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Chap 1. Điện hóa học

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PHN II IN HA HC Ni dung Tnh cht ca dung dch in ly(DDL) S chuyn vn in tchtrong DDL Pin v in cc ng hc ca qu trnh in ha Chng 1: Chng 2: Chng 3: Chng 4: 1.1. Dung dch cc cht in ly (DDL) Chng 1: Tnh cht ca DDL H2O HCl H3O++ Cl- T cao NaCl Na++ Cl- Cht in ly (cht in phn): l cc cht c th to ra cc dung dch ion v hn hp nng chy c cha cc ion. V d: mui, axit, baz l dung dch c tnh cht: ng nht v tnh cht ho l mi im trong th tch Thnh phn c th thay i lin tc trong mt gii hn Tng im si Gim im kt tinh Tng p sut thm thu: t =i.C.R.T (do s phn ly cht in ly thnh ion nn lm tng s ht phn t trong dung dch Dung dch cht in ly m s s m dd ddc K i T c K i TT R C i T R n i V, 2 , 2 = A = A = H = Hp sut thm thu: t =i.C.R.T Ap suat tham thau: Ap suat can phai tac dung len dung dch e dung moi nguyen chat khong the tham thau qua mang ban tham sang dung dch .1.1. Dung dch cc cht in ly (DDL) (tt) S in ly (electrolysis): l s phn ly ca cc cht in ly trong dung dch. l mt qu trnh thun nghch Hng s in ly (KD, electrolyte constant): Chng 1: Tnh cht ca DDL ++ v vv + v +z zA M A M| | | || | + +v vvv+=A MA . MKz zD1.1. Dung dch cc cht in ly (DDL) (tt) phn ly (o, electrolytic capacity): Chng 1: Tnh cht ca DDL o =S phn t phn ly S phn t ban u + + +vv+v + v v + vv vo o= ) ( ) .() 1 (. nK) ( ) 1 (0Do ln ( 1) : Cht in ly mnh:Mui, acid v c, baz mnh o nh ( 0) : Cht in ly yu:acid hu c, acid yu, baz yu Khi v+ = v- = 1 o o=1. nK20Dn0 = s mol cht in ly (ban u) Chto HCl0.926 H2SO4 0.60 H2C2O4 0.50 HF0.09 CH3COOH0.014 H2CO3 0.0017 HCN0.0001 KOH0.90 Ba(OH)2 0.77 NH3 0.014 KCl0.862 MgCl2 0.765 K2SO4 0.722 MgSO4 0.449 in ly o ca mt s cht in ly trong dung dch nc 0.1 N S phn ly ca cht tan thnh ion thng xy ra yu trong dung mi c cc yu (c b) v xy ra mnh trong dung mi c cc mnh (c ln) in ly tng khi nng dung dch gim v ngc li in ly gim khi nng dung dch tng.(tng nng tng tng tc)CN 0.10.050.010.0050.001 o0.0140.0190.0420.0600.124 S thay i in ly ca axit axetic H s Van tHoff i H s Van tHoff i: l t s gia s phn t sau phn ly (ion + phn t) so vi s phn t trc khi phn ly (s phn t ho tan) v = v+ + v- 00 0 0 0n. n . . n . ) . n n (io v + o v + o = +i = 1 + (v-1).o 11 i v= oH s vant Hoff ca mt s cht 1.1. Dung dch cc cht in ly (DDL) (tt) nh hng ca dung mi: nh lut Coulomb: Chng 1: Tnh cht ca DDL 22 1. . . 4.rq qfc t=q1, q2 : in tch ca 2 ion r: khong cch gia 2 ion c : thm in mi: c= c0.D c0= 8.85x10-12 J.C/m (F/m) D: thm in mi tng i D nh f ln ion tri du t kt hp li hay khng to ra ion o nhCng vi D, bn cht ca cht in ly v dung mi cng nh hng:V d: Mt s cht ha tan v phn ly trong HCN (D=96) yu hn trong H2O (D=81) ChtcChtc HF84.00Aniline7.00 Nc80.00Acit axetic 6.19 Ethanol25.20Clorofoc4.72 Amoniac25.00Ete ietyl4.10 Axetone20.70Benzen2.30 SO2 lng 15.35CF4 2.23 Hng s in mi ca mt s cht c trng quan trng ca dung mi l hng s in mi c. c ph thuc c cc, cu to v kch thc ca phn t dung mi. c cc cng ln, c cng ln. Da vo c c th d on c kh nng ca dung mi i vi s ha tan, phn ly 1.2. Nc Chng 1: Tnh cht ca DDL O H H 104,523o 0.276 nm -Momen lng cc: 1,84 0.02D -Lien ket cua moi phan t nc trung bnh ln hn 2. Xung quanh mot phan t nc co t nhat 4 phan t khac bao boc (cau truc t dien eu). -Nc a ket tinh dang luc phng. Nc co cau truc gia tinh the, con goi la bang ba chieu, hay claster. Cac claster bi trong nc t do. Nc v tnh cht ca n1.3. Phn loi cht in ly Chng 1: Tnh cht ca DDL C 3 loi cht in ly MNH Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. YU Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. KHNG IN LY Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ng ma, glucose, urea H2CO3, CH3COOH, NH4OH HCl, H2SO4, NaOH, KOH Cht in ly thc, th: da vo bn cht lin kt trong phn t, tinh th. Lin kt ion: cht in ly thc (true electrolyte): v d: NaCl Tng tac gia ion trong tinh the va phan t dung moi lng cc + chuyen ong nhiet cua ion lien ket mang b t ion lien ket vi dung moi, tach khoi mang. Cc loi khc: cht in ly th (potential electrolyte): v d: HCl Tng tac lng cc - lng cc + chuyen ong nhiet Ph hy phan t co cc thanh ion hydrat hoa ion tng tac nc tao ion H3O+ (oxonium, hydronium). * Dung mi khc nc: - Dung mi proton: Cho H+, to lin kt hydro gia DM v cht tan c th ha tan, phn ly, solvat ha cc ion. - Dung mi aproton: C nhng cp e- dng chung c th solvat cc cation, to cc dung dch ion. 1.4. S solvat ha cc ion Chng 1: Tnh cht ca DDL Qua trnh tac dung cua ion vi cac phan t dung moi to tap hp tng oi ben chac goi la solvat hoa. Dung moi nc: hydrat hoa. So cua cac phan t nc rang buoc trong hydrat (phc nc) goi la so cua hydrat hoa (so phoi t). Vi dung moi nc: (quan trong nhat) dung dch cua cation kim loai kiem, kiem tho tao ra do tng tac tnh ien ion lng cc.Tng tac nay phu thuoc: ien tch, ban knh, khoi lng cation 1.4. S solvat ha cc ion Nguyen nhan cua s ien ly: S solvat hoa, hydrat hoa giai phong nang lng tinh the ran b pha v, lien ket hoa hoc b pha v. lp hydrat hoa gan: lien ket cho nhan, do con obitan trong cua cation va cap ien t ghep oi cua nc. lp hydrat hoa xa: cac phan t dung moi nh hng khong hoan toan, do lien ket long leo hn. Cac ligand nc khong co nh trong phc nc ma co the thay oi v tr. Ligand nay ra i, ligand khac vao the. Dung dch l tng v dung dch thc Dung dch l tng: Cation v anion khng tng tc vi nhau. Dung dch thc: c s tng tc tnh in. Cc ion ngc du tp hp gn nhau - - - - - + + + + + + - - - - - - + + + + + + - Shaded region: Solvation cage L tng Thc 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL NaCl Na+ + Cl- Th ha hc + + =Cl NaNaCl oCloNaoNaCl + + = NaCloNaCl NaCla RT ln + = i ioi ioi im RT a RT ln ln + = + = + + = salt + + ++ + + = + m RT m RT a RTo osaltosaltln ln lnH s hot trung bnh ion ( ) + + ++ + = + m m RT a RTo osaltosaltln lno o osalt + + = ( ) + += m m RT a RTsaltln ln2 += i vi NaCl, m+ = m- 2 2= m asaltTi sao khng s dng nng mol m s dng nng molan ??? MgCl2 Mg2+ + 2 Cl

+ + ++ + + = + Cl0Cl Mg0Mgsalt0saltln 2 2 ln ln2 2m RT m RT a RT2 2 0Cl0Mgsalt0saltln 2 ln2 + + + + = + + m m RT a RT+ + = 2 2saltm m aTa c m+ = m v m- = 2m ( )312+ = nh ngha:( )3 3 32salt4 2 = = m m m a ++ z zy xyB xA B A( ) ( ) ( ) ( ) ( )( )( )yBxAy x y xm y x ym xm m m a a += = =yBxAyB BxA AyBxA( )( ) y x+ =/ 1yBxA ( )( ) ( ) y x y x y xm y x a a++= yBxA( )3 3 3 3 2 1 2Cl4 2 1- = =+ m m a aMgCho MgCl2 : h s hot trung bnh ion l i lng quan trng, c trng cho s sai lch nng dung dch mui so vi khi trng thi l tng 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL H s hot trung bnh ion (tt) *Ch :- Tnh m NaCl trong dung dch c mNaCl = 0.1 v mCaCl2 = 0.3 : m NaCl = (0.11 . 0.71 )1/2 = 0.27 - H s hot trung bnh biu din cc loi nng khc nhau: ac =c.C hocax =x.x ( ) ( ) + + = v v ln . . ln . .0RT m RTmVi: v.RT.ln() : Th ha d - thc o tng tc gia cc ion vinhauvviphntdungmivc cc phn t dung mi vi nhau v.RT.ln() 0, tc = 1; dung dch = ly tng Dung dch v cng long: 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL H s hot trung bnh ion (tt) Th ha:Trng thi chun Trng thi chun l trng thi m m = = 1 tt c cc khong nhit v p sut = 0 nhit bt ktrng vi dung dch v cng long. ngha vt l ca xc nh s khc bit gia dung dch thc v dung dch l tng C hai loi lc tc dng vo cc ion trong dung dch: tng tc vi phn t dung mi tng tc tnh in vi cc ion khc Khi pha long dung dch tng khong cch gia cc ion gim tng tc Dung dch v cng long tng tc solvat ho l ch yu dung dch v cng long ch l gn ng l tng, xem nh nng lng solvat ho khng ph thuc nng . Lc ion (I) Lc ion ca dung dch l mt na tng tch s nng ca mi ion vi bnh phng s in tch (ha tr) ca tt c ion c trong dung dch Im = 0.5Emi .Zi2 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL Nu biu din theo nng mol Ci : Ic = 0.5ECi .Zi2

Trong :i: k hiu tt c cc ion trong dung dch mi hayCi :lnngthc(khngkphn khng phn ly) Lc ion (I) (tt) nh lut thc nghim Lewis v Randall: Cht in ly mnh, vng nng long, cc dung dchccnglcionhshottrungbnh nh nhau 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL =ii iz m I2210.010 m HCl: mH+ = mCl- = 0.01 mol/kg, ( ) ( ) ( )1 -2 2kg mol 01 . 0 1 01 . 0 1 01 . 021 = + + = I0.10 m Na2SO4 Cc phng php xc nh Phng php nghim lnh: o h nhit kt tinh ca dung dchAT vi cc nng thp bit, ri xc nh m 02 / 1STC=1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL Phng php da vo ha tan: Khi bit tch s tan T, xc nh ha tan S0, ri tnh c PT Gibbs-Duhem: n1lna1 + n2lna2 = 0 21lnRT dTa d =mKT da d) (ln2A=(*) Phng php nghim lnh xc nh 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL PT Sreder : Vi: n2= m ; n1= 1000/M1 ; AT=|T0-T| h nhit kt tinh T2=T02 K= R.T02M1/(1000) : hng s nghim lnh Do:a2= av=v.mv (v+v+. v-v-) Nn (*) tr thnh: t : Do :d(ln) = -dj j.dlnm Ly tch phn 2 v: mKT dm d dv) (ln lnA= +dmKmTmKT ddjmKTj2) (1v v vA+A = A =dm m j jm} =0) / ( ln Phng php nghim lnh xc nh (tt) 1.5. Tnh cht nhit ng ca DDLChng 1: Tnh cht ca DDL j/m m m 0 }mdm m j0) / (