tham luận 1

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www.daykemquynhon.ucoz.com Trung tm bi dng kin thc vn ha cho hc sinh cp 2 + 3

1000B Trn Hng o Tp Quy Nhn3

I . CHUYN GING DY HA HC

VN DNG L THUYT HA HC PHN TCH TRONG GING DYNI DUNG CHUN AXIT-BAZ TRNG CHUYN,

PHC V BI DNG HC SINH GII QUC GIA, QUC TI. CHUN N AXIT- N BAZ

.PGS.TS.o Th Phng Dip-Khoa Ha hc, trng HSP H Ni

Ths.on Th Kim Dung-Trng chuyn Lng Vn Ty, Ninh bnh

M UHa hc phn tch ni chung v phn ng axt- baz ni ring c vai tr to ln, chim mt

v tr quan trng trong qu trnh ging dy mn ho hc trng trung hc ph thng (THPT),c bit i vi cc trng chuyn v luyn thi hc sinh gii Quc gia, Quc t. Trong nhng

nm gn y c mt s cng trnh nghin cu vic vn dng l thuyt ha hc phn tch [1], vphn ng oxi ha-kh [2], phn ng axit-baz [3], phn ng to thnh hp cht t tan [4] trongging dy hc sinh trng chuyn v bi dng hc sinh gii Quc gia. Nhng cc cng trnhtrn mi ch tp trung i su vo tnh ton cn bng ion trong dung dch, m cha cp nchun dung dch, l mt trong nhng ni dung c trong chng trnh thi hc sinh gii Qucgia [5], thi chn i tuyn Quc gia d thi Olympic Ha hc Quc t ca cc nc trn thgii [6], [7], [8], [9], [10] v c c trong thi Olympic ha hc Quc t hng nm [11], [12]. Nidung ny cng c a vo trong chng trnh sch gio khoa (SGK) 12 nng cao [13]nhng vi thi lng qu t v ni dung kin thc cng ht sc n gin. rt ngn khongcch gia ni dung kin thc c hc cc trng chuyn v ni dung thi Olympic Quc gia,

Quc t, cn thit phi trang b cho c gio vin (GV) v hc sinh (HS) nhng kin thc nng caongang tm chng trnh i hc, nhng vn m bo mc hp l, ph hp vi trnh hcsinh ph thng. y chnh l mt trong nhng ni dung chng ti xin c trnh by trong thngbo ny vi phm vi vchun n axitn baz .

NI DUNGT lu chun axit-baz c a vo trong mt s thi HSG Quc gia v Quc t

vi ni dung ht sc phong ph v a dng: T php chun n axit, n baz n cc phpchun a axit, a baz; T ni dung tnh pH ti cc thi im chun , tnh nng cc chttham gia phn ng chun v tnh th tch ca thuc th cn dng n yu cu tnh sai s chun v chn ch th thch hp; T ni dung v ng cong chun n yu cu xc nh thnh

phn hn hp phn tch trn c s dng ng cong chun v.vNh vy, ni dung thiOlympic Quc gia, Quc t c mt s chnh lch kh xa so vi chng trnh ha hc ca cctrng chuyn. Chnh v th cn thit phi trang b kin thc chun axit-baz cho HS mt cchhp l cc em nm c bn cht ca cc qu trnh ha hc xy ra trong dung dch, t gipcho cc em c phng php gii n gin ph hp vi trnh ca HS. Sau y l mt s v dminh ha v chun n axit, n baz.

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V d 1: Chun 100,00 ml dung dch NaOH n mt mu ch th phenolphtalein th tiuth ht 48,00 ml dung dch HCl 5,00.10-3M. Tnh chnh xc nng dung dch NaOH[14].

Phn tch:i vi hc sinh ph thng (HSPT), khi nim v chun axit-baz cn kh mi m, v

vy GV cn ging gii hc sinh (HS) thy c phn ng chun axit-baz thc cht l phn

ng trung ha. V bi ton trn c th t vn nh sau: Nh vi git phenolphtalein vo 100,00ml dung dch NaOH cha bit nng . Thm tip dung dch HCl 5,00.10-3M cho ti khi mt muca phenolphtalein th phi dng ht 48,00 ml dung dch HCl. Tnh chnh xc nng dung dchNaOH.

Trc ht GV cn phn tch HS thy c nu lng axit cho vo trung ha va vi100,00 mlNaOH th h t n im tng ng (T) v pHT = 7,00. Nhng y php chun (hay qu trnh trung ha) li kt thc ti thi im ch th phenolphtalein i mu. Cn c vokhong chuyn mu ca ch th v da vo th t chun , GV hng dn HS xc nh cti im cui chun , pHc = 8,00 > pHT = 7,00 (dng trc im T), tc l sau phn ng cnd NaOH.

tnh chnh xc CNaOH, HS c th s dng phng trnh bo ton in tch, hoc nu ckin thc v ha hc phn tch ni chung hay kin thc v chun axit - baz ni ring [15],[16], th cc em c th tnh sai s ca php chun . Cn c vo gi tr sai s tnh c (dnghoc m) hiu chnh, tnh chnh xc th tch tng ng (VT) l th tch dung dch HClcn dng trung ha ht 100,00 mlNaOH. T HS s tnh c chnh xc nng NaOH.

Tuy nhin i vi HSPT cha c khi nim v phng trnh sai s chun axit baz, vvy GV cn vn dng kin thc v chun axit - baz mt cch hp l hng dn cc em giibi ton ny m khng cnphi s dng n phng trnh sai s(qu nng i vi cc em), vcng khng nn dng phng trnh bo ton in tch (nng v ngha vt l hn l t duy hahc):

t CNaOH = Co. V pHc = 8,00 > 7,00, do cn d NaOH theo phn ng:H+ + OH- H2O

C148

48.10.5 3 o

100.

148

C

C 03

o100. 5.10 .48

148

C

pH = 8,00 [OH-] = 10-6 ? [H+] = 10-8 c th b qua s phn li ca nc [OH -

] ; -OHC (d), ngha l: 10-6 =

3o100. 5.10 .48

148

C

Co = 2,4015.10-3M.

Nh vy vi cch gii ny cc em hiu c bn cht ca cc qu trnh xy ra trong dungdch, nm c hin tng ha hc, gip pht trin t duy ha hc.Cng cn lu rng: thng thng khi cn nh gi mc phn li ca nc, cc em

thng so snh [H+] hoc [OH-] vi gi tr 10-7. Trong bi ton ny ta khng th ni 10-6 10-7v nh vy kh c th thuyt phc cc em chp nhn [OH-] ; -OHC (d). Nhng nu so snh

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[OH-] = 10-6 ? [H+] = 10-8, c ngha l s phn li ca H2O c th b qua, do s chp nhn[OH-] ; -OHC (d) l hp l.

Nh vy t gi tr pH chuyn mu ca ch th khi thc hin phn ng trung ha cho phpnh gi c nng ca cht tham gia phn ng. Trn c s ny gio vin c th nng mc

phc tp ca bi ton nu s dng hai ch th s cho php xc nh nng ca c cht phn tchv cht chun.V d 2: Chun 50,00 ml dung dch HCl bng dung dch NaOH n i mu

phenolphtalein (pT=9,00) th phi dng 25,00 ml NaOH. Nu chun n xut hin mu vngca metyl da cam (pT=4,40) th phi dng 24,50 ml NaOH. Tnh nng HCl v NaOH[14].

Phn tch:Tng t v d 1 gio vin nn hng dn cho cc em phn tch bn cht cc qu trnh xy

ra trong dung dch: khi trung ha n pH = 9,00 > 7,00 mi trng baz d NaOH; ngcli khi chun n pH = 4,40 < 7,00 mi trng axit d HCl, t cc em d dng tnhc nng cc cht m khng phi s dng phng trnh bo ton in tch - ch thin v cngc ton hc, nng v ngha vt l m khng ch n t duy ha hc.

T 2 v d trn ta thy cc bi ton u gii hn trong phm vi l php chun u ktthc cc gi tr pH m ti c th b qua qu trnh phn li ca H2O (pH = 8,00; 9,00; 4,40).Nhng trn thc t c th trung ha n gi tr pH bt k, do gio vin cn khai thc thmtrng hp nng cao ny.

V d 3: Chun 100,00 ml dung dch NaOH bng dung dch HCl 0,010 M. Nu lng HCltiu th l 40,00 ml th pH ca dung dch thu c bng 10,00. Tnh th tch dung dch HCl 0,010M cn phi cho vo 100,00 ml dung dch NaOH trn trong qu trnh chun pH ca hnhp thu c bng 7,50.

Phn tch: tnh c th tch dung dch HCl cn cho vo dung dch NaOH, trc ht cc em phi

tnh c nng NaOH (CNaOH) t d kin th nht ca bi. Ging nh v d 1, c rt nhiucch tnh CNaOH, nhng GV nn hng dn cc em xt cc qu trnh xy ra trong h trnhkhng phi s dng phng trnh sai s hoc phng trnh bo ton in tch:

T gi tr pHh = 10,00 sau khi thm 40,00 mlHCl, cc em s thy c sau khi trung ha,OH- cn d v [OH-] ; -OHC (d). T cc em d dng tnh c CNaOH = 4,14.10

-3 (M)

Sau khi tnh c CNaOH, cc em phi cn c vo gi tr pH ti thi im dng chun xc nh thnh phn ca h: pH = 7,5 > 7,0 mi trng baz, d NaOH v [H+] = 10-7,5 [OH-] = 10-6,5 S phn li ca H2O khng th b qua, do [OH

-] = CNaOH d + [H+] (tnh theo

cn bng phn li ca nc hoc tnh theo iu kin proton - KP). T s tnh c th tch

dung dch HCl: VHCl = 41,40 ml.V d 4: Chun 25,00 ml dung dch HCOOH ht 12,50 ml dung dch NaOH 0,100 M.

Tnh pH ca dung dch trc khi chun v sau khi thm NaOH vi th tch:1/ 10,00 ml; 2/ 12,45 ml; 3/ 12,50 ml; 4/ 13,00 ml.Phn tch:

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y l dng bi tp tnh pH ca dung dch sau khi trung ha axit yu (hoc baz yu) bngbaz mnh (hoc bng axit mnh). Dng ton ny c s dng rt nhiu trong cc thi hcsinh gii v kt hp c nhiu tnh hung t n gin n phc tp. a v dng bi quenthuc HS phi vit c phn ng chun v xc nh c thnh phn gii hn mi thiim dng chun , t d dng tnh c pH ca h.

V d 5: C th chn metyl da cam, metyl hoc phenolphtalein lm ch th thch hp chophp chun HCl bng NaOH 0,10 M c khng, bit rng trung ha ht 20,00 ml dungdch HCl trn cn 20,00 ml dung dch NaOH 0,10 M.

Phn tch:Trc ht GV cn nu nguyn tc chung ca vic chn ch th sao cho pH chuyn mu r

nht ca ch th (hay cn gi l ch s chun pT ca cht ch th) cng gn vi pH T cng tt.Nh vy HS cn xc nh c pHT cng nh ch s chun ca 3 cht ch th cho trongphp chun HCl bng NaOH. Nhng nu ch cn c vo nguyn tc trn, tc l nu so snh gitr pTmetyl da cam = 4,40; pTmetyl = 6,20 v pTphenolphtalein = 9,00 vi pHT = 7,00 th cng lm HSch dm chn metyl c pT = 6,20 7,00 lm ch th thch hp cho php chun trn.

Chnh v vy GV cn linh hot hng dn cc em da vo th tch thuc th (NaOH) tiu th khis dng cc ch th khc nhau chn ch th thch hp: Nu ti thi im chuyn mu ca ch thm th tch thuc th tiu th xp x th tch tng ng ( VT = VNaOH = 20,00 ml) th c thchn c ch th cho php chun ang xt.

Nh vy cn c vo gi tr pH chuyn mu ca cc ch th ti im dng chun , vi CHCl

=0,10.20

0,1020

! M cc em d dng tnh c VNaOH dng:

Vi ch th metyl da cam, pTmetyl da cam = 4,40 sau khi trung ha CH+

(d) = [H+] = 10-

4,40.

-

-

4,40 H

H

0,1.20 0,1

1020

V

V

-

19,984V ! (ml) VT dng c ch th metyl da cam.

Tng t c th chn phenolphtalein cho php chun ny, bi v th tch thuc thtiu th l -

20,004V ! (ml) VT.

Nu dng metyl , c pT = 6,20 7,00 ngoi lng d axit ti im cui chun ,phi k n lng H+ do nc phn li ra: [H+] = 10-6,20 = CH+(d)+ [OH

-]

Hay:-

-

7,80 6,20OH

OH

0,1.20 0,110 10

20

V

V

!

-

19,9998V ! (ml) VT. Vy s dng ch th metyl

l hon ton hp l v cho sai s nh nht trong s 3 ch th trn.

V d 6: C th dng phenolphtalein lm ch th cho php chun 25,00 ml dung dchC6H5COOH 0,10 M bng dung dch NaOH 0,20 M c khng, nu sai s cho php khng vtqu 0,1%?

Phn tch:So vi v d 5, bi ton ny phc tp hn do y l php chun n axit yu bng baz

mnh, nn pHT { 7,00. c th chn ch th theo nguyn tc chung, cn tnh pHT. i vi HS

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trng chuyn, vic xc nh pHT l hon ton c th thc hin c. Xut pht t phn ngchun : C6H5COOH + OH

- C6H5COO- + H2O, cc em d dng tnh c th tch NaOH

tiu th ti T: VT =25.0,1

12,50,2

ml! . T xc nh c pHT theo cn bng thu proton

ca C6H5COO-

c nng C = 3

2,0

5,1225

25.1,0! M: pHT = 8,5 9,00. Do c th chn c

phenolphtalein lm ch th cho php chun trn.Tuy nhin khng phi lc no cng chn c ch th c pT pHT, hn na khc vi v d

5, y vic chn ch th thch hp c khng ch bi sai s cho php, do ngoi nguyn tctrn, GV cn gii thch HS nm c khi nim bc nhy chun (BNC) v vi s xuthin BNC trn ng cong chun cho php m rng phm vi chn ch th: c th chn bt kch th no c ch s chun pT nm trong BNC ng vi sai s cho php. Ngha l chnc ch th thch hp, c th tnh BNC ng vi sai s cho trc hoc tnh sai s i vi tng

ch th theo phng trnh sai s: q = - w 0

0 a

( )K C C h

h

h CC K h

.

Tuy nhin c hai cch ny u khng ph hp vi HSPT do cc em khng bit phng trnhsai s. Chnh v vy GV cn hng dn cc em tnh c gi tr pH u bc nhy (pH) vpH cui bc nhy (pHc) ng vi gi tr th tch thuc th cho vo thiu (V: th tch u bcnhy) v tha (Vc: th tch cui bc nhy) 0,1% so vi th tch tng ng, t chn nhngch th c pH < pT < pHc.

u bc nhy: V = VT.99,9% = 12,5.99,9% = 12,4875 mlCui bc nhy: Vc = VT.100,1% = 12,5.100,1% = 12,50125 ml

T hai gi tr th tch ny cc em s tnh c gi tr pH v pHc tng ng:Ti V =12,4875 ml d C6H5COOH:

C6H5COOH + OH-

C6H5COO-

+ H2OC

4875,34

1,0.25

4875,37

4875,12

C 6,67.10-5 _ 0,0666 tnh pH theo h C6H5COOH 6,67.10

-5Mv C6H5COO- 0,0666M:

pH = 4,2 + lg 510.67,6

0666,0

7,20. Nhng vi pH = 7,20 7,00 cn phi nh gi chnh

xc theo KP v phi k n cn bng phn li ca H2O. Nhng nu ch cn so snh pH vipTphenolphtalein = 9,00 th c th chp nhn c gi tr gn ng ny.

Tng t vi Vc = 12,5015 > VT d OH-

COH- = 610.666,6

50125,371,0.252,0.50125,12 !

pHc = 9,80

Vy BNC = 7,20 9,80 c th chn c phenolphtalein lm ch th cho php chun trn v 7,20 < 9,00 < 9,80.

V d 7: Chun 50,00 ml dung dch NH3 ht 30,00 ml dung dch HCl 0,25 M1/ Tnh nng dung dch NH3.

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2/ Tnh pH ti thi im chun c 50% cht phn tch (pH1/2) v pHT.3/Cn c vo pHT, c th chn c phenolphtalein, metyl da cam, metyl lm ch th thch

hp cho php chun trn khng?4/C th chn nhng ch th c gi tr pT bng bao nhiu cho php chun ny vi sai s cho

php l

0,1%.

5/ Nu chn ch th metyl da cam th sai s chun l bao nhiu?Phn tch:y l bi ton tng hp gm tt c cc dng: Tnh nng cc cht phn ng; Tnh pH ti

cc thi im chun ; Chn ch th thch hp cho php chun ; Tnh sai s chun v tnhBNC.

Da vo phn ng chun , HS d dng tnh c CHCl = 0,15M. Trn c s , khi chun c 50% lng NH3, HS s tnh c pH1/2 l pH ca h m.

T thnh phn ti T l NH 4 vi C =

3050

50.15,0

, HS d dng tnh c pHT = 5,134 t

c th chn metyl c pT = 5,00 5,314 lm ch th cho php chun ny.

tr li c cu 4, cc em phi tnh c BNC, tc l tnh khong pH ng vi th tchthuc th cho vo thiu v tha so vi VT l 0,1% tng t nh v d 6.

nh gi sai s ca ch th metyl da cam, GV cn hng dn cho HS phi tnh VHCl cndng chun 50,00 mlNH3 trn n i mu metyl da cam ti pH = 4,00, t tnh c sais theo s sai lch gia VHCl tiu th v VT m khng cn phi tnh theo phng trnh sai s: q =

T

T

V V

V

T thnh phn ca h ti pH = 4,00 gm NH 4 to thnh vi +

4NHC =

V50

50.15,0Mv H+ d

vi +H 0, 25 0,15.5050VC

V! M, cc em s tnh c V

HCl = 30,03 ml.

Vy q = T

T

30, 03 30.100 0,10%

30

V V

V

! !

Tuy cch gii ny di hn nhng ph hp vi trnh ca HS, gip cc em pht trin ct duy ha hc.

KT LUNTrn c s phn tch ni dung chun axit baz ca chng trnh chuyn ho chng ta

thy c v tr, vai tr ca Ha hc phn tch trong vic hnh thnh kin thc c bn v chun axit-baz cho hc sinh trng chuyn, cng nh thy c mi quan h mt thit, hu c giachng trnh chuyn ha, chng trnh thi hc sinh gii Quc gia, Quc t vi Ha hc phn tch.T vn dng linh hot, hp l l thuyt Ha hc phn tch ni chung v l thuyt chun axit-baz ni ring trong ging dy ha hc trng chuyn, phc v bi dng hc sinh giiQuc gia, Quc t.

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TI LIU THAM KHO[1] Dng Th Lng. Vn dng l thuyt ha hc phn tch gii cc bi ton cn bng ion

trong dung dch-Bi dng hc sinh gii chuyn ha. Lun vn thc s khoa hc ha hc.Thi nguyn, 2007.

[2] L Th Ngc H. Tm hiu vic vn dng l thuyt phn ng oxi ha kh trong ging dy ha

hc ph thng qua h thng bi tp, thi i hc, thi hc sinh gii quc gia . Lun vnthc s khoa hc ha hc. H Ni, 2003.

[3] Nguyn Th Hin. Phn loi, nh gi tc dng, xy dng cc tiu ch, cu trc cc bi tp vphn ng axit-baz phc v cho vic bi dng hc sinh gii quc gia. Lun vn thc s khoahc ha hc. H Ni, 2003.

[4] Vng B Huy. Phn loi, xy dng tiu ch cu trc cc bi tp v hp cht t tan phc vcho vic bi dng hc sinh gii quc gia. Lun vn thc s khoa hc ha hc. H Ni, 2006.

[5] thi HSG Quc gia bng A, bng B t nm 1994 n nm 2009.[6] thi chn i tuyn thi Olympic quc t vng 2 t nm 2005 n nm 2009.[7] Bi tp chun b Olympic Quc t t nm 2002 n nm 2006.

[8] Trn Thnh Hu, Nguyn Trng Th, Phm nh Hin. Olympic Ha hc Vit Nam v Quct, tp II. NXB Gio dc, 2000.

[9] Nguyn Trng Th. Olympic Ha hc Vit Nam v Quc t, tp III. NXB Gio dc, 2000.[10] o Qu Triu, T B Trng (Hong Minh Chu, o nh Thc hiu nh). Olympic Ha

hc Vit Nam v Quc t, tp IV. NXB Gio dc, 2000.[11] Hong Minh Chu, Ng Th Thn, H Th Dip, o nh Thc (hiu nh ting c), Trn

Thnh Hu, Nguyn Trng Th, Phm nh Hin. Olympic Ha hc Vit Nam v Quc t, tpV. NXB Gio dc, 2003.

[12] thi Olympic Quc t t nm 2002 n nm 2009.[13] L Xun Trng, Nguyn Hu nh, T Vng Nghi, nh Rng, Cao Th Thng. Ha hc

12 nng cao. NXB Gio dc, 2008.[14] Nguyn Tinh Dung.Bi tp ha hc phn tch. NXB Gio dc,1982.[15] Nguyn Tinh Dung.Ha hc phn tch phn III. Cc phng php nh lng ha hc (ti

bn ln th 4). NXB Gio dc, 2007.[16] o Th Phng Dip, Vn Hu. Gio trnh Ha hc phn tch. C s phn tch nh

lng ha hc. NXB i hc S phm, 2007.

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S DNG PHNG PHP TH XC NH BC PHN NGV TNH HNG S CN BNG

Mai Chu PhngTHPTChuyn Lam sn-Thanh Ha

I.1. C s l thuyt.

I.1.1. Phn ng bc mt.A p sn phm

Phng trnh ng hc dng tch phn:

C = Co.e-kt (1)oCkt=ln

C(2)

Trong : k: hng s tc phn ngCo: nng ban u ca cht tham gia phn ng A.C : nng cht A ti thi im t.t : thi gian phn ng.

T phng trnh (1) v (2) nhn thy nu ta kho st s ph thuc nng cht phn ng theothi gian nh phng trnh (1) th vic xc nh hng s tc phn ng k l kh khn v y lhm m, cn nu kho st phng trnh (2) th thy phng trnh (2) chnh l phng trnh tuyn

tnh bc nht dng y=ax, trong oCy=ln

C, a=k, x=t. Do , da vo s liu thc nghim ta

kho st hmoCln =f(t)

Cph thuc tuyn tnh theo thi gian vi dc ca th xc nh

c hng s tc phn ng k=tgE

Hnh 1: th biu din s ph thuc logarit nng cht tham gia phn ng theo thi gian.

I.1.2. Phn ng bc hai.

I.1.2.1. Trng hp 1: Nng ban u ca hai cht tham gia phn ng khc nhau.

ln C

C0

0 t

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A + B p sn phmt=0 01C

02C 0

t C1 C2 xPhng trnh tc phn ng dng tch phn:

01 2

0 0 02 1 2 1

C .C1kt= lnC -C C .C (3)

Suy ra:0

0 02 22 1 0

1 1

C Cln =(C -C )kt+lnC C

(4)

T phng trnh (4) c 2

1

Cln

Cph thuc tuyn tnh vo thi gian v ng biu din s ph thuc

ny c dng y=ax + b vi dc ca th bng 0 02 1tg =(C -C )k v ct trc tung ti

0201

ClnC

.

Suy ra:0 0

2 1

tgk=

C -C

Hnh 2: th biu din s ph thuc logarit nng hai cht tham gia phn ng theo thi gian. I.1.2.2. Trng hp hai: Nng ban u ca hai cht tham gia phn ng bng nhau a=b hay

0 02 1C =C .

Phng trnh tc phn ng dng tch phn:0

1 1= +ktC C

(5)

Phng trnh (5) biu din nghch o ca nng cht tham gia phn ng ti thi im t phthuc tuyn tnh theo thi gian. ng biu din

1= (t)

Cc dng y=ax+b, c dc bng

k=tg , ct trc tung ti0

1C

ln 1

2

C

C

ln

0

1

02

C

C

0 t

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Hnh 3: th biu din s ph thuc nghch o nng cht tham gia phn ng theo thigian.Nh vy, trong phng php th ngi ta th cc s liu thc nghim vo dng tch

phn ca phng trnh ng hc tm s ph hp. Trc ht gi nh phn ng c bc n. Sau t dng tch phn ca phng trnh ng hc tm mt hm ca nng cht tham gia phn ngph thuc tuyn tnh theo thi gian. V ng biu din ca hm ny da vo cc kt qu thcnghim. S thng hng ca cc im biu din chng t tnh ng n ca bc phn ng ginh, t cng xc nh c hng s tc phn ng da vo dc ca th

I.1.3. Cch dng th biu din s ph thuc tuyn tnh.

C th xy dng th dng y=ax, y=ax+b da trn phn mm Microso t O ice Excel,hoc dng th trn giy li.

I.1.3.2. Dng th trn phn mm Microsoft Office Excel.

Bc 1: M mn hnh Microso t O ice Excel.Bc 2: Nhp s liu theo ct cc gi tr x, y tng ng ca phng trnh dng y=ax hay y=ax+b.Bc 3: Bi en hai ct s liu, sau ln lt vo Insert/Chart/XY(Scatter)/Chart sub-type/Next/Next/Finish.Bc 4: Tch phi chut vo vng khng gian trng trong th, vo Chart Options, nhp cchbiu din trc honh Value (X) axis, cch biu din trc tung Value (Y) axis.

Bc 5: Tch phi chut vo mt im thc nghim, voAdd Trendline/Linear/Display equation on chart/Ok c ng thng biu din hm ph thucv phng trnh hi quy. T thu c h s gc ca th, do tnh c hng s tc phn ng.

Vic dng th trn phn mm Microso t O ice Excel, sau my tnh hi quy li thuc ng chun trn c s phng php bnh phng ti thiu.

C

1

0

1

C

0 t

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I.1.3.1. Dng th trn giy.

Bc 1: Gi thit phn ng tun theo ng hc bc no .Bc 2: Chuyn sang phng trnh ng hc dng tch phn.Bc 3: K bng tnh s liu biu din s ph thuc thch hp ca nng theo thi gian.

Bc 4: Dng th biu din s ph thuc dng thch hp ca nng theo thi gian bng cchv cc im thc nghim. Sau k ng chun.Ch : i vi dng th trn giy li cn ch chn t l xch trn giy sao cho ph hp, gcto khng nht thit phi l im khng. Sau khi dng c cc im thc nghim th dngng chun khng phi l ng i qua nhiu im thc nghim, m ng chun l ng tiu khi tng bnh phng khong cch t cc im thc nghim n ng chun l nh nht, nnk ng chun da vo khong cch ti cc im thc nghim.Bc 5: Dng mt tam gic vung bt k c cnh huyn nm trn ng chun (u mt ca cnhhuyn tam gic ny khng nn trng vi bt c im thc nghim no cho kt qu thu ckhch quan).Bc 6: Tnh h s gc tgE da trn vic o di hai cnh gc vung.

Ch : di hai cnh gc vung ny phi tng ng vi t l xch c chia trn trc honhv trc tung, khng phi l di o c trn thc hay trn giy li.Ngoi cch dng th c th kim tra li tnh ng n ca phng php bng cch so

snh vi cch tnh theo phng php bnh phng ti thiu hay phng php trung bnh.Phng php ny c th p dng v m rng cho vic xc nh cc thng s khc nh

nng lng hot ho, cc gi tr nhit ng, h s hp th trong o trc quang.

II.2. Bi tp vn dng.

Bi 1:(Trch thi Olympic ho hc quc t ln th 41-Anh 2009)[8]Bin tnh protein.

i vi cc phn t protein nh, phn ng khng bin tnh c th biu din theo cn bng sau:Protein bin tnh Protein khng bin tnhTa c th gi s rng cc phn ng bin tnh protein ch xy ra qua mt giai on. Phn ng bintnh protein l phn ng bc nht, hng s tc ca phn ng ny c th xc nh bng cchtheo di cng hunh quang khi lm bin tnh mt mu protein khng bin tnh ban u(thng thng bng cch thay i pH ca dung dch). Bt u t dung dch protein khng bintnh c nng 1,0 M, ngi ta lm bin tnh protein v o c nng ca cc protein khngbin tnh 25 oC theo thi gian:

Thi gian (ms) 0 10 20 30 40

Nng (M) 1 0,64 0,36 0,23 0,14

V mt th thch hp, t tnh hng s tc ca phn ng bin tnh protein k 250CBi gii:Do phn ng bin tnh ca protein l phn ng bc nht nn c phng trnh

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ng hc dng tch phn l:o

kt ln

Trong : k l hng s tc phn ng bin tnh protein.0: nng protein khng bin tnh ban u.: nng protein khng bin tnh ti thi im t.

T s liu thc nghim ta c bng s liu sau:Thi gian(ms) 0 10 20 30 40Nng ( ) 1,00 0,64 0,36 0,23 0,14

lno

C

C

0,446 1,022 1,470 1,966

Dng cc im thc nghim c to (t, lno

C

C), sau k ng chun.

10 20 4030 t(ms)0

0,5

1,0

1,5

ln(Co/C)

2,0

A

C

B

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Dng tam gic vung bt k ABC. Da vo t l xch chia trn trc honh

v trc tung tnh c0,975

0,050519,3

CBtg = =

AB!

Kim tra li kt qu dng th trn giy vi kt qu dng th trn phnmm Microso t O ice Excel

y = . x - .

R

= .

.

.

.

t(ms)

ln(C/C)

T th ta thy h s gc k=0,0501 ms-1 hay k=50s-1. Nh vy kt qu thu c bng cch dng th trn giy hon ton ph hp.

Bi 2:(Trch trong bi tp chun b Olympic ho hc quc t ln th 40 Hungari 2008)[7]Phn ng ca axeton vi brom sinh ra bromaxeton.a)Vit phng trnh ha hc ca phn ng nu nh axeton d.

Khi nghin cu c ch phn ng th ngi ta thc hin mt s th nghim ng hc sau 25 Ctrong dung dch nc bng cch o nng ca Br2 bng phng php trc quang. ng congng hc tng ng c ghi li khi nng u ca cc cht l [Br2]0 = 0.520 mmol/dm

3,[C3H6O]0 = 0.300 mol/dm

3, v [HClO4]0 = 0.050 mol/dm3.

t(min) 0 2 4 6 8 10 12 14[Br2] ( mol/dm

3) 520 471 415 377 322 269 223 173t(min) 16 18 20 22 24 26 28 30[Br2] ( mol/dm

3) 124 69 20 0 0 0 0 0b)Tc nhn phn ng ht trong th nghim ny lc)Bc ca phn ng i vi tc nhn ny lThi im m xy ra bc gy v tnh cht (characteristic break point) trn ng cong

ng hc c gi l thi gian phn ng v n c o trong dung dch nc 25 C. Bng saucho ta mt vi gi tr thi gian phn ng trong mt vi th nghim khc nhau ( ch pht, chgiy)

2 03

[Br ]

mmol/dm 3 6 0

3

[C H O]

mmol/dm

4 03

[HClO ]

mmol/dm

thi gian phn ng

0.151 300 50 5560.138 300 100 244

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0.395 300 100 7320.520 100 100 30370.520 200 100 15130.520 500 100 6090.520 300 200 455

0.520 300 400 228d)Xc nh bc phn ng ng vi ba cht tham gia.e)Vit biu thc tc tng qut.Mt phng php in ha khc cho php ta s dng nng Br2 b hn. mt ng congng hc vi nng cc cht u [Br2]0 = 1.80 mol/dm

3, [C3H6O]0 = 1.30 mmol/dm3, v

[HClO4]0 = 0.100 mol/dm3 c cho bng sau:

t(s) 0 10 20 30 40 50 60 70[Br2] ( mol/dm

3) 1.80 1.57 1.39 1.27 1.06 0.97 0.82 0.73t(s) 80 90 100 110 120 130 140 150[Br2] ( mol/dm

3) 0.66 0.58 0.49 0.45 0.39 0.34 0.30 0.26f)Tc nhn phn ng ht trong th nghim ny l

g)Bc ca phn ng i vi tc nhn ny lBi gii:a) Phn ng

b) Br2 l tc nhn gii hn (cht phn ng ht)c) th ng hc ca phn ng (ng cong ng hc)

ng cong ng hc lc ny l mt ng thng nn qu trnh ny l qu trnh bc 0 ca brom

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d) Do qu trnh l bc 0 i vi Br2 v tt c cc cht tc nhn khc u lng d nn gi tr tc l mt hng s trong mi th nghim. N c th c tnh mt cch n gin: v = [Br2]o / tphnnnVi tphn ng l thi gian phn ng. S ph thuc gia tc phn ng vo tc nhn phn ng htc th c kho st trc tip t cng thc. V th vn tc nh l mt hm ca nng axeton

trong mi trng axit (nng khng i 0,100 mol/dm

3

) cho:

th l mt ng thng, phn ng c bc 1 ng vi axeton.V th nh l mt hm ca nng axit vi nng axeton gi khng i (0,300 mol/dm3)cho:

th l mt ng thng, phn ng c bc 1 i vi [H+]

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e) v= ka[C3H6O][H+]

f) Br2 l tc nhn gii hn (cht phn ng ht)g) ng cong ng hc:

ng ny khng hn l ng thng nn qu trnh ny khng phi bc 0. Kim tra vi phn ngbc 1 th thy c kh nng bng cch xy dng mt s ph thuc vo logarit. th c dng iunh sau:

Cc im c th thng hng vi nhau. Nh vy qu trnh ny l bc 1 vi bromMt phng php khc l xc nh thi gian bn hu t nhng cp gi tr nng cho. T rt ra c s phn hu Br2 tun theo quy lut phn ng bc 1.

Bi 3: nghin cu ng hc ca cc phn ng gia ion Br- v ClO-:Br- + ClO- p BrO- + Cl-

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ngi ta trn ln 100 ml dung dch NaClO 0,1M vi 48 ml dung dch NaOH 0,5M v 21 ml ncct. Hn hp ny c t vo my iu nhit 298K. Sau cho 81ml dung dch KBr 1% (cng 298K) vo hn hp trn. Sau mi khong thi gian xc nh (t) ngi ta ly mu v xc nhlng BrO- trong mu. Kt qu thu c nh sau:t(pht) 0 3,65 7,65 15,05 26,00 47,60 90,60

2

-BrOC .10 (M) 0 0,0560 0,0953 0,1420 0,1800 0,2117 0,2367 Nng ca NaClO v KBr trong hn hp phn ng t=0 ln lt bng 0,003230M v0,002508M. pH cu dung dch l 11,28. Xc nh bc v hng s tc phn ng.Bi gii:Gi s phn ng c ng hc bc hai. Phng trnh ng hc ca phn ng l:

- -Br ClO.v=k.C C

Br- + ClO- p BrO- + Cl-t=0 01C

02C 0 0

t 1C 2C x x

- -

0

BrBr BrO=C -CC - - -

0

ClO ClO BrO=C -CC Ta c bng s liu sau:t(pht) 0 3,65 7,65 15,05 26,00 47,60 90,60

-

3

Br.10C 2,508 1,948 1,555 1,088 0,708 0,391 0,141

-

3

ClO.10C 3,230 2,670 2,227 1,810 1,430 1,113 0,863

-

-

ClO

Br

lnCC

0,253 0,315 0,381 0,509 0,703 1,046 1,812

Dng th c to cc im l (t,-

-

ClO

Br

lnCC

)

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Dng tam gic vung bt k ABC. Da vo t l xch chia trn trc honh

v trc tung tnh c0,66

0,017438

CBtgAB

!

- -

1 10 0 3

l

r

tg 0,0174k 24,100( . )- (3,230 2,508).10

M ph

! !

C th kim tra li kt qu da vo dng th trn Microso t O ice Excel hocgii theo phng php trung bnh thy kt qu hon thon ph hp.

20 40 8060

t(pht)

0

0,4

0,8

1,2

1,6

A

C

B

2,0

100

-

-

l

r

ln

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Bi tp vn dng.1. thu c oxi ngi ta phn hu 15cm3 dung dch H2O2 vi s c mt ca cht xc tc. Thtch oxi thu c theo thi gian nh sau:t(pht) 2 4 6 8 gV(ml)O2 1,3 2,36 3,36 3,98 6,18

p s: Bc nht, k=0,124 ph-1

2. ng hc ca phn ng gia natri thiosun at vi propylbromua:2- - -S O +RBr RS O +Br 2 3 2 3p , c nghin cu 37,3

0C. Nng ca 2-2 3S O sau mi khong

thi gian t c xc nh bng phng php chun th tch. Th tch dung dch I2 nng0,02572N dng chun hn hp phn ng (10,02ml) cc thi im khc nhau c chodi y:t(s) 0 1110 2010 3192 5052 7380 11232 g

2Iv (ml) 0 37,63 35,20 31,90 29,86 28,04 26,01 22,24

Nng ca 2-2 3S O ban u l 0,100M. Xc nh hng s tc phn ng.

p s: k=0,0016

36

M

-1

.s

-1

3. Etylen oxit b nhit phn theo phng trnh:

O (k) p CH4(k) + CO(k) 687,7K p sut chung ca hn hp trong phn ngbin i theo thi gian nh sau:t(pht) 0 5 7 9 12 18P.105(N.m-2) 0,155 0,163 0,168 0,172 0,178 0,188P(mmHg) 116,51 120,56 125,72 128,74 133,23 141,37Hy chngt rng phn ng phn hu etylen oxit l bc nht v tnh hng s tc ca phn ng.4. Dng phng php th hy xc nh bc ca phn ng:(CH3)3CBr(aq) + H2O(l) p (CH3)3COH(aq) + H

+ + Br-t cc d kin thc nghim sau y 298Kt(s) 0 15000 35000 55000 95000 145000[(CH3)3CBr](mol.l-1)

0,0380 0,0308 0,0233 0,0176 0,0100 0,00502

p s: bc nht.5. i vi phn ng x phng ho etyl axetatCH3COOC2H5 + OH

- CH3COO- + C2H5OH

thi im ban u t=0 hn hp phn ng cha este v xt vi nng bng nhau 0,05M. Phnng c theo di bng cch mi thi im t ngi ta ly 10ml hn hp phn ng ri chun lng xt cn li bng dung dch HCl 0,01M. Kt qu thu c nh sau:t(pht) 4 9 15 24 37 53VHCl(ml) 44,1 38,6 33,7 27,9 22,9 18,5Xc nh bc phn ng, hng s tc phn ng v thi gian na phn ng.

p s: Bc hai, k=0,647M-1.pht-1

6. Styren phn ng vi axit hipoclor cho ta clohirin C6H5-CHOH-CH2Cl. Cht ny chuynthnh epoxit trong mi trng kim:

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dU = HQ = P .dV dU = Q - dVP.2

1

m(U = Q - dVP.2

1

*Nhit ng tch: Nu h bin i V = const p dV = 0(U = QV QV l 1 hm trng thi.

*Nhit ng p:Nu h bin i P = const th:

dVP.2

1

= P . dV2

1

= P. V2 - P. V1

(U = U2 - U1 = QP - P. V2 + P .V1 QP = (U2 + P.V2) - (U1 + P .V1)t U + P.V = H = entanpi = hm trng thi

QP = H2 - H1 = (H = s bin thin entanpi ca h.*Nhit phn ng:Xt 1 h kn trong c phn ng: aA + bB p cC + dDNhit phn ng ca phn ng ny l nhit lng trao i vi mi trng khi a mol A phn ngvi b mol B to ra c mol C v d mol D T = const.- Nu phn ng c thc hin P = const th nhit phn ng c gi l nhit phn ng ng p

QP = (H- Nu phn ng c thc hin V = const th nhit phn ng c gi l nhit phn ng ngtch QV=(U* Quan h gia QP v QVQP = (H = ((U + PV)P = (U + P. (V (H = (U + P . (V = (U + (n .RTQP = QV + (n .RT ( (n = n kh sp - n kh p )Khi (n = 0 QP = QV hay (H = (U

U = QV = n .CV . T( (

H = QP = n .CP . T( ( *Nhit dung mol ng p (CP) l nhit lng cn cung cp lm 1 mol cht nng thm 1otrong iu kin ng p (m trong qu trnh khng c s bin i trng thi).

* Tng t vi CV: (H = 2

1

.T

T

P dTC ; (U = 2

1

.T

T

T dTC

CP, CV l hm ca nhit .

Vi 1 mol kh l tng: CP =T

H

(

(; CV =

T

U

(

(

M (U = (H - P. (V CP =T

H

(

(=

T

U

(

(+

T

VP

(

(.= CV + R

Q, W: Khng phi l hm trng thiQV = (U; QP = (H QV, QP l hm trng thi ch ph thuc vo trng thi u v trngthi cui ca h m khng ph thuc vo qu trnh bin i l thun nghch hay khng thunnghch.9) NH LUT HESS:(H ((U) ca 1 qu trnh ch ph thuc vo trng thi u v trng thicui ca h m khng ph thuc vo ng i.(Hp = (Hs (sn phm) - (Hs (cht u) = (Hc (cht u) - (Hc (sn phm)

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10) NH LUT KIRCHHOFF:

(

(

(

n1 n2 n3C n42

H2

HaHb

(n1 n2 n3C n4

1

H1

heo nh lut Hess: (H2 = (Ha (H1 (Hb

M:

(Ha = 2

1

)...(21

T

T

PP dTCnCn bA = - 2

1

)...( 21

T

T

PP dTCnCn BA

(Hb = 2

1

)...( 43

T

T

PP dTCnCn C

(H2 = (H1 + 2

1

)].()..[( 2143T

T

PPPP dTCnCnCnCn !"#C = (H1 + (2

1

.T

T

P dTC

- (H1 thng c xc nh iu kin chun: (HoT = (H

o298 + (

T

oP dTC

298

.

Vi (CoP = Co

P(sp) - CoP(tham gia)

CoP l nhit dung mol ng p iu kin chun (1atm).- Trong khong hp ca nhit c th coi (CoP = constTh: (H2 = (H1 + (CP.(T2 -T1)

(HoT = (Ho298 + (C

oP (T - 298)

11) ENTROPI (S)- Trong s bin i thun nghch v cng nh T = const h trao i vi mi trng mt lng

nhit HQTN th s bin thin entropi trong qu trnh ny l: dS =T

QTNH

S l hm trng thi (J/mol.K)

- Nu s bin i l bt thun nghch th dS >T

QTNH

- V l hm trng thi nn khi chuyn t trng thi 1 sang trng thi 2 bng bin thin thun

nghch hay bt thun nghch th S2 - S1 = (S = 2

1 T

QTNH

((STN = (SBTN)12) NGUYN L II CA NHIT NG HC:

dS uT

QH

- Trong h c lp HQ = 0. nn:+ dS = 0: trong h c lp entropi ca h khng i nu xy ra qu trnh thun nghch.

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+ dS > 0 : trong h c lp, qu trnh t xy ra (BTN) theo chiu tng entropi ca h v tng choti khi t gi tr max th h s t trng thi cn bng.* Entropi l thc o hn n ca h: hn n ca 1 h hay 1 cht cng ln khi h hay cht gm nhng ht v s dao ng ca cc ht cng mnh (khi lin kt gia cc ht cng yu).

: S S S

S S S$

2(k)%

2(k)%

3 (k)

$

2%

(r)$

2%

(l) $2

%

(h)

S l 1 i lng dung .13) S BIN THIN S TRONG QU TRNH BIN I TRNG THI CA CHT:Khi cht nguyn cht nng chy hoc si P = const th:

T = const (S = 2

1 T

QH=

T

H(

(H = nhit bin thin trng thi = Ln/c hoc Lh14)(S TRONG QU TRNH GIN N NG NHIT KH L TNG:Xt n mol kh l tng gin n th tch t V1p V2 t

o = const. V ni nng ca kh l tng ch

ph thuc nhit nn trong s bin i ny:(U = QTN + WTN = QBTN + WBTN = 0

QTN = - WTN = nRT. ln1

2

V

V( = -(- P. (V) = dV

V

nRT.

2

1

).

T = const (S =T

QTN = nRln1

2

V

V= n.R.ln

2

1

P

P

15) S BIN THIN ENTROPI CA CHT NGUYN CHT THEO NHIT .- Qu trnh P = const: un nng 1 cht nguyn cht t T1p T2, khng c s chuyn pha:

(S =

2

1

T

T

TN

T

QHVi HQ = HQP = dH = n.CP.dT

(S =T

dTCn

T

T

P..2

1

* Trong khong nhit hp, coi CP = const (S = n.CP.ln1

2

T

T

- Qu trnh: V = const (S = n .CV.ln1

2

T

T

16) ENTROPI TUYT I* Nguyn l III ca nhit ng hc:

- Entropi ca cht nguyn cht di dng tinh th hon chnh 0(K) bng 0: S (T = 0) = 0* Xut pht t tin trn ta c th tnh c entropi tuyt i ca cc cht cc nhit khcnhau.VD: Tnh S ca 1 cht nhit T no , ta hnh dung cht c un nng t 0(K) p T(K)xt P=const. Nu trong qu trnh un nng c s chuyn pha th:

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(S = (ST - (S(T = 0) = ST = !

(5

1iiS

ST =T

dTCn

T

Ln

T

dTCn

T

Ln

T

dTCn

T

T

hPS

ST

T

lPnc

ncT

rP

S

S

nc

nc

........ )()(0

)(

1

Gi tr entropi c xc nh P = 1 atm = const v nhit T no cgi l gi tr entropi chun, k hiu l S0T, thng T = 298Kp S

0298

17) S BIN THIN ENTROPI TRONG PHN NG HO HC:+ Khi phn ng thc hin P = const, T = const th: (S =

&

S(sp) -&

S(t/g)+ Nu iu kin chun v 250C th: (S0298=

&

S0298(sp) -&

S0298(t/g)+ V S ca cht kh >> cht rn, lng nn nu s mol kh sn phm (sp) > s molkh tham gia th (S > 0 v ngc li. Cn trong trng hp s mol kh 2 vbng nhau hoc phn ng khng c cht kh th (S c gi tr nh.18) TH NHIT NG

(Sc lp = (S h + (S mt 0a)Th ng p G:

Xt h xy ra s bin i P, T u khng i trong qu trnh ny mi trng nhn ca h mtnhit lng (Hmt do h to ra p(Hmt = - (H h = - (H

p(S mt = -T

H(

+ iu kin t din bin ca h:

p(S c lp = (S h -T

H(> 0 p(H T. (S < 0

+ H trng thi cn bng khi (H T. (S = 0+ t G = H TS nhit , P khng i th qu trnh xy ra theo chiu c

(G = (H T. (S < 0

V t ti trng thi cn bng khi (G = 0.b) Th ng tch: (Nng lng Helmholtz)Nu h bin i iu kin T, V khng i nhit ng tch m mi trng nhn ca cc h l

(Umt p(Smt = -T

Umt(

p iu kin t din bin ca h trong qu trnh ng nhit, ng tch l(F = (U T. (S < 0

V t trng thi cn bng khi (F = 0Trong : F = U TSV H = U + PV p G = H TS = U TS + PV p G = F + PV

+ i vi qu trnh T,P = const p(G = Wmax+ i vi qu trnh T, V = const p(S = WmaxTM LI :* Qu trnh ng p: P = const- Cng:HWP = - P.dV = -n.R.dT p WP = - P. (V = - nR(T

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B - H THNG CC CU HI V BI TP PHN NHIT HA HC :Bi 1:Cho 100 g N2 0

oC, 1atm. Tnh Q, W, (U, (H trong cc bin i sau y c tin hnh thunnghch nhit ng:a) Nung nng ng tch ti P = 1,5atm.

b) Gin ng p ti V = 2V ban u.c) Gin ng nhit ti V = 200ld) Gin on nhit ti V = 200lChp nhn rng N2 l kh l tng v nhit dung ng p khng i trong qu trnh th nghim vbng 29,1J/mol.KGii

a) V = const W = ! 0.dVP (U = QV = n VC .(T = ( PC - R).(T2 T1) .n

= ( PC - R).(1

2

P

P-1).T1.n = (29,1 - 8,314).( )1

1

5,1 .273,15 = 14194,04(J)

b) Vo = )(804,22.28

100 l! V= 2Vo = 160 (l)

W = -P. (V = -1(160 80) = -80 (l.at) = -80 .101,33 = -8106,4(J)

QP = (H = PC .n .(T =

11

1

2 ..1,29.28

100TT

V

V= 29,1.

28

100(2.273,15 273,15) = 28388,1(J)

(U = Q + W = 28388,1 = 8106,4 = 20281,7(J)c) T = const p(U = 0; (H = 0

W = - 2

1

.V

dVn

4

T = - nRT.ln1

2

V

V

W = -28

100 .8,314 .273,15.ln80200 = -7431,67(J)

(U = Q + W = 0 Q = -W = 7431,67(J)d) Q = 0 (S = const)Theo PT poisson: T1. V

11

K = T2 . V1

2K

p T2 = T1.(2

1

V

V)K-1 Vi 4,1

314,81,29

1,29}

!

!!

P

P

V

P

CR

C

C

CK

W = (U = n. VC (T2 T1) = 28

100(29,1-8,314).(189,33 -273,15) = 6222,4(J)

(H = n PC .(T = 28100 .29,1(189,33 273,15) = - 8711,3(J)

Bi 2:Tnh oSH 298,( ca Cl

-(aq). Bit:

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(a):2

1H2 +

2

1Cl2(k)p HCl(k)

oSH 298,( = -92,2(kJ)

(b): HCl(k) + aq p H+

(aq) + Cl-(aq)

oSH 298,( = -75,13(kJ)

(c):

2

1H2 + aq p H

+(aq) + e

oSH 298,( = 0

Gii:

Ly (a) + (b) (c) :2

1Cl2 + e + aq = Cl

-(aq)

oSH 298,( = - 167,33(kJ)

Bi 3:Tnh hiu ng nhit ca phn ng:

3Fe(NO3)2(aq) + 4HNO3(aq)p 3Fe(NO3)3(aq) + NO(k) + 2H2O (l)Din ra trong nc 25oC. Cho bit:

Fe2+(aq) Fe3+

(aq) NO3-(aq) NO(k) H2O(l)

oSH 298,( (kJ/mol) -87,86 - 47,7 -206,57 90,25 -285,6

Gii:Phng trnh ion ca phn ng:3Fe2+(aq) + 4H

+(aq) + NO3

-(aq)p 3Fe

3+(aq) + NO(k) + 2H2O (l)

(H=3. oSH 298,( (Fe3+

,aq)+oSH 298,( (NO)+2.

oSH 298,( (H2O(l))3.

oSH 298,( (Fe

2+,aq)-

oSH 298,( (NO3

-, aq)

= 3.(-47,7) + 90,25 + 2.(-285,6) + 3.87,6 + 206,57 = -153,9(kJ)Bi 4:1) So snh (H, (U ca cc phn ng: CnH2n + H2p CnH2n+2

2) Khi t chy hon ton 2 anome E v F ca D glucoz mi th 1 mol p sut khngi, ngi ta o c hiu ng nhit ca cc phn ng 500K ln lt bng:-2790,0kJ v - 2805,1kJ

a) Tnh (U i vi mi phn ng.

b) Trong 2 dng glucoz, dng no bn hnGii:1) (H = (U + P. (V = (U + (n.RTPhn ng trn c: (n = 1-2 = -1 (H = (U RT (H < (U2) C6H12O6 + 6O2p 6CO2 + 6H2O(U(E) = (H(E) - (n.RT = - 2799 6.8,314.10

-3.500 = -2824(kJ)(U(F) = (H(F) - (n.RT = - 2805,1 6.8,314.10

-3 .500 = -2830 (kJ)oH

E( = 6. o COSH )( 2( + 6.

oOHSH )( 2( -

oSH )(E(

oHF( = 6.o

COSH )( 2( + 6.o

OHSH )( 2( -oSH )(F(

oSH )(E( -oSH )(F( =

oHF( -oH

E( = -2805,1 + 2799 = -6,1(kJ)

oSH )(E( 139,82(kJ/mol)

O3 c cu trc vng kn rt khng bn cu trc ny khng chp nhn c.

Bi 6:Entanpi sinh tiu chun ca CH4(k) v C2H6(k) ln lt bng -74,80 v -84,60 kJ/mol. Tnh entanpitiu chun ca C4H10(k). Bin lun v kt qu thu c. Cho bit entanpi thng hoa ca than chv nng lng lin kt H- H ln lt bng: 710,6 v - 431,65 kJ/mol.Gii:* (1) C than ch + 2H2 (k)p CH4(k)

oCHSH 4,( =-74,8kJ

(2) Cthan ch

p C(k)

oth

H( = 710,6 kJ

(3) H2 (k)p 2H (k) lkH( = 431,65 kJ

Ly (1) [(2) + 2.(3)] ta c:C(k) + 4H(k) p CH4(k)

oCHtungSH 4,/,( = -1648,7(kJ/mol)

Nng lng lin kt trung bnh ca lin kt C H l:4

1(-1648,7) = - 412,175 (J/mol).

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* (4) 2C than ch + 3H2p C2H6(k)o

KHCSH ),( 62( = -84,6 (kJ/mol)

Ly (4) [2 .(2) + 3.(3)] ta c:2C(k) + 6H (k)p C2H6(k)

oHCtungSH

62,/,( = -2800,75 (kJ/mol)

Coi EC H trong CH4 v C2H6 nh nhau th:

E C- C = =1800,75 6(- 412,175) = -327,7(kJ/mol)* Coi EC-H; EC- C trong cc cht CH4, C2H6, C4H10 u nh nhau th:o

HCtungSH104

,/,( = 3. EC- C + 10.EC- H = 3.(- 327,7) + 10( -412,75) = -5110,6 (kJ/mol)

* (5) 4C(k) + 10 H(k)p C4H10(k)o

HCtungSH104

,/,( = -5110,6 (kJ/mol)

Ly (2). 4 + (3).5 + (5) ta c:

4Cthan ch + 5H2(k)p C4H10(k)o

HCSH104

,( = -109,95(kJ/mol)

* Kt qu thu c ch l gn ng do coi E lk(C C), Elk(C- H) trong mi trng hp l nh nhau.V v vy s khng tnh r c oSH( ca cc ng phn khc nhau.

Bi 7: Tnh (Ho ca cc phn ng sau:1) Fe2O3(r) + 2Al(r)p 2Fe(r) + Al2O3(r) ( 1)Cho bit o OFeS rH )(32,( = -822,2 kJ/mol;

oOAlS r

H)(32,

( = -1676 (kJ/mol)

2) S(r) +2

3O2(k)p SO3(k) (2)

Bit (3) : S(r) + O2(k)p SO2(k)oH298( = -296,6 kJ

(4): 2SO2(k) + O2(k)p 2SO3(k)oH298( = -195,96 kJ

T kt qu thu c v kh nng din bin thc t ca 2 phn ng trn c th rt ra kt lun gGii:1) opuH )1(( =

oOAlS r

H)(32,

( - o OFeS rH )(32,( = -1676 + 822,2 = - 853,8(kJ)

2) opuH )2(( =o

puH )3(( +2

1 opuH )4(( = -296,6 -

2

1.195,96 = -394,58 (kJ)

KL: Hai phn ng (1), (2) u to nhit mnh. Song trn thc t 2 phn ng khng t xy ra.Nh vy, ch da vo (H khng khng nh chiu ca 1 qu trnh ho hc (tuy nhin trongnhiu trng hp, d on theo tiu chun ny l ng).

Bi 8:1) Tnh hiu ng nhit ng tch tiu chun ca cc phn ng sau 25oC.

a) Fe2O3(r) + 3CO(k) p 2Fe(r) + 3CO2(k)

o

H298( = 28,17 (kJ)b) Cthan ch + O2(k)p CO2 (k)

oH298( = -393,1(kJ)

c) Zn(r) + S(r)p ZnS(r)oH298( = -202,9(kJ)

d) 2SO2(k) + O2(k)p 2SO3(k)oH298( = -195,96 (kJ)

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!( oH1000oH298( + dTCP,

1000

298 ( =

oH298( + dTT)10.541,311,32(1000

298

3

= oH298( + )210.541,311,32(

21000

298

3 TT

= - 46,2.103 +31,541 .10-3.21 (10002 -1982) 32,1(1000 298)= - 54364,183 (J/mol)

Khi tng hp 17 kg NH3 th nhit lng to ra l:

Q =17

17000.(-54364,183 .10-3) = -54364,183 (kJ)

Bi 11:Tnh nng lng mng li tinh th BaCl2 t 2 t hp d kin sau:

1) Entanpi sinh ca BaCl2 tinh th: - 859,41 kJ/molEntanpi phn li ca Cl2: 238,26 kJ/mol

Entanpi thng hoa ca Ba: 192,28 kJ/molNng lng ion ho th nht ca Ba: 500,76 kJ/molNng lng ion ho th hai ca Ba: 961,40 kJ/moli lc electron ca Cl : - 363,66 kJ/mol2) Hiu ng nhit ca qu trnh ho tan 1 mol BaCl2 vo g mol H2O l: -10,16kJ/mol.Nhit hirat ho ion Ba2+ : - 1344 kJ/molNhit hirat ho ion Cl- : - 363 kJ/molTrong cc kt qu thu c, kt qu no ng tin cy hn.Gii:

Ba(r) + Cl2(k) BaCl2(tt)

Hth(Ba) Uml

Ba(k) + 2Cl (k) Ba2+ + 2Cl-

I1(Ba) + I2(Ba)

2. ACl

Uml = H - Hth (Ba) - H - I1(Ba) - I2(Ba) - 2ACl

= - 859,41 - 192,28 - 238,26 - 500,76 - 961,40 + 2 .363,66

= - 2024,79 (kJ/mol)

HS(BaCl2, tt)

Hpl(Cl2)

o(

( (

S(BaCl2, tt)pl(Cl2)

( ((o

(

2) BaCl2 (tt) Ba(aq) + 2Cl(aq)

- Uml +H2O

H1 H2

Ba2+ + 2Cl-

Hht(BaCl2)

((

2+ -

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P

V

V

V

3 3

10,K = =

32

K

K1-53

53

3

T@

73,15 .(10)-0,4 108,74 (K)

U W . (108,74 - 73,15) .101,33 9148,6(J)73,16

10 .10(

V2 = = 39,81 (l)1 .273,110.10.108,74

P2 .T1

P1 .V1.T2 ~~

c) Q = 0 (U = W

m n. VC (T2 T1) = -Png .(V2 V1) = -P2 .

1

1

2

2

P

nRT

P

nRT

m n.2

3R(T2 T1) = -nR.1

10112 TT m T2 = 0,64T1

V2 64(l)A

.0. 0.0,64 A

2 . A

A .V1. 2 =

(U = W = -Png(V2 V1) = -1(64 10) = -54(l.atm)= -54(l.atm) .101,33 .J/l.atm = - 5471,82 (J)

Bi 13 :

Phn ng sau: Ag +2

1Cl2 = AgCl

Xy ra di p sut 1 atm v 25oC to ra 1 nhit lng l 126,566 kJ. Nu cho phn ng xy ra trong 1 nguyn t ganvani P, T = const th ho nng s cchuyn thnh in nng v sn ra cng W = 109,622 kJ.Hy chng t rng trong c 2 trng hp trn, bin thin ni nng ca h vn ch l mt, cnnhit th khc nhau v tnh gi tr bin thin ni nng .Gii:- Do U l hm trng thi nn (U = U2 U1 = const, cho d s bin i c thc hin bng cchno. V vy (U trong 2 trng hp trn ch l mt.- V (U = Q + W = Q + W - P(V = Q + W - (n.RTDo (nRT = const; (U = constNn khi W (cng c ch) thay i th Q cng thay i

- (U = (H - (nRT = -126,566 +2

1. 8,314 .298,15.10-3 = - 125,327 (kJ)

Bi 14:

Tnh cng ca s bin i ng nhit thun nghch v bt thun nghch 42g kh N2 300K khi:a) Gin n t 5atm n 1atm.b) Nn t 1atm n 5atm.(Kh c coi l l tng). So snh cc kt qu v rt ra kt lun.Gii:

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a) * WTN = -1

2

1

22

1

2

1

lnlnP

PnRT

V

VnRT

V

dVnRTPdV !!!

WTN =28

42.8,314 .300. ln

5

1= -6201,39(J)

*WBTN = - Png . (V = -Png(V2 V1) = -Png

1

2

11 VP

VP = - P2.V1

1

2

1

PP

= - P2.

!

1

2

2

1

1

11.P

PnRT

P

P

P

nRT= -

28

42.8,314 .300

5

11 = -2993,04 (J)

b) WTN = nRTln1

2

P

P=

28

42.8,314 .300.ln

1

5= 6201,39(J)

WBTN = - Png. (V= -Png(V2 V1) = -Png

12 P

nRT

P

nRT

= -nRT.P2

1211 PP = -nRT

1

21 P

P= - 2842 .8,314 .300

1

51 = 14965,2 (J)

KL: - Cng m h thc hin (sinh) trong qu trnh bin thin thun nghch t trng thi 1 ntrng thi 2 bng cng m h nhn khi t trng thi 2 v trng thi 1. Cn trong qu trnh binthin bt thun nghch th cng h sinh nh hn cng h nhn.- Trong s bin thin thun nghch th h sinh cng ln hn trong qu trnh bin thin bt thunnghch.

Bi 15: Phn ng: C6H6 +2

15O2(k)p 6CO2(k) + 3H2O

300K c QP QV = 1245(J). Hi C6H6 v H2O trong phn ng trng thi lng hay hiGii:

QP QV =(nRT = 1245(J) p(n =300.314,8

1245= 0,5

H2O v C6H6 phi th hi th (n = 0,5

Bi 16:Tnh nhit lng cn thit nng nhit ca 0,5 mol H2O t -50

oC n 500oC P = 1atm. Bit nhit nng chy ca nc 273K l Lnc = 6004J/mol,

nhit bay hi ca nc 373K l Lh = 40660 J/mol.o

hOHPC ),( 2= 30,2 + 10-2T(J/molK) ; o rOHPC ),( 2 = 35,56(J/molK);

olOHPC ),( 2

= 75,3(J/molK)

Gii:

H2O(r) H2O(r) H2O(l) H2O(l) H2O(h) H2O(h)(500oC)

H1 H2 H3 H4 H5

-50oC 0oC 0oC 100oC 100oC

( (( ( (

(Ho = ho

lPnco

rP LndTCnLndTCnH ......373

273

)(

273

223

)(

5

1

!( + 773

373

)( .. dTCno

hP

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- S bin thin entropi c tnh theo CT: (S = nRln1

2

V

V

2NS( = 0,2 .8,314.ln

V

V

2,0= 2,676J/K

2HS( = 0,5.8,314.ln VV5,0 = 2,881J/K

3NHS( = 0,3.8,314.ln

V

V

3,0= 3,003J/K

(S =2N

S( +2H

S( +3NH

S( = 8,56(J/K)

* Qu trnh khuch tn kh l tng l ng nhit nn (H = 0(G 273 = (H T. (S = -273.8,56 = -2336,88(J)Bi 20: Trong cc phn ng sau, nhng phn ng no c (S > 0; (S < 0 v (S { 0 t.

C(r) + CO2(k) p 2CO(k) (1)

CO(k)

+2

1O

2(k)p CO

2(k)(2)

H2(k) + Cl2(k)p 2HCl(k) (3)S(r) + O2(k)p SO2(k) (4)

Gii:Phn ng (1) c (n kh = 2 -1 = 1 > 0 p(S > 0

Phn ng (2) c (n kh = 1 -1-2

1< 0 p(S

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o(S298,p- = S

o - So - So298(H2O)298(C2H4)298(C2H5OH)

= 282 - 219,45 - 188,72 = - 126,17(J/K)G = H - T. S

H298,p- = G298,p- + T. S298,p-

= -8,13 + 298(- 126,17 .10-3) = - 45,72866(kJ)H298,p- < 0 phn ng to nhit

( ( (

( ( (

(

o

o

o o

Bi 22: Mt mol kh l tng n nguyn t 300K v 15atm gin n ti p sut 1atm. Sgin n c thc hin bng con ng:a) ng nhit v thun nghch nhit ng.b) ng nhit v khng thun nghch.c) on nhit v thun nghch.d) on nhit bt thun nghch.Trong cc qu trnh bt thun nghch, s gin n chng li p sut 1atm. Tnh Q, W, (U, (H, (Stpcho mi trng hp.Gii:a) T = const p(U = 0 ; (H = 0

WTN = - 2

1

PdV = - nRTln1

2

V

V= -nRTln

2

1

P

P

WTN = -1(mol).8,314 (J.mol-1K-1) .ln

1

15.300(K) = -6754,42(J)

Q = -W = 6754,42(J)Qu trnh gin n thun nghch: (Stp = (Smt + (Sh = 0b) T = const p(U = 0 ; (H = 0

WBTN = -Png(V2 - V1) = -P2(2P

nRT-

1P

nRT) = nRT(

1

2

P

P- 1) = 1. 8,314.300.(

15

1- 1)= -2327,92(J)

pQBTN = -W = 2327,92(J)(Stp = (Smt + (Sh

(Sh(BTN) = (Sh(TN) =T

QTN = !(

T

WU2

1 T

dTnCV + dVV

nR2

1

= nRln1

2

V

V= nRln

2

1

P

P

= 1.8,314 .ln1

15= 22,515(J/K)

(Smt = T

Qmt

= - T

Qhe

= 300

92,2327

= -7,76(J/K) (Stp = 22,515 - 7,76 = 14,755 (J/K)( Qu trnh gin n ny t xy ra)c) on nhit Q = 0on nhit thun nghch Theo Poisson T.VK- 1 = const

M V =P

nRT T.

1

K

P

nRT= const p TK .P1-K = const

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T1 . 1 T2 . 2

T1T2

=P2

P1

K K

KT1

T2

P2

P1=

K

KE - K

E -

KE -

KE -

T2 = T1 .P2

P1KK1-

Vi kh l tng n nguyn t th CV =2

3R; CP =

2

5R

K = =CPCV

53

1-=

1-= -0,4

2 = 300.-0,4

= 101,55( )

K

K

53

53151

(U = W = nCV( 2- 1) = 1.2

3.8,314.(101,55 - 300) = -2474,87(J)

(H = nCP( 2- 1) = 1.2

5.8,314 .(101,55 - 300)= - 4124,78(J)

(SF N =

T

Q= 0

d) on nhit Q = 0on nhit, khng thun nghch khng p dng c PT poisson(U = W m nCV. (T = -Png. (V

m n.2

3.R(T2 - T1) = -P2(

2

2

P

nRT-

1

1

P

nRT)

m2

3(T2 - 300) = -( T2 -

1

2

P

P.T1)

m2

3T2 - 450 = -T2 +

15

1.300

p V2 =2

2

PnRT =

1188.082,0.1 = 15,416(l)

V1 =2

2

P

nRT= 1,64(l)

(U = W = 1.2

3 .8,314.(188- 300) = -1396,752(J)

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(S v tr = (Smt + (S h = 04) Cc na phn ng:Anot: C3H8 + 26OH

-p 3 23CO + 17H2O + 20e

Catot: O2 + 2H2O + 4e p 4OH-

Phn ng tng cng:

C3H8(k) + 5O2(k) + 6OH-(aq)p 3 2 )(3 aqCO + 7H2O(l)

S pin:(-) Pt, C3H8(1atm)/KOH(5M), K2CO3(1M)/ O2(1atm), Pt (+)

0puH( = 3(-677,14) + 7.(-285,83) + 103,85 - 5.0 - 6(-229,99) = -2548,44(KJ)0

puS( = 3.(-56,9) + 7.69,91 - 269,91 - 5.205,138- 6(-10,74) = -912,43(KJ)0

puG( =0

puH( = T.0

puS( = -2548,44 + 298,15.912,43.10-3 = - 2276,399(KJ)

0puE = - nF

G 0(=

96485.20

2276399= 1,18(V)

p E = E0 - 56

32

3

283..][

][lg20

0592,0

OHC PPOH

CO

= 1,18 - 20

0592,0lg(5)-6 = 1,19(V)

p P = E .I = 1,19 .0,1 = 0,119(W)

Bi 29: Tnh bin thin entropi khi chuyn 418,4J nhit t vt c t0 = 150oC n vt c t0 =50oC.Gii:Qu trnh bin i trn l khng thun nghch c coi nh gm 3 qu trnh bin thin thunnghch:1) Vt 150oC truyn nhit thun nghch T = const.

(S1 = T

Q

= 15,273150

4,418

= - 0,989(J/K)2) H bin thin on nhit t 150oC n 50oC(S2 = 03) Vt 50oC nhn nhit thun nghch T = const

(S3 = -T

Q=

15,27350

4,418

= 1,295(J/K)

Do S l hm trng thi nn:(SBTN = (STN = (S1 + (S2 + (S3 = 0,306(J/K)

Bi 30: Bit -15oC, Phi(H2O, l) = 1,428 (torr)

-15oC, Phi (H2O,r) = 1,215(torr)Hy tnh (G ca qu trnh ng c 1 mol H2O(l) thnh nc -15

oC v 1atm.Gii:

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1 o , 1 mol HG l 1 o , 1mol HG (r)

( )

GBTN= ?

(1)(Qu trnh TN oHG hi, bo ho

nm cn bng v i HG

(l))

1 o , 1mol HG1,4 Torr

( )1 o , 1mol HG (h)

1, 1 Torr

(

(1), (3) l qu trnh chuyn pha thun nghchp(G1 = (G3 = 0

p(G = (G2 = nRTln1

2

P

P= 1.8,314. 258,15 ln

428,1

215,1= -346,687(J)

Bi 31: C 1 mol O2 nguyn cht 25oC, 2atm, 1 mol O2 nguyn cht 25

oC, 1atm1 mol O2 25

oC trong khng kh trn mt t (P = 1atm, O2 chim 21% V khng kh)- So snh gi tr hm G ca 1 mol O2 trong 3 trng hp trn hn km nhau bao nhiu J . T rt ra kt lun: Kh nng phn ng ca O2 trong mi trng hp trn cao hay thp hn so vitrng hp khcGii:* G0 l hm Gibb ca 1 mol O2 1atm- 1 mol O2, 1atm, 25

oC p 1 mol O2, 2atm, 25oC

(G0) (G1)

(G1 = G1 - G0 = nRTln

1

2

P

P= 1. 8,3145 .298,15.ln

1

2 = 1718,29(J)

G1 > G0.

- Gi G2 l hm Gibb ca 1mol O2 25oC trong khng kh (0,21 atm)1mol O2, 25

oC, 1atm p 1 mol O2, 25oC, 0,21atm

(G0) (G2)(G2 = G2 - G

0 = 1. 8,3145 .298,15.ln1

21,0 = -3868,8(J)

G2 < G0.

Vy:G2(1mol O2, 25

oC, 0,21atm) < G0(1 mol O2, 25oC, 1atm) < G1(1 mol H2O, 25oC, 2atm)

- 1 cht c hm G cng cao th cng km bn 1 mol O2 25oC, 2atm c kh nng phn ng cao

nht cn 1 mol O2 nm trong khng kh th b nht c kh nng phn ng km nht.

Bi 32: Nhit ho tan ((Hht) 0,672g phenol trong 135,9g cloro om l -88J v ca 1,56gphenol trong 148,69g cloro om l -172J.Tnh nhit pha long i vi dung dch c nng nh dung dch th 2 cha 1 mol phenol khipha long n nng ca dung dch th nht bng cloro om.Gii:

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Bi 35:a) Tnh cng trong qu trnh t chy 1 mol ru etylic kc v 25oC.

b) Nu H2O dng hi th cng km theo qu trnh ny l bao nhiuGii:

a) C2H5OH(l) + 3O2 (k)p 2CO2 (k) + 3H2O (l)(n = -1

W = -Png . (V = -Png.ngP

RTn.(= R.T = 8,314.29815 = 2478,82 (J)

b) Nu H2O dng hi th: (n = 2. W = - (n. RT = -2 .8,314 .298,15 = - 4957,64(J)

Bi 36: Tnh (S, (G trong qu trnh gin khng thun nghch 2 mol kh l tng t 4lt n20 lt 54oC.Gii:V S, G l cc hm trng thi nn (S, (G khng ph thuc vo qu trnh bin thin l thunnghch hay bt thun nghch m ch ph thuc vo trng thi u v trng thi cui. V vy:

(S = nRln1

2

V

V= 2. 8,314.ln

4

20= 26,76 (J/K)

T = const p(H = 0; (U = 0p(G = (H - T. (S = 0 -(273,15 + 54) .26,76 = - 8755,1 (J)

Bi 37: Mt bnh c th tch V = 5(l) c ngn lm 2 phn bng nhau. Phn 1 cha N 2 298K v p sut 2atm, phn 2 298K v p sut 1atm. Tnh (G, (H, (S ca qu trnh trn ln 2kh khi ngi ta b vch ngn i.Gii:

T = 298K ; Vb (N2) = Vb(O2) =25 (l)

(S = (S(N2) + (S(O2) =2N

n .Rln1

2

V

V+

2On Rln

1

2

V

V=

RT

VP NN 22 . Rln5,2

5+

RT

VP OO 22 . Rln5,2

5

= 2ln..

2222

T

VPVP OONN = 0,0174(l.at/K) = 0,0174 .101,325 = 1,763 (J/K)

- Qu trnh ng nhit p(H = 0(G = (H - T. (S = - 298 .1,763 = - 525,374 (J)

Bi 38: Cho cc d liu sau y 298KCht 0SH( (kJ/molK) S

0(J/molK) V(m3/mol)

Cthan ch 0,00 5,696 5,31.106

Ckim cng 1,90 2,427 3,416.10-6

1) 298K c th c mt phn rt nh kim cng cng tn ti vi than ch c khng2) Tnh p sut ti thiu phi dng c th iu ch c kim cng 298K

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0puS( =

0

)(2 aqZnS +

0)(rCuS -

0)(rZnS -

0

)(2 aqCuS = -106,5 + 33,3 - 41,6 + 98,7 = -16,1 (J/K)

0puG( =

0H( - T. 0S( = -216,79 + 298,15 .16,1.10-3= -211,99(kJ)

(Uo = QP =0

puH( = -216,79 (kJ)

W = 0; qu trnh BTN; W = 0

b) * 0puG( = -211,99 (kJ) 0 Qu trnh l bt thun nghch m phn ng t xy ra.c) Khi thc hin phn ng trn TN trong pin in th cc gi tr (H0, (S0, (G0, (U0 khng thayi do H, S, G, U l cc hm trng thi nn khng ph thuc qu trnh bin i l thun nghch

hay bt thun nghch nhng cc gi tr Q, W th thay i.C th: Wtt = 0; Wmax = (G0 = -211,99(kJ)Q = T. (S = 298,15 .(-16,1) = - 4800,215 (J)

(Smt =T

Qmt =T

Qhe = 16,1 (J/K) (S v tr = (Smt + (Sh = 0

Epin = -nF

G 0(=

96485.2

211990} 1,1(V)

Bi 42:i vi nguyn t anien 15oC ngi ta xc nh c sc in ng E = 1,09337V v h

s nhit ca sc in ng

T

E

x

x= 0,000429 V/K. Hy tnh hiu ng nhit ca phn ng ho

hcGii:

(G = - nEF T

G

x

x= - nF.

T

E

x

x= - (S (S = nF .

T

E

x

x

(H = (G + T. (S = nF.(T.T

E

x

x- E)

(H = 2. 96485 .(298,15.0,000429 - 1,09337) - - 187162,5(J)Bi 43: Cho phn ng ho hc: Zn + Cu2+p Zn2+ + Cuxy ra mt cch thun nghch ng nhit, ng p 25oC trong nguyn t Ganvani.Sc in ng ca nguyn t o c l 1,1V v h s nhit ca sc in ng l

PT

E

x

x= 3,3.10-5 (V/K).

a) Tnh hiu ng nhit Q, bin thin Gipx(G v bin thin entropi (S ca phn ng ho hc cho.b) Tnh Qtn ca qu trnh

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PHN CU TO CHTTrng THPT chuyn Thi Bnh

A- MT S VN CHUNG V L THUYT

I. cc c trng v cu to phn t:Mt phn t hnh thnh c v tn ti bn nh kt qu ca tng tc gia cc ht nhn velectron dn n mt nng lng h cc tiu (nng lng ny ca phn t phi thp hn nnglng ca h ban u). Trong phn t c s phn b v tr tng i gia cc ht nhn nguyn tnn c c hnh dng khng gian ca phn t vi di lin kt v gc xc nh.1) Nng lng lin ktNng lng lin kt gia hai nguyn t A v B l nng lng cn thit va ph v lin kt hay nng lng to ra khi hai nguyn t A v B trng thi c bn kt hp vi nhau. Tuynhin nng lng lin kt l su ca cc tiu nng lng trn c cong th nngTh d: phn ng H2 2 H cn nng lng bng 436 kJ.mol

-1. Phn t H2 bn vng nn khi chohai nguyn t H kt hp vi nhau: 2 H H2 to ra mt nng lng bng 436 kJ.mol-1. Nh vy

hai gi tr nng lng bng nhau v gi tr v ngc nhau v du. Quy c rng nng lng linkt c du dng bin lun rng lin kt cng bn th nng lng lin kt cng ln??? nn EH-

H = 436 kJ.mol-1.

Trong phn t c nhiu lin kt th nng lng lin kt c tnh trung bnh.2) di lin kt

di ca mt lin kt trong phn t l khong cch trung bnh gia hai ht nhn nguyn t tora lin kt khi phn t trng thi nng lng thp nht. di lin kt thng c k hiu ld.

Phng php ph vi sng hay phng php nhiu x electron thng c dng xc nh di lin kt. Tr s di lin kt trong khong t 0,74 (phn t H 2) n 4,47 (phn t CS2);thng thng trong khong 1,0 2,0 i vi lin kt gia hai nguyn t ca cc nguyn t chu

k 2, 3, 4. di ca mt lin kt no thng gn ng l mt hng s trong cc phn t khc nhau.Chng hn lin kt n C-C trong hu ht cc phn t hirocacbon khng lin hp vo khong1,53-1,54. Trong C6H6 (benzen) di lin kt gia hai nguyn t C cnh nhau bng 1,40 .Tr s ny nm trong khong di mt lin kt C-C l 1,54 v di mt lin kt i C=C l1,34. di lin kt cng nh, lin kt cng bn.

Bn knh lin kt: T cc s liu c th thy rng di lin kt dAB xp x bng 1/2(dAA + dBB)vi dAA, dBB l di lin kt A-A, B-B tng ng. Chng hn, coi A l Cl, B l Cl; bit dCl-Cl= 1,99A, vy dC-Cl = 1/2(dC-C + dCl-Cl) = 1/2(1,54 + 1,99) = 1,765. Tr s thc nghim cho bitdC-Cl = 1,766. Do ngi ta coi 1/2dAA l bn knh lin kt hay bn knh cng ho tr rA canguyn t .

3) Gc lin kt :Gc lin kt l gc to bi hai na ng thng xut pht t mt ht nhn nguyn t i qua haiht nhn ca hai nguyn t lin kt vi nguyn t .

Cc trng hp in hnh v gc lin kt theo nh ngha trn l:- Phn t thng, gc lin kt bng 180o (2T); chng hn C2H2, CO2,- Phn t c gc, gc lin kt khc 180o, chng hn BF3 hay C2H4 c gc 120

o, H2O c gc104,5o,

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- Phn t t din, gc lin kt bng 109o28, chng hn CH4,Trong mt s trng hp, ngi ta ch n gc c to ra t 4 nguyn t hay 2 mt phng,

l gc nh din hay gc xon (hay gc vn). Di y l hnh nh mt s phn t cho thy chngc kch thc ring.

4) Cc dng lin kt ho hcXt mt cch i cng, lin kt ho hc c bn dng:- Lin kt cng ho tr (hay lin kt nguyn t)- Lin kt ion (hay lin kt in ho tr)- Lin kt kim loi- Lin kt hiro, tng tc Van de Van; gi chung l tng tc yu.

Thc t khng c ranh gii r rt gia cc dng lin kt . Tuy nhin, thun li khi xem xt,ngi ta vn cp ring tng dng , hai dng u thng c cp n nhiu hn.

II. Quy tc bt t(Octet):

T s phn tch kt qu thc nghim v cu to ho hc ca cc phn t, nm 1916 nh hohc Cxen (Kossel) v Liuyx (Lewis) a ra nhn xt m ngy nay gi l quy tc bt t (hayquy tc octet): Khi to lin kt ho hc, cc nguyn t c xu hng t ti cu hnh lp ngoicng bn vng ca nguyn t kh tr vi 8e.

Cn lu l quy tc ch p dng c cho mt s gii hn cc nguyn t, ch yu l ccnguyn t chu k 2. Quy tc bt t (octet) th hin trong tng dng lin kt c th. Thngthng trong lin kt ion, sau khi cho nhn electron lp v ngoi cng c s electron nhcc nguyn t kh him. Thc t quy lut y ch ng cho a s cc trng hp nguyn t nhmA. (Hc vin ly th d v nhng trng hp khng tun theo quy tc bt t).

III. Thuyt liuytx(Lewis) (nm 1916):

1. Ni dung ca thuyt:

Trong phn t c to ra t nguyn t cc nguyn t phi kim, lin kt ho hc gia hainguyn t c thc hin bi cp (i) e dng chung, nh m mi nguyn t u c ccu hnh lp ngoi cng bn vng ca nguyn t kh tr vi 8e.

Electron ca mi nguyn t c th tham gia c lin kt l e ho tr. i e to lin kt phic spin i song.

V d: Phn t Cl2 c lin kt gia hai nguyn t Cl c thc hin nh cp e gp chung.cp e ny l cp e lin kt, c k hiu oq hay qo, cc e cn li c gi l e khng lin kt.

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a) Phn loi lin kt cng ho trCn c vo v tr cp e dng chung so vi ht nhn nguyn t tham gia lin kt, ngi ta chia

lin kt cng ho tr thnh hai loi:- Lin kt cng ho tr khng phn cc (hay khng c cc): i e dng chung gia khong

cch hai ht nhn nguyn t. l lin kt trong cc phn t n cht nh Cl 2, Br2... (trng

hp hiu m in e 0,4).- Lin kt cng ho tr c cc (hay phn cc): i e dng chung lch v pha nguyn t canguyn t c tnh phi kim mnh hn (hay c m in ln hn). l lin kt ho hc trongcc phn t hp cht nh H2O, NH3, CH4,... (hiu m in trong khong 0,40 z 1,70).

b) Tnh nh hng khng gian ca lin kt cng ho trLin kt cng ho tr c tnh nh hng khng gian. Trong lin kt cng ho tr, cp e dng

chung (hay cp e lin kt) c phn b khong khng gian gia hai ht nhn tham gia linkt.

Lin kt cng ho tr c tnh cht bo ho. Chng hn trong hp cht gia Cl vi H, ch c 1nguyn t H lin kt vi 1 nguyn t Cl to thnh HCl; khng th c nhiu hn mt nguyn tH lin kt vi mt nguyn t Cl. Do vy s nguyn t lin kt vi mt nguyn t cho trc b

hn ch bi ho tr ca nguyn t .Bi tp: Cho cc nguyn t H, F, Cl, Br, I.1) Hy vit CTPT ca cc cht c to ra t cc nguyn t cho.2) Trong s cc cht nu cht no c lin kt khng c cc, c cc Hy ch r v tr ca

cp electron lin kt trong mi cht.2. Cng thc cu toLiuytx(Lewis):

Biu din lin kt v cu to phn t kh trc quan1. Cng thc:Mi du chm biu th mt electron. Hai chm hay mt vch ch mt cp electron trong

nguyn t hay phn t. Cc electron ny l cc electron ho tr. Cng thc ho hc ch r th tlin kt gia cc nguyn t v cc k hiu ch s phn b electron ho tr c gi l cng thc

Lewis (do Lewis xng). Thng thng cc cp electron lin kt vit bng du vch,electron khng lin kt biu din bng chm. Cng thc Lewis khng ch dng cho cc hp chtc lin kt cng ho tr m dng c cho cc hp cht c lin kt ion.

Bi tp:Vit cng thc Lewis cho cc phn t : a) nit, b) nc, c) Canxi clorua.2. Cch vit cng thc Lewis:

a) Cc khi nim cn dng:+ Nguyn t trung tm v phi t: Trong mt cng thc ho hc, c nguyn t trung tm l

nguyn t cn nhiu e nht to c cu hnh tm electron (octet) lp ngoi cng ca n(hay nguyn t c s oxi ho cao nht); cc nguyn t khc v c cp electron khng lin ktca nguyn t trung tm c gi l phi t. V d: trong phn t NH3, nguyn t trung tm l

N, phi t gm 3H v 1 cp e khng lin kt ca N ( v ho tr). Trong phn t HCN, nguynt trung tm l C, phi t gm 1H v 1N ( y khng c cp e khng lin kt v ho tr).+ Li ca nguyn t: Phn li ca mt nguyn t (khi nguyn t ny l thnh phn ca mt

cng thc ho hc c xt) gm ht nhn v cc electron cc lp bn trong. V d: Xt linguyn t ca cc nguyn t trong HCN ta c: li nguyn t N gm ht nhn v hai e phnlp 1s2; li nguyn t C gm ht nhn v 2 electron phn lp 1s2; li nguyn t H ch gm htnhn, thc t H thng c coi l trng hp ngoi l.

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+ in tch:- in tch li nguyn t: l s n v in tch ca nguyn t khi ta b cc electron lp ho

tr i nn l mt s nguyn dng, c tr s bng s e ho tr vn c ca nguyn t .- in tch hnh thc ca mt nguyn t = (in tch ca li nguyn t - tng s e ring ca

nguyn t tng s e to lin kt c nguyn t tham gia/2).

V d: Xc nh in tch hnh thc ca N trong NH3, NH4+

- Trong NH3: T cu to Lewis, ta thy:in tch li ca N l 5S e khng lin kt ca N l 2Tng s e to lin kt c N tham gia l 6 (hay c 3 lin kt)

Vy in tch hnh thc ca N = 5 2 6/2 = 0- Trong NH4

+: Xt tng t nh trn, ch N khng cn e khng lin kt v N tham gia 4 linkt vi 4 H.

Vy in tch hnh thc ca N = 5 0 8/2 = +1y chnh l in tch ca c nhm NH4

+.b) Cc bc vit cu to Lewis: HCN

Bc 1: Vit cng thc cu to s b ca cht da vo ho tr ca cc nguyn t v gi thitrng ch c lin kt n c hnh thnh. Nu cha bit th t lin kt gia cc nguyn t, hydng gi thit vit th t .

y ta c: H : C : N (a) hay H : N : C (b)Bc 2: gi n1 l tng s e ho tr ca cc nguyn t.- Thng thng da vo cu hnh e ca cc nguyn t

H: 1s2p 1eC: 1s2 2s2 2p2p 4eN: 1s2 2s2 2p3p 5e

Vy n1 = (1 + 4 + 5) e = 10 eCh : Nu cng thc l:

+I

on m: 1 n v in tch m do c cng thm 1e vo tng trn.+ Ion dng: 1 n v in tch dng do tr i 1e t tng trn.HCN l phn t trung ho nn khng p dng phn ny.Bc 3: Tm cng thc Lewis (gn ng)- gi n2 l tng s e to lin kt trong cng thc a ra bc 1. S e cn li khng tham

gia lin kt n3 = n1 n2- S e cn ly to bt t cho nguyn t m in nht trong cng thc ban u bng n4.Khi p dng ba bc trn cho HCN. n2 = 4e, vy n3 = n1 n2 = 6e.Trong (a), N m in hn C nn phi to bt t cho N. Trong cng thc ban u N mi c 2e,

n cn 6e na mi thnh 8 e. Nh vy n4 = 6e.Bc 4: Tm cng thc Lewis ng

- Tm s e cn li, k hiu n5 = n3 n4+ Nu n5 = 0: tnh in tch hnh thc mi nguyn t trong cng thc va vit bc 3.+ Nu n5{ 0: chnh l s e cn dng to bt t cho nguyn t trung tm.

Ch : Vic ny ch c thc hin khi nguyn t trung tm l nguyn t ca nguyn t thucchu k 3 tr i.

Sau tnh li in tch hnh thc cho mi nguyn t trong cng thc va vit.p dng: Vi HCN c n3 = n4 = 6e nn n5 = 0.

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IV. Thuyt sc y gia cc cp electron:

1. M hnh s y gia cc cp electron v ho tr:Mi lin kt cng ho tr gia hai nguyn t c to thnh nh cp electron lin kt hay cp

electron dng chung. i e lin kt phn b trong khong khng gian gia hai ht nhn nguynt to ra lin kt . Lin kt cng ho tr l lin kt c nh hng khng gian lm cho phn tc hnh dng nht nh c trng cho phn t v cho cht cho.Nhiu nguyn t sau khi gp chung e to lin kt cn c cc e khng lin kt. Chng hn

trong N NH3 ngoi 3 cp electron lin kt vi 3 nguyn t H, cn c 1 cp e khng lin kt.Cc cp electron d lin kt hay khng lin kt ny s y nhau do cng tch in m.

Trong phn t AXn, A l nguyn t trung tm, X l phi t; n l s phi t X c trong AXn.Nu A cn c m cp e khng lin kt, mi cp c k hiu l E, ta c k hiu AX nEm. Mhnh VSEPR xt s phn b khng gian gia A vi X, vi E. Coi nguyn t trung tm A cdng cu. Tm ca hnh cu l ht nhn nguyn t A v cc electron phi ho tr bn trong (li),v qu cu l cc e lp ngoi cng (e ho tr). Mi cp e ho tr chim mt khong khng gianno ca qu cu.Nh vy, mt mc nht nh, hnh dng ca phn t ph thuc vo khong khng gian

chim bi cc e ho tr ca nguyn t trung tm A. Hnh dng phn t ph thuc ch yu vo sphn b cc cp e hay cc m my e ho tr ca nguyn t A.

2. Ni dung ca thuyt sc y gia cc cp e ho tr (VSEPR)Vo nhng nm 1940, N. Sidgwick, H. Powell a ra thuyt sc y gia cc cp electron

ho tr v sau c cc nh bc hc khc, trong c R. Gillespie, b sung v hon chnh.+ Cu hnh cc lin kt ca nguyn t (hay ion) ph thuc vo tng s cp electron ho tr

lin kt hay khng lin kt ca nguyn t.+ Cc obitan c cc cp e ho tr c phn b u nhau v cch nhau xa nht c lc y

nh nht gia chng.C s khng tng ng gia cp e lin kt v cp e khng lin kt. i e lin kt chu lc

ht ng thi ca hai ht nhn nguyn t A v X to ra lin kt nn chuyn ng ch yu vng khng gian gia hai ht nhn. Trong khi , cp e khng lin kt ch chu lc ht ca htnhn A nn c th chuyn ng ra xa hn. Kt qu l cp e khng lin kt chim khong khnggian rng hn so vi khong khng gian chim bi cp e lin kt.

+ Obitan c cp electron khng lin kt chim khng gian ln hn so vi obitan cha cpelectron lin kt v th sc y gia cc cp electron lin kt gim hn so vi cp khng linkt. Th d gc lin kt trong cc phn t CH 4, NH3 v H2O tng ng bng 109

o28, 107o v104,5o do cc phn t c s cp electron khng lin kt bng 0, 1 v 2.

+ Khng gian ca cp electron lin kt s gim nu m in ca cc nguyn t lin kt ln.gc lin kt trong NF3 ch l 102

o so vi 107o ca NH3. Tng t gc lin kt tgim trong dy:PI3 (102

o), PBr3 (101,5o), PCl3 (100,3

o) v PF3 (97,8o).

a) M hnh s y gia cc cp electron lin kt: AXn vi n = 2 p 6n=2: hai cp e c phn b trn ng thng. Phn t thng nh BeH2. Gc lin kt

XAX bng 180o.n=3: ba cp e c phn b trn ba nh ca tam gic u. Phn t c hnh tam gic u,

phng. Gc XAX bng 120o. V d: BF3, AlCl3,...n=4: bn cp e c phn b bn nh t din u, tm l A. Phn t c hnh t din u.

Gc XAX bng 109o28. V d: CH4, NH4+,...

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te3 =2

1(s - px+ py - pZ)

te4 =2

1(s - px - py + pZ)

s =T4

1 ; px =T43 cos U cos N ;

py =T4

3cos U sin N ; pZ =

T4

3cos U

- Lai ho sp2 c p dng gii thch lin kt ho hc trong cc anken CnH2n (nu2), benzen,... Cc phn t c gc lin kt bng 120o.

d) Lai ho sp3- Lai ho trong AO - 2s t hp tuyn tnh vi 3 AO - 2p to ra 4 AO mi c cng nng

lng c gi l lai ho sp3.- C th hnh dung qu trnh lai ho sp3 nh sau:

- Hnh dng ca mi AO lai ho sp3 cng tng t nh hnh dng AO lai ho sp, sp2 va xt. 4AO-sp3 hng ra 4 nh ca t din u m tm ca t din l nguyn t (chnh xc l ht nhn

nguyn t) c cc AO lai ho. Do lai ho sp3

c gi l lai ha t din. AO-sp3

c khiu l te. Vy ta c 4 AO-sp3 l te1, te2, te3, te4.

- Lai ho sp3 c p dng gii thch lin kt ho hc trong cc ankan CnH2n+2 (nu1), inhnh l trong CH4. Ngoi ra n cng c p dng trong cc trng hp khc nhau H2O,NH4

+,... cc phn t c gc lin kt gn bng 109o28Ngoi ra cn c cc kiu lai ho sp3d2 hay sp2d vi s tham gia ca cc obitan d.S dng quan nim lai ho c th gii thch v hnh dng cc phn t kh ng.Kiu laiho

Gc gia cc obitan lai ho Dng phn t V d

spsp2

sp3dsp2sp3d2

180o120o

109o2890o90o

ng thngTam gic u

T din uHnh vungBt din u

BeH2, BeCl2, ZnCl2, CO2BF3, NO3

-, CO32-

CH4, CCl4, NH4+, ClO4-, SO42-,PO4

3-PtCl4, CuCl4

2-, Ni(CN)42-

SF6, AlF63-, SiF6

2-Kiu lai ho cc obitan ca nguyn t cho thy s obitan lai ho c to nn ng thi l s

phi tr ti a ca nguyn t . Bng di y h thng li kh nng lai ho cc obitan canguyn t cc nguyn t v s phi tr ti a m nguyn t c:

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Nguyn t chu k Kiu lai ho v s phi tr (vit trong dungoc)

IIIIIIV

V v VI

sp(2), sp2(3), sp3(4)sp3(4), dsp3(5), d2sp3(6), sp3d2(6)sp3(4), dsp3(5), d2sp3(6), sp3d2(6)

d2

sp3

(6), d2

sp3

(7)Da vo bng , c th hiu c cng thc cc dy hp cht v anion sau y ca cacnguyn t thuc cng nhm trong bng tun hon:

Nhm IVA CH4 SiF62- GeF6

2- SnF84- PbF8

4-CO3

2- SiO44- GeO4

4- SnO68- PbO6

8-Nhm VIA H2SO4 H6TeO6Nhm VIIA HClO4 H5IO6Tuy nhin, gii thch hnh dng ca phn t, ngoi s lai ho, cn vn dng thm mt s

gi thit khc na. V d nh i vi nhng phn t sau y, cc nguyn t trung tm u cng mt kiu lai ho sp3 ca cc obitan, s bin i ca gc ho tr c gii thch nh sau:

Gc ho tr gim xung v vai tr ca s trong s lai ho sp3 gim xung

Gc linkt trongcc phn t

H2O (105o)

NH3 (107o)

CH4 (108o28)

H2S (92o)

PH3 (94o)

H2Se (91o)

AsH3 (92o)

H2Te (90o)

SbH3 (90o)

B. Cu to cc phn t n gin

1. Phn t O2Nguyn t O c cu hnh electron: 2s2 2p4Hai nguyn t O lin kt vi nhau bng hai cp electron chung:

Cng thc cu to phn t vi lin kt cp gia hai nguyn t O ph hp nng lng ca linkt l 494 kJ/mol v di ca lin kt l 1,21A v gii thch c hu ht tnh cht ca oxi trt tnh.

Oxi trng thi kh, lng hay rn u c tnh thun t. T tnh o c cho thy s c mtca hai electron c thn trong phn t O2. Bi vy gii thch tnh cht ngi ta buc phigi thit thm rng lin kt cng ho tr cng c th c to nn nh 3 electron gi l lin ktba electron, ngha l phn t O2 c cu to:

Trong ngoi lin kt cng ho tr bnh thng c to nn bng cp electron (vch lin)cn c hai lin kt c to nn nh ba electron (vch ri). Lin kt ba electron cn c gi llin kt mt electron v thc t trong ba electron ch c mt electron c dng chung gia hainguyn t:

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Tuy nhin s lin kt gia hai nguyn t O cng ch l hai. (Thuyt obitan phn t gii thchmt cch n gin tnh thun t ca O2 bng s tn ti ca hai electron c thn trn haiobitan phn t phn lin kt)

2. Phn t N2Nguyn t N c cu hnh electron 1s2 2s2 2p3

Hai nguyn t N lin kt vi nhau bng ba cp electron chung:

Hai obitan 2px ca hai nguyn t N che ph nhau to thnh lin kt W.

Cp obitan 2py v cp obitan 2pz ca hai nguyn t N che ph nhau, theo tng cp mt, tothnh hai lin kt T.

Nh vy, lin kt trong phn t l lin kt ba. Lin kt ba ny c nng lng rt ln (942kJ/mol) nn phn t N2 rt bn. Nit kh tham gia phn ng nhit thng.

3. Phn t NOTrc y ngi ta cho rng phn t NO c cu to:

Ngha l trong phn t c lin kt i.Nhng thc nghim cho thy rng phn t ny c mmen lng cc rt b (Q = 0,15D) v

di ca lin kt N-O l 1,44A, ngha l trung gian gia di lin kt i trong ion NO - (1,18A)v ca lin kt ba trong ion NO+ (1,06A). Vy bc ca lin kt trong NO khng th bng 2 mbng 2,5, ngha l phn t c cu to:

Trong ngoi hai lin kt cng ho tr bnh thng (c to nn nh cp electron chung)cn c mt lin kt ba electron na (ion NO+ c mt trong cc hp cht NOCl, NOClO4, ionNO- c mt trong NaNO).

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Mt obitan lai ho c mt cp electron t do v mi mt obitan lai ho cn li c mt electronc thn che ph vi obitan 2p ca hai nguyn t O cng c electron c thn to thnh lin ktW:

Obitan 3p khng lai ho ca S c electron c thn che ph vi obitan 2p khc ca nguyn tO (gi s bn tri ca hnh v) c electron c thn to thnh mt lin kt T v mt obitan 3dkhng lai ho ca S c electron c thn che ph vi obitan 2p khc ca O (gi s bn phihnh v) c electron c thn to thnh mt lin kt T na.

8. Phn t NO2Phn t NO2 c dng gp khc gn ging nh O3 v SO2

S to thnh cc lin kt trong phn t c m t tng t nh i vi phn t O3 v SO2,

ngha l NO2 c cng thc cu to:

hay cng thc vi lin kt T khng nh ch

9. Phn t SO3

Phn t SO3 c dng hnh tam gic u, nguyn t S nm trng tm ca tam gic v banguyn t O nm nh:

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Ba obitan lai ho ca S, mi obitan c mt electron c thn che ph vi obitan 2p electronc thn ca ba nguyn t O to thnh ba lin kt cng ho tr. Ngoi ra mt obitan 3p v haiobitan 3d khng lai ho ca S, mi mt c mt electron c thn che ph vi obitan 2p cn lic electron c thn ca ba nguyn t O to thnh ba lin kt T. Nh vy trong phn t SO3,nguyn t S c ho tr su.

10.Phn t NH3Phn t NH3 c dng hnh chp tam gic, nguyn t N nh v ba nguyn t H nh ca

tam gic u:

Trong phn t NH3, nguyn t N trng thi lai ho sp3:

Mt obitan lai ho c cp electron khng lin kt, cn ba obitan lai ho khc, mi obitan cmt electron c thn che ph vi obitan 1s c electron c thn ca ba nguyn t H to thnhba lin kt cng ho tr:

Vy phn t NH3 c cng thc cu to:

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11.Ion NH4+

Ion NH4+ c dng hnh t din u, nguyn t N nm trung tm v bn nguyn t H nm

nh ca t din:

Trong ion NH4+, s to thnh ba lin kt cng ho tr bi ba cp electron chung gia N v H

xy ra tng t nh trong phn t NH3. Ch khc y l obitan lai ho c cp electron cheph vi obitan 1s trng electron ca H+ to thnh lin kt cho nhn. Tuy nhin c bn lin kt N- H u ging nhau v u l lin kt cng ho tr to nn bi cp electron:

12.Phn t H2SPhn t H2S c dng gp khc ging cc phn t O3, SO2:

Trong phn t H2S, nguyn t S trng thi lai ho sp3. Hai obitan lai ho, mi obitan c mt

cp electron khng lin kt cn hai obitan lai ho cn li, mi mt c mt electron c thn che

ph vi obitan 1s c electron c thn ca hai nguyn t H to thnh hai lin kt cng ho tr :

13.Phn tCO2Phn t CO2 c dng ng thng, nguyn t C gia hai nguyn t O:

trong phn t CO2, nguyn t C trng thi lai ho sp v c cu hnh electron trng thikch thch:

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3. Xc nh xem phn t no l phn cc v khng phn cc. Gii thch kt qu chnCu 3:Cho hai nguyn t A v B c tng s ht l 65 trong hiu s ht mang in v khngmang in l 19. Tng s ht mang in ca B nhiu hn ca A l 26.a) Xc nh A, B; vit cu hnh electron ca A, B v cho bit b 4 s lng t ng vi electron

sau cng trong nguyn t A, B.

b) Xc nh v tr ca A, B trong HTTH.c) Vit cng thc Lewis ca phn t AB2, cho bit dng hnh hc ca phn t, trng thi lai hoca nguyn t trung tm

d) Hy gii thch ti sao phn t AB2 c khuynh hng polime hoCU 41. Vit cng thc cu to Lewis, nu trng thi lai ha v v dng hnh hc ca cc phn t sau:

(a) B2H6 (b) XeO3 (c) Al2Cl6Gii thch v sao c Al2Cl6 m khng c phn t B2F6

2. Trnh by cu to ca cc ion sau: O 2 , O2

2 theo thuyt MO (cu hnh electron, cng thccu to). Nhn xt v t tnh ca mi ion trn.

3. So snh v gii thch ngn gn phn cc (momen lng cc) ca cc cht sau: NH 3, NF3,

BF3.4. Ha tan 2,00 gam mui CrCl3.6H20 vo nc, sau thm lng d dung dch AgNO3 v lcnhanh kt ta AgCl cn c 2,1525 gam. Cho bit mui crom ni trn tn ti di dng phccht.

4.1. Hy xc nh cng thc ca phc cht .4.2. Hy xc nh cu trc (trng thi lai ha, dng hnh hc) v nu t tnh ca phc

cht trn.CU 5: Cho biet trang thai lai hoa cua ngt trung tam va dang hnh hoc cua cacphan t sau :

H2O , H2S , H2Se , H2Te .- Hay sap xep theo chieu tang dan o ln goc lien ket va giai thch s

sap xep o.- Tai sao ieu kien thng H2O the long,con H2S , H2Se , H2Te thekh

- Hay sap xep theo chieu tang dan tnh kh cua cac chat tren.Giai thch.Cu 6

1. X, Y l hai phi kim. Trong nguyn t X, Y c s ht mang in nhiu hn s htkhng mang in ln lt l 14 v 16.Hp cht A c cng thc XYn, c c im:

- X chim 15,0486% v khi lng- Tng s proton l 100- Tng s ntron l 106

a. Xc nh s khi v tn nguyn t X, Y. Cho bit b bn s lng t ca e cui cngtrn X, Yb. Bit X, Y to vi nhau hai hp cht l A, B. Vit cu trc hnh hc v cho bit trng

thi lai ho ca nguyn t trung tm ca A, B.c. Vit cc phng trnh phn ng gia A vi P2O5 v vi H2O

Vit cc phng trnh phn ng gia B vi O2 v vi H2O

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103o15

2.3. Nhng phn t no sau y c moment lng cc ln hn 0 BF3, NH3, SiF4, SiHCl3,SF2, O3.

Cho bit Zp = 15, ZAs = 33, ZO = 16, ZF = 9, ZCl = 17, ZB = 5, ZN = 7, ZSi = 14, ZS = 16.Cu 10Cho b bn s lng t ca electron cht cng trn nguyn t ca cc nguyn t A, X, Z nh sau:

A: n = 3, l = 1, m = - 1, s = -1/2X: n = 2, l = 1, m = - 1, s = -1/2Z: n = 2, l = 1, m = 0, s = +1/2

1 Xc nh A, X, Z.2 Cho bit trng thi lai ho v cu trc hnh hc ca cc phn t v ion sau: ZA 2, AX2, AX3

2-,AX4

2-.3 Bng thuyt lai ho gii thch s to thnh phn t ZX. Gii thch v sao ZX c moment lngcc b. Gii thch s hnh thnh lin kt trong phn t phc trung ho Fe(CO)5 bng thuyt VB.4 Gii thch v sao AX3

2- li c kh nng ho tan A to thnh A2X32-.

CU 11(1) Cho bit s bin i trng thi lai ho ca nguyn t Al trong phn ng sau v cu

to hnh hc ca AlCl3, AlCl5

4 .AlCl3 + Cl

5 p AlCl 54

(2) Biu din s hnh thnh lin kt phi tr trong cc trng hp sau:(o): Sn phm tng tc gia NH3 v BF3.(b): Sn phm tng tc gia AgCl vi dung dch NH3.(3): Gii thch s khc nhau v gc lin kt trong tng cp phn t sau:(a)

S OCl Cl Cl Cl

(b)O O

F F Cl Cl

Cu 121. Cu hnh electron ngoi cng ca nguyn t X l 5p5. T s ntron v in tch ht

nhn bng 1,3962. S ntron ca X bng 3,7 ln s ntron ca nguyn t thuc nguynt Y. Khi cho 4,29 gam Y tc dng vi lng d X th thu c 18,26 gam sn phmc cng thc l XY. Hy xc nh in tch ht nhn Z ca X v Y v vit cu hnhelectron ca Y tm c.

2. Hy cho bit trng thi lai ha v dng hnh hc ca hp cht XCl3.3. Bn knh nguyn t Cobalt l 1,25. Tnh th tch ca n v ca tinh th Co nu

trong 1 trt t gn xem Co kt tinh dng lp phng tm mt.

Cu 13A, B l 2 nguyn t k tip nhau trong cng mt chu k ca bng tun hon trong B c tng slng t ( n + l ) ln hn tng s lng t ( n + l ) ca A l 1. Tng s i s ca b 4 s lng tca electron cui cng ca cation A a l 3,5.

a)Xc nh b 4 s lng t ca electron cui cng trn A, B.

111o103o

111o

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b)Vit cu hnh electron v xc nh tn ca A, B.Cu 14: 1) Cho cc cht sau y:

CO2 , SO2 , C2H5OH, CH3COOH, HIHy cho bit cht no c nhit si cao nht Gii thch

2) Dng thuyt ni ha tr, hy cho bit c cu lp th (biu din bng hnh v) v trng

thi lai ha ca nguyn t trung tm ca cc phn t v ion sau:H2SO4 , [Ni(CN)4]2- , ICl3 , XeF4

Cu 15:a. Hy cho bit cu trc hnh hc , kiu lai ha ca cc phn t : SF6 , XeF2 , OF2b. Da vo cu hnh electron ca uran [Rn]5 36d17s2. Hy cho bit hai hp cht X,Y ca uran

vi lo , cho bit ti sao c c 2 hp cht ny . Hon thnh phn ng sauClF3 + A p B + Cl2

c. Ti sao ozon tan nhiu trong nc , nhng oxi t tan trong ncCu 16:Nguyn t ca nguyn t phi kim A c electron cui cng c b 4 s lng t tha mn m + l = 0v n + ms = 3/2 ( quy c cc gi tr m t thp n cao )

1. Xc nh s hiu nguyn t, gi tn nguyn t A. Vit cng thc electron, cng thc cu toca phn t A2. Kim chng s lin kt v tnh cht thun t ca A2 bng cu hnh electron caphn t.2.Ion A3B

2- v A3C2- ln lt c 42 v 32 electron

2.1. Tm 2 nguyn t B v C ( s hiu nguyn t, tn, k hiu )2.2. Dung dch mui ca A3B

2- v A3C2- khi tc dng vi axit clohidric cho kh D v F tng ng.

- M t dng hnh hc ca phn t D v E.- Nu phng php ha hc phn bit D v E.- Kh no trong 2 kh c th kt hp vi O2 Ti saoCu 171. So snh bn knh ca cc ht sau: Al, Al3+, Na, Na+, Mg, Mg2+, F-, O2-.

2. Trong s cc cu trc c th c sau y:a) Ca ICl4(-):

I

Cl Cl

Cl Cl

..

..

I

Cl Cl..

Cl

Cl

tham luận 1

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