BI 3: XC TC NG TH - PHN NG PHN HY H2O2I Mc ch th nghim: Xc nh hng s tc , chu k bn hy v nng lng hot ha ca phn ng phn hy H2O2 vi Cu2+ l cht xc tc. II C s l thuyt: H2O2 t phn hy theo phng trnh sau: H2 O2 H2 O + O 2 Tc phn ng tng nhanh khi c cc cht xc tc Pt, mui ca cc kim loi chuyn tip, Phn ng tin hnh theo 2 giai on:1. 2. HOOH HOOH + 2H+ 2 H2O2 O2 + 2H+ ( chm ) ( nhanh ) 2 H2O

2H2O + O2

Tc tng qut ca phn ng c xc nh bi giai on 1 v do phn ng xy ra theo bc 1. III Dng c v ha cht: 1. Dng c: Bnh tam gic 100ml: Pipet: B iu nhit: 2. Ha cht: Dung dch H2O2 30% Dung dch CuSO4 10% Dung dch H2SO4 10% Dung dch KMnO4 0,1N IV Tin trnh thc hnh:1. Thc hin phn ng 300C

3 1 1

Ly 1ml dung dch H2O2 v 19ml nc ct cho vo mt bnh nn; ly 10ml dung dch CuSO4 10% cho vo mt bnh nn khc.

c c hai bnh ny vo b iu nhit 300C trong 25 pht. Trn chung hai bnh li vi nhau, lc vi ln trn u hn hp, vn hn hp trong bnh iu nhit. Sau ly 2ml hn hp phn ng cho vo mt bnh nn c cha sn 3ml H2SO4 10% ri chun vi KMnO4 0,1N. Khi bt u nh git KMnO4 u tin th ghi thi gian, l thi im t = 0. Ghi s ml KMnO4 dng. Sau 5, 10, 15, 20, v 30 pht k t thi im t = 0 li ly 2ml em chun .2. Thc hin tng t 400C

V Kt qu thc nghim:t (pht) V (KMnO4) (ml) C (H2O2) (N) V (KMnO4) (ml) C (H2O2) (N) 0 1.8 0.09 2.4 0.12 5 0.9 0.045 1.4 0.07 10 0.8 0.04 1.3 0.065 15 0.6 0.03 1.2 0.06 20 0.5 0.025 1 0.05 30 0.4 0.02 0.8 0.04

300C 400C

Cch tnh C(H2O2) nh sau: Theo nh lut ng lng: CHCH 2 O2 = CKMnO 4 .VKMnO 4 VH 2 O22 O2

.VH 2 O2 = C KMnO 4 .VKMnO 4

1. Tnh hng s tc : Th nghim xy ra cc phng trnh phn ng sau: 2 H2O2CuSO4

O2

+

2H2O

(1)

2KMnO4 + 5 H2O2 + 6H+

K+ + Mn2+ + 5O2 + 8H2O2 ( 2 )

Ta thy phng trnh (1) l phn ng phn hy H2O2 l phng trnh bc nht nn hng s tc c tnh theo cng thc sau:1 C( H 2O2 ) 1 C0( H 2O2 ) K = ln = . ln t C0 ( H 2O2 ) t C( H 2O2 )30 C 400C0

K5 0.1386 0.1078

K10 0.0811 0.0613

K15 0.0732 0.0462

K20 0.064 0.0438

K30 0.0501 0.0366

Hng s tc trung bnh: K = 300C: Ktb = 0.0814 ph-1 400C: Ktb = 0.0591 ph-1

K 5 + K10 + K15 + K 20 + K 30 5

2. Tnh nng lng hot ha ca phn ng:E = R ln K T2 1 1 0.0591 1 1 4 = 8.314 ln = 2.81 10 T 0.0814 313 303 K T1 2 T1

3. Tnh chu k bn hy: 300C: t1 / 2 = 400C: t1 / 2 =ln 2 = 8.52 ( s ) K ln 2 = 11 .7( s ) K

BI 4: XC NH BC PHN NGI. MC CH Trong bi ny chng ta xc nh bc ring v bc chung ca phn ng:Fe3+ IFe2+ 1/2 I2

II. C S L THUYT xc nh bc ring theo ca Vant-hoff. Gi s vn tc u ca phn ng c xc nh bng phng trnh: (1) Trong , ( ( l nng u ca l nng u ca l bc ring phn ca Ly logarit ca phng trnh (1) ta c: lg = lgk + lg + lg (2) v , chng ta s dng phng php vi phn

Nu tin hnh mt chui cc th nghim vi nhng nng u ca nhau trong khi nng ca mt chui nng ca c gi khng i, ta c th xc nh c khng i, chng ta s xc nh c +

khc . V

. T tnh

c bc tng qut ca phn ng n =

Chui th nghim th nht cho ta dng nh sau ca phng trnh (2) lg =A+ lg( = - lg( . thng s dng phng trnh ta c ng thng vi gc nghing

Xy dng th lg vi tg=

. Tng t c th xc nh

xc nh vn tc thi im u kinh nghim sau:

l nng mol ca t l thi gian phn ng

sinh ra mi thi im t

, l nhng hng s nghim Ly o hm = ti t=o, ta c: (3) ta xc nh c hng s thc nghim t

Nh vy xy dng th

y c th tnh c vn tc u ca phn ng (3). l nng mol ca sinh ra. Lng ti thi im t, c th xc nh thng qua nng

c th c kim sot bng cch cho vo hn hp phn ng

dung dch Na2S2O3 vi nhng lng bit trc chnh xc km theo h tinh bt. Khi ton b lng Na2S2O3 tc dunhj ht vi c sinh ra trong qu trnh phn

ng th lng mu xanh. Nng

mi sinh ra trong hn hp phn ng cng vi h tinh bt s cho

ti mi thi im t c xc nh theo biu thc: =

: nng ng lng ca : ton b th tch dung dch cho n thi im t. : nng ca Fe2+ c cho vo hn hp phn ng cho n thi im t. : th tch ca hn hp phn ng ( qui v th tch u khng i). III. TH NGHIM1. Xc nh bc ring theo Fe3+:

c cho vo hn hp phn ng

Dng ng ht cho vo 4 bnh tam gic cc ha cht vi lng chnh xc theo thnh phn bng sau: Dung dch Fe(NO3)3M/60,ml HNO3 0,1M,ml KNO3 1M,ml H2O,ml Bnh 1 10 10 40 20 Bnh 2 20 10 30 20 Bnh 3 30 10 20 20 Bnh 4 40 10 10 20

Tin hnh th nghim ln lt vi tng bnh. Cho vo bnh 1 vi git h tinh bt v ng 20ml dung dch KI M/40. Ghi nhn thi im bt u phn ng v khng tt ng h trong sut th nghim ca tng bnh. Dng buret cho mt lng nht nh 0,01N (V1).

Khi dung dch c mu xanh tr li, ghi nhn thi im t1- l lc I2 sinh ra tc dng vi tng lng trc tip trn buret ). - V2 v ghi nhn thi im t2 xut hin li mu xanh (V2 l hin din trong hn hp phn ng cho thi im t2 , V2 c

Tip tc theo di tin trnh phn ng nh vy cho n khi khong 8 im V1- t1. Lng nh vo va sao cho khong cch gia hai ti lin tip

khong 1 pht ri. Ton b th nghim vi 1 bnh khng qu 10-15 pht m phi t nht 8 im. Lm nh vy vi cc bnh cn li.2. Xc nh bc ring theo

:

Dng ng ht cho vo 4 bnh tam gic cc ha cht vi lng chnh xc theo thnh phn bng sau: Dung dch KI M/40, ml HNO3 0,1M, ml H2O, ml KNO3, ml Bnh 1 10 10 27,5 32,5 Bnh 2 20 10 20 30 Bnh 3 30 10 12,5 27,5 Bnh 4 40 10 5 25

Cho vo bnh 1 vi git tinh bt v 20ml dd Fe(NO3)3M/60, lc mnh, ghi thi im bt u phn ng. Cc bc tip theo c tin hnh tng t nh phn xc nh bc ting theo Fe3+ . IV. KT QU Chui th nghim 1: Bnh 1:(C0)I =1,667x10-3

t,giy 115 100 100 110 110 118 100 100 2.6 3.0 3.4 3.9 4.4 4.9 5.3 5.8

ml 2.6x10-4 3 x10-4 3.4 x10-4 3.9 x10-4 4.4 x10-4 4.9 x10-4 5.3 x10-4 5.8 x10-4

(104) 0.38 0.33 0.29 0.26 0.23 0.20 0.19 0.17 8.7x10-3 0.01 0.01 9.1 x10-3 9.1 x10-3 8.5 x10-3 0.01 0.01

Chui th nghim 1 bnh 1 T biu trn ta c c gi tr = 19.297 Bnh 2: (C0)I =3,33 x10-3

t,giy 50 50 60 55 59 61 62 60 2.3 3.0 3.8 4.5 5.3 6.0 6.8 7.5

ml 2.3 x10-4 3.0 x10-4 3.8 x10-4 4.5 x10-4 5.3 x10-4 6.0 x10-4 6.8 x10-4 7.5 x10-4

(104) 0.44 0.33 0.26 0.22 0.19 0.17 0.15 0.13 0.02 0.02 0.017 0.018 0.017 0.016 0.016 0.017

Chui th nghim 1 bnh 2 T biu trn ta c c gi tr = 57.552

Bnh 3: (C0)I =5x10-3 t,giy 32 30 32 30 35 30 35 35 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 ml 3.5 x10-4 4.0 x10-4 4.5 x10-4 5.0 x10-4 5.5 x10-4 6.0 x10-4 6.5 x10-4 7.0 x10-4 (104) 0.29 0.25 0.22 0.20 0.18 0.17 0.15 0.14 0.031 0.033 0.031 0.033 0.028 0.033 0.028 0.028

Chui th nghim 1 bnh 3 T biu trn ta c c gi tr = 11.353 Bnh 4: (C0)I = 6,67 x10-3

t,giy 50 40 40 40 45 50 49 40 5.9 6.6 7.3 8.0 8.8 9.6 10.5 11.4

ml 5.9 x10-4 6.6 x10-4 7.3 x10-4 8.0 x10-4 8.8 x10-4 9.6 x10-4 10.5 x10-4 11.4 x10-4

(104) 0.17 0.15 0.14 0.13 0.11 0.10 0.95 0.88 0.02 0.025 0.025 0.025 0.022 0.02 0.02 0.025

Chui th nghim 1 bnh 4 T biu trn ta c c gi tr = 11.782

th xc nh bc ring phn ca I-

bc ring phn ca I- l 0.29 Chui th nghim 2: Bnh 1: (C0)Fe3+ = 2.5 x10-3

t,giy

ml

(104)

60 65 60 60 64 65 60 65

2.6 3.2 3.8 4.4 5.0 5.6 6.2 6.8

2.6 x10-4 3.2 x10-4 3.8 x10-4 4.4 x10-4 5.0 x10-4 5.6 x10-4 6.2 x10-4 6.8 x10-4

0.38 0.31 0.26 0.23 0.20 0.18 0.16 0.15

0.017 0.015 0.017 0.017 0.016 0.015 0.017 0.015

Chui th nghim 2 bnh 1 T biu trn ta c c gi tr = 22.727 Bnh 2: (C0)Fe3+ =5 x10-3

t,giy 120 120 120 110 110 100 100 110 2.8 3.3 3.8 4.3 4.8 5.3 5.8 6.4

ml 2.8 x10-4 3.3 x10-4 3.8 x10-4 4.3 x10-4 4.8 x10-4 5.3 x10-4 5.8 x10-4 6.4 x10-4

(104) 0.36 0.30 0.26 0.23 0.20 0.19 0.17 0.16 8.3 x10-3 8.3 x10-3 8.3 x10-3 9.1 x10-3 9.1 x10-3 0.01 0.01 9.1 x10-3

Chui th nghim 2 bnh 2 T biu trn ta c c gi tr = 76.433 Bnh 3: (C0)Fe3+ = 7,5 x10-3

t,giy 60 60 55 55 59 58 55 55 4.4 5.4 6.2 6.0 6.8 7.6 8.2 9.0

ml 4.4 x10-4 5.4 x10-4 6.2 x10-4 6.0 x10-4 6.8 x10-4 7.6 x10-4 8.2 x10-4 9.0 x10-4

(104) 0.23 0.19 0.16 0.17 0.15 0.13 0.12 0.11 0.017 0.017 0.018 0.018 0.017 0.017 0.018 0.018

Chui th nghim 2 bnh 3 T biu trn ta c c gi tr = 35 Bnh 4: (C0)Fe3+ = 0,01

t,giy 30 30 25 30 30 28 29 29 2.7 3.7 4.7 5.8 6.8 7.6 8.6 9.7

ml 2.7 x10-4 3.7 x10-4 4.7 x10-4 5.8 x10-4 6.8 x10-4 7.6 x10-4 8.6 x10-4 9.7 x10-4

(104) 0.37 0.27 0.21 0.17 0.15 0.13 0.12 0.10 0.03 0.03 0.04 0.03 0.03 0.04 0.03 0.03

Chui th nghim 2 bnh 4 T biu trn ta c c gi tr = 2.6667

Biu xc nh bc ring phn ca Fe3+

bc ring phn ca Fe3+ l 0.97 bc chung ca phn ng l n=0.97+0.29=1.26

BI 5: KHO XC TC THY PHN ESTER XC NH NNG LNG HOT HA CA PHN NGI- MC CH Kho st tc phn ng thy phn ester nhiu nhit t xc nh nng lng hot ha ca phn ng.I- C S L THUYT

Nng lng ha ha Ea da vo phng trnh Arrhenius:lg k = Ea + lg k 2,303 RT

o k nhiu nhit T khc nhau, v ng biu din lgk theo l Ea v tung l lgk* 2.303 R

1 , h s gc T

Khi bit Ea ta phng tnh k2 T2 nu bit c k1 T1. T (1) suy ra:lg Ea 1 k2 1 = T k1 2,303 R 1 T2

Acetate etyl: CH3COOC2H5 hay RCOOR RCOOR + H2O RCOOH + ROH

Nu c xc tc ion H+ th phn ng xy ra nhanh hn RCOOR + H2O RCOOH + ROH

Do nng ca acid CH3COOH v ru C2H5OH hin din rt t nn phn ng thun nghch coi nh khng ng k. Phng trnh ng hc:

d [ RCOOR '] = k 1[RCOOR' ] dt

y l hm bt mt nn: lg ( a x ) = vi: a l nng u ca acid acetat etyl (a - x) l nng thi im t

k1 t + lg a 2,303

- Gi V l th tch dung dch NaOH cn chun 10ml mu lc phn ng xy ra hon ton. - Gi V0 l th tch dung dch NaOH cn chun 10ml mu lc mi bt u phn ng. - Gi Vt l th tch dung dch NaOH cn chun 10ml mu thi im t. V th tch t l vi nng nn:lg (V Vt ) = k1t + lg (V V0 ) 2,303

V ng biu din lg(V-Vt) theo t, suy ra k t h s gc ca ng biu din. II- TIN TRNH TH NGHIM1. Thc hin phn ng 30oC

Cho 5ml ester acetat etyl vo bnh nh mc 100ml, 100ml dung dch HCl 0.2N vo cc 100ml ri c hai vo b iu nhit. Khong 10 pht, ly cc HCl cho vo bnh nh mc ti ng vch. Ghi nhn thi im t0. Lc u hn hp vi ln ri vo my iu nhit, sau ht nhanh 10ml cho vo bnh nn 250ml c cha sn 50ml nc ct c ngm lnh trc ngng phn ng. Ghi nhn thi im t1.

nh phn acid bng dung dch NaOH 0.1N vi ch th phenolphtalein n khi xut hin mu hng nht. Ghi nhn th tch t1 thi im t1. Sau 10 pht lp li s nh phn nh trn 6 ln na, t2 l thi gian tnh t t0, tng t i vi cc im cn li. Kt qu: t ( pht) 5 15 25 Vt (ml) 16 17 18 V 67.2 V - Vt Vt ) 51.2 50.2 49.2 1.709 1.700 1.692 lg(V -

V ng biu din lg(V - Vt) theo t

V ng biu din lg(V-Vt) Ta c: VT1 = 16 (ml)

VT2 = (5000*d)/M = (5000*0.901)/88 = 51.2 Vy V = VT1 + VT2 = 51.2 + 16 = 67.2 (ml) T phng trnh (*) v th suy ra k1/2.303 = -0.0011 v lgk* = 1.713 k1 = 2.5333 x 10-3 v k* = 512. Thc hin phn ng 40oC

Lm tng t nh trn nhng 40oC. Kt qu: t ( pht) 5 15 25 Vt (ml) 19.6 21.4 22.6 V 70.8 V - Vt Vt ) 51.2 49.4 48.2 1.709 1.694 1.683 lg(V -

V ng biu din lg(V - Vt) theo t

Ta c: VT1 = 19.6 (ml)

VT2 = (5000*d)/M = (5000*0.901)/88 = 51.2 Vy V = VT1 + VT2 = 51.2 + 19.6 = 70.8 (ml) T phng trnh (*) v th suy ra k2/2.303 = -0.001 v lgk* = 1.714 K2 = 2.303 x 10-3 v k* = 51 Tnh Ea: t phng trnh lgEa 1 k2 1 = T k1 2,303 R 1 T2

vi T1 = 303, T2 = 313, R = 8.314 Ea = 1982958.214