Tom Tat Ly Thuyet 10

Tm tt l thuyt ho hc v c 10

Chng 1 : NGUYN TI. Thnh phn cu to ca nguyn t Kt lun : thnh phn cu to ca nguyn t gm: Ht nhn nm tm nguyn t gm cc ht proton v ntron V nguyn t gm cc electron chuyn ng xung quanh ht nhn Electron me= 9,1094.10-31 kg qe= -1,602.10 -19 C k hiu l eo qui c bng 1Proton Ht proton l 1 thnh phn cu to ca ht nhn nguyn t,mang in tch dng, k hiu p m= 1,6726.10 -27 kg q= + 1,602.10 -19 C k hiu eo, qui c 1+ Ntron Ht ntron l 1 thnh phn cu to ca ht nhn nguyn t, khng mang in , k hiu n.Khi lng gn bng khi lng proton II.Kch thc v khi lng ca nguyn t 1- Kch thc Nguyn t cc nguyn t c kch thc v cng nh, nguyn t khc nhau c kch thc khc nhau. n v biu din A(angstron) hay nm(nanomet) 1nm= 10 -9 m ; 1nm= 10A 1A= 10 -10 m = 10 -8 cm 2- Khi lng Khi lng nguyn t rt nh b, biu th khi lng ca nguyn t, phn t, p, n, e dng n v khi lng nguyn t, k hiu u (vc) 1u = 1/12 khi lng 1 nguyn t ng v cacbon-12 1u = 19,9265.10 -27 kg/12 = 1,6605.10 -27kg III-Ht nhn nguyn t 1. in tch ht nhn Proton mang in tch 1+, nu ht nhn c Z proton th in tch ca ht nhn bng Z+ Trong nguyn t : S n v in tch ht nhn = S p = S e V d : nguyn t Na c Z = 11+ ngt Na c 11p, 11e 2. S khi L tng s ht proton v ntron ca ht nhn A=Z+N V d 1: Ht nhn nguyn t O c 8p v 8n A = 8 + 8 = 16 V d 2: Nguyn t Li c A =7 v Z = 3 Z = p = e = 3 ; N = 7 - 3 =4 Nguyn t Li c 3p, 3e v 4n IV- Nguyn t ha hcTrang 1


1.nh ngha Nguyn t ha hc l nhng nguyn t c cng in tch ht nhn V d : Tt c cc nguyn t c cng Z l 8 u thuc nguyn t oxi, chng u c 8p, 8e 2.S hiu nguyn t S n v in tch ht nhn nguyn t ca 1 nguyn t c gi l s hiu nguyn t ca nguyn t (Z) 3.K hiu nguyn t S khi

A ZS hiu nguyn t V d :23 11

X

Na

Cho bit nguyn t ca nguyn t natri c Z=11, 11p, 11e v 12n (23-11=12) V - NG V Cc ng v ca cng 1 nguyn t ha hc l nhng nguyn t c cng s proton nhng khc nhau v s ntron, do s khi ca chng khc nhau V d : Nguyn t oxi c 3 ng v1 6 8

O,

1 7 8

O,

1 8 8

O

Ch : - Cc nguyn t ca cng 1 nguyn t c th c s khi khc nhau - Cc ng v c tnh cht ha hc ging nhau VI- Nguyn t khi v nguyn t khi trung bnh ca cc nguyn t ha hc 1- Nguyn t khi Nguyn t khi ca 1 nguyn t cho bit khi lng ca nguyn t nng gp bao nhiu ln n v khi lng nguyn t V khi lng nguyn t tp trung nhn nguyn t nn nguyn t khi coi nh bng s khi (Khi khng cn chnh xc) V d : Xc nh nguyn t khi ca P bit P cZ=15, N=16 Nguyn t khi ca P=31 2- Nguyn t khi trung bnh Trong t nhin a s nguyn t ha hc l hn hp ca nhiu ng v(c s khi khc nhau) Nguyn t khi ca nguyn t l nguyn t khi trung bnh ca cc ng v .A= aX + bY 100

X, Y: nguyn t khi ca ng v X, Y a,b : % s nguyn t ca ng v X, Y V d : Clo l hn hp ca 2 ng v35 17

Cl chim 75,77%A=

v

35 17

Cl

chim 24,23% nguyn t khi trung bnh ca clo l:75 ,77 24 ,23 + 35 .5 100 100

VII- Cu hnh electron nguyn t 1.S chuyn ng ca cc electron trong nguyn t:Trang 2


-Cc electron chuyn ng rt nhanh trong khu vc xung quanh ht nhn nguyn t khng theo nhng qu o xc nh to nn v nguyn t. - Trong nguyn t: S e = s p = Z 2.Lp electron v phn lp electron a.Lp electron: - trng thi c bn, cc electron ln lt chim cc mc nng lng t thp n cao (t gn ht nhn ra xa ht nhn) v xp thnh tng lp. - Cc electron trn cng mt lp c mc nng lng gn bng nhau Th t lp 1 2 3 4 5 6 7 Tn lp K L M N O P Q b.Phn lp electron: - Cc e trn cng mt phn lp c mc nng lng bng nhau - Cc phn lp c k hiu bng ch ci thng : s, p, d, f, - S phn lp = s th t ca lp V d: + Lp th nht (lp K,n=1) c 1 phn lp :s + Lp th hai (lp L,n=2) c 2 phn lp : s, p + Lp th ba (lp M,n=3) c 3 phn lp :s, p, d + Lp th t (lp N,n=4) c 4 phn lp: s, p, d, f - Cc electron phn lp s gi l electron s, tng t ep, ed, c. Obitan nguyn t : L khu vc khng gian xung quanh ht nhn m xc sut c mt electron l ln nht ( 90%) k hiu l AO. Trn 1 AO ch cha ti a 2 electron c gi l electron ghp i Nu trong 1AO cha 1 lectron c gi l e c thn Nu trong AO khng cha e c gi l AO trng. - Phn lp s c 1 AO hnh cu. - Phn lp p c 3 AO hnh s 8 ni cn i. - Phn lp d c 5 AO hnh phc tp. - Phn lp f c 7 AO hnh phc tp. 3.S electron ti a trong mt phn lp , mt lp: a.S electron ti a trong mt phn lp : Phn Phn Phn Phn lp s lp p lp d lp f S e ti a 2 6 10 14 2 6 10 Cch ghi S p d f14 - Phn lp s electron ti a gi l phn lp electron bo ha.

b. S electron ti a trong mt lp : Lp

Lp K

Lp LTrang 3

Lp M

Lp N


Th t n=1 n=2 n=3 n=4 Sphnlp 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f S e ti a ( 2n2) 2e 8e 18e 32e - Lp electron s e ti a gi l lp e bo ha. 14 Th d : Xc nh s lp electron ca cc nguyn t : 7 N 4.Cu hnh electron nguyn t a.Nguyn l vng bn - Cc e trong nguyn t trng thi c bn ln lt chim cc mc nng lng t thp n cao. - Mc nng lng ca : 1s2s2p3s3p4s3d5s4d5p6s4f5d6p7s5f6d... - Khi in tch ht nhn tng ln s xut hin s chn mc nng lng gia s v d hay s v f. + Lp : tng theo th t t 1 n 7 k t gn ht nhn nht +Phn lp: tng theo th t s, p, d, f. b. Nguyn l pauli: Trn 1obitan nguyn t cha ti a 2 electron v c chiu t quay khc chiu nhau xung quanh trc ring ca mi electron. c. Qui tc hun : Trong cng mt phn lp cc electron in vo cc obitan sao cho s lectron c thn l ln nht. e. Cu hnh electron ca nguyn t: - Cu hnh electron ca nguyn t: Cu hnh electron ca nguyn t biu din s phn b electrron trn cc phn lp thuc cc lp khc nhau. - Quy c cch vit cu hnh electron : + STT lp e c ghi bng ch s (1, 2, 3. . .) + Phn lp c ghi bng cc ch ci thng s, p, d, f. + S e c ghi bng s pha trn bn phi ca phn lp.(s2 , p6 ) - Mt s ch khi vit cu hnh electron: + Cn xc nh ng s e ca nguyn t hay ion. ( s e = s p = Z ) + Nm vng cc nguyn l v qui tc, k hiu ca lp v phn lp ... + Qui tc bo ho v bn bo ho trn d v f : Cu hnh electron bn khi cc electron in vo phn lp d v f t bo ho ( d10, f14 ) hoc bn bo ho ( d5, f7 ) - Cc bc vit cu hnh electron nguyn t Bc 1: in ln lt cc e vo cc phn lp theo th t tng dn mc nng lng. Bc 2: Sp xp li theo th t cc lp v phn lp theo nguyn tc t trong ra ngoi. Bc 3: Xem xt phn lp no c kh nng t n bo ho hoc bn bo ho, th c s sp xp li cc electron cc phn lp ( ch yu l d v f ) V d: Vit cu hnh electron nguyn t cc nguyn t sau + H( Z = 1) + Ne(Z = 10) + Cl(Z = 17) 1s22s22p63s23p5 + Fe, Z = 26, 1s22s22p63s23p63d64s2 + Cu ( Z = 29); Cr ( Z = 24) -Cch xc nh nguyn t s, p, d, f: + Nguyn t s : c electron cui cng in vo phn lp s. Na, Z =11, 1s22s22p63s1Trang 4


+Nguyn t p: c electron cui cng in vo phn lp p. Br, Z =35, 1s22s22p63s23p64s23d104p5 Hay 1s22s22p63s23p63d104s24p5 + Nguyn t d: c electron cui cng in vo phn lp d. Co, Z =27, 1s22s22p63s23p64s23d7 Hay 1s22s22p63s23p63d74s2 + Nguyn t f: c electron cui cng in vo phn lp f c. Cu hnh e nguyn t ca 20 nguyn t u(sgk) d. c im ca lp e ngoi cng: -i vi nguyn t ca tt c cc nguyn t, lp ngoi cng c nhiu nht l 8 e. - Cc electron lp ngoi cng quyt nh n tnh cht ho hc ca mt nguyn t. +Nhng nguyn t kh him c 8 e lp ngoi cng (ns2np6) hoc 2e lp ngoi cng (nguyn t He ns2 ) khng tham gia vo phn ng ho hc . +Nhng nguyn t kim loi thng c 1, 2, 3 e lp ngoi cng. Ca, Z = 20, 1s22s22p63s23p64s2 , Ca c 2 electron lp ngoi cng nn Ca l kim loi. +Nhng nguyn t phi kim thng c 5, 6, 7 e lp ngoi cng. O, Z = 8, 1s22s22p4, O c 6 electron lp ngoi cng nn O l phi kim. +Nhng nguyn t c 4 e lp ngoi cng c th l kim loi hoc phi kim. Kt lun: Bit cu hnh electron nguyn t th d on tnh cht ho hc nguyn t. PHNG PHP GII MT S BI TON CHNG 1 I-Mt s im lu khi gii ton chng nguyn t. Trong nguyn t ta lun c: - S e = s p - S n = S A s p - p n 1,5p hay P N 1,5Z - n,p,e thuc tp s nguyn dng. ( sau chng ta bin i bt ng thc t kim tra nghim ) II- Mt s bi ton v d 1. Bi ton v cc ht: xut nhiu cch gii, chn cch gii hay V d 1: Mt nguyn t c tng s cc loi ht l 13 . Hy xc nh s lng tng loi ht trong nguyn t. V d 2: Tng s ht trong ht nhn nguyn t l 9. Hy xc nh s lng tng loi ht trong nguyn t. V d 3: Tng s ht trong nguyn t bng 115, s ht mang in nhiu hn s ht khng mang in l 25. Xc nh s ht e ca nguyn t .

V d 4: Ion M3+ c cu to bi 37 ht. S ht mang in nhiu hn s ht khng mang in l 9. a. Xc nh s lng tng ht trong M . b. Vit cu hnh electron v s phn b cc e vo cc AO.Trang 5


2. Bi ton v ng v : xut nhiu cch gii, cch gii hay V d 1: Trong t nhin ng c 2 ng v 63Cu chim 73 % v 65Cu chim 27%. Xc nh khi lng nguyn t trung bnh ca ng. V d 2: Trong t nhin ng c 2 ng v 63Cu chim 73 % v A Cu. Xc nh s khi A bit khi lng nguyn t trung bnh ca ng bng 63,54. V d 3: Trong t nhin ng c 2 ng v X Cu chim 73 % v Y Cu. Xc nh X,Y bit khi lng nguyn t trung bnh ca ng bng 63,54 v s khi ca ng v th hai ln hn ng v th nht 2 n v. V d 4: Trong t nhin ng c 2 ng v 63 Cu v 65 Cu. Xc nh % ca ng v th nht bit khi lng nguyn t trung bnh ca ng bng 63,54 . V d 5: Ion M+ v X2- u c cu hnh electron : 1s22s22p63s23p6. a. Vit cu hnh e ca M v X. b. Tnh tng s ht mang in trong hp cht c to bi 2 ion trn.

Chng 2 : BNG TUN HON CC NGUYN T HA HC V NH LUT TUN HON CC NGUYN T HO HCI- BNG TUN HON CC NGUYN T HA HCTrang 6


1. Nguyn tc sp xp : * Cc nguyn t c xp theo chiu tng dn ca in tch ht nhn nguyn t. * Cc nguyn t c cng s lp electron trong nguyn t c xp thnh mt hng. * Cc nguyn t c cng s e ha tr trong nguyn t c xp thnh mt ct. 2. Cu to bng tun hon: a- nguyn t: S th t ca nguyn t ng bng s hiu nguyn t ca nguyn t . b- Chu k: Chu k l dy cc nguyn t m nguyn t ca chng c cng s lp electron, c xp theo chiu in tch ht nhn tng dn. S th t ca chu k trng vi s lp electron ca nguyn t cc nguyn t trong chu k . * Chu k nh: gm chu k 1, 2, 3. * Chu k ln : gm chu k 4, 5, 6, 7. c- Nhm nguyn t: l tp hp cc nguyn t m nguyn t c cu hnh electron tng t nhau , do c tnh cht ha hc gn ging nhau v c xp thnh mt ct. d- Khi cc nguyn t: * Khi cc nguyn t s : gm cc nguyn t nhm IA v IIA Nguyn t s l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp s. * Khi cc nguyn t p: gm cc nguyn t thuc cc nhm t IIIA n VIIIA ( tr He). Nguyn t p l nhng nguyn t m nguyn t c electron cui cng c in vo phn lp p. * Khi cc nguyn t d : gm cc nguyn t thuc nhm B. Nguyn t d l cc nguyn t m nguyn t c electron cui cng c in vo phn lp d. * Khi cc nguyn t f: gm cc nguyn t thuc h Lantan v h Actini. Nguyn t f l cc nguyn t m nguyn t c electron cui cng c in vo phn lp f. II-S BIN I TUN HON MT S TNH CHT CA CC NGUYN T 1. Cc nguyn t nhm A: nguyn t s v p * S th t nhm = s electron ha tr = s electron lp ngoi cng. * S bin i tun hon v cu hnh electron lp ngoi cng ca nguyn t cc nguyn t khi in tch ht nhn tng dn chnh l nguyn nhn ca s bin i tun hon tnh cht ca cc nguyn t. 2. Cc nguyn t nhm B: nguyn t d v f. ( kim loi chuyn tip). * Cu hnh electron nguyn t c dng : (n1)da ns2(a=1 10) * S electron ha tr = s electron lp n + s electron phn lp (n1)d nhng cha bo ha. * t S = a + 2 , ta c : - S 8 th S = s th t nhm. - 8 S 10 th nguyn t nhm VIII B. 3. S bin i mt s i lng vt l: a S bin i bn knh nguyn t khi in tch ht nhn tng : * Trong cng chu k : bn knh gim. * Trong cng nhm A : bn knh tng. b S bin i nng lng ion ha th nht ca cc nguyn t nhm A: Khi in tch ht nhn tng : * Trong cng chu k nng lng ion ha tng. * Trong cng nhm, nng lng ion ha gim. Nng lng ion ha th nht (I1) ca nguyn t l nng lng ti thiu cn tch electron th nht ra khi nguyn t trng thi c bn. ( tnh bng Kj/mol)Trang 7


4. m in: ca mt nguyn t l i lng c trng cho kh nng ht electron ca nguyn t khi to thnh lin kt ha hc. Khi in tch ht nhn tng: trong cng chu k, m in tng. trong cng nhm, m in gim. 5. S bin i tnh kim loiphi kim: a Trong cng chu k, khi in tch ht nhn tng: * tnh kim loi gim, tnh phi kim tng dn. b trong cng nhm A, khi in tch ht nhn tng: * tnh kim loi tng, tnh phi kim gim dn. 6. S bin i ha tr: Trong cng chu k , khi in tch ht nhn tng , ha tr cao nht vi oxi tng t 1 n 7, ha tr i vi hidro gim t 4 n 1. Ha tr i vi hidro= s th t nhm ha tr i vi oxi Cng thc phn t ng vi cc nhm nguyn t ( R : l nguyn t ) R2On : n l s th t ca nhm. RH8-n : n l s th t ca nhm. Nhm IA IIA IIIA IVA VA VIA Oxit R20 RO R2O3 RO2 R2O5 RO3 Hirua RH4 RH3 RH2 7. S bin i tnh axit-baz ca oxit v hidroxit tng ng: a Trong cng chu k , khi in tch ht nhn tng : tnh baz gim , tnh axit tng . b Trong cng nhm A, khi in tch ht nhn tng : tnh baz tng, tnh axit gim. * Tng kt : N.L ion Bn knh m Tnh Tnh Tnh ha (I1) n.t(r) in kim loi Phi kim baz Chu k (Tri sang phi) Nhm A (Trn xung ) 8. nh lut tun hon cc nguyn t ho hc. Tnh cht ca cc nguyn t v n cht cng nh thnh phn v tnh cht ca cc hp cht to nn t cc nguyn t bin i tun hon theo chiu tng ca in tch ht nhn nguyn t.

VIIA R2O7 RH

Tnh axit

III. QUAN H H GIA V TR NGUYN T V CU TO NGUYN T. 1.Mi quan h cu hnh v v tr trong HTTH.

Trang 8

Tm tt l thuyt ho hc v c 10-

Tog so n e Nguye to hoa p n s c Nguye to hoa f n d c So ngoacug e i n So p e l

-

Stt nguye to n Thuo nho A c m Thuo nho B c m Stt cu nho a m Stt chu k

Ca hnh e u nguye t n

-

V d : Xt i vi nguyn t P ( Z = 15)-

Ca hnh e u nguye t n

Tog so n e Nguye to hoa p n s c Nguye to hoa f n d c So ngoacug e i n So p e l

: 16 ne Stt nguye to n n :16 : P ne thuo nho A n c m : : 6e ne thuo nho VIA n c m : 3 l ne thuo chu k 3 p n c

2. Quan h h gia v tr nguyn t v tnh cht ca nguyn t. V tr nguyn t suy ra: Thuc nhm KL (IA, IIA, IIIA) tr B v H. Ho tr trong h/c oxit cao nht v trong h/c vi hiro. H/C xit cao v h/c vi hiro. Tnh axit, tnh baz ca h/c oxit v hiroxit. V d: Cho bit S th 16: Suy ra: S nhm VI, CK3, PK Ho tr cao nht vi xi 6, vi hiro l 2. CT oxit cao nht SO3, h/c vi hiro l H2S. SO3 l xit axit v H2SO4 l axit mnh. 3.So snh tnh cht ho hc ca mt nguyn t vi cc ng/t ln cn. a.Trong chu k theo chiu tng ca in tch ht nhn, c th v: Tnh kim loi yu dn, tnh phi kim mnh dn. Tnh baz, ca oxit v hiroxit y dn, tnh axit mnh dn. b. Tong nhm A, theo chiu tng ca in tch ht nhn, c th: Tnh kim loi mnh dn, tnh phi kim yu dn. Theo chu k : Tnh phi kim Si< P< S Theo nhm A: Tnh phi kim As < P< N 4. Lu khi xc nh v tr cc nguyn t nhm B . a. Nguyn t h d : (n-1)dansb vi a = 1 10 ; b = 1 2 + Nu a + b < 8 a + b l s th t ca nhm . + Nu a + b > 10 (a + b) 10 l s th t ca nhm. + Nu 8 a + b 10 nguyn t thuc nhm VIII B b. Nguyn t h f : (n-2)fansb vi a = 1 14 ; b = 1 2 + Nu n = 6 Nguyn t thuc h lantan. + Nu n = 7 Nguyn t thuc h actini. (a + b) 3 = s th t ca nguyn t trong h V d : Z = 62 ; n = 6, a = 6, b = 2 6 + 2 3 = 5 , thuc th 5 trong h lantan. PHNG PHP GII MT S BI TON CHNG II A. Phng php v qui tc h tr:Trang 9


- Qui tc tam xut. - Phng php t n s v gii cc phng trnh. - Phng php gi tr trung bnh. A,x mol, MA m x.M A +y.MB M A 0 : Thu nhit * Lu : H = H 2 H 1 nu H < 0 : To nhit III. Nhng ch quan trng a. Cn bng ho hc l cn bng ng Ngha l ti thi im cn bng c thit lp khng c ngha l phn ng dng li m vn xy ra nhng tc ca phn ng thun bng tc phn ng nghch. ( vt=vn).Trang 28


b.Khi bin i h s trong phng trnh ho hc biu din cn bng ho hc th hng s cn bng cng bin i theo. Th d : 2A + B C + D Kcb 4A + 2B 2C + 2D K'cb = (Kcb)2 IV . Cu hi v bi tp 1. Cho mt mu vi nng 10g vo 200ml dung dch HCl 2M. Tc phn ng thay i nh th no nu: a. Nghin nh vi trc khi cho vo ? b. dng 100ml dung dch HCl 4M ? c. tng nhit ca phn ng ? d. Cho thm vo 500ml dung dch HCl 4M ? e. Thc hin phn ng trong nghim ln hn ? 2. Cho H2 + I2 2 HI. Vn tc phn ng thay i th no khi nng ca hiro tng gp hai ln. 3. Tc ca phn ng tng ln bao nhiu ln khi nhit ca phn ng tng t 20 oC 80oC. Bit c tng 10oC th tc tng ln: a. 2 ln b. 3 ln 4. Cho phn ng tng hp NH3 N2 + 3H2 2NH3 H < 0 . Cn tc ng nhng yu t no thu c nhiu NH3 nht ? 5. Cn bng ca phn ng sau s chuyn dch v pha no khi: Tng nhit ca h. H p sut ca h . Tng nng cc cht tham gia phn ng. a) N2 + 3H2 2 NH3 + Q. CaO + CO2 Q. b) CaCO3 c) N2 + O2 2NO + Q. d) CO2 + H2 H2O + CO Q. e) C2H4 + H2O C2H5OH + Q. f) 2NO + O2 2NO2 + Q. g) Cl2 + H2 2HCl + Q. 2SO2 + O2 Q. h) 2SO3 6. Cho 2SO2 + O2 2SO3 + 44 Kcal. Cho bit cn bng ca phn ng chuyn dch theo chiu no khi: a. Tng nhit ca h. b. Tng nng ca O2 ln gp i . 7. Cn bng phn ng CO2 + H2 CO + H2O c thit lp t0C khi nng cc cht trng thi cn bng nh sau: [ CO2] = 0,2 M; [H2] = 0,8 M ; [CO] =0,3 M; [H2O] = 0,3 M. a) Tnh hng s cn bng ? b) Tnh nng H2, CO2 ban u. 8. Cho phn ng PCl5 (k) PCl3 (k) + Cl2 (k)Trang 29


C hng s cn bng 503oC l 33,33mol/lit . Tnh nng cn bng ca cc cht bit nng ban u ca PCl5 l 1,5M v Cl2 1M 9. Cho phn ng thun nghch N2 + O2 2NO c hng s cn bng 2400oC l Kcb = 35.10-4 Bit lc cn bng nng ca N2 v O2 ln lt bng 5M v 7M. Tnh nng mol/lit ca NO lc cn bng v nng N2 v O2 ban u. 10. Xt cn bng : Cl2 (k) + H2 (k) 2HCl (k) a. nhit no hng s cn bng l 0,8 v nng cn bng ca HCl l 0,2M. Tnh nng ca Cl2 v H2 lc ban u, bit rng lc u lng H2 ly gp 3 ln Cl2. b. Nu tng p sut ca h th c nh hng g n cn bng khng ? ti sao ? 11. Cho cn bng 2A(k) B(k) + C(k) a. nhit no Kcb = 1/729. Tnh xem c bao nhiu % A b phn hu. b. Tnh hng s cn bng ca phn ng cng nhit trn khi c vit A(k) 1/2B(k) + 1/2 C(k) B(k) + C(k) 2A(k) 12. Xt cn bng sau : CaCO3 (r) CaO(r) + CO2(k) H > 0 Cn bng s chuyn dch nh th no khi bin i mt trong cc iu kin sau - Tng nhit - Thm lng CaCO3 - Ly bt CO2 - Tng p sut chung bng cch nn th tch ca h gim xung. 13. Trong qu trnh sn xut gang , xy ra phn ng Fe2O3(r) + 3CO(r) 2Fe (r) + 3CO2 (k) H > 0 C th dng nhng bin php g tng tc phn ng ? H > 0 14. Xt cn bng CO(k) + H2O(k) CO2(k) + H2 Bit rng nu thc hin phn ng gia 1 mol CO v 1 mol H2O th trng thi cn bng c 2/3 mol CO2 c sinh ra. Tnh hng s cn bng cca phn ng ?

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Tom Tat Ly Thuyet 10