CNGI CNG

GV: L Minh ThnhGV: L Minh ThnhGV: L Minh ThnhGV: L Minh Thnh

1

CCCCuuuu hnh electron nguyn thnh electron nguyn thnh electron nguyn thnh electron nguyn t&&&&

Tnh tuTnh tuTnh tuTnh tun hon ha hn hon ha hn hon ha hn hon ha hcccc

Khi nim spin electron:

S lng t spin electron: ms.

Gi tr ca ms=, i din cho 2 chiu quay

B s lng t y m t 1 e trong nguyn t

l: (n, , m, ms).

Cu hi: b s (3, 1, -1, +) m t e no?

HD: phn tch ln lt v tr

3

T tnh (magnetic property) l mt tnh cht ca vt li

hng ng di s tc ng ca mt t trng.

T tnh c ngun gc t lc t, lc ny lun i lin v

lc in nn thng c gi l lc in t.

Lc in t l mt trong bn lc c bn ca t nhin.

lin h gia lc t, lc in, v ngun gc ca chng

c cho bi h phng trnh Maxwell.

Lc in t sinh ra khi cc ht tch in chuyn ng v

nh cc in t chuyn ng trong dng in, hoc trnquan im lng t th lc in t gy ra bi chuyn ng

qu o v spin ca in t trong nguyn t.

T tnh c th phn ra lm cc loi: st t, phn st t,

ferri t, thun t, nghch t. Thng thng khi ta ni mt

vt liu c t tnh, tc l m ch vt liu c tnh st t, phn

st t hoc ferri t.

5

Tnh thun t: l tnh cht b ht v t trng c

cc cht (n/t hoc ion to nn n c e c thn

Tnh nghch t: l tnh cht b y bi t trng

ca cc cht (n/t, ion to nn n ko c e c th

Nguyn nhn tnh thun t, nghch t:

Cu hi: Nguyn t, ion no sau y

thun t, nghch t: Na+, Cl, O, Fe2+..

HD: vit cu hnh e cho nt, ion trn


13/63

Cht thun t Cht nghch t Cht st t

St t: l cc cht c t tnh mnh, hay kh nng hng ngmnh di tc dng ca t trng ngoi, m tiu biu l Fe.

Tn gi "st t" c t cho nhm cc cht c tnh cht tging vi st. Cc cht st t c hnh vi gn ging vi cc chtthun t c im hng ng thun theo t trng ngoi, (vd nh st (Fe), cban (Co), niken (Ni), gali (Gd)... l cccht st t in hnh) 7

Hnh nh ca cht thun, nghch t v st t:

H qu: ..

Cu hi: S e ti a mi obitan (m ), mi phn

lp () v mi lp (n) l bao nhiu?

Cu hi: Vi cu hnh electron obitan sau y, h

chra b s lng t tng ng vi cc electron

tng ng? 2s

4p

Ni dung: trong mt nguyn t ko th tn ti ng

thi 2 e c chung b bn s lng t (n, , m, m

Ni dung nguyn l Aufbau (quy tc (n+)):

Electron c xp vo cc phn lp theo chiu

tng dn ca gi tr (n+).

Trng hp gi tr (n+) bng nhau, th u tin

phn b vo gi tr no c n nh hn trc.

9

H qu: a ra th t nng lng ca cc phn lp

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p.

Th t nnglng ca cc

phn lp ph

thuc vo 2 s

lng t n v .

Quy tc

Klescopski

minh ha nguyn

l Aufbau:

Khi nim cu hnh electron:

Biu din cu hnh electron thng theo 2 cch:

Cu hnh spdf, dng b k hiu: na.

Cu hnh obitan ( lng t), dng cc vung, in e

11

biu din c cu hnh obitan quy tc

Hund (quy tc bi cc i): cc e in vo mt phn

lp sao cho tng spin ca chng l cc i.

H qu quy tc Hund: a ra tnh thun

t, nghch t ca nguyn t, ion, phn t

ChChChCh quan tr quan tr quan tr quan trng khing khing khing khi vivivivit ct ct ct cu hnh electron:u hnh electron:u hnh electron:u hnh electron:

Vit theo nng lng trc, sau a cc ph

lp v ng v tr ca lp.

Trng hp xut hin phn lp d9, f13 d10,

hin tng gi bo ha. Cn nu xut hin p

lp d4, f6 d5, f7 hin tng gi na bo h

Cu hnh electron ca ion dng (hoc ion m)

c vit bng cch, sau khi vit cu hnh

electron ca nguyn t, ta tr i (hoc thm v

cu hnh nguyn t .


14/63

13

Cu hi: hy vit cu hnh electron theo 2 cch

cho cc nguyn t sau: A (Z=15); B (Z=24)?

A: 1s2 2s2 2p6 3s2 3p3

B: 1s2 2s2 2p6 3s2 3p6 4s2 3d4

1s2 2s2 2p6 3s2 3p6 3d4 4s2

1s2 2s2 2p6 3s2 3p6 3d5 4s1

Cu hi: hy vit cu hnh electron ca ion A2- v

B3+ bit: A (Z=16); B (Z=27)?

a. a. a. a. nh lunh lunh lunh lut tut tut tut tun honn honn honn hon

Ni dung: Tnh cht ca cc

n cht, thnh phn v tnh

cht ca cc hp cht tng

ng ca cc nguyn t bin

i tun hon theo chiu tng

ca THN.

H qu:.....

15Dmitri Ivanovich MendeleevDmitri Ivanovich MendeleevDmitri Ivanovich MendeleevDmitri Ivanovich Mendeleev

b. Gib. Gib. Gib. Gii thii thii thii thiu mu mu mu mt st st st s ddddng bng bng bng bng HTTHng HTTHng HTTHng HTTH

17


15/63

19

a. Bn knh nguyn t v ion

BKNT l na khong cch gia hai nguyn t gn nhau

nht trong n cht.

BKNT cn ph thuc kiu lai ho, kiu mng tinh th.

BK ion cng tng t BKNT, ngi ta coi tng bn knh

cation v anion bng khong cch gn nht gia cation

v anion trong tinh th ion.

21

c im bn knh ion:

Cation c hnh thnh do nguyn t mt

electron, v vy bn knh cation < bn knh n.t

Anion c hnh thnh do nguyn t nhn

electron, v vy bn knh anion > bn knh n.t.

b. Nng lng ion ha

Khi nim: l nng lng cn thit tch mt e ra khi

nguyn t pha kh trng thi c bn.

K hiu: IE, hoc I..... n v o......

Ch :

Tnh kim loi ca nguyn t cng ln th IE cng nh.

Mt nguyn t trung ha, tch ln lt cc e th 1, 2, 3

th gi tr nng lng ion ha l....

Nng lng tch e ha tr nh hn rt nhiu so vi e li.23

c. i lc electron

Khi nim: l nng lng ca qu trnh nhn thm m

ca nguyn t pha kh trng thi c bn.

K hiu: EA, hoc E..... n v o...

Ch :

Kh nng ht e ca n.t cng ln th EA cng m.

Mt nguyn t trung ha, nhn ln lt e th 1, 2, 3

gi tr i lc electron ln lt l...


16/63

d. Quy lud. Quy lud. Quy lud. Quy lut bit bit bit bin thin cc tnh chn thin cc tnh chn thin cc tnh chn thin cc tnh chtttt

25

R

EA, IE

Tham kho thm quy lut bin i hnh 8.11; 8.12

(v bn knh); hnh 8.13 (v IE); hnh 8.14 (v EA).

Cu hi:

So snh ba nt: 13Al, 15P v 9F theo cc tnh cht sau

a) Th t tng dn bn knh nguyn t ca chng

b) Nguyn t no c nng lng ion ha ln nht?

c) Nguyn t no c i lc electron m hn, Al hay F?

HD:

a. F < P < Al, ti v.

b. Al < P < F, ti v.

c. F m hn Al, ti v.

CuCuCuCu hhhhi:i:i:i:

Sp xp cc ion 11Na+, 12Mg2+, 9F- v 8O2- theo

chiu bn knh tng dn? Chn p n ng?

a. O2- < F- < Na+ < Mg2+

b. F- < Mg2+ < O2- < Na+

c. Na+ < Mg2+ < F- < O2-

d. Mg2+ < Na+ < F- < O2-

27

Bi sau: Chng 9

BTBTBTBT chng 8chng 8chng 8chng 8

3, 13, 17, 25, 27, 29, 31, 47, 51, 53

29


17/63

HA HC

I CNGGV: L Minh Thnh

112/02/2011

L

C 9

12/02/2011

9.1. E

3

Khi nim e ha tr: l cc e

lp ngoi cng + s e phn lp

st ngoi cng cha bo ha.

Khi nim electron li: l cc e cn li ngoi e ha tr.

Cu hi: Hy xc nh s e ha tr cho mi nt sau y:

13A: 1s2 2s2 2p6 3s2 3p3

28B: 1s2 2s2 2p6 3s2 3p6 3d8 4s2

12/02/2011

Ch khi biu din k hiu Lewis:

o in e ln lt xung quanh theo 4 hng sau tip tc

o Vi cc nguyn t phn nhm ph:

K hiu Lewis cho cc nguyn t: k hiu ca nguyn t l i

din cho ht nhn + cc electron li, cn cc electron ha tr

c biu din bng cc du chm t u n bn pha xu

quanh k hiu nguyn t.

12/02/2011

9.2.

Khi nim lin kt ha hc: .

S phn loi lin kt ha hc (gia cc nguyn t):

Theo l thuyt c in: 3 loi: ion, cng ha tr, kim loi.

Theo l thuyt hin i: 2 loi: xch ma, pi.

Ngoi ra cn c lin kt lin phn t: l loi lc lin kt gia cc

phn t vi nhau. V d: lin kt hir,

Nguyn tc chung ca vic to thnh lin kt ha hc thuyt c

in l tun theo quy tc bt t: 512/02/2011

9.3. L

Khi nim: l lin kt ha hc c bn cht lc ht tnh igia hai ion mang in tch tri du.

S hnh thnh:

1e

Na + Cl Na+ + Cl NaCl

Na ([Ne]3s1) + Cl ([Ne]3s23p5) Na+ ([Ne]) + Cl ([Ne]3s2312/02/2011

1e


18/63

c im lin kt ion:

o thng l lin kt gia cc nguyn t nguyn t phi kim vi

cc nguyn t nguyn t kim loi.

o in tch ion tham gia lin kt cng ln, th lc ht cng mnh,

lin kt cng bn vng.

o kch thc ca ion, hay khong cch gia cc ion tri du

cng ln, th lc ht gia cc ion cng gim, cng km bn.

Nng lng mng tinh th ion ph thuc vo:o Qu trnh ion ha, nng lng ca n th hin bng IE.

o Qu trnh nhn thm e, nng lng ca n th hin bng EA

o Qu trnh to phn t t cc ion th kh, th hin bng Ecp ion712/02/2011

9.4. L

Khi nim: l lin kt c hnh thnh gia cc nguyn

bng mt hay nhiu cp in t (electron) chung.

S hnh thnh:

Cl Cl

CT e

Cl-Cl

CTCT

Cl Cl+

H Cl H-Cl HCl+H

12/02/2011

c im lin kt cng ha tro Thng l lin kt gia cc n.t nguyn t phi kim vi nhau.

o Nu lin kt gia 2 nguyn t bng 1 cp electron dng

chung lin kt gi l lin kt n. Cn nu 2 nguyn

t bng 2, 3 cp electron dng chung lin kt gi l

lin kt i, ba.

912/02/2011

Ch khi so snh cc tnh cht ca hp cht ion v cngha tr v dng tn ti, tonc,, t

os, tnh tan, tnh dn in, E

hyth chng tun theo quy lut l:

oHp cht ion thng c tonc,, to

s, tnh tan, tnh dn i

Ephn hy cao hn so vi h/c cng ha tr.

oTrong cc hp cht CHT, cht c phn cc lin kt

cng ln th cht c tonc,, to

s, tnh tan trong nc (

phn cc) cng ln.

12/02/2011

12/02/2011 11

Cu hi: Hy so snh nhit nng chy, tnh tan trong

nc ca cc cht sau y v xp theo chiu tng dn cc

tnh cht : HCl, NaCl, CaCl2, Cl2, NH3.

HD:

oPhn loi 4 cht trn xem chng thuc loi ion hay CHT.

oCng l h/c CHT th xt xem phn cc cc lk ntn.

oCng l h/c ion th xem xt ln in tch ion, kch

thc ion so snh.

oKt qu ng l: .

9.5. C L :

,

12/02/2011

Cch v cu trc Lewis cho 1 cht: (tr.419/T1)

1. Xc nh tng s electron ha tr ca phn t, ion .

2. Chn nguyn t trung tm thng l nguyn t c i lc e

nh nht, cc nguyn t cn li lm nguyn t xung quanh.

3. t 1 cp electron vo gia mi cp nguyn t . Sau t

cc cp electron cn li quanh cc nguyn t xung quanh (t

H) chng c bt t.

4. Chuyn mt (hoc mt vi cp e khng lin kt) ca nguyn

ngoi vo lm bt t cho nguyn t trung tm.


19/63

CH

H

H

H

CH4 PF3

PF

F

F

..: ..

....

..

..:

: :

CO2

CO O

..:

..:

....CO O:

.... :

CH3COO

..

CO

O

:

C

H

H

H

:

:

:

H2SO4

1312/02/2011 12/02/2011

Ch khi v cu trc Lewis: ....oMt hp cht c th c 1 hoc nhiu cu trc Lewis khc nhau.

oCc cht khc nhau, nhng c cng tng s e ha tr (cc cht

ng e) th thng c cu trc Lewis ging nhau.

oCc nguyn t C, N, O, F lun tun theo quy tc bt t.

oXem thm Ch 9.1 trang 423.

1512/02/2011

9.6. Ni dung quy tc bt t: ....

Cc trng hp ngoi l:

oNguyn t Hidro trong mi hp cht ch cn 2e.

oTrng hp c t hn 8 electron ha tr: VD: hp cht ca Bo

oTrng hp c nhiu hn 8 electron ha tr: VD: hp cht ca

oTrng hp c s l electron: VD: hp cht NO2,..

12/02/2011

9.7. H Thuyt sc y cp electron ha tr (VSEPR): cp e lk

v cp e cha lk trong v ha tr ca nt lun y nhau v

phn b sao cho cng xa cng tt.

ng dng: d on hnh dng phn b khng gian ca

cc nt trong phn t, trong ion...

1712/02/2011

S phn b cc cp e ( lk + cha lk) ca nt trung tm

to nn hnh hc cp electron.

S phn b cc nguyn t c mt xung quanh nt trung

tm to nn hnh hc phn t.12/02/2011

2 cp e 3 cp e 4 cp e 5 cp e 6 cp e

ngthng

180O

Tam gic

120O

T din u

109,5O

Lng thptam gic

120O; 900

Bt din

90O

H qu ca thuyt VSEPR:

o d on hnh hc cp e ca nguyn t trung tm trong phn t

ion. theo m hnh di y (ch d vo s cp e quanh nt

o d on hnh hc phn t ca c phn t, ion. phi kt hp

qu ca hnh hc cp e trn vi s lng cc nt c xq nt


20/63

12/02/2011 19

2 cp e3 cp e 4 cp e

5 cp e 6 cp e

Hnh hc cp e theo thuyt VSEPR

C , :

1. Tnh tng s electron ha tr ca phn t, ion .

2. V cu trc Lewis cho pt, ion

3. m s cp e ( lk v cha lk) quanh nguyn t trung tm,

cc lin kt bi (i, ba) cng coi nh 1 cp e lin kt.

4. Xc nh hnh hc cp e theo quy tc VSEPR, ch da vo tn

cp e quanh n.t trung tm. (c th v ra cho d hnh dung)

5. Da vo hnh hc cp e, ng thi da vo s nguyn t xun

quanh nguyn t trung tm, kt lun v hnh hc phn t, (

ra gc lin kt nu c yu cu).12/02/2011

Ch v hnh hc cp e v hnh hc phn t:

D on hnh hc cp e ch cn cn c vo tng s cp e

xung quanh nguyn t trung tm (tnh c cp e lk v cp e

cha lin kt).

D on hnh dng ca phn t phi da vo s nguyn t

xung quanh nguyn t trung tm, c kt hp hnh hc cp e.

Hnh hc cp e v hnh hc phn t u coi lin kt i, ba

nh cc lin kt n.

2112/02/2011 12/02/2011

12/02/2011 23

Cu hi: hy xc nh hnh hc cp e v hnh hc phn t c

cc phn t v ion sau: CO2, O3, SO32, PO4

3

Cu hi: hy xc nh hnh hc cp e v hnh hc phn t c

cc trng hp tng qut sau:

12/02/2011


21/63

9.8.

in tch chnh thc ca n.t trong p.t =

( ).

2512/02/2011

Trng hp c nhiu cng thc Lewis, v d: N2O, BF3 tcng thc no m in tch m nm trn nt m in lnht s l cng thc ph hp nht.

N N O

N N O==

N N O

2 +1 +1 1 +1 0 0 +1 1

12/02/2011

9.8..

Trong hp cht ion: phn cc lin kt c k hiu (+) v () cvit ngay cnh k hiu nguyn t .

Trong lin kt cng ha tr phn cc, s phn cc c ch ra bi khiu (+) v () vit cnh k hiu nguyn t

C th s dng m in d on khuynh hng ca s phncc lin kt bng cch tnh

2 1 =

Gi tr delta X cng ln lin kt cng phn cc mnh.

phn cc ca lin kt c biu din bng mi tn hng tnguyn t m in yu sang nguyn t m in mnh.

2712/02/2011

9.8..

S phn b in tch trn nguyn t theo hai quy tc sau

o Cc electron s phn b theo cch no sao cho in tch t

mi nguyn t trong phn t gn gi tr 0 nht.

o Nu c mt in tch m xut hin, n s nh v trn nguyn

c m in ln nht.

12/02/2011

9.9. Khi nim: l trng thi ca phn t trong c mt u mang

in tch m, u cn li mang in tch m.

2912/02/2011

Cc bc d on phn cc phn t:

1. Tnh tng e ha tr.

2. V cu trc Lewis.

3. Xc nh hnh hc cp e.

4. Xc nh hnh hc phn t.

5. Xc nh phn cc mi lin kt.

6. Kt lun v s phn cc ton p.t theo quy tc hp lc.

S phn cc phn t thng c o bng m men

lng cc ( ): l i lng c trng cho mc ph

cc ca phn t

= | | (Debye, D)

: di lng cc (cm)

: ln in tch

1 D = 3,34 1030 C.m

M men lng cc cng ln, phn t cng phn cc

:

= 1

3012/02/2011


22/63

3112/02/2011

9.10. C

Bc lin kt: l s cp e dng chung gia 2 nt trong 1 p

di lin kt: l khong cch gia 2 ht nhn ca 2 nt

tham gia lk.oLin h: bc lin kt cng ln, th di lin kt tng ng cn

nh, lin kt cng bn.

Nng lng lin kt: l n/l cn ph v lk trong pt

oLin h: nng lng lin kt cng ln th lin kt cng bn

S mi lin kt ca XYtrong phn t hoc ion

S cp e dng chung lin kt X vi YBc lin kt = (9.2)

o12/02/2011

Cu hi: Xc nh bc lin kt, so snh nng lng lin ktca cc lin kt NO trong phn t v ion: N2O, NO2

+, NO2

v NO3.

N N O

Bc lin kt: 1; 2; 1,5 v 1,3.

di lin kt :

3312/02/2011

Bi sau: Chng 10

BT chng 9:

11, 13, 19, 21, 33, 35, 45, 47, 51, 53.

L :

12/02/2011

BeH2

CO2

BH3

CO32

3512/02/2011

CH4 PO43-

NH3 SO32-

12/02/2011


23/63

HA HC


12/12/2011

Chng 10

Lin kt v cu to phn t:

S lai ha obitan v obitan phn t2/12/2011

10.1. Obitan v10.1. Obitan v cccc ll thuyt v lin ktthuyt v lin kt

Khi nim v obitan nguyn t, obitan phn t.

Phn loi obitan nguyn t (AO): s, p, d, f

Phn loi obitan phn t (MO): ,

Hai thuyt hin i gii thch lin kt ha hc da trn

obitan :

thuyt lin kt ha tr (thuyt VB, valent bond)

thuyt obitan phn t (thuyt MO, molecular orbital

32/12/2011

10.2. Thuyt lin kt ha tr (thuyt VB)10.2. Thuyt lin kt ha tr (thuyt VB)10.2.a. M hnh xen ph obitan

Lun im chnh ca thuyt VB: lin kt ha hc

sinh ra do s xen ph cc obitan nguyn t .

Khi nim xen ph AO: s an xen hai hay nhiu

vng obitan vo nhau to thnh khu vc c mt

electron cao hn gia hai hay nhiu nguyn t.

Cc hnh thc xen ph:

Xen ph trc: to lin kt xichma .

Xen ph bn: to lin kt pi hoc lin kt delta2/12/2011

-Xen ph trc: l s xen ph trong vng xen ph nm

trn trc lin kt, l ng ni tm 2 nguyn t tham gia

lin kt.

- Cc kiu xen ph trc:

s-s s-p p-p

+ Cc lin kt n trong cc phn t u l lin kt xichma

+ Xen ph trc to ra lin kt xichma , l lin kt bn

vng nht, bi vng xen ph nm trn ng ni tm gia

2 ht nhn ca nguyn t.

52/12/2011

-Xen ph bn: l s xen ph trong vng xen ph

2 bn trc lin kt, hoc song song vi trc lin kt (

ni tm 2 nguyn t tham gia lin kt ).

+ Xen ph bn to ra lin kt pi v lin kt delta ,

lin kt ny km bn hn so vi lin kt xichma do v

xen ph xa hai ht nhn hn.

- Cc kiu xen ph bn:

p-p p-d d-d

d-d

62/12/2011


24/63


25/63

Cch gii thch lin kt ha hc theo thuyt VB, c kt

hp thuyt lai ha:

1. Vit cu hnh e nguyn t tham gia lin kt, xc nh

obitan cha e c thn.

2. V cu trc Lewis cho phn t .

3. Xc nh hnh hc cp e cho nguyn t trung tm.

4. Ch nh kiu lai ha cho nguyn t trung tm m phhp vi hnh hc cp e .

5. Tin hnh xen ph to lin kt gia cc obitan

thun khit v obitan lai ha.132/12/2011

V d: gii thch lin kt trong NH3 theo VB.

o Cu hnh e: 7N: 1s22s22p3.

o Cu trc Lewis:

o Hnh hc cp e ca N l t din.

o Vy N lai ha sp3, l s t hp.

o S xen ph trong NH3 nh sau:

V d: gii thch lin kttrong CH4 (xem kt qu)

142/12/2011

V d: gii thch lin kt trong BF3 theo VB. Cu hnh e: 5B: 1s22s22p1.

9F: 1s22s22p5

Cu trc Lewis:

Hnh hc cp e ca B l tamgic

Vy B lai ha sp2, t hp v

S xen ph trong BF3 nh sau:

Cu hi: hy gii thch lin kt

trong cc phn t NO3- theo VB?

152/12/2011

Mt s hnh

nh lin kt

trong phn t

c gii

thch theo

thuyt VB:spsp 2p

z

2pz

Be FF 162/12/2011

10.2.c. Lin kt n v lin kt bi theo thuyt VB

Phn loi:

Lin kt n: do s 1 xen ph trc to thnh, ch gm lin

kt xichma .

Lin kt i: do 1 xen ph trc v 1 xen ph bn to thnh

(1 + 1).

Lin kt ba: do 1 xen ph trc v 2 xen ph bn (1 + 2).

Nhn xt:

Lin kt pi ch hnh thnh khi c s xen ph bn ca cc

obitan p, d khng tham gia lai ha.

Mt cu trc Lewis c lin kt bi th nguyn t tham

gia lin kt phi c lai ha sp2 hoc sp.172/12/2011

Lin kt n

Lin kt i

Lin kt ba

182/12/2011


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10.2.d. ng phn cis-trans

Nhn xt:

Cc lin kt n d dng xoay quanh trc ca n, to

ra cc cu dng khc nhau ( ngay nhit phng).

Cc lin kt i khi xoay quanh trc ca n, i nng

lng cao hn nhiu, v khng th xy ra nhit

phng, do tn ti 2 dng khc nhau trong khng

gian xut hin ng phn cis-trans.

Kt lun: s sp xp khc nhau ca cc nguyn t xung

quanh mt phng cha lin kt i trong khng gian, to

thnh ng phn cis-trans.192/12/2011

10.3. Thuyt obitan phn t (thuyt MO)10.3. Thuyt obitan phn t (thuyt MO)

10.3.a. Nguyn l c bn ca thuyt

Tng s cc MO thu c lun bng tng s cc AO th

gia t hp.

Obitan lin kt (MO) c nng lng thp hn obitan g

v obitan phn lin kt (MO*) c nng lng cao hn.

Cc electron ca phn t c in vo cc MO theo t

t nng lng tng dn, theo nguyn l loi tr Pauli v

quy tc Hund.

Cc AO kt hp to MO hiu qu nht khi n c m

nng lng bng nhau2/12/2011

10.3.b. S hnh thnh cc MO v MO* C 2 AO (s-s hoc p-p) khi t hp vi nhau, s to thnh 1

MO lin kt v 1 MO phn lin kt (MO*).

t tn cho cc MO v MO* to ra theo 2 dng xen ph (

)

212/12/2011 2/12/2011

- Cch v cho cc phn t ng hch dng X2.

2p

2p

*2p*2p

2s

*2s

1s

*1s

AO AOMO

1s

2p

2s

E

1s

2p

2s

23

2p

*2p

- Ch : i vi O2v F2th 2p c nng lng thp hn 2p2/12/2011

10.3.c. Cc bc vit cu hnh e cho phn t, ion

1) Vit cu hnh e cho cc nguyn t tham gia lin kt.

2) V gin nng lng ca qu trnh t hp cc AO to ra cc MO v MO*.

3) t tn cho cc MO v MO* to ra.

4) in electron theo th t tng dn nng lng, tng s

in vo cc MO v MO* bng tng s e cc AO c.

5) Kt lun cu hnh electron ca phn t t gin trn

6) Bc lin kt = (tng s e MO tng s e MO*).

7) Kt lun v s tn ti phn t, v cc lin kt c trong

phn t 2/12/2011


27/63

- Cch gii thch lin kt theo thuyt MO: v d cho He2

+ Cu hnh e cho phn t He2: (1s)2(*1s)2.

+ Bc lin kt cho phn t He2 = ( 2 - 2 ) = 0

1s

*1s

He HeHe2E

1s1s

25

+ V gin MO cho phn t He2.

+ Cu hnh e cho nguyn t He: 1s2

2/12/2011

-- Cch gii thch lin kt theo thuyt MOCch gii thch lin kt theo thuyt MO:: v d chov d cho phn tphn t

1s

2

2s

E

2p

2p

*2p*2p

2s

*2s

1s

*1s

+ Bc lin kt = (10 - 4) = 3 lk ba

N NN2

+ Cu hnh e cho N2: (1s)2(*1s)2(2s)2(*2s)2 (2p)2(2p)2 (2p

1s

2p

2s

+ Cu hnh e cho nguyn t N: 1s2 2s2 2p3

2/12/2011

Gin MO: cho phn tLi2.

Cu hnh e cho phn tLi2:

(1s)2(*1s)2(2s)2

Bc lin kt cho phn tLi2 = 1

27

-- Cch gii thch lin ktCch gii thch lin kt theotheo MOMO:: v d chov d cho phnphn tt LiLi22

2/12/2011

Cu hi: Hy vit cu hnh electron ca O2+, N2+. T m t bc lin kt v t tnh ca chng?

+ Cu hnh e cho N2: (1s)2(*1s)2(2s)2(*2s)2(2p)2(2p)2 (2p)2

+ Cu hnh e cho O2: (1s)2(*1s)2(2s)2(*2s)2 (2p)2(2p)2(2p)2(2p)1(

+ T suy racu hnhca ion:

Cu hnh e cho O2+: (1s)2(*1s)2(2s)2(*2s)2 (2p)2(2p)2(2p)2(2p)

Cu hnh e cho N2+: (1s)2(*1s)2(2s)2(*2s)2(2p)2(2p)2 (2p)1

+ T cuhnh ca ion suyra bclin ktv ttnh ca ion :

C hai ion O2+, N2+ u c e c thn, nn chng thun t.

Ion O2+ c bc lk = 2,5 ; cn ion N2+ c bc lk cng = 2,5.

2/12/2011

10.4. So snh hai thuyt VB v MO

Thuyt VB.

cung cp cho ta mt bc tranh m hnh v cu to phn

t v lin kt.

c bit hu ch cho phn t khi xem xt nhiu nt.

m t tt v lin kt i vi cc phn t trng thi c

bn, hoc trng thi c nng lng thp nht.

Thuyt MO.

c s dng khi cn mt bc tranh v s lng linkt

rt cn thit nu chng ta mun m t phn t trng

thi kch thch c nng lng cao

i vi mt vi phn t nh NO v O2, thuyt obitan

phn t l thuyt duy nht m t c lin kt tht s

trong chng.292/12/2011

Bi sau: Chng 12 (Tp II)

BT chng 10:

1, 3, 5, 11, 15, 17, 23, 29, 31, 41.

Trng thi kh

2/12/2011


28/63

HA HC


12/12/2011

TRANGTRANGTRANGTRANGTRANGTRANGTRANGTRANG THAI KHTHAI KHTHAI KHTHAI KHTHAI KHTHAI KHTHAI KHTHAI KH2/12/2011

..

1 atm = 760 mmHg =101,3 kPa =1,013 bar = 760 torr 1 bar =105 Pa

3

a. p sut kh

Khi nim: lc tc dng ca cc phn t kh ln thnh

bnh cha trn 1 n v din tch

Biu thc:

n v o:

Mi lin h quy i:

P, KP N/2

,

H

2/12/2011

b. Nhit Khi nim: l tnh cht vt l ca vt cht, (l thang o

"nng" v "lnh). Vt cht c nhit cao hn th nng hn

n v o:

Mi lin h quy i:

o T (K) = t (oC) + 273,15

o T (oF) = T (K)*1,8 459,67 = T (oC)*1,8 + 3,2.

Ch : Trong h o lng quc t, nhit c o bng

n v K.

Trong i sng Vit Nam v nhiu nc, n c o

bng C (1 C bng 274,15 K). Trong i sng nc Anh, M v mt s nc, n

o bng F (1 F bng 255,927778 K).

C K F

2/12/2011

5

C : H L F C?

HD: 43 F 22,1oC

2/12/2011

..

12.2..

N :

B : P 1/ V

( )

H :(12.1)1 2

2 1

P V

P V=

Robert Boyle (1627 -

C : B?2/12/2011


29/63

12.2..

N : ,

,

.

B : V ~ T

(n v P khng i)

H :1 1

2 2

V T

V T=

(12.2)

Jacques Charles (1746-1823)

7

Cu hi: Mt qu bng c bm y kh heli n th tch 45lt

nhit phng (25oC). Lm lnh qu bng n -10oC, th

tch mi ca qu bng l bao nhiu? Coi p sut khng i2/12/2011

12.2.c. nh lut kh tng qut:

Ni dung: l s kt hp nh lut Boyle v nh lut Charl

Biu thc: V ~ T v V ~ 1/P (n khng i)

H qu :

Phng trnh ny c gi l nh lut kh tng qut, ha

nh lut kh kt hp.

1 1 2 2

1 2

P V P V

T T= (12.3)

Cu hi: Ta c mt bnh tr 22 lt cha kh Heli p sut l

atm v nhit 31oC. Ta c th bm c bao nhiu qu

bng, mi qu 5 lt khi p sut bn ngoi l 755 mmHg v n

l 22oC?2/12/2011

12.2.d. nh lut Avogadro

Ni dung: nhng th tch kh bng nhau cng mt iu

kin nhit v p sut s c s ht bng nhau.

Biu thc: V ~ n

(T v P khng i)

H qu : trong phn ng cht kh, tnh theo th

tch hoc s mol u c.

1 1

2 2

V n

V n=

92/12/2011

12.2.e. nh lut kh l tng

Ni dung: l s kt hp ca 3 nh lut: Boyle, Charles, v

Avogadro.

Biu thc: V ~ n/(T.P) PV = nRT (12.4).

Ch : R = 0,082 (atm.l/mol.K) khi P (atm)

R = 8,413 (J/mol.)

R = 1,987 (cal/mol.)

R = 62,32 (mmHg.l/mol.) khi P (mmHg)

2/12/2011

Kh l tng: l mt loi cht kh tng tng cha cc ht ging nhau c

kch thc v cng nh so vi th tch ca khi kh v khng tng tc vi nhau,

chng ch va chm n hi vi tng bao quanh khi kh.

Phn loi kh l tng:

kh l tng c in, tun th thng k Maxwell-Boltzmann

kh l tng lng t tun th thng k Bose.

kh l tng lng t tun th thng k Fermi .

Kh thc: l cht kh c k n kch thc ca phn t v th nng tng tc

gia chng.

c im kh thc: Khi kh thc trng thi p sut cao v nhit thp,

lc tng tc gia cc ht trong kh (cc phn t hay nguyn t) c nh hng

ng k trong cc tnh cht ca kh112/12/2011

12.2. f. Khi lng ring ca kh

Khi nim: l mt c tnh v mt ca vt cht , l i

lng o bng thng s gia khi lng m ca vt cht

(nguyn cht) v th tch V ca vt.

Biu thc:

H qu: tnh khi lng mol phn t khi bit P, V, T, m ca .

Cu hi: Khi lng ring ca mt kh cha xc nh l 5,0

g/lt 15oC v 745 mmHg. Hy tnh khi lng mol phn t

m P.Md

V R.T= =

(12.5)

PVn

RT=

mM (g / mol)

n=

2/12/2011


30/63

..

13

D: ,

,

.

AA

h

nX =n

B :

P = P1+ P2+ P3+ (12.6)

H :

:

PA= XA.P (12.8)

John Dalton (1766-1844)2/12/2011

Cu hi: Khi trn 71 gam kh Clo vi 64 gam kh oxi. Nu

sut ton phn ca hn hp l 800 mmHg.

a) p sut ring phn ca mi kh l bao nhiu?

b) Nng phn mol ca mi kh l bao nhiu?

HD:

Xc nh xem c phn ng ha hc xy ra ko?

Tnh s mol kh clo, s mol kh oxi, t tnh s mol hh p dng cng thc tnh p sut ring phn, nng ph

mol:

P= X.P A

A

h

nX =

n2/12/2011

..

15

12...

Cht kh c cu to t cc phn t ring r, c kch thc rt

nh so vi khong cch gia chng.

Cc phn t kh chuyn ng hn lon khng ngng chuyn ng

ny cng nhanh th nhit cht kh cng cao.

Khi chuyn ng hn lon cc phn t kh va chm vo nhau v va

chm vo thnh bnh v khng mt nng lng.

Mi phn t va chm vo thnh bnh tc dng ln thnh bnh mt

lc khng ng k, nhng v s phn t kh va chm vo thnhbnh tc dng ln thnh bnh mt lc ng k, lc ny gy ra p

sut ca cht kh ln thnh bnh.2/12/2011

12...

: E= (.2)

ng nng trung bnh ca mt tp hp nhiu phn t kh:

T thc nghim, ng nng trung bnh lin h vi nhit theo

thc:

Tc bnh phng trung bnh:

2

d

1E mv

2=

d

3E R.T

2=

(R = 8,3145 J/.K

2 3RTC vM

= = (12.9) M: /

C: /2/12/2011

S

2/12/2011 17

S

2/12/2011


31/63

..

19

Khi nim khuych tn:l s ha trn cc phn t ca

hai hoc nhiu kh do chuyn ng hn lon ca cc

phn t kh vo nhau thnh mt hn hp ng nht.

Khi nim phng lu: l s chuyn ng ca mt kh

thng qua khe h rt nh t bnh cha ny sang bnh

cha khc trong iu kin p sut thp.

T

N2

H2

2/12/2011

H qu: tc phng lu ca mt kh t l nghch vi c

bc hai ca khi lng phn t kh (nh lut Graha

Biu thc nh lut Graham:

T 1

T 22

1

M

M

Thomas Graham (1805-182/12/2011

21

Cu hi: Cho hai bnh kh cha ln lt l nguyn t Heli v

phn t N2 25 oC.

a) Hy tnh tc C ca 2 kh trn.

b) Bnh kh no c tc phng lu nhanh hn?

HD:

S dng cng thc:

Vi MHe = 4.10-3kg/mol, MN2 = 28.10-3 kg/mol, gi tr

R = 8,3145 J/mol.K, T = 25 + 273, ta thu c kt qu l..

S dng cng thc ca Graham so snh tc phng

lu 2 kh trn

2 3RTC vM

= =

2/12/2011

Bi sau: Chng 13

BT chng 12

1, 11, 23, 27, 31, 39, 41, 47, 59, 65.

,

2/12/2011

232/12/2011


32/63


33/63

8

Cu hi: Hy so snh nng lng

hirat ca qu trnh ha tan NaCl,

CsCl v MgCl2 vo dung mi nc.

Gii thch?

Cu hi: Gia F- v Cl-, ion no cnng lng hirat ha m cao

hn? Gii thch ngn gn ti sao.

13.2.b. Lin kt gia: lng cc bt bin - lngcc bt bin

9

Xut hin khi ha tan hp cht lng cc vo mt d

mi lng cc khc. VD: ha tan kh amoniac

nc.

So snh v bn lin kt:

lin kt ionlng cc > lin kt lng cc-lng c

lin kt lng cc-lng cc cm ng

Dng gii thch cc tnh cht v nhit si, n

lng bay hi, kh nng ha tan .ca cc cht.

2/12/201

10 2/12/2011

Quy lut: phn cc ca phn t lng cc cng ln th li

kt lin phn t cng bn, nng lng bay hi cng

cao (Hbh) v im si cng cao(Tos).

Cc cht cng tnh phn cc th d ha tan vo nha

cht phn cc d ha tan vo dung mi phn cc

Cu hi: Hy so snh cc lin kt lin phn t xut

khi ha tan kh H2S vo dung mi nc?

Cu hi: Hy so snh kh nng ha tan ca kh oxi,

hiroclorua v kh amoniac vo dung mi nc?

11 2/12/201

13.2.c. Lin kt gia: lng cc bt bin - lngcc cm ng

12

Xut hin khi ha tan hp cht lng cc vo mt dungmi khng cc hoc ngc li. VD: ha tan kh Cl 2 vo

nc.

So snh v bn lin kt:

lin kt lng cc lng cc > lin kt lng cc-cm

ng > lin kt cm ng-cm ng

Xut hin hin tng phn t ang l khng cc, do nh

hng ca phn t c cc tr thnh c cc tm thi (do

cm ng), hin tng ny gi l lng cc ha (hay phn

cc ha)2/12/2011

Quy lut:

Khi lng phn t cng ln th kh nng phn cc c

cao (cng d b phn cc cm ng).

Kh nng ha tan ca kh (m phn t khng phn c

l thun vi khi lng phn t ca chng.

Cu hi: Hy so snh kh nng ha tan ca kh O

kh Hiro, kh Metan v kh Clo vo dung mi nc13 2/12/201


34/63

13.2.d. Lin kt gia: lng cc cm ng- lngcc cm ng (lc phn tn London)

14

Xut hin khi ha tan hp cht khng cc vo mt dung

mi khng cc. VD: ha tan kh Cl2vo benzen lng.

So snh v bn lin kt:

L loi lin kt yu nht trong cc loi lin kt lin phn t

Xut hin hin tng phn cc tm thi (b ng) do s

chuyn ng lin tc ca m my e trong phn t.

2/12/2011

13.2.e. Lin kt hidro

15

Khi nim: l lin kt gia nguyn t H ca mi lin k

X-H v Y; trong X, Y l nhng nguyn t c m

in cao (O, N, F); cn Y thng c cp e cha lin k

Bn cht lin kt hiro l mt loi lin kt lng cc

lng cc c bit.

So snh v bn lin kt:

lin kt ion-lng cc> lin kt hiro > lin kt lng c

lng cc Lin kt hidro c th xut hin gia cc phn t vi n

hoc trong mt phn tn nh hng n Tos, Tonc,

nng ha tan v gy nn tnh cht bt thng ca n

v khi lng ring.2/12/201

Cu hi: Ch ra cc loi

lin kt lin phn t c

th c khi ha tan ng

thi kh HCl, HF v O2

vo dung mi nc?16 2/12/2011 17

13.2.f. Tng kt cc loi lk lin phn t Cu hi: chng ta va hc my loi lin kt lin ph

t? K tn cc loi ?

Cu hi: bn cc loi lin kt c xp theo t

t gim dn nh th no?

lin kt ionion > lin kt ionlng cc > lin kt hi

> lin kt lng cc-lng cc > lin kt lng c

lng cc cm ng > lin kt cm ng-cm ng

2/12/201

13.3. Mt s tnh cht ca cht lng

18

13.3.a. S bay hi

Khi nim s bay hi:

Nng lng ca qu trnh bay hi: Hbh.

Khi nim ngng t:

c im: nng lng ca qu trnh bay hi ph thuc

cht ch vo cc lin kt lin phn t c trong cht lng

.

Cu hi: So snh nng lng bay hi ca cc cht H2O,

HI, HF, HCl, HBr?2/12/2011

13.3.b. p sut hi bo ha

19

Khi nim: l p sut hi cn bng ca gia pha lng

pha hi khi n c thit lp trong h kn, trong h lc

tc bay hi bng vi tc ngng t.

c im: p sut hi bo ho (ti nhit cho trc)

cng cao th kh nng bay hi ca hp cht cng ln.

p sut hi bo ha tun theo pt kh l tng:

P.V = n.R.T

Phng trnh Clausius-Clapeyron a ra quan h p s

hi bo ha v nhit bay hi:o

bhbh

Hln P C

RT

= +

(13.1)

2/12/201


35/63

Cu hi: Hy so snh nhit si ca ietylete,

etanol v nc p sut bo ha 400mmHg?20 2/12/2011

13.3.c. im si

21

Khi nim: l nhit m ti cht lng c p su

bo ha bng p sut bn ngoi.

c im: im si ca cng mt cht s cng thp n

p sut bn ngoi ca qu trnh si cng thp..

13.3.d. Nhit v p sut ti hn

Nhit ti hn l nhit ca mt cht ti thi im t

hn (thi im m s phn chia gia cht lng v dng

bin mt)

p sut ti hn l p sut hi bo ha ti thi im ti

Cht tn ti trng thi ny gi l cht lng siu ti h2/12/201

13.4. Trng thi rn

22

Lin kt ha hc trong cht rn thng c lin kt ion, lin

kt kim loi, lin kt cng ha tr, lin kt hiro, lin kt

lin phn t khc

Dng cu trc: mng tinh th ion, tinh th nguyn t, tinh

th phn t, mng polime

Khi nim mng tinh th: l mt s sp xp c bit ca

cc nguyn t, phn t hoc ion trong mt mng li

khng gian 3 chiu theo cc quy lut nht nh.

2/12/2011

n v: l hnh khi khng gian nh nht, c tnh ixng theo s sp xp ca cc nguyn t, ion, phn t

trong tinh th .

c im: nu ta lp li n v theo mi hng(b

cch tnh tin) th n s chim y khng gian v s

nn ton tinh th

S ng gp ca 1 qu cu no vo n v l:

Qu cu trong lng ca n v ng gp 1 qu

Qu cu b mt ca n v ng gp qu

Qu cu cnh ca n v ng gp qu

Qu cu gc ca n v ng gp qu23 2/12/201

Khi sp xp cc qu cu ng dng trong t nhin, chng

thng to ra 3 loi hnh c bn sau:

Lp phng n gin = sc.

Lp phng tm khi = bcc.

Lp phng tm mt = fcc.

C 3 loi hnh ny u c cc qu cu nh, ngoi ra

cn c loi hnh nh: hcp, ccp24

2/12/2011

Lp phngn gin (sc)

Lp phngtm khi (bcc)

Lp phngtm mt (fcc)

- S qu cu c mt trong mt n v ca mi loi hnh:

Dng cu trc S qu cu c trong mi n

Lp phng n gin 1 = 8*()

Lp phng tm khi 2 = 1 + 8*()

Lp phng tm mt 4 = 6*() + 8*()2/12/201


36/63

Vi hp cht ion:

cu trc ph thuc kch thc

cc ion v thnh phn cu to

ca hp cht.

Cu trc ca a s l sc hoc

fcc ca cc anion, cn cc

cation t trong hc mng.

Cc loi hc mng: Hc t din: 4 qu

cu bao quanh.

Hc bt din: 8 qu

cu bao quanh

Hc btdinHc t

din

26 2/12/2011

Cch suy ra CTPT ca hp cht t 1 t bo cho

trc (bng hnh v hoc m t bng li):

V ng cu trc ca t bo (theo m t)

Tnh s ion (+) v (-) trong mt t bo

Rt gn t l nguyn t (hoc ion) ri kt lun cng thc

Cu hi: Xc nh cng thc

ng cho hp cht to nnt cc ion A v B theo hnh

v bn. Gii thch?

27 2/12/201

Cu hi: Xc nh cng thc ng cho hp cht tonn t cc ion A v B theo hnh v bn. Gii thch?

HD: S ion A = 8.1 = 8 ion.S ion B = 8. + 6. = 4 ion

28 2/12/2011

13.5. Gin trng thi ca nc

29 2/12/201

30

Gin trng thi l hnh minh ha mi lin h gia

cc trng thi ca vt cht (r,l,k) cc iu kin nhit

v p sut khc nhau.

Cc ng cong trong gin l iu kin tn ti

trng thi cn bng gia hai pha.

Cc im khng nm trn ng cong ch ra iu

kin ti tn ti 1 pha duy nht no .

im giao ca 2 ng cong l im ba, im xc

nh iu kin c cn bng ca c 3 trng thi vt

cht. 2/12/2011

Bi sau: Chng 14

Dung dch v tnh cht ca d

BT chng 13:

5, 7, 9, 13, 15, 25, 27, 39, 49, 53

31

2/12/201


37/63

HA HC I CNG

G: L Mih Thh

2/12/2011 1

DUNG DCH &

TNH CHT DUNG DCH

CHNG 14

2/12/2011

14.1. M

1) Nng mol/l (CM): l i lng biu din s mol cht

tan c trong mt lt dung dch.

o Biu thc:

o n v o: (mol/l) hoc (M)

2) Nng phn trm (C%): biu din t l phn trm ca

khi lng cht tan so vi khi lng dd.

o

Biu thc: o n v o: (%)2/12/2011 3 2/12/2011

3) Nng molan (Cm): l i lng biu din s mol cht tac trong mt kg dung mi.

oBiu thc:

on v o: (m) hoc (molan)

4) Nng phn mol (Xi): l t s v s mol ca cht cn tnh

chia cho tng s mol cc cht c trong dd.

oBiu thc:

on v o:

5) Nng phn triu (Cppm): l t s v khi lng ca cht

tnh chia cho khi lng dung mi.

oBiu thc:

on v o: (ppm) hoc (mg/l).

2/12/2011 5

6) Nng ng lng (CN): l i lng biu din s ng

lng cht tan c trong mt lit dung dch.

oBiu thc:on v o: (N)

Ch .

oTng nng phn mol (Xi) ca tt c cc cht trong h = 1.

XA+ XB+ XC + XD + = 1

oS ng lng c tnh l:

ong lng mt cht c tnh l:

2/12/2011

7) Lin h gia cc loi nng hc.

oGia C% v CM:

oGia CNv CM:

C :Nu ha tan 10 g ng (C12H22O11) vo mt c

nc (250 g). Hi nng mol/l, nng phn mol, n

molan v nng phn trm. Cho DH2O= 1 g/ml.

HD: nct, Vdd CM; Cm; mdm; ndm Xct,Xdm;

mdm, mct C%.


38/63

2/12/2011 7

C :Cho dung dch H2SO40,5M. Hy tnh nng phn

mol, nng molan, nng ng lng v nng

phn trm. Cho Ddd= 1,1 g/ml.

HD: xt Vdd= 1 lt;

Vdd ; CMnctmct;

Ddd; Vdd mdd mdm ndm Xct,Xdm;

nct

; mdm

Cm

;

n*; CM CN;

mct; mdd C%

2/12/2011

Cc khi nim khc lin quan n dung dch.

o a (S): l g ch a khi ch a c b

i lg ch khg a g dg dch.

n v o ca tan: (M, mol/lt, g/lt, mg/lt).

oDung dch bo ha:

oDung dch qu bo ha: trong lng cht tan nhiu h

so vi trong lng dung dch bo ha.

oDung dch cha bo ha:

14.2. C a a14.2.a. S a a a

Qi l hg dg: Cc ch gig ha ( h cc) h

ha a ha.

VD:

oEtanol lng ha tan mi t l trong trong nc, v chng

u l cht phn cc, hn na chng to ra c lk Hidro

lin phn t.

oOctan C8H18 lng v cloroform CCl4lng u khng ha tan

trong nc, v chng l cc cht khng phn cc, dmi nc

phn cc.

oOctan C8H18li cng ha tan trong CCl4 vi mi t l ,v ....2/12/2011 9 2/12/2011

14.2.b. S a a Qi l cc ch gig ha h ha a ha cg c

dg ch ic ha a ch g lg.

Qui lut trn i lc khng hiu qu, nhng vn c dng cho c

cht cha ion. VD: AgCl ko tan trong nc

Vic to thnh lin kt hidro gia phn t dung mi vi phn t c

tan lun lm tng kh nng ha tan. Cc cht c lk hidro thng c

nhm OH, NH, FH v to lk hidro vi nc

VD:

o It, I2, mt cht rn khng phn cc, ha tan hn ch trong n

nhng ha tan tt trong dung mi khng phn cc nh CCl4

.

o ng sucro, mt cht rn phn cc, khng tan trong dung m

khng phn cc, nhng tan tt trong nc

2/12/2011 11

14.2.. N a a.

Xt qu trnh ha tan cht rn tinh th ion vo nc:

oBan u cc nh th cht rn b tch thnh cc ion n l(thnh pha kh). Nng lng cn cung cp ph v lin kt

gia cc ion l Hnh th.

oSau , cc ion ny b hyrat ha, tc l to ra l.k gia ion v

cc pt nc. Nng lng ca qa trnh ny l Hhyrat ha.

oNng lng ca qu trnh tng c gi l nhit ca dung

dch v l tng ca hai nng lng thnh phn:

Hdung dch= H nh th+ Hhyrat ha

Hodung dch= [Ho

f,cht sn phm] [Ho

f, cht ban u]2/12/2011

Nhn xt:

oi lng Hnh thv Hhyrat ha thng c gi tr m.

oa s qu trnh ha tan cc mui l ta nhit, ch c mt s t

trnh l thu nhit.

oC H nh thv H hyrat hau b nh hng bi kch thc v

in ch ca ion. Ion c kch thc nh hn s c th c nng

lng mng li nh th H nh thln hn v nng lng son

Hhyrat haln hn.

C : Trong cc mui sau, mui no c nhit hirat ha cao nh

Li2SO4, Na2SO4, K2SO4, Cs2SO4?

C : Sp xp tan cc lng sau y trong nc theo th t

dn: C6H6, Br2, C2H5OH, CH3COOH?


39/63

14.3. C a a :

14.3.a. a :

Tun theo nh lut Henry: ha a ca 1 ch kh

g ch lg l h i ig ca kh .

Biu thc: Si= kH.Pi n v o ca P (mmHg v S (mol/l).

2/12/2011 13

Ch : h l He ch ghi

g g g h ch a

khg h g ha hc i

dg i. 2/12/2011

14.3.b. N a a

Tun theo nguyn l Le Chatelier: Khi c ha i ca b

h hg c bg, h h ha i h

hg a ch l gi h hg ca c g .

Kh + dung mi lng dung dch bo ha H


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2/12/2011 19

p sut thm thu () l p sut thy tnh c to ra bi

2 mi trng c nng khc nhau..

Biu thc: = C.R.T (14.10)

(R = 0,082057 l.atm/K.mol, th c n v l atm)

2/12/2011

14.4.. X KLPT (M) a a a

ct

cht tan s

s dm

mM K .

T .m=

cht tan

ct

dm

mM K .

T .m=

cht tanct

dd

mM

.V=

C :Alee l hicacb c cg hc hc ghi C

Ha a 0,640 g ih h ch g 99,0 bee, dg dch h

c c i i l 80,23C. Xc h cg hc h ca al

M lieile, ch d h bi, c khi l

1,40 g c ha a h g bee a 100 l dg

h h ca dg dch c bg 1,86 Hg 25

Th khi lg l h g bh ca h ch lie .

2/12/2011 21

14.5. T a a a , ba

14.5.a. T b a .

gim p sut hi bo ha, tng nhit si, gim nhit

ng c v p sut thum thu ca dung dch cht in li ln

hn so vi ca dung dch cht khng in li c cng nng

molan.

Ng h: ..

a ra h s Vant Hoff (i):

2/12/2011

14.5.b. C Cc cng thc cho dd in ly s l phng trnh Raoult v

Vant Hoff hiu chnh ( nhn thm h s Vant Hoff (i)

o P = P P

= . X. P

o T, =T T = . K. C,

o T, = T T = . K. C,

o = .C.R.T

Ch : (i) gn st cc gi tr nguyn (2, 3, ) ch khi dung

dch rt long.

VD: gi tr ca i= 2 i vi NaCl, i = 3 i vi Na2SO4, i =

i vi AlCl3v i = 5 i vi Al2(SO4)3.

2/12/2011 23

C : Trong cc dung dch sau, dung dch no c nhit

ng c thp nht: MgCl20,05m (A); ng saccaroz 0,15m

(B); nhm sunfat 0,05m (C) v etylenglicol 0,2m (D)? HD:

o Tnh gim nhit ng c cho tng dung dch.

T, A = 3. K. 0,05 T, B = 1 . K. 0,15

T, C = 5 . K. 0,05 T, D= 1 . K. 0,2

(cng mt dung mi, nn cng mt gi tr K )

o Tnh nhit ng c: T dd = T,li+ T dm .

o Vy dung dch c nhit ng c

thp nht l:..

C a

Bi sau: Chng 15

BT chng 14

3, 17, 21, 27, 31, 35, 39, 49, 51, 59..

2/12/2011


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2/12/2011 1

HA HHA HHA HHA HCCCCI CI CI CI CNGNGNGNG

GV: L Minh Thnh

2/12/2011

15

2/12/2011 3

15.1.

Tc ca phn ng ho hc : l s bin thin nng ca

cc cht trong mt n thi gian.

Xt p/: aA + bB cC + dD, biu thc n tc trung bnh

n ca n tc: (mol/l.thi gian)

Tc tc thi ca phn ng c c nh bng bin

thin nng ti mt thi im (t).

pu

C 1 [A] 1 [B] 1 [C] 1 [D]V

t a t b t c t d t

= = = = + = +

2/12/2011

Biu thc tc tc thi:

Ch : tc ring phn ca cht tham gia (hoc to

thnh) trong phn ng c khc i khi nim tc

Biu thc tc ring phn ca mi cht l:

Tc tc thi c tnh bng dc ca ng thngtip tun i ng cong (C, t) ti thi im t

dt

dCv AA =

dt

dCv BB =

dt

dCv CC +=

dt

dCv D +=

dt

dC

ddt

dC

cdt

dC

bdt

dC

adt

dCv DCBApu

1111+=+====

2/12/2011 5 2/12/2011

Cu hi: ng saccaro phn hu thnh flucto gluco

trong mi trng ait. th biu th s bin thin ca nng

theo thi gian c a ra di :

a) H c nh tc

bin i ca nng

ng saccaro sau 2 gi

u tin?

b) H c nh tc

bin i nng ng

saccaro ca 2 gi cui

cng. H c tnh tc

tc thi ti t = 4 gi?


42/63

2/12/2011 7

15.2.

iu kin c bn phn ng xy ra: cc phn t cht phn

ng phi va chm vi nhau.

Bn cnh , tc phn ng ph thuc vo cc k nh:

Nng cht tham gia phn ng: nng cht t/gia tng

Nhit : nhit phn ng tng

Cht xc tc: ty vo xc tc dng hay m

Din tch tip xc (vi p c cht rn tham gia)

V d: C6H6 + HNO3 C6H5NO2 + H2O

Nng cht tham gia phn ng: .

Nhit : nhit phng (25oC)

Cht xc tc: (Cu(NO3)2.3H2O trn bentonit)2/12/2011

15.3.

Biu thc ca nh lut tc dng khi lng:

aA + bB sn phm Vp = k.[A]m.[B]n

S m m v n khng nht thit phi l h s t lng tron

phng trnh phn ng (a v b), n c th l s dng,

m hoc 0.

Phng trnh trn cn gi l phng trnh ng hc ca

phn ng (hoc phng trnh tc ca p). Bc ca phn ng chnh l s m ca nng cht phn

trong biu thc tc phn ng.

Bc phn ng ca A l m, ca B l n, ca c phn ng l

(m+n)

2/12/2011 9

Cu hi: Cho phn ng: 2NO(k) + Cl2 (k) 2NOCl (k). C

biu thc vn tc l: V = k.[NO]3/2.[Cl2]2/3. Xc nh bc

ring phn v bc ton phn?

HD:

o Bc ring phn ca NO l:

o Bc ring phn ca Cl2 l:

o Bc ton phn ca c phn ng trn l: .

2/12/2011

Hng s tc k l h s t l c lin quan vi h s v

nng ca cc cht ti mt nhit xc nh (cn g

tc ring ca phn ng).

n v o ca k ph thuc vo bc ton phn ca p:

o Phn ng bc 1 hng s k c n v l: thi gian-1

o Phn ng bc 2 hng s k c n v l: l/mol.thi gia

o Phn ng bc 0 hng s k c n v l: mol/l.thi gia

V d: phn ng: CO (k) + NO2(k) CO2(k) + NO (k

o Vn tc phn ng: v = 6,8.10-8 (mol/l.h)

o Hng s tc : k = 1,9 (l/mol.h)

o Vy phn ng trn c bc l 2.

2/12/2011 11

Biu thc tc phn ng c c nh bng thc nghim

: Cho phn ng: aA + bB cC + dD

H c nh biu thc tc phn ng?

Vp= k.[A]m.[B]n

TN Nng u ([A]0) Nng u ([B]0) Tc ban u (0)

1 1.00 M 1.00 M 1.25 102 M/s

2 1.00 M 2.00 M 2.5 102 M/s

3 2.00 M 2.00 M 2.5 102 M/s

2/12/2011

15.4. :

Phng trnh tc theo thi gian ca phn ng bc 1 sn phm) s l:

Phng trnh tc theo thi gian ca phn ng bc 2

sn phm; hoc A + B sn phm ) s l:

Phng trnh tc theo thi gian ca phn ng bc 0

sn phm) s l:

[ ]

[ ]t

0

Aln kt

A= (15.1)

[ ] [ ]t 0

1 1kt

A A = (15.2)

[ ] [ ]0 t

A A kt = (15.3)


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2/12/2011 13

T cc biu thc ca pt tc trn, a ra cch xc nh

bc phn ng m, n v hng s tc k theo pp th:

Thi gian bn hu (t1/2) ca phn ng l thi gian cn thit

nng cht phn ng gim mt na so vi nng

ban u.

phn ng bc 1, d dng tnh c:

Bc p Biu thc tc phn ng

Biu thc bin ica tc p

thng thng

0 [A]t= kt + [A]o = a + b [A]t thi gian

1 ln[A]t= kt + ln [A]o = a + b ln[A]t thi gian

2 1/[A]t= +kt + 1/[A]o = a + b 1/[A]t thi gian

kt

693,02/1 =

2/12/2011

Cu hi: S liu ca phn ng phn hu N2O5trong dung m

45oC c a ra trong bng di y?

Hy v th tm, bc ca p. v hng s tc ca p

HD:

biu din s bin thin [N2O5] theo t.

biu din s bin thin ln[N2O5] theo t

biu din s bin thin 1/[N2O5] theo t.

mt trong 3 th trn s l ng thng bc p k

25, (/) , ()

2,08 3,07

1,67 8,77

1,36 14,45

0,72 31,28

2/12/2011 15

Cu hi: ng saccaro b phn hu trong mi trng axit tothnh glucoz v fructoz. l phn ng bc mt vi hng

s tc k = 0,21 (h-1) 25 oC. Nu nng ban u ca

ng saccaro l 0,010 (mol/l), hy tnh nng ca n cn li

sau 5,0 (h)?

HD:

Biu thc tnh ca p/ bc 1:

Theo bi ta c [sacca]o = 0,01 M, k = 0,21 (h -1), t=5 (h)

lp vo biu thc trn, gii pt logarit mt n bc nht ta tnh

c [sacca]5h =

[ ]

[ ]t

0

Aln kt

A=

2/12/2011

15.5. 15.5.)

Thut a chm ca phn ng nu ra ba iu kin cn c:

c a chm ra gia cc pt tham gia;

c mc nng lng ln trong cc pt tham gia;

hng a cham phi thun li cho ic ph to lk.

15.5.)

Khi tng nhit thng lm tng tc phn ng do nhit

lm tng t l s phn t c nng lng t qua hng r

nng lng hot ho.

nh lut Vant Hoff: khi nhit tng 10o, tc p thn

tng ln t 2 4 ln.

2/12/2011 17

15.5.c) Yu t nng lng hot ha

Nng lng hot ha (Ea) l nng lng d ti thiu m h

phn ng cn phi c so vi nng lng trung bnh ca h ban

u phn ng c th xy ra.

Nu gi tr Ea cng ln, th phn ng cng kh xy ra, v s

pt tham gia t ti gi tr Ea cng t i, dn n s va chm

hiu qu khng nhiu.

Ngc li, Ea cng nh, phn ng cng d xy ra.

2/12/2011

15.5.d) Yu t xc tc

Xc tc l cc cht c kh nng lm thay i tc phn

ho hc. Phn loi xc tc: xc tc dng v xc tc m.

C ch chung ca cht xc

tc dng l bin mt phn

ng t giai on (mi giai

on c nng lng hot ha

rt cao) thnh mt phn ng

c nhiu giai on (vi mi

giai on c nng lng hot

ha thp hn nhiu)


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2/12/2011 19

15.5.e) Yu t bc phn ng

Bc phn ng c xc nh l s lng cc phn t (hoc

ion, nguyn t, gc t do) ca cht phn ng.

Bc phn ng cng ln th xc sut va chm hiu qu ca

ng thi s phn t tham gia p cng thp, p cng kh

xy ra:

Phn ng bc 1, bc 2 rt thng gp trong t nhin

cng nh trong cuc sng, l cc p/ tng i d xy ra

P bc 3: s va chm cng mt lc ca ba phn t rt

kh xy ra tr khi mt trong cc phn t c nng

ln.

P bc 4 tr ln hu nh l khng th xy ra c2/12/2011

15.5.f) S tng hp cc yu t: Phng trnh Arrheni

S ph thuc ca tc phn ng (v) vo cc yu t

(nng lng, tn sut va chm, nhit , nh hng h

hc ca cc va chm) c tm tt trong biu thc ca

Arrhenius.

aE / R Tk A.e = (15.5)

Ea 1

ln k ln AR T

= +

22 1

1 2 1

k Ea 1 1ln k ln k ln

k R T T

= =

(15.7

2/12/2011 21

Bi sau: Chng 16

CN BNG HA HC

B 15:

5, 11, 15, 19, 25, 27, 33, 63, 65, 71.

2/12/2011


45/63

HA HC


12/13/2011

16

2/13/2011

16.1.

Khi nim v phn ng thun nghch: .

Khi nim v cn bng ha hc: l 1 trng thi ca phn

ng thun nghch m vn tc phn ng thun bng

vn tc phn ng nghch.

c im ca trng thi cn bng ha hc:

+ l mt trng thi ng:

+ c tnh linh ng:.

+ khng tha i theo thi gian nu .

+ c th xc lp theo 2 chiu, 32/13/2011

16.2.

Xt p/ : a A (dd) + b B (dd) cC (dd)+ dD (dd)

Phn ng thun c: vt= kt. [A]a.[B]b.

Phn ng nghch c: vn= kn. [C]c.[D]d.

Khi t ti trng thi cn bng th vt= vnnn t s kt/kn=

c gi l hng s cn bng (theo nng ):

c d

C a b

[C] .[D]

K [A] .[B]=

2/13/2011

Trong trng hp phn ng c s tham gia ca cht kh, c

th tha nng bng p sut ring phn.

Xt phn ng: a A (k) + b B (k) cC (k)+ dD (k)

Hng s cn bng KP theo p sut

Lin h gia KCv KP: (vi n=c+d-a-b)

:Khi no th gi tr Kpv gi tr Kclun bng nhau

trong mt phn ng ha hc bt k?5

c d

C DP a b

A B

[P ] .[P ]K

[P ] .[P ]=

( )p cK K RT

=

2/13/2011

c im ca hng s cn bng Kchoc Kp:

1. N ,

2. G K

.

3. K .

4. N

,

5. K >> 1: , K


46/63

7

2:H thit lp cc biu thc hng s cn bng KCv KPca cc phn ng sau:

a) N2(k) + O2(k) 2NO(k)

b) C2H5OH(dd)+ CH3COOH(dd) CH3COOC2H5(dd)+ H2O(l)

c) C(r)+ O2(k) CO(k)

d) AgCl(r) Ag+

(dd)+ Cl(dd)

e) AgCl(r)+ 2NH3(dd) Ag(NH3)2+

(dd)+ Cl(dd)

3:H so snh quan h gia K1, K2, K3ca cc p sau:

a) N2 (k) + 3H2 (k) 2 NH3 (k) K1.

b) N2 (k) + /2 H2 (k) NH3 (k) K2.

c) 2 NH3 (k) N2 (k) + 3H2 (k) K3.2/13/2011

Trong trng hp phn ng khng trng thi cn bn

th t s ca nng sn phm chia cho cht tham gia,

c gi l t s phn ng.

Xt phn ng: a A (dd) + b B (dd) cC (dd) + dD (dd

T s phn ng Q:

( .)

c d

C D

a b

A B

C .CQ

C .C=

Mi lin h Q v K Chiu ca phn ng

Q < K Chiu thun

Q = K P/ t trng thi cn bng

Q > K Chiu nghch2/13/2011

16.3.

K phn ng tng nhiu phn ng = K1.K2.K3

K phn ng hiu ca phn ng (1) tr i (2) = K1/K2.

Khi nhn h s cn bng ca phn ng ln n ln

Kmi=(Kc)n.

Khi phn ng c vit ngc li Kmi= 1/(Kc).

Mi bi ton c s dng n hng s cn bng K u s

dng bng ICE tnh ton.

92/13/2011

:Cho phn ng: 2SO2(k) + O2(k) 2 SO3 (k) 1000K. Cho 1 mol SO2v 1 mol O2vo 1 bnh 1 lt phn

ng. trng thi cn bng c 0,8 mol SO3to ra. Tnh K?

[cn bng] = E 0,2 0,6 0,8

= (0,8)2

/(0,2)2

.(0.6) = 26,67

2SO2 (k) + O2 (k) 2 SO3 (k)

[ban u] = I 1,00 1,00 0

[phn ng] = C -0,8 -0,4 +0,8

2/13/2011

11

:Cho 2 mol H2v 3 mol I2vo bnh kn 1,0L 600oC.

Tnh s mol cc cht khi cn bng, bit K= 50,0?

E (M) 2-x 3-x 2x

[I2] = 1,21M ; [H2] = 0,21M; [HI] = 3,58M

x = 1,79

H2 (k) + I2 (k) 2HI (k)

I (M) 2 3 0

C (M) -x -x +2x

2/13/2011

16.4.

Cc u t thng gp: nhit , nng , p sut

S nh hng ca cc u t trn u tun theo mt q

lut chung, l tun theo ngun l chun dch cn b

Le Chartelier:

Cch xt cc nh hng n mt phn ng:

: H

: .

: ,

2/13/2011


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16.4.)

Xt phn ng: A + B C + D c H < 0

Chiu thun l chiu ta nhit.

Nu ta tng nhit ca h cn bng s chun dch

theo chiu chng li s tng nhit chun dch theo

chiu lm gim nhit chun dch theo chiu nghch.

16.4.)

Xt phn ng: A + B C + D

Nu ta tng nng ca cht C (cht sn phm) cn

bng s chun dch theo chiu chng li s tng nng

n chun dch theo chiu lm gim nng cht C

chun dch theo chiu nghch. 132/13/2011

16.4.)

Xt phn ng: aA (k)+ bB (k) cC (k)+ dD (k)c n = c+da

Nu n < 0, phn ng l gim p sut.

Nu ta tng p sut ca h cn bng s chun dch t

chiu chng li s tng p sut n chun dch theo

chiu lm gim p sut chun dch theo chiu thun

:Nu p c n = 0 , th p sut nh hng ntn?

16.4.)

i vi phn ng pha kh, th tch h gim i th p su

h tng ln, nh hng ca n ging nh p sut

i vi phn ng dung dch.

2/13/2011

15

: trng thi cn bng c 0,2 mol H2v 0,3 mol I2v0,1 mol HI trong bnh kn 1,0L 600oC. Thm vo h 0,5mol

HI, h tnh s mol cc cht khi cn bng mi c thit lp?

E (M) 0,2+x/2 0,3+x/2 0,6-x

H2 (k) + I2 (k) 2HI (k)

cb (M) 0,2 0,3 0,1

C (M) +x/2 +x/2 -xI (M) 0,2 0,3 0,1+0,5

(0,6x)2/(0,2+x/2).(0,3+x/2) =

x = 2/13/2011 16

Cu hi: Cho phn ng: N2 + 3H2 2NH3, c H < 0a) Thm xc tc, cn bng chuyn dch theo chiu no?

b) Thm NH3, cn bng chuyn dch theo chiu no?

c) Gim nhit , cn bng chuyn dch theo chiu no?

d) Tng p sut h, cn bng chuyn dch theo chiu no?

e) Tng th tch h phn ng ln (khng lm thay i s mol

cht), cn bng chuyn dch theo chiu no?

f) Trong cc yu t sau: nhit , nng , p sut, xc tc,

tch, din tch tip xc, bc phn ng ch ra yu t m g

nh hng ln:

tc phn ng?

cn bng ha hc?2/13/2011

17

Cu hi: Vit phn ng m t cn bng ha hc xy ra sau y

(hnh v). T phn ng , hy ch ra chiu chuyn dch cn

bng khi:

a) Thm NH4Cl vo dung dch hn hp Na2CrO4 v Na2Cr2O7.

b) Thm Na2CO3 vo dung dch hn hp Na2CrO4 v Na2Cr2O7.

c) Thm KCl dung dch hn hp Na2CrO4 v Na2Cr2O7.

2/13/2011

Bi sau: Chng 17

BT chng 16:

1, 5, 9, 17, 23, 25, 27, 29, 45, 51.

2/13/2011


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HA HC


12/13/2011

TTTTnh cht cu?a axit baznh cht cu?a axit baznh cht cu?a axit baznh cht cu?a axit baz

Chuong 17Chuong 17Chuong 17Chuong 17

2/13/2011

17.1. K Khi nim axitbaz ca Arrhenius :

Axit: l cht khi ha tan trong nc phn ly ra ion H3O+ (H+).

Baz: l cht khi ha tan trong nc phn ly ra ion OH

u im v nhc im ca thuyt c:

S in ly ca axit baz trong nc l mt qu trnh cn bng,

v c c trng bi hng s in ly.

VD: HA + H2O H3O+ + A. Ka=

B + H2O BH+ + OH. Kb=

32/13/2011

S phn loi axitbaz mnh

Axit, baz mnh: l axit baz c in ly =0,750,95 (K>

Axit, baz yu: l axit baz c in ly =0,050,75 (K


49/63

Khi nim v cp axitbaz lin hp: l mt cp axit/baz m

baz c to ra t axit tng ng, v ngc li.

Xt p: HA + H2O H3O+ + A.

(axit) (baz) (axit) (baz)

Theo chiu thun: cp axit/baz lin hp l: HA/A.

Theo chiu nghch: cp axit/baz lin hp l: H3O+/H2O.

C : Cho p: H2O + H2O

H3O+

+ OH

. Hy k tn cpaxit/baz lin hp trong phn ng trn?

C : Cho p: CO32 + H2O HCO3 + OH. Hy k tn cp

axit/baz lin hp trong phn ng trn?72/13/2011

17.3. S H

Qu trnh t ion ha ca H2O :

H2O (l)+ H2O (l) H3O+

(dd)+ OH

(dd) Kc=?

c im:

Kwgi l tch s ion ca nc.

Kw

= [H3O+].[OH] = 1,0.1014 ( 25oC)

Trong nc nguyn cht, trung tnh: [H3O+] = [OH] = 10

Mi trng axit: [H3O+] > [OH] .

Mi trng baz: [H3O+] < [OH].

+ -

3c 2

2

[H O ][OH ]K

[H O]= Kw= [H3O+] [[OH] = Kc[H2O]2

2/13/2011

Khi nim v pH, pOH: l mt i lng c trng cho nng mol/lt ca ion H+ hoc OH trong dung dch.

H qu:

pH = a [H+] = 10a v ngc li.

pOH = b [OH] = 10b v ngc li.

pKw= pH + pOH = 14,00 ( 25oC)

ngha ca ch s pH: xc nh mi trng ca dung dch (axit,baz, trung tnh).

pH = lg[H3O+] pOH = lg[OH]

92/13/2011 2/13/2011

2/13/2011 11

17.4. H Hng s cn bng ca cc axit, baz trong qu trnh in ly tro

dung mi, c gi l hng s axit, baz.HA (dd)+ H2O (l) H3O

+(dd)+ A

(dd)

B (dd)+ H2O (l) BH+

(dd)+ OH

(dd)

c im:

Gi tr Ka, Kbcng ln th axit, baz cng mnh v ngc

Gi tr ca Ka, Kbch ph thuc vo nhit v bn cht axb

Axit cng mnh th baz lin hp ca n cng yu v ngc l

Vi 1 cp axit baz lin hp ta c mi quan h gia Kav Kb:

- +

b

[OH ][BH ]K

[B]=

- +

3a

[A ][H O ]K

[HA]=

Ka .Kb= Kw = 1,0.1014

2/13/2011


50/63

Ch s axit pKa, ch s baz pKb: l mt dng khc ca hng s

axit, baz biu din theo hm logarit.

c im.

Gi tr ca pKa, pKbcng ln, axit hoc baz cng yu.

Vi 1 cp axit baz lin hp ta c mi quan h gia pKav

pKb:

C : Cho axit yu HX v HY c ch s baz pKbca baz lin

hp vi chng ln lt l 3 v 5. Hi axit no mnh hn?

pKa= lgKa pKb= lgKb

pKa + pKb= pKw = 14

132/13/2011

17.5. D Quy lut: axit mnh + baz mnh axit yu + baz yu.

V d:

HCl (dd)+ H2O (l) H3O+

(dd) + Cl

(dd)

Tnh axit > H3O+ Tnh baz > Cl Tnh axit < HCl Tnh baz < H2O

CH3COOH (dd)+ H2O (l) CH3COO

(dd)+ H3O+

(dd)

Tnh axit < H3O+ Tnh baz < CH3COO

Tnh baz > H2O Tnh axit > CH3C

C : P: CH3

COOH(dd)

+ HCO3

(dd) CH3

COO(dd)

+H2

CO

D on chiu phn ng?

Bit H2CO3 (Ka = 4,2.107 ), CH3COOH (Ka = 1,8.10

5) v CH3CO

(Kb = 5,6.1010), HCO3

(Kb = 2,4.108).

2/13/2011

17.6. M , Quy lut:

Mui to bi axit mnh v baz mnh cho mt trung tnh.

Mui to bi axit mnh v baz yu cho mt axit.

Mui to bi axit yu v baz mnh cho mt baz.

Mui to bi axit yu v baz yu cho mt ty thuc vo

tng quan gia ax yu v bz yu .

Nguyn nhn: do s thy phn ca ion cht yu c mt trong

dung dch vi dung mi nc, lm tng lng ion H3O+ hoc

OH trong dung dch mui v to mi trng

152/13/2011

17.6.. D Xut hin khi mui c cha anion ca axit mnh (Cl-, B

, NO3-, ClO4-) v cation ca baz mnh (Na+, K+, Li+, R

Cs+, Ca2+, Sr2+, Ba2+):

Nguyn nhn: C cation v anion ca mui u khng

phn ng thy phn vi H2O to ra H+ hoc OH-.

VD: dung dch mui cha NaCl:

NaCl (r) Na+ (dd) + Cl- (dd)

Ion Na+ v Cl- u khng thy phn trong nc khng c

thay i no pH = 7,00 mt trung tnh

2/13/2011

17.6.. D

Xut hin khi mui c cha anion ca axit mnh (Cl-, Br-, I-

, NO3-, ClO4-) v cation ca baz yu (NH4+; Fe3+) hoc

mt s mui axit nh NaH2PO4 :

Nguyn nhn: do s thy phn ca cation ca baz yu

vi H2O to ra nhiu H+ :

VD: NH4Cl NH4+ + Cl-

NH4+(dd) + H2O (l) NH3 (dd) + H3O+(dd)..

VD: FeCl3 + 6H2O [Fe(H2O)6]3+ + 3 Cl-

[Fe(H2O)6]3+(dd) + H2O (l) [Fe(H2O)5(OH)]2+(dd) + H3O+(dd)172/13/2011

17.6.. D

Xut hin khi mui c cha anion ca axit yu (F-, NO2

PO43-, CH3COO- ) v cation ca baz mnh (Na+; K+

Nguyn nhn: do s thy phn ca anion ca axit yu

H2O to ra nhiu OH- hn.

VD: NaCN Na+ (dd) + CN- (dd)

CN- (dd) + H2O (l) HCN (dd) + OH- (dd)..

VD: K2CO3 2K+ + CO32-

CO32-(dd) + H2O (l) HCO3-(dd) + OH-(dd)

HCO3-(dd) + H2O (l) H2CO3(dd) + OH-(dd)2/13/2011


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17.6.. D Xut hin khi mui c cha anion ca axit yu (F, NO2

, PO43,

CH3COO ) v cation ca baz yu (NH4

+; Fe3+) :

Nguyn nhn: c cation v anion ca mui u thy phn, nn

dd c th l axit hoc baz, ph thuc vo gi tr Kav Kbca

cc ion.

VD: NH4CN (dd) NH4+

(dd) + CN

(dd)

CN

(dd)+ H2O (l) HCN (dd) + OH

(dd)

NH4+

(dd)+ H2O (l) NH3 (dd)+ H3O+

(dd)

Ka (NH4+)= 5,7.1010; Kb(CN)=2,5.10

5 mt baz

Quy lut: Nu Ka> Kb, dung dch mui c tnh axit; nu Ka< Kb,dung dch mui c tnh baz. 192/13/2011

C : Trong cc dung dch mui sau y: LiCl, FeCl3,

NH4NO3, CH3COOK, C6H5ONa, KOOCCOOK, Cs2SO4,

Al2(SO4)3, nc Javel, soa, Kaliperrmanganat. dung d

no cho mi trng:

a) Trung tnh.

b) Axit

c) Baz

d) Khng xc nh c

2/13/2011

17.7. C D 1: A + sp.

V d:

HCl (dd)+ NaOH (dd) NaCl (dd) + H2O (l)

H3O+

(dd)+ OH

(dd) H2O (l)

Cc phn ng dng ny u c hng s cn bng v cng

ln, nn c coi nh xy ra hon ton.

Cc ion Na+ v Cl khng thy phn trong dung dch sau p

C : Tnh pH ca dung dch thu c sau khi trn400ml HCl 0,2M vi 100 ml NaOH 1M.

212/13/2011

D 2: A + sp.V d:

HCOOH (dd)+ NaOH (dd) NaHCOO (dd) + H2O (l)

HCOOH (dd)+ OH

(dd) HCOO

(dd) + H2O (l) K = K

Khi xem xt mi trng dung dch (tnh pH) phi xem x

quan h lng ban u v sp 3 trng hp.

C : Tnh pH ca dung dch thu c sau khi trn

400ml HCOOH 0,2M vi 100 ml NaOH 1M.

C : Tnh pH ca dung dch thu c sau khi trn400ml HCOOH 0,3M vi 100 ml NaOH 1M.

2/13/2011

D 3: A + sp.

V d:

HCl (dd)+ NaHCO3 (dd) NaCl (dd) + H2CO3 (dd)

H3O+

(dd)+ HCO3

(dd) H2CO3 (dd) + H2O (l) K = K/K

Tng t dng 2, trong dng ny 3 trng hp.

D 4: A + sp.

V d:

CH3COOH (dd)+ NH3 (dd) CH3COONH4 (dd)

CH3COOH (dd)+ NH3 (dd) CH3COO

(dd) + NH4+

(dd)

pH ca dung dch ph thuc vo c Kav Kb.23

a btong

w

K .KK

K=

2/13/2011

17.8. K, K H Gi tr ca Kav Kbc xc nh bng thc nghim.

Khi tnh ton, cc gi tr lp vo biu thc Kav Kbu l cc gnng mol/l ca cc ion v cht thi im cn bng.

D: Dung dch HNO20,50M c pH = 1,72. Tnh Ka?

pH = 1,72 [H+] = x = 101,72 = 0,019 M.

HNO2 H+ + NO2-

[ban u] 0,50M 0 0

[phn ng] -x +x +x

[cn bng] (0,50-x) x x

2 24

a

x xK 7,1.10

0,5 x 0,5

= =

2/13/2011


52/63

D: Tnh pH ca dung dch NH30,1 M. Bit Kb= 1,8.105?

x = 1,3.103M = [OH] pH = 11,11

D: Tnh pH ca dung dch HCOOH 0,010M v tnh % ion

ha ca HCOOH? Bit Ka = 1,6.104.

%ion ha = (s mol HCOOH in ly) / (s mol HCOOH b)25

NH3 + H2O NH4+ + OH-

[ban u] 0,10M 0 0

[phn ng] -x +x +x

[cn bng] (0,10-x) x x

2 254

b

3

[NH ][OH ] x xK 1,8.10

[NH ] 0,1 x 0,1

+

= = =

2/13/2011

D: Tnh pH ca dung dch H3PO4 5,0M? Bit cc Kaca

l Ka1= 7,2.103; Ka2= 6,2.10

8; Ka3= 4,8.1013;

Do Ka3


53/63

HA HC


12/13/2011

C

C 18

2/13/2011

18.1.

Khi nim ion ng dng: l in hm h m ng i

in c m ng h ang kh .

nh hng ca ion ng dng tun theo nguyn l chuyn

dch cn bng Le Chatelier.

Tnh ton trong bi ton ca ion ng dng, thng s

dng bng ICE lm.

C : Cho cb: CH3COOH+ H2O H3O+ + CH3COO

.

Nu s chuyn dch cb khi:

a) Thm CH3COONa b) Thm HCl c) Thm NH4Cl 32/13/2011

: Tnh pH ca dd thu c khi cho 200 dd HCl 1M vo300ml dd CH3COOH 1M. Bit Ka=1,8.105.

Hng dn:

Tnh li nng mol/L ca CH3COOH v HCl.

Lp bng ICE vi cn bng ca axit:

Dng Ka m x m nng H3O+

m pH:

CH3COOH + H2O H3O+ CH3CO

[ban u] 0,6 M 0,4 M 0

[phn ng] +x -x -x

[cn bng] (0,6+x) (0,4-x) -x

2/13/2011

18.2. D

Khi nim: l m dng dch hn h gm ai mi

ng ng ca n (hc ba mi ng ng ca n).

c im dd m:

Cha axit yu v baz lin hp vi n (hoc baz yu v axit

lin hp ca n).

C kh nng chng li s thay i ca pH khi thm vo dung

dch mt lng nh H+ hay OH.

V d: dd cha ng thi NH4Cl v NH3

dd cha ng thi CH3COONa v CH3COOH.

52/13/2011

: Tnh pH dd m thu c khi cho 200 mlNaOH 1

tc dng vi 300ml CH3COOH 1M. C Ka=1,8.105

Hng dn:Tnh s mol ca NaOH, CH3COOH NaOH thiu dd sau p

CH3COOH v CH3COONa l mt dd m.

Tnh li nng mol/L ca CH3COOH v CH3COONa sau p.

Lp bng ICE vi cn bng ca axit:

CH3COOH+ H2O H3O+ + CH3COO dng Ka=

(Hoc lp bng ICE vi cn bng ca baz:

CH3COO+ H2O OH + CH3COOH dng Kb=)

Tnh nng H3O+ (hoc OH) ri tnh pH ca dd sau p. 2/13/2011


54/63

Biu thc tnh pH cho dd m: (ptHenderson Hasselbalch) :

= + ( /)

dng ch dd m gm ai mi ca n

= + ( /)

dng ch dd m gm ba mi ca n.

: Tnh pH dd m thu c khi cho 200 mlNaOH 1M tcdng vi 300ml CH3COOH 1M. C Ka=1,8.10

5.

Hng dn:


[CH3COOH] = 0,2M [CH3COONa] = 0,4M

Ta c: pH = pKa+ log ([0,4]/[0,2]) = 4,74 + log2 = 72/13/2011

: Tnh pH dd m mi thu c khi cho 100 ml HC

1M vo mt dd m c sn 700ml CH3COOH 1M v

CH3COONa 0,8M. C Ka=1,8.105 .

Hng dn:

Coi ton b lng HCl cho thm vo s tc dng ht CH3COO

CH3COONa + HCl CH3COOH + NaCl


Lp bng ICE vi cn bng ca axit, tnh nng H3O+ ri tnh

ca dd sau p :

CH3COOH+ H2O H3O+ + CH3COO

dng Ka=

(Hoc dng phng trnh Henderson Hasselbalch tnh pH2/13/2011

18.3. C Chun l phng php dng xc nh hm lng ca mt

cht c trong mt dung dch hoc xc nh tnh cht vt l ca

mt phn t.

Dung dch tiu chun l dung dch bit chnh xc nng ca

n, dng dung dch ny xc nh nng cc dung dch khc.

Dung dch chun l dung dch cha bit nng . Ta dng

dung dch tiu chun xc nh n.

im tng ng l thi im khi dung dch tiu chun v

dung dch chun tc dng va vi nhau. Ngi ta xc

nh im tng ng bng cht ch th mu. 92/13/2011

Cht ch th mu (cho chun axitbaz) l cht c mu scthay i theo gi tr pH ca dung dch..

ng cong chun : l th quan h gia pH v th tch

dung dch axit hoc baz thm vo.

Cc bi ton chun thng gp (5 dng):

Chn ai mnh i ba mnh.

Chn ai i ba mnh.

Chn ba bng ai mnh.

Chn ai a bc bng ba mnh.

Chn ba a bc bng ai mnh.

Biu thc dng tnh pH v nng cc cht trong h, cho

trng hp sau l : phng trnh Henderson Hasselbalch. 2/13/2011

112/13/2011 2/13/2011


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: Xt qu trnh chun 100 ml dd HCl 0,1M bng dd

NaOH 0,1M.

Dd i chn l NaOH 0,1M ( n be); dd chn l HCl

(ch ng bnh am gic).

Ch ch h m c h l m, hc P.P.

Phng nh : H+ + OH H2O.

Ban khi cha chn : H = 1.

Ti im ng ng: H = 7.

Sa im ng ng (khi NaOH d 10ml): H = 11,67.

h bi din ng cng chn : 132/13/2011 2/13/2011

: Xt qu trnh chun 100 ml dd CH3COOH 0,1Mbng dd NaOH 0,1M. Bit Ka=1,8.10

5 .

Dd i chn l NaOH 0,1M ( n be); dd chn l

CH3COOH (ch ng bnh am gic).

Ch ch h m l P.P.

Phng nh : CH3COOH + OH CH3COO

+ H2O.

Ban khi cha chn : = 2,87

Ti im ng ng: H = 8,72.

Sa im ng ng (khi NaOH d 10ml): H = 11,67.

h bi din ng cng chn : 152/13/2011 2/13/2011

Khi nim cht ch th mu axitbaz: l cht c mu sc

thay i theo gi tr pH dung dch.

Cc cht ch th thng l cc axt hoc baz hu c yu m

dng phn t hoc ion chng c mu khc nhau.

Phenlhalein, C20H14O4. Tinh h nh m ng, khng mi,

khng , nc= 259 263C. an ng nc, d an ng eanl

hc ee, khng m H < 9; m hng H > 9; dng ng

hc (lm hc ), ng cng nghi, ng h hm nhm.

HInd(komu) H+ +Ind- (muhng)

Gi (lim) l gi c m dng dch eanl hc nc i

ch m ch c a (ngnh hc cng inh gia

nm) Rccella Dendgaha, c m gc ban l m

m (nn cn c gi l gi m), c dng ng

ngnh ha hc h, kim nghim H.

172/13/2011

Phenolphtalein

2/13/2011

Bài giảng Hoá học đại cương - Lê Minh Thành

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