bài tập trắc nghiệm hóa đại cương

8/21/2019 bi tp trc nghim ha i cng

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TTRRNNGGIIHHCCBBCCHHKKHHOOAATTHHNNHHPPHHHHCCHHMMIINNHHKKHHOOAACCNNGGNNGGHHHHAAHHCCVVDDUUKKHH

BBMMNNCCNNGGNNGGHHVVCC

GIO TRNH BI TP

HA I CNG

9 - 2004Cc khiu vthut ngtrong gio trnh bi tp ny tun theo gio trnh ha i cngxut bn nm 2002 cugio sNguyn nh Soa

PHN 1 : CU TO CHTA. BI TP TN

Bi 1.1:Cbao nhiu ocbitan nguyn ttrong phn lp lng tl = 2 ca lp M? Gi tn vvccocbitan nguyn t.

Bi 1.2:Hy vit cc slng tl, mlvtnh selectron cthctrn lp N trong nguyn t.


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Bi 1.3:Da vo trt tphn bcc mc nng lng cho bit cu to lp velectron nguyn tca ccnguyn t: S (Z = 16), Ti ( Z = 22) vNd ( Z = 60).

Bi 1.4:Vit cu hnh electron vvcc ocbitan nguyn t lp ngi cng ca: Si ( Z =14, chu k IIIphn nhm IVA), Fe ( Z = 26, chu kIV, phn nhm VIIIB), Ag ( Z = 47, chu kV, phn nhmIB) vAt ( Z = 85, chu kVI, phn nhm VIIA).

Bi 1.5:Xc nh vtr( , chu k, phn nhm), hvtn nguyn t, cc mc oxy ha dng cao nhtvm thp nht ca nhng nguyn tccu hnh electron nguyn tnhsau:

- 1s22s22p63s23p63d54s2- 1s

22s

22p

63s

23p

63d

104s

24p

64d

105s

25p

3 -

1s22s

22p

63s

23p

63d

104s

24p

64d

104f

145s

25p

65d

106s

1

Bi 1.6:Vit cu hnh electron ca cc ion Ag+, Ti2+, Ti4+, Mn2+, Fe2+, Se2-vBr-. Nhng nguyn tvion no ccu hnh ging ion Br-?

Bi 1.7:Tnh ha trvsoxy ha ca cc nguyn ttrong nhng hp cht sau:H2O, H2O2, HClO4, Hg2Cl2, CBr4, Al4C3, CaH2, H2S vNa2S2O3.

Bi 1.8:Phn tch sto thnh lin kt (kiu, bc), cu hnh khng gian (dng hnh hc, gc ha tr) cacc phn tsau y bng phng php lin kt ha tr(LH) : F2, HBr, H2Te ( HTeH = 90

0)

NF3( FNF = 1020), CCl4 ( ClCCl = 109

05), CS2( SCS = 180

0), NO2( ONO = 132

0; bc lin

kt = 1,5), NO2-( ONO = 115

0; bc lin kt = 1,5).

Bi 1.9:Phn tch sto thnh phn tN2vCO bng cc phng php lin kt cng ha tr(LH) vocbitan phn t(OP). Tso snh cc c trng lin kt vlha tnh ca N2vCO.

Bi 1.10:So snh di, bn, bc lin kt OO trong dy O22-

- O2-- O2- O2

+. Nhn xt vttnh

bn, tnh oxy ha ca chng.B. BI TP TRC NGHIM

CHNG 2 : CU TO NGUYN T2.1 Cu to ht nhn nguyn t

2.1Trong cc pht biu cho sau y, cc pht biu ng l:1) Cc nguyn tccng in tch ht nhn Z vc skhi A khc nhau c gi lcc ng v.2) Ht nhn nguyn tca cc ng vca mt nguyn tcsntron khc nhau.3) Nguyn tlng ca mt nguyn ttrong bng hthng tun hn ltrung bnh cng ca nguyntlng ca cc ng vtheo tltn ti trong tnhin.4) Trng vcnhiu nht ca mt nguyn tX, cc ng vkhc u lnhng ng vphngx.a) 1 b) 1,2 c) 1,4 d) 1,2,3

2.2 Khi lng nguyn tca ng v2H gm:a) Khi lng ca 1 proton + 1 ntron b) khi lng ca electronc) khi lng ca electron + 1 ntron d) khi lng ca 1 proton

2.3 Chn pht biu ng vtnh cht ca cc ng vca cng 1 nguyn t:a) Cc ng vca cng mt nguyn tthging nhau vtt ccc tnh cht l, ha hc.b) Cc nguyn tccng in tch ht nhn, cskhi nhnhau c gi lcc ng v.c) Cc ng vccng sproton vcng sntron.

d) ng vchim cng mt trong bng hthng tun hn cc nguyn t.2.4 Pht biu no di y lng

a) Cc nguyn tccng in tch ht nhn, cskhi nhnhau c gi lng v.b) Vi mi nguyn t, slngproton trong ht nhn nguyn tlcnh,song cthkhcnhau vsntron, lhin tng ng v.c) Cc nguyn tcskhi nhnhau, song sproton ca ht nhn li khc nhau c gi lcc cht ng v.d) Cc ng vca cng mt nguyn tthging nhau vtt ccc tnh cht l, ha hc.

2.5 Chn p n ng vy nht.


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1) ng vgm cc nguyn tccng bc snguyn t(Z) nhng cskhc nhau v skhi lng(A).

2) Nguyn tlng ca mt nguyn tltrung bnh cng ca cc nguyn tlng ca cc ng vtheo tlca cc ng vny trong thin nhin.

3) Khc nhau duy nht vccu gia cc ng vlcsntron khc nhau.4) Trng vcnhiu nht ca mt nguyn t, cc ng vkhc u lnhng ng vphng x.

a) Chc1 ng b) Chc1 v2 ng c) Chc1 v4 ng d) 1, 2 v3 ng.2.2. Cu to lp vnguyn t

2.2.1 Mhnh nguyn tBorh vquang phnguyn t2.6 Chn pht biu sai vkiu nguyn tBohr p dng cho nguyn tHidro hoc cc ion ging Hidro

(ion chc1 electron)a) Khi chuyn ng trn qy o Bohr, nng lng ca electron khng thay i.b) Electron khi lng m, chuyn ng vi tc v trn qy o Bohr bn knh r, c ln ca

momen ng lng bng: mvr = nh/2.c) Electron chthu vo hay pht ra bc xkhi chuyn tqy o ny sang qy o khc.d) Bc xpht ra c bc sng bng: = IE- EcI/h.

2.7 di sng ca bc xdo nguyn thidro pht ra tun theo hthc: 1/= RH(1/n121/n22). Nun1= 1 vn2= 4, bc xny ng vi schuyn electron:

a) Tquo 4 xung quo 1, bc xthuc dy Lyman.b) Tquo 1 ln quo 4, bc xthuc dy Lyman.c) Tquo 1 ln quo 4, bc xthuc dy Balmer.d) Tquo 4 xung qo 1, bc xthc dy Balmer.

2.8 Bc xcbc sng cc tiu ca nguyn tHidro pht ra khi electron t:a) Vcc (n = ) ri xung qy o 1 (n = 1). b) quo 1 ln quo 2.c) Quo 1 ln vcc. d). quo 2 xung quo 1

2.2.2 Lp velectron theo chc lng tCc slng tvocbitan nguyn t2.9 Chn pht biu sai:

1) Cc AO lp n bao gicng cnng lng ln hn AO lp (n-1).2) Slng tphl xc nh dng vtn ca ocbitan nguyn t.3) Slng ttmlccc gitrtn n n.4) Slng tphccc gitrt0 n n-1.

a) Cu 1 v2 sai. b) Cu 1 v3 sai.c) Cu 1, 2 v3 sai. d) Cu 1, 3 v4 sai.

2.10 Cc pht biu sau u ng tr:a) Slng tchnh n cgitrnguyn dng vgitrti a l7.b) Slng tphl (ng vi mt gitrca slng tchnh n) lun lun nhhn n.c) Nng lng electron vkhong cch trung bnh ca electron i vi ht nhn nguyn t tng

theo n.

d) Cng thc 2n2cho bit selectron ti a cthctrong lp electron thn ca mt nguyn ttrong bng hthng tun hn.

2.11 Slng tchnh n vslng tphl ln lt xc nh:a) Snh hng vhnh dng ca ocbitan nguyn t.b) Hnh dng vsnh hng ca ocbitan nguyn t.c) Nng lng ca electron vsnh hng ca ocbitan nguyn t.d) Nng lng ca electron vhnh dng ca ocbitan nguyn t.

2.12 Slng tmlc trng cho:a) Dng ocbitan nguyn t b) Kch thc ocbitan nguyn tc) Snh hng ca ocbitan nguyn t d) Tt cu ng


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2.13Chn pht biu sai:Slng ttmla) c trng cho snh hng ca cc AO trong khng gian.b) Cho bit slng AO trong mt phn lpc) Cgitrbao gml , , 0 , , l.d) c trng cho nng lng ca cc phn lp.

2.14 Chn pht biu sai:a)

Slng tchnh n cthnhn gitrnguyn dng (1,2, 3) , xc nhnng lng electronkch thc ocbitan nguyn t; n cng ln th nng lng ca electron cng cao, kch thcocbitan nguyn tcng ln. Trong nguyn ta electron, nhng electron ccng gitrn lpnn mt lp electron vchng ccng gitrnng lng.

b) S lng tphl c thnhn gitrt0 n n-1. Slng tphl xc nh hnh dng cam my electron vnng lng ca electron nguyn t. Nhng electron ccng gitrn vllp nn mt phn lp electron vchng cnng lng nhnhau.

c) Slng ttmlcthnhn gitrtl n +l. Slng ttc trng cho snh hngca cc ocbitan nguyn ttrong ttrng.

d) Slng tspin c trng cho thuc tnh ring ca electron vchchai gitr1/2 v+1/2.2.15 Chn cung: AO l:hm sng mttrng thi ca electron trong nguyn tc xc nh bi ba slng tn, l vml.bmt cmt electron bng nhau ca m my electron.qy o chuyn ng ca electron trong nguyn t.c trng cho trng thi nng lng ca electron trong nguyn t.Khong khng gian bn trong cc electron ca nguyn tchuyn ng.

a) 1 v5 b) 1 , 2 v3 c) 1 d) cnm cu u ng.2.16 Chn pht biu sai:

a) Slng ttmlccc gitrtn n nb) Slng tphl ccc gitrt0 n n1c) Slng tchnh n xc nh kch thc ca ocbitan nguyn td) Slng tphl xc nh cu hnh vtn ca ocbitan nguyn t

Cc quy tc xy dng lp velectron nguyn t2.17 Thuyt chc lng tcho nguyn tkhng chp nhn iu no trong 4 iu sau y (chn cu

sai):a) trng thi cbn, cc electron ln lt chim cc mc nng lng tthp n cao.b) Trong mt nguyn t,ct nht 2 electron ccng 4 slng t.c) Slng tphl xc nh tn vhnh dng ca orbital nguyn t.d) Trong mi phn lp, cc electron sp xp sao cho selectron c thn lti a.

2.18 Sphn bcc electron trong nguyn tCacbon trng thi bn l:

1s2 2s

2 2p

2

t cstrn:a) Nguyn lvng bn Paoli vquy tc Hund.b) Nguyn lvng bn Paoli, nguyn lngoitrPaoli, quy tc Hund vquy tc Cleskovxkic) Nguyn lvng bn Paoli, nguyn lngoi trPaoli vquy tc Hund.d) Cc quy tc Hund vCleskovxki.

2.19Trng thi ca electron lp ngi cng trong nguyn tcZ = 30 c c trng bng cc slng t:

a) n = 3, l = 2, ml= -2, ms= +1/2 b) n = 4, l = 0, ml= 0, ms= +1/2 v-1/2c) n = 3, l = 2, ml= +2, ms= -1/2 d) n = 4, l = 0, ml= 1, ms= +1/2 v-1/2


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2.20 Nhng bba slng tno di y lnhng bc chp nhn:1) n = 4, l = 3, ml= -3 2) n = 4, l = 2, ml= +3

3) n = 4, l = 1, ml= 0 4) n = 4, l = 0, ml= 0

a) 1,3,4 b) 1,4 c) 2,3,4 d) 3,4

2.21 Chn tt ccc bba slng tc chp nhn trong cc bsau:1) n = 4, l = 3, ml= -3 2) n = 4, l = 2, ml= +3

3) n = 4, l = 1, ml= 2 4) n = 4, l = 0, ml= 0

a) 1,3,4 b) 1,4 c) 2,3,4 d) 3,42.22 Tn cc ocbitan ng vi n = 5, l = 2; n= 4, l = 3; n =3, l = 0 ln lt l:

a) 5d, 4f, 3s b) 5p, 4d, 3s c) 5s, 4d, 3p d) 5d, 4p, 3s2.23 Ocbitan 3pxc xc nh bi cc slng tsau

a) chcn n , l , m b) Chcn n , mc) Chcn l , m d) n , l , m , s

2.24 Trong cc nguyn t vion sau, tiu phn no ccu hnh electron lp ngi cng l3s23p6a) X (Z = 17) b) X ( Z = 19) c) X

-( Z = 17) d) X

+( Z = 20)

2.25 Cho bit selectron ti a vslng tchnh n ca cc lp lng tL vN:a) lp L :18 e, n = 3; lp N : 32 e, n = 4b) lp L : 8 e, n = 2; lp N : 32 e, n = 4c) lp L : 8 e, n = 2; lp N : 18 e, n = 3d) lp L : 18 e, n = 3; lp N : 32 e, n = 5

2.26 Electron cui ca nguyn tS (Z = 16) cbcc slng tsau (quy c electron in vo ccocbitan theo thtmlt+l nl)a) n = 3, l = 2, ml= -2, ms= +1/2 b) n = 3, l = 2, ml= +2, ms= -1/2

c) n = 3, l = 1, ml= -1, ms= +1/2 d) n = 3, l = 1, ml= +1, ms= -1/2

2.27 Chn slng tt(ml) thchhp cho mt electron trong mt nguyn tcslng tchnhbng 4, slng tocbitan bng 2 vslng tspin bng1/2.

a) -2 b) 3 c) -3 d) -4

2.28 Cu hnh electron ha trca ion Co3+( Z = 27 ) trng thi bnh thng l:a) 3d

6(khng celectron c thn) b) 3d44s2( celectron c thn)

c) 3d6(celectron c thn) d) 3d44s2( khng celectron c thn)2.29 Xc nh cu hnh electron ha trca nguyn tcsthttrong bng hthng tun hn l47.

a) 4d10

5s25p

1 b) 4d

95s

2 c) 4d

105s

1 d) 4d

10.

2.30 Cu hnh electron ha trca ion Fe3+(Z= 26) trng thi bnh thng l:a) 3d

44s

1 b) 3d

34s

2 c) 3d

6 d) 3d

5

2.31 Cng thc electron ca Cu2+ trng thi bnh thng l:a) 1s

22s

22p

63s

23p

63d

94s

0 b) 1s

22s

22p

63s

23p

63d

74s

2

c) 1s22

s22p

63s

23p

63d

84s

1 d) 1s

22s

22p

63s

23p

63d

104s

0

2.32 Ocbitan 1s ca nguyn tH cdng hnh cu, ngha l:a) Xc sut gp electron 1sca H ging nhau theo mi hng trong khng gian.b) Khong cch ca electron 1s n nhn H lun lun khng i.

c) electron 1s chdi chuyn ti vng khng gian bn trong hnh cu y.d) C3 u ng.

2.33Chn pht biu ng. Trong cng mt nguyn t1) ocbitan 2s ckch thc ln hn ocbitan 1s.2) nng lng ca electron trn AO 2s ln hn nng lng ca electron trn AO 1s.3) xc sut gp electron ca AO 2pxln nht trn trc x.4) nng lng ca electron trn AO 2pzln hn nng lng ca electron trn AO 2pxa) Chccc cu 1 , 2 , 3 ng. b) C4 cu u ng.c) Chccc cu 2 , 3 , 4 ng. d) chccc cu 3 , 4 ng.


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2.34 Cc electron ccng slng tchnh chu tc dng chn yu nht l:a )Cc electron f b) Cc electron s c) Cc electron p d) Cc electron d

CHNG 3: NH LUT TUN HN, H THNG TUN HN CC NGUYN T HAHC

VCU TO NGUYN T3. 1 Cu trc electron ca nguyn tvhthng tun hn cc nguyn t

3.1 Hy chn trong cc pht biu di y cpht biu no sai :1) in tch ht nhn nguyn tca bt knguyn tno vtrsbng sthtca nguyn t

trong bng hthng tun hn.2) Tnh cht ca n cht, thnh phn vtnh cht cc hp cht bin thin tun hn theo chiu tng

ca in tch ht nhn.3) Trong bng hthng tun hn, phn nhm VIIIB cha phi lphn nhm cha nhiu nguyn

tnht.4) Chu klmt dy cc nguyn t, mu lmt kim loi kim vkt thc lmt khhim.a) 1 b) 3 c) 2 d) Khng cpht biu no sai

3.2 Chn pht biu saisau y vbng hthng tun hn cc nguyn tha hc:a) Cc nguyn tcng 1 phn nhm chnh ctnh cht tng tnhau.b) Cc nguyn ttrongcng chu kctnh cht tng tnhau.c) Cc nguyn ttrong cng mt phn nhm chnh ctnh khtng dn ttrn xung.d) Cc nguyn ttrong bng h thng tun hn c sp xp theo thttng dn in tch ht

nhn cc nguyn t.3.3 Chn cung:

"Sthtca phn nhm bng tng selectron lp ngi cng". Quy tc ny:a) ng vi mi phn nhm.b) Sai vi mi phn nhm.c) ng vi cc phnnhm chnh, trHidro phn nhm 7A vHeli.d) ng vi cc phn nhm phtrphn nhm VIIIB.

3.4 Trong chu k4, nguyn tno trng thi cbn c3 electron c thn:a) V, Ni, As b) V, Co, Br c) V, Co, As d) Mn, Co, As

3.5Vtrtrong bng hthng tun hn ca nguyn tccu hnh electron 1s22s22p63s23p63d54s2l:a) chu k4, phn nhm VIIB, 23 b) chu k4, phn nhm VIIB, 25c) chu k4, phn nhm VIIA, 25 c) chu k4, phn nhm VB, 25

3.6 Fe (Z = 26), Co (Z = 27) vNi (Z = 28) thuc phn nhm VIIIB nn c:a) Cu hnh electron ha trging nhau.b) Selectron ha trbng sthtnhm.c) Selectron ca lp electron ngi cng ging nhau.d) Selectron ha trging nhau.

3.7 Chn pht biu saivcc nguyn tphn nhm VIA :a) CthcsOxy ha cao nht l+6.

b) SOxy ha m thp nht ca chng l-2.c) as cc nguyn tlkim loi.d) Cu hnh lp ngi cng lns2np4.

3.8 Chn pht biu ng. Cu hnh electron ca hai nguyn tthuc phn nhm VIB vVIA ca chu k4 ln lt l:

1) 1s22s

22p

63s

23p

63d

44s

2 2) 1s

22s

22p

63s

23p

63d

54s

1

3) 1s22s

22p

63s

23p

63d

104s

24p

4 4) 1s

22s

22p

63s

23p

63d

104s

14p

5

a) 1, 3 b) 2, 3 c)1, 4 d) 2, 4

3.9 Chn pht biu ng. Nguyn tno di y khng thuc hd:


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a) Sn ( Z = 50 ) b) V ( Z = 23 ) c) Pd ( Z = 46 ) d) Zn ( Z = 30 )

3.10 Da vo cu hnh electron ngi cng l4d105s2, hy xc nh vtrca nguyn ttrong bng hthng tun hn :

a) Chu k5 , phn nhm IIA, 50 b) Chu k4, phn nhm IIB, 48c) Chu k5, phn nhm IIB, 48 d) Chu k5, phn nhm IIB, 50

3.11Chn pht biu ng. Cc electron ha trca:a)nguyn tBr (Z = 35) l4s24p5 b) Nguyn tSn (Z = 50) l3d24s1c) Nguyn tTi (Z = 22) l5s2 d) Nguyn tSr (Z = 38) l4d105s2

3.12 Cho cc nguyn t: Ca (Z = 20), Fe (Z = 26), Cd (Z = 48), La (Z = 57), cc ion ccu hnh lp velectron ging cc khtrgn nl:

a) Ca2+

, Cd2+

c) Ca2+

, Cd2+

b) La3+

, Fe3+

d) Ca2+

, La3+

3.13 Chn pht biu ng:a) SOxy ha dng cc i ca mt nguyn tlun bng vi sthtca phn nhm ca nguyn t

.b) SOxy ha dng cc i lun bng vi selectron lp ngi cng ca nguyn t.c) SOxy ha dng cc i lun bng selectron trn cc phn lp ha trca nguyn t.d) SOxy ha dng cc i cacc nguyn tphn nhm VA bng +5.

3.14 Nguyn tccu hnh lp ngi cng l3d54s1cvtrtrong bng hthng tun hn vcc tnhcht c trng nhsau:

a) Chu k4, phn nhm VIB , 24 , phi kim loi, soxy ha dng cao nht 6+.b) Chu k4, phn nhm VIB , 24, kim loi, soxy ha dng cao nht 6+, soxy ha m thpnht 1-.c) Chu k4, phn nhm VIB, 24, kim loi, soxy ha dng cao nht 6+.d) Chu k4, phn nhm VB, 24, kim loi, soxy ha dng cao nht 6+.

3.15 Phn nhm cm in ln nht trong bng hthng tun hn l:a) Phn nhm IIIA b) Phn nhm VIIAc) Phn nhm VIA d) Phn nhm IA

3.16Nguyn tA ccu hnh electron phn lp cui cng l4p3. A phi:a) thuc phn nhm IIIA, csoxy ha dng cao nht +3 vkhng csoxy ha m.b)

thuc phn nhm IIIB, csoxy ha dng cao nht +3 vcsoxy ha m thp nht -3.c) thuc phn nhm VB, csoxy ha dng cao nht +5 vcsoxy ha m thp nht -3.d) thuc phn nhm VA, csoxy ha dng cao nht +5 vcsoxy ha mthp nht -3.

3.17 Nguyn tccu hnh lp electron lp ngi cng l4s2cvtra) phn nhm IIA. b) Ctnh kim loi mnh.c) Csoxi ha +2 lbn nht. d) C3p n trn u cha chc ng.

3.18 Chn trng hpng:Nguyn tA chu kIV, phn nhm VIA. Nguyn tA c:

a) Z = 34, lphi kim. b) Z = 24, lkim loi.c) Z = 24, lphi kim. d) Z = 34, lkim loi.

3.19 Chn trng hp ng:Nguyn tB chu kIV, phn nhm VIIB . Nguyn tB c:

a) Z = 25 , lkim loi. b) Z = 24, lkim loi.c) Z = 26, lphi kim loi. d) Z = 25, lphi kim loi.

3.2 Sthay i tnh cht ca cc nguyn ttrong hthng tun hn3.20 Trong cc pht biu di y, pht biu no sai.

Trong cng mt chu ktheo thtttri qua phi, ta c:1) Slp electron tng dn . 2) Tnh phi kim loi gim dn.3) Tnh kim loi tng dn. 4) Tnh phi kim loi tng dn.


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a) 1,2,4 b) 4 c) 1 d) 1,2,3

3.21 Chn pht biu sai.a) Trong mt phn nhm chnh, m in gim dn ttrn xung di.b) Trong mt phn nhm ph, bn knh nguyn ttng u ttrn xung di.c) Trong mt chu knh(trkhhim), bn knh nguyn tgim dn ttri qua phi.d) Tnh kim loi gim dn, tnh phi kim loi tng dn ttri qua phi trong mt chu knh(trkh

him).3.22 Trong mt phn nhm chnh ca hthng tun hn, tnh oxy ha ca nguyn tkhi i ttrn

xung di bin thin theo chiu:a) Tng dn. b) Gim dn. c) Khng i. d) Khng xc nh c.

3.23 Trong mt phn nhm phca hthng tun hn, tnh kim loi ca cc nguyn tkhi i ttrnxung di bin i nhsau:a) Khng i. b) Tng dn. c) Gim dn. d) Khng xc nh c.

3.24 Chn pht biu ng:a) Trong cng chu k, bn knh nguyn tthuc phn nhm chnh tng dn tu n cui chu k.b) Trong mt chu kngn, m in tng dn ttri qua phi.c) Cc nguyn tnhm IA ddng nhn thm 1 to anion.d) Trong bng hthng tun hn, chu kIII cphn nhm ph.

3.25 Bn knh ion ca cc nguyn t phn nhm VIA ln hn bn knh ion ng electron ca ccnguyn tphn nhm VIIA (cng chu k) do cc nguyn tphn nhm VIA c:

a) Khi lng nguyn tnhhn. b) in tch ht nhn nguyn tnhhn.c) Ai lc electron nhhn. d) m in nhhn.

3.26 Chn pht biu ng:a) Bn knh ion lun nhhn bn knh nguyn t.b) Cc ion ca nhng nguyn tnm trong cng mt chu kthcbn knh bng nhau.c) Trong chui ion ng electron(cc ion cselectron bng nhau), ion csoxy ha ln hn ckch

thc nhhn.d) Trong mt chu k, khi i ttri sang phi, bn knh ca nguyn tng sau lun nhhn bn knh

ca nguyn tng trc.3.27 Chn pht biu ng. Dy nguyn tCa (Z = 20), Al (Z = 13), P (Z = 15), K (Z = 19) c bn knh

R tng dn theo dy :a) RP< RAl< RCa< RK b) RP< RAl< RK< RCac) RAl< RP< RK< RCa d) RK< RCa< RP< RAl

3.28 Nng lng ion ha ca nguyn thydro lnng lng phi cung cp a electron t:a) Tng 1 ( n = 1) ln tng 2 b) Tng 1 ln tng 7c) Tng 1 ra vcc d) Tvcc xung tng 1

3.29 Chncu sai. Sthay i nng lng ion ha thnht (I1) ca cc nguyn ttrong cc phn nhmtheo chiu tng sthtnguyn tc gii thch nhsau:

a) Trong phn nhm chnh, I1gim do stng hiu ng chn.b) Trong phn nhm ph, I1 tng do s tng in tch ht nhn v hiu ng xm nhp ca cc

electron ns.c) Trong phn nhm ph, I1gim do sgim hiu ng xm nhp ca cc electronns.d) Trong phn nhm chnh, I1gim do stng kch thc nguyn t.

3.30 Chn trng hpng:So snh nng lng ion ha thnht I1ca N (Z = 7) vO (Z = 8):

a) I1(N)< I1(O)vtrong mt chu k, khi i ttri sang phi I1tng dn.b) I1(N)> I1(O)vN ccu hnh bn boha phn lp 2p.c) I1(N)I1(O)velectron cui cng ca N vO cng thuc phn lp 2p.d) Khng so snh c.


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3.31 Cho cc nguyn tha hc sau: Ne ( Z = 10), Na (Z = 11) vMg ( Z = 12) . Chn pht biu ng:a) I1 (nng lng ion ha thnht) ca Mg nhhn I1ca ca Ne.b) I1ca Mg nhhn I1ca Na.c) I2( nng lng ion ha thhai) ca Na nhhn I2ca Ne.d) I2ca Mg ln hn I2ca Na.

3.32 Chn trng hp ng. Nng lng ion ha thnht (I1) ca cc nguyn tccu trc electron1s

22s

22p

4(1) , 1s

22s

22p

3(2), 1s

22s

22p

6(3) v1s22s22p63s1(4) tng theo chiu:

a) 1 2 3 4 b) 3 2 1 4c) 4 1 2 3 d) 4 3 2 1

3.33 Chn cu ng. Ai lc electron ca nguyn t:a) lnng lng pht ra (-) hay thu vo (+) khi kt hp electron vo nguyn tthkhkhng b

kch thch.b) lnng lng cn tiu tn kt hp thm electron vo nguyn ttrung ha.c) tng u n trong mt chu kttri qua phi.d) ctrsbng nng lng ion ha thnht ( I1) ca nguyn t.

3.34 Chn pht biu ng:a) m in ca mt kim loi ln hn m in ca mt phi kim loi.b) Trong mt phn nhm chnh, m in tng dn ttrn xung di.c)

Trong mt chu k, kim loi kim cm in nhnht.d) Ssai bit gia hai m in ca A vB cng ln thlin kt AB cng t phn cc.

3.35 Da vo m in:Nguyn t H C N Om in 2,1 2,5 3,0 3,5

Trong 4 ni cng ha trn sau, ni no bphn cc nht?a) CH b) NH c) OH d) CO

CHNG 4: LIN KT HA HC VCU TO PHN T4.1 nng lng lin kt, di lin kt, gc ha tr

4.1 Chn cu sai. Lin kt ClO trong dy cc ion ClO-, ClO2-, ClO3

-vClO4

-cdi tng ng

1,7; 1,64; 1,57 v1,42 A0. Ty suy ra theo dy ion cho:a) bn ion tng dn b) Nng lng lin kt tng dn.c) Tnh bn ca cc ion gim dn. d) Bc lin kt tng dn.

4.2 Chn pht biu sai:1) di lin kt lkhong cch gia hai ht nhn nguyn ttrong lin kt (n vangstrom ).2) Nng lng lin kt lnng lng cn tiu tn phvlin kt (n vJ/mol hay cal/mol)3) Gc ha trlmt i lng c trng cho tt ccc loi phn t.4) Mi loi lin kt ha hc u cbn cht in.5) phn cc mt phn tbng tng phn cc ca cc lin kt ctrong phn t.a) 1, 3, 5 b) 3,5 c) 3, 4, 5 d) khng cpht biu no sai.4.2 Lin kt ion

4.3 Chn cu saitrong cc pht biu sau vhp cht ion:a) Nhit nng chy cao. b) Phn ly thnh ion khi tan trong nc.c) Dn in trng thi tinh th. d) Dn in trng thi nng chy.

4.4 Lin kt ion ccc c trng cbn khc vi lin kt cng ha trl:a)Tnh khng bo ha vkhngnh hng. b) Ckhng phn cc cao hn.c) Cmt trong a shp cht ha hc. d) Cu a vb u ng.

4.3 Lin kt cng ha tr4.3.1 Phng php lin kt ha tr(VB)

4.5Cng ha trcc i ca nguyn tc quyt nh bi:


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a) Socbitan nguyn tha tr. b) Selectron ha tr.c) Selectron ha trc thn trng thi kch thch. d) Tt cu ng

4.6 Hp cht no di y ckhnng nhhp:a) NO2 b) SO2 c) O3 d) CO2

4.7 Chn pht biu sai:a) Lin kt cng ha trkiu lkiu lin kt cng ha trbn nht.b) Lin kt cng ha trc hnh thnh trn 2 cch: Cho nhn vghp i.c)

Lin kt llin kt c hnh thnhtrn cssche phca cc orbital nguyn tnmtrn trc ni 2 ht nhn.

d) Snh hng ca lin kt cng ha trc quyt nh bi slai ha ca nguyn ttrungtm tham gia to lin kt.

4.8 Chn pht biu sai. Theo lthuyt lin kt ha tr(VB):a) Lin kt cng ha trhnh thnh do skt i ca 2 electron cspin tri du, y csph

ca hai ocbitan nguyn t.b) Lin kt cng ha trcng bn khi mc phca cc ocbitan nguyn tcng ln.c) S lin kt cng ha tr ca mt nguyn t trong mt phn tbng socbitan ha tr ca n

tham gia che ph.d)

Nitc5 lin kt cng ha trtrong hp cht HNO3.4.9 Chn pht biu ng:

a) Lin kt cng ha trnh chllin kt hai electron nhiu tm.b) Lin kt cng ha trlun ctnh phn cc mnh.c) Lin kt cng ha trnh chllin kt hai electron hai tm.d) Trong lin kt cng ha trcc electron lca chung phn tvchng lun thp vi nhau thnhcc orbital phn t.

4.10 Theo thuyt lin kt ha tr(thuyt VB), selectron ha trca N vslin kt cng ha trti amN cthto thnh trong cc hp cht ca nln lt l:

a) 3, 3 b) 5, 4 c) 5, 5 d) 5, 3

4.3.2 Thuyt lai ha vcu hnh phn t4.11 Theo thuyt lai ha, cc orbital tham gia lai ha cn phi ccc iu kin:

a) Cc orbital ging nhau hn tn vnng lng.b) Cc orbital chnh dng hn tn ging nhau.c) Cc orbital cnng lnggn nhau vcmt electron ln.d) Cc orbital lai ha lun nhn tt ccc trc ta lm trc i xng.

4.12 Chn pht biu ng:Theo thuyt lai ha cc orbitan nguyn tta c:a) Slai ha thng khng clin hn hnh hc phn t.b) Lai ha sp c thchin do sthp mt orbitan s vmt orbitan p (ca cng mt nguyn t)

, kt qa xut hin 2 orbitan lai ha sp phn bi xng di mt gc 1800.c) Lai ha sp2c thc hin do sthp mt orbitan s v2 orbitan p (ca cng mt nguyn t)

kt quxut hin 3 orbitan lai ha sp2phn bi xng di mt gc 109,280.d) Lai ha sp3c thc hin do sthp mt orbitan s v3 orbitan p (ca cng mt nguyn t)

kt quxut hin 4 orbitan lai ha sp3phn bi xng di mt gc 1200.4.13 Slai ha sp3ca nguyn ttrung tm trong dy ion: SiO4

4-- PO4

3-- SO4

2-- ClO4

-gim dn do:

a) Schnh lch nng lng gia cc phn lp electron 3s v3p tng dn.b) Kch thc cc nguyn ttrung tm tham gia lai ha tng dn.c) Nng lng cc ocbitan nguyn t(AO) tham gia lai ha tng dn.d) Tt cu sai

4.14 Bn orbital lai ha sp3ca phn tCH4cc im:


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a) Hnh dng ging nhau nhng nng lng vnh hng khng gian khc nhau.b) Hnh dng vnng lng ging nhau nhng nh hng khng gian khc nhau.c) Hnh dng, nng lng v nh hng khng gian hn tn ging nhau vi gc lai ha l

109o28.

d) Nng lng bng nhau, hnh dng vnh hng khng gian khc nhau.4.15 Trong ion NH2

-, kiu lai ha ca nguyn tnitvdng hnh hc ca ion NH2

-l:

a) sp3vgc b) sp2vtam gic phng

c) sp vthng hng d) sp2vgc4.16 Cho bit Nittrong phn tNF3trng thi lai ha sp

3, vy phn tNF3cc im :

a) Cu hnh tam gic phng, gc ha tr120ob) Cu hnh tdin, gc ha tr109o28.c) Cu hnh thp, phn cc.d) Cu hnh thp, khng ccc.

4.17 Trong cc tiu phn sau, tiu phn ccu trc tdin u l:a) NH4

+ b) SF4 c) XeF4 d) SO2Cl2

Bit N (Z=7), S (Z=16), Xe (Z=54)4.18 Chn pht biu ng vcu hnh phn tNH3:

a) Cu hnh tam gic phng, phn cc. b) Cu hnh tdin u, phn ccc) Cu hnh tam gic phng, khng phn cc. d) Cu hnh thp tam gic, phn cc.

4.19 Trng thi lai ha ca cc nguyn tC theo thtttri qua phi ca phn t CH2= C = CHCH3l:

a) sp ,sp , sp , sp b) sp , sp , sp , spc) sp , sp , sp , sp d) sp , sp , sp , sp

4.20 Chn pht biu ng. Phn tCH3CH2CH3cc im:a) 3 nguyn tC u khng lai ha. b) 3 nguyn tC u lai ha sp2c) 3 nguyn tC u lai ha sp. d) 3 nguyn tC u lai ha sp3

4.21 Sp xp cc hp cht cng ha trsau theo chiu tng dn gc lin kt:1. CH4 2. NH3 3. H2O

a) 1, 2, 3 b) 2,1, 3 c) 3, 2,1 d) 3, 1, 2

4.22

Chn pht biu ng:a) CO2vSO2u ccu trc thng hng.b) CH4vNH4

+u ccu trc tdin u.

c) CO32-

vSO32-

u ccu trc phng.d) H2O vBeCl2u ccu trc gc.

4.23Phn tSO2cgc ha trOSO = 11905 ccc c im cu to l:

a) Dng gc, bc lin kt 1,33, clin kt khng nh ch3 tm.b) Dng gc, bc lin kt 1,5, clin kt khng nh ch3 tm.c) Dng tam gic, bc lin kt 1, khng clin kt .d) Dng gc, bc lin kt 2, clin kt 2 tm.

4.24Nhng phn tno trong scc phn tsau y cmomen lng cc bng khng? H2, H2S , CO2, NH3, H2O , SO2(cho bit H (Z = 1), C (Z = 6) , O (Z = 8) , N (Z = 7) vS (Z = 16))a) H2, H2S b) CO2, NH3 c) H2O , SO2 d) H2, CO2

4.3.3 Thuyt ocbitan phn t(MO)4.25 Chn pht biu ngtheo phng php MO:1) Phng php Ocbitan phn tcho rng trong phn tkhng cn tn ti ocbitan nguyn t, thay vo

y lcc ocbitan phn t.2) Phn tlthp thng nht ca cc ht nhn nguyn tv, trng thi c c trng bng hm

ssng phn t.3) Cc ca cc nguyn tchchu lc tc dng ca ht nhn nguyn t.


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4) Cc orbital phn tc to thnh do sthp tuyn tnh cc orbital nguyn t, sMO to thnhbng sAO tham gia thp.

a) 1,2 v3 b)1,2 v4 c) 2 v4 d) 1 v24.26 Chn pht biusaivphng php MO

a) Cc electron trong phn tchu nh hng ca tt ccc ht nhn nguyn ttrong phn t.b) Cc electron phn btrong phn ttheo cc quy tc nhtrong nguyn ta electron (trquy tc

Cleskovxki).

c)

MO lin kt cnnglng ln hn AO ban u.d) Ngi MO lin kt vphn lin kt cn cMO khng lin kt.

4.27 Da theo thuyt ocbitan phn t(MO) trong cc phn t H2, H2-vH2

2-phn tno clin kt

bn nht, phn tno thun t, phn tno khng tn ti (cho kt qa theo thttrn)a) H2, H2

2-, H2

- b) H2, H2

-, H2

2-

c) H22-

, H2-, H2 d) H2

-, H2, H2

2-

4.28 Bit in tch ht nhn ca Be, F, N vLi ln lt l4, 9, 7 v3. Phn tno khng tn ti trongthc t?a) N2 b) Li2 c) Be2 d) F2

4.29 Chn cu ng. Sthm electron vo ocbitan phn tphn lin kt dn n hqa:a) Gim di vtng nng lng lin kt.b)

Tng di vgim nng lng lin kt.c) Gim di vgim nng lng lin kt.d) Tng di vtng nng lng lin kt.

4.30 So snh bc lin kt trong N2, CO vCN-:

a) Trong CO ln nht b) Trong CN-ln nht c) Trong N2ln nht d) Bng nhau4.31 di lin kt trong cc tiu phn NO, NO+vNO-tng dn theo tht:

a) NO < NO-< NO b) NO < NO < NO

- c) NO

-< NO < NO d) NO < NO < NO

-

4.32 Chn trng hp ng:bn lin kt trong cc tiu phn NO, NO+, NO-tng dn theo tht:

a) NO < NO < NO- b) NO < NO

-< NO c) NO

-< NO < NO d) NO < NO < NO

-

4.33 Theo thuyt MO, bc ln kt ca CO, CN-vNO+ln lt l:a) 1 ; 2 ; 3. b) bng nhau vbng 3.

c) 2 ; 2,5 ; 3 d) Bng nhau vbng 2 .4.34Bit cacbon cZ bng 6 vNitcZ bng 7. Cu hnh electron ca ion CN-l:(z ltrc lin kt)

a) (s)2(s*)2(z)2(x,y)4 b) (s)2(s*)2 (x)2(z)2(y)2c) (s)2(s*)2(x,y)4 (z )2 d) (s)2(s*)2(x,y)4 (z )1(x*)1

4.35 Chn pht biu ng. Theo phng php obitan phn t, cu hnh electron phn tBN l(z ltrclin kt ):

a) (slk)2(s*)2(xlk)2(ylk)1(zlk)1 b) (slk)2(xlk)2(s*)2(ylk)2c) (xlk)2(ylk)2(slk)2(s*)2 d) (slk)2(s*)2(zlk)2(xlk)1(ylk)1

4.36 Chn pht biu chnh xc:a)

O2cbc lin kt bhn O2+ b)

Nng lng lin kt ca O2+ ln hn O2c) Lin kt O2cdi ln hn O2 d) Cc phtbiu trn u ng

4.4 Sphn cc vsbphn cc ca ion4.37 Sp cc cation Na+, Al3+, Cs+vMg2+theo stng dn phn cc ca chng :

a) Na+< Cs

+< Mg

2+< Al

3+ b) Cs

+< Na

+< Mg

2+< Al

3+

c) Al3+

< Mg2+

< Na+< Cs

+ d) Mg

2+< Al

3+< Na

+< Cs

+

4. 38 Sp xp cc hp cht VCl3, VCl2, VCl4vVCl5theo stng dn tnh cng ha trca lin kta) VCl2< VCl3< VCl4< VCl5 b) VCl5< VCl4< VCl3< VCl2

b) VCl3< VCl4< VCl2< VCl5 d) VCl4< VCl2< VCl3< VCl5


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4.39 Hy sp xp cc phn tsau y theo chiu tng dn bphn cc ca ion m:1. NaF 2. NaCl 3. NaBr 4. NaI

a) NaF , NaBr , NaI , NaCl b) NaI , NaBr , NaCl , NaF

c) NaF , NaCl , NaBr , NaI d) NaF , NaCl , NaI , NaBr

4.5 Cc loi lin kt khc4.40 Trong cc cht sau: HF, NH3vH2S cht no clin kt hydro

a) ChcHF b) ChcNH3 c) HF, NH3 d) cba cht4.41 Chn pht biu ng:

a) Lin kt gia hai phi kim loi lun lun llin kt cng ha tr.b) Lin kt gia hai kim loi llin kt ion.c) Lin kt gia kim loi vphi kim loi lun lun llin kt ion.d) Hp cht no ccha O vN u cho c lin kt hydro.

4.42 Chn pht biu ng:a) Hp cht ccha F, O lun lun cho lin kt hydrob) Lin kt hydro lin phn tlm tng nhit si ca hp cht.c) Hp cht to c lin kt hydro vi nc lun lun ha tan vi nc theo bt ktlno.d) Lin kt hydro chckhi hp cht thrn.

4.43 Theo thuyt min nng lng kim cng khng dn in v:a) Trong tinh thkim cng min ha trc in y electron, cn min cm cln hn 3 eV.b) Cmin cm gia min ha trvmin dn ca kim cng.c) Lin kt gia cc nguyn tC trong tinh thkim cng llin kt cng ha trbn vng.

d) Sche phcp i gia cc ON lai ha sp3ca cc nguyn tC lm cho min ha trca kimcng bo ha.

4.6 nh hng ca bn cht lin kt vcu trc phn tn cc tnh cht vt lca cc cht4.44 Chn pht biu sai:

a) I2rn dthng hoa vI2cmng tinh thcng ha tr.b) NaCl khnng chy vNaCl cmng tinh thion.c) Kim cng rt khnng chy vkim cng cmng tinh thcng ha tr.d) ng dn in rt tt vng cmng tinh thkim loi.

4.45 Chn pht biu ng:a) Cacbon graphit khng dn in vnlmt phi kim loi.b) Tinh thNaCl dn in vnccha cc ion.c) Kim cng khng dn invvng cm cnng lng ln hn 3eV.d) Cht bn dn lcht cmin dn vmin ha trche phnhau.

4.46 Trong cc khCO2, SO2, NH3vHe, khkhha lng nht l:a) He b) CO2 c) NH3 d) SO2

4.47 Sp cc cht sau NH3, H2S vH2O theo thtnhit si tng dn:a) H2S < H2O < NH3 b) H2S < NH3< H2O c) NH3< H2S < H2O d) NH3< H2O < H2S

4.48

Cc cht HCl, HBr, H2vBaCl2cnhit si gim dn trong dy:a) HCl > BaCl2> HBr > H2 b) H2> HCl > BaCl2> HBr

c) HCl > HBr > BaCl2> H2 d) BaCl2> HBr > HCl > H2

4.49 Trong dy H2O, H2S, H2Se, H2Te, (O, S, Se, Te ccu hnh electron ha trln lt l2s22p

4,

3s23p

4, 4s

24p

4, 5s

25p

4), nhit si cc cht bin thin nhsau:

a) Tng dn tH2O n H2Teb) Gim dn tH2O n H2Tec) Nhit si ca H2O >H2S < H2Se < H2Te (nhit si ca H2S thp nht)d) Nhit si ca H2O < H2S > H2Se > H2Te (nhit si ca H2S cao nht)


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4.50 CaCl2vCdCl2u lcc hp cht ion. Cc ion Ca2+

(lp vngi cng 3s23p6) vCd2+(lp vngi cng 4s24p64d10) ckch thc xp xnhau. Chn pht biung:

a) Nhit nng chy ca hai hp cht xp xnhau vchng c cu to tcc ion cin tch vkch thc xp xnhau.

b) Nhit nng chy ca CaCl2lnhn ca CdCl2vCaCl2ctnh ion ln hn.c) Nhit nng chy ca CaCl2nhhn ca CdCl2vCaCl2 nhhn CdCl2.d) Nhit nng chy ca CaCl2nhhn ca CdCl2vCa

2+ckhnng phn cc mnh hn Cd2+.

4.51 Chn pht biu ng:a) SO2tan trong nc nhiu hn CO2vSO2ckhi lng phn tln hn CO2.b) SO2tan trong nc nhiu hn CO2vphn tSO2cmoment lng cc khckhng, CO2c

moment lng cc bng khng.c) SO2vCO2u t tan trong nc vchai u lhp cht cng ha trmnc chha tan

c cc hp cht ion.d) SO2vCO2u tan nhiu trong nc vu ccha lin kt phn cc.

4.52 Chn pht biu ng:a) Chchp cht ion mi tan trong ncb) Cc hp cht cng ha tru khng tan trong ncc) Cc hp cht cnng lng mng tinh th(U) nh, khtan trong ncd)

Cc hp cht cng ha trphn tnhvto c lin kt hidro vi nc thtan nhiu trongnc

4.53 Chn pht biu sai:a) Etylamin (C2H5NH2) v ru etylic u tan nhiu trong nc do to c lin kt hydro vnc.b) Toluen (C6H5CH3) lmt hidrocacbon nn t tan trong ncc) C2H5-O-C2H5lphn tphn cc nn tan nhiu hn C6H14d)

Cht to lin kt hidro vi nc cthtan trong nc theo bt ctlno.4.54 Chn pht biung:

Xt cc hp cht dng H2X ca cc nguyn tphn nhm VIA: O, S, Se, Te.a) H2Te cnhit nng chy cao nht vckhi lng phn tln nht.b) H2O cnhit nng chycao nht vclin kt hydrogen.c) Chng cnhit nng chy xp xnhau vccu trc phn ttng tnhau.d) Khng so snh c vphn cc ca chng khc nhau.

4.55 Sp cc cht sau y: C6H14, CH3-O-CH3vC2H5OH theo thttan trong nc tng dn:a) CH3-O-CH3< C6H14< C2H5OH b) C6H14< C2H5OH < CH3-O-CH3

c) C2H5OH < CH3-O-CH3< C6H14 d) C6H14< CH3-O-CH3< C2H5OH

4.7 Cc cu hi tng hp

4.56 Trong cc hp cht sau : AlCl3, BCl3, KCl vMgCl2, hp cht no ctnh cng ha trnhiu nhtvhp cht no ctnh ion nhiu nht? (Cho bit Al (Z = 13) , B (Z= 5) , K (Z = 19) , Mg (Z = 12)p n sp theo thtcu hi)

a) AlCl3; KCl b) BCl3 ; KCl c) KCl ; BCl3 d) MgCl2; AlCl3

4.57 Chn trng hpng:Trong cc loi lin kt sau, lin kt no cnng lng lin kt nhnht:

a) Ion b) Cng ha tr c) Van der Waals d) Hydrogen4.58 Chn pht biu ng:


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1) Lin kt cng ha trllin kt mnh nht do nto ra c cc hp cht ccng cao nhnhkim cng.

2) Lin kt Van Der Waals tn ti cbn trong mt phn thu hn (vd: C2H5OH).3) Lin kt cng ha tryu hn lin kt ion do cc hp cht ion ccng ha trcao u km

bn vcnhit nng chy khthp. Vd: FeCl2cnhit nng chy 672oC, nhit s

1026oC, trong khi FeCl3cnhit nng chy 307,5

oC vnhit si 316oC.

4)Tt ccc loi hp cht ha hc c to thnh t t nht mt trong ba loi lin kt mnh l ioncng ha trvkim loi.

a) 1 v4 b) 1 , 2 v4 c) 3 v4 d) 44.59Trong cc cht H2, RbF, NaCl vNH3, cht no c% tnh ion cao nht, cht no c% tnh ion thp

nht trong lin kt (cho kt qa theo thttrn):a) H2, RbF b) RbF, H2 c) NaCl, NH3 d) RbF, NH3

4.60 Chn pht biu sai:a) Soxy ha lmt i lng quy c vi githit nguyn tnhn hn hoc cho hn electron ha

trc thn hoc bkch thch n trng thi c thn.b) Cng ha trcc i camt nguyn tbng socbitan ha trtham gia lai ha.c) Lin kt ion ctnh khng bo ha, tnh khng nh hng vtnh ccc.d) Lin kt cng ha trccc tnh cht : nh hng, bo ha, ccc hoc khng ccc.

4.61 Chn cu saitrong cc pht biu sau vhp cht ion:a) Cnhit nng chy cao.b) Dn in trng thi tinhth.c) Phn ly thnh ion khi tan trong nc.d) Dn in trng thi nng chy.

4.62 Loi lin kt no lchyu trong hp cht CH3OH.a) Lin kt ion. b) Lin kt cng ha tr. c)Lin kt hidro. d) Lin kt kim loi.

4.63 Chn cu sai:a) NaCl clin kt ion b)Ngi lin kt ion, KCl cn clin kt Vander Waalsc) HCl clin kt cng ha tr d) NH3clin kt hidro lin phn t

4.64 Chn pht biu saitrong cc pht biu sau:a) Cc lin kt Hidro vVan der Waals llin kt yu, ni phn t.b) Cc lin kt cng ha trv ion cbn cht in.c) Lin kt hidro lin phn tslm tng nhit si ca cht lng.d) Lin kt kim loi llin kt khng nh ch.

4.65 Trong 4 hp cht sau BaF2, CaCl2, CF4, HF, hp cht mlin kt ctnh ion cao nht la) CaCl2 b) BaF2 c) CF4 d) HF

4.66 Trong cc lin kt cng ha trsau H-F, H-Br, H-I, H-Cl, lin kt t bphn cc nht la) H-F b) H-I c) H-Cl d) H-Br.

4.67Nguyn tA ccu hnh lp cui cng l2s22p6. Chn pht biu sai:a) A lnguyn ttrvmt ha hc iu kin khquyn.b) A lcht rn iu kin thng.c) A chu k2 vphn nhm VIII A.d) Lnguyn tcui cng ca chu k2.

4.68 Chn pht biu ng:1) Lc tng tc Van der Waals gia cc phn ttrung ha c gii thch bng ba hiu ng: Hiu

ng nh hng, hiu ng cm ng vhiu ng khuych tn.2) m in khng phi lmt hng snguyn tmphthuc nhiu yu tnhtrng thi ha tr,

soxy ha ca nguyn t, thnh phn ca cc hp cht cho nn, mt cch cht chta phi nim in ca mt nguyn ttrong nhng iu kin cthxc nh.


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3) Do clin kt hydro nn nc ccu trc c bit, tng i xp nn tkhi nh, nn nc nhhn nc lng.a) 2 b) 1, 2 c) 1, 3 d) cbacu u ng.

4.69 Ngc li vi NaCl, LiI tan nhiu trong ru, tan t trong nc, nhit nng chy thp. Ldo lv:

a) Lin kt trong phn tLiI mang nhiu c tnh cng ha tr, tri li lin kt trong phn tNaCmang nhiu c tnh ion.

b)

Ion Li+cbn knh nhhn ion Na+, trong khi ion I-cbn knh ln hn ion Cl-.c) Nng lng mng li tinh thLiI ln hn nng lng mng li tinh thNaCl.d) Chai ldo a vb u ng.

P N PHN 1:CU TO CHT

Cu 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10p n d a d b d d a a b a

Cu 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20p n d c d a c a b b b a

Cu 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30p n b a a c b c a c c d

Cu 2.31 2.32 2.33 2.34p n a a a b

Cu 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10p n d b c c b c c b a c

Cu 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20p n a d d c b d d a a d

Cu 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30p n b b c b b c a c c bCu 3.31 3.32 3.33 3.34 3.35

p n a c a c cCu 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

p n c b c a a a c d c bCu 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20

p n c b a c a c a d a dCu 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30

p n c b d d b c b c b dCu 4.31 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39 4.40

p n b c b c a d b a c ccu 4.41 4.42 4.43 4.44 4.45 4.46 4.47 4.48 4.49 4.50

p n a b a a c a b d c bCu 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 4.60

p n b d d b d b c d b bCu 4.61 4.62 4.63 4.64 4.65 4.66 4.67 4.68 4.69

p n b b b a b b b d d


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PHN 2: LTHUYT DIN RA CC QUTRNH HA HC

A. BI TP TNBi 2.1:Cho phn ng : H2 (k) + 1/2O2(k) = H2O (l). Hy xc nh :

a) o

vUo

ca phn ng 25o

C.b) o100oC,Cho bit nhit dung ng p trong khong nhit t25 n 100oC i vi H2(k), O2(k) vH2O (l) l28,9; 29,4 v75,5 J/mol.Cho bit : o298,tt ca H2O (l) bng68,32 kcal/molp s: a)68,32 kcal ; -67,43 kcal . b)67,75 kcalBi 2.2:Tnh hiu ng nhit ca cc phn ng trung ha:

HCl (dd) + NaOH (dd) = NaCl (dd) + H2O (l) (1)

HCl (dd) + KOH (dd) = KCl (dd) + H2O (l) (2)

Cho bit nhit to thnh ca cc ion trong nc nhsau (kJ/mol) :ion H

+.aq Na

+.aq K

+.aq OH

-.aq Cl

-.aq H2O (l)

o298,tt 0,0 -240 -252 -230 -167 -286

Cho bit ti sao nhit trung ha gia cc axit mnh vbazmnh u ccng gitrnhnhau? p s: -56 kJBi 2.3:Ha tan 1 mol CuSO4, 1mol CuSO4. H2O hoc 1mol CuSO4.5H2O trong 800 mol nckm theo sgii phng hay thu vo mt lng nhit tng ng l15,90; -9,33 v2,80 kcal.Hy tnh hiu ng nhit ca cc qa trnh:

CuSO4 CuSO4.H2O (1)CuSO4.H2O CuSO4.5H2O (2)CuSO4 CuSO4.5H2O (3)


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p s: -6,57 kcal ; -12,13 kcal v18,70 kcalBi 2.4:Hy xc nh nng lng lin kt trung bnh ca mt ni C-H trong phn tCH4, cho bit nhitthng hoa ca grafit bng 170,9 kcal/mol, nhit phn li ca khhydro bng 103,26 kcal/mol vhiu ngnhit ca phn ng sau: C (gr) + 2H2(k) =CH4(k) , o298 = -17,89 kcal

p s: 98,83 kcalBi 2.5: Tnh thay i entropi khi t nng 1 nguyn tgam cadimi tt1= 25

oC n t2= 727

oC.

Cho bit cadimi c: Nhit nng chy 321oC, nhit nng chy l1460 cal/ntg, nhit dung nguyn tng p thrn l: Cpr = 5,46 + 2,47.10-3T (cal/ntg.K), nhit dung nguyn tng p thlng l:Cpl = 7,13 cal/ntg.K.

p s: 10,72 cal/ntg.K

Bi 2.6: Cho phn ng : NH3(k) + HCl (k) = NH4Cl (r). Hy da vo cc gitrnhit tothnh tiu chun, entropi tiu chun vthng p tiu chun cho trong bi camt scht :

a) Tnh o298, So298, Go298ca phn ng.b) Tnh Go298,ttca NH4Cl (r)c) Tcc kt qa thu c cthrt ra nhng kt lun g?Cho bit :

Nhit to thnh tiu chun 25oC (kcal/ mol) ca NH3(k), HCl (k) vNH4Cl (r) ln lt l: -11,00 ; -22,24 v-75,38Entropi tiu chun 25oC (cal/mol.K) ca NH3(k), HCl (k) vNH4Cl (r) ln lt l: 45,95 ; 46,04 v22,6Thng p to thnh tiu chun 25oC (kcal/mol.K) ca NH3(k) vHCl (k) ln lt l: -4,00 v22,74p s: a) -42,14 kcal; -68,22 cal/; -21,46 kcal b) -48,21 kcalBi 2.7: Cho phn ng: CH4(k) + 2H2O (k) = CO2(k) + 4H2(k)

a) Tnh o298, So298, Go298ca phn ng.Tnh o1000, So1000, Go1000ca phn ng, cho bit nhit dung phn tng p trung bnh trongkhong nhit 273K n 1000oK ca cc khCH4, H2O, CO2vH2ln lt l: 8,54; 8,03; 8,87 v6,89cal/mol.K.So snh khnng vchiu hng ca phn ng trn cc nhit 298K v1000K.

Cho bit:Nhit to thnh tiu chun 25oC (kcal/mol) ca cc khCH4, H2O vCO2ln lt l: -17,89; -57,80-94,10.

Entropi tiu chun 25oC (cal/mol.K) ca cc khCH4, H2O,CO2vH2ln lt l: 45,50; 45,13; 51,10v31,21.Bi 2.8: Cho phn ng thun nghch: H2(k) + I2(k) 2HI (k)

Xc nh sgam HI c to thnh vhiu sut ca phn ng theo lthuyt khi cho 2g hydro v254g iot phn ng trong bnh kn cdung tch 3 lt 699K, bit hng scn bng KPca phn ng nhit ny l54,5.

p s: 201,37g ; 78,7%Bi 2.9:Xc nh hng scn bng KPca phn ng: N2O4(k) = 2NO (k) 25

oC, cho bit phn li

ca N2O4nhit ny vp sut 1 atm l0,185. Tnh phn li ca N2O4khi p sut chung l10atm.

p s: KP= 0,141 ; = 0,059Bi 2.10:Phn ng thun nghch: CO (k) + Cl2 (k) = COCl2(k) , c thc hin trong bnh kn nhitkhng i. Nng ban u ca CO vCl2 bng nhau vbng 0,4 ptg/lit.


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Tnh hng scn bng KCca phn ng, bit rng khi ht trng thi cn bng thchcn 50% lngCO ban u.Sau khi cn bng c thit lp ta thm 0,1 ptg CO vo 1 lt hn hp. Tnh nng cc cht lc cnbng mi c thit lp.

p s: a) KC= 5 , b) 0,27 (CO) ; 0,17 (CO2); 0,23 (COCl2) (ptg/lit)Bi 2.11:Hng stc ca phn ng n phn tbng 8.10-3pht-1. Hy tnh thi gian cho nng ban u ca cht phn ng gim i 1/4

p s: 36 phtBi 2.12: 35oC khN2O5tinh khit cha trong bnh kn cp sut 0,100 atm. Hng stc caphn ng bc mt phn hy N2O5 thnh N2O4vO2bng 1,34.10

-4giy-1. Tnh p sut ring phn ca

oxy sau 10 pht vsau 1 gi.p s: sau 10 pht : P = 0,004 atm; sau 1 gi: P = 0,019 atm

Bi 2.13: Hi etyl clorua bnhit phn theo phn ng : C2H5Cl C2H4 + HCl. y lphn ngbcmt vchng stc phn ng :trong = 1,6.1014sec-1, E*= 59,5 kcal/mol. Tnh:

Hng sk 700K% etyl clorua bphn hy nhit ny sau 10 phtNhit my phn ng ctc ln hn 2 ln

p s: a) 4,24.10-5sec-1, b) 2,53%, c) 712KBi 2.14:Xc nh bc ca phn ng chuyn etylen thnh buten vhng stc k 427oC, bit rngnhit ny khi nng ca etylen l0,1 mol/lit thtc phn ng bng 7,6.10-6mol/lit.sec, cn khinng etylen l0,01 mol/lit thtc phn ng chcn l7,6.10-8mol/lit.sec.

p s: Bc 2; k = 7,6.10-4sec-1Bi 2.15: Ngi ta ha tan mt khi lp phng kim loi ha tr2 ccnh bng 1 cm vo axit. Hy vitphng trnh tc phn ng ha hc xy ra. Xc nh tc phn ng tng ln bao nhiu ln nu phnchia khi lp phng kim loi ban u thnh nhng khi lp phng ccnh bng 0,1 cm ri mi chotc dng vi axit?

p s: v = k.S.CH; 10 ln

B. BI TP TRC NGHIM

CHNG 6: HIU NG NHIT CA CC QUTRNH HA HC6.1 Cc khi nim cbn ca nhit ng hc vnhit ng ha hc. nh lut thnht ca nhitng hc.6.1Chn pht biu sai:a) Hclp lhkhng ctrao i cht, khng trao i nng lng di dng nhit vcng vi mitrng.b) Hkn lh khng trao i cht vcng, song cthtrao i nhit vi mi trng.c) Hon nhit lh khng trao i cht vnhit, song cthtrao i cng vi mi trng.d) Hhlhkhng brng buc bi hn chno, cthtrao i cht vnng lng vi mi trng.6.2 Xt phn ng NO(k) + 1/2O2(k) NO2(k) o298= -7,4 kcal. Phn ng c thc hin trong bnhkn cthtch khng i, sau phn ng c a vnhit ban u. Hnhthl:

a) Hclp b) hkn& ng th c) Hkn & dth d) hclp vng th6.3 Chn sai:

RT

E

ek

*

.


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a) Nguyn lI nhit ng hc thc cht lnh lut bo tn nng lng.b) Nhit chcthttruyn tvt thcnhit cao sang vt thcnhit thp.c) Hiu ng nhit ca mt phn ng llng nhit tora hay thu vo ca phn ng .d) bin thin entanpi ca mt qutrnhkhng thay i theo nhit.6.4Trong mt chu trnh , cng hnhn l2 kcal. Tnh nhit mhtrao i :

a) -2 kcal b) +4 kcal c) +2 kcal d) 06.5 Trong iu kin ng tch, phn ng pht nhit lphn ng c:

a) Cng A < 0 b) U < 0 c) H < 0 d) U > 06.6Chn qutrnh ng:

Xt phn ng: NO (k) + O2 (k) = NO2(k)Phn ng c thc hin trong bnh kn cthtch khng i, sau phn ng c a vnhit

ban u. Qutrnh nhthlqutrnh :a) ng p , ng nhit b) ng tchc) ng tch, ng nhit d) ng p , ng tch

6.7 Mt hcni nng tng ( U2> U1) , khi i ttrng thi 1 sang trng thi 2 trong iu kin ng p.Bit rng trong qutrnh bin i ny htonhit (< 0) , vy h:

a) Sinh ra cng b) Nhn cng

c) Khng trao i cng d) Khng thdn c6.8 Sbin thin ni nng U khi mt hthng i ttrng thi thnht (I) sang trng thi thhai (II) bng nhng ng i khc nhau ctnh cht sau:

a) Khng i do nhit Q vcng A u khng thay i.b) Thay i do nhit Q vcng A thay i theo ng i.c) Khng thay i vbng Q - A theo nguyn lbo tn nng lng.d) Khng thtnh c do mi ng i cQ vA khc nhau.

6.9Mt hthng hp thu mt lng nng lng di dng nhit l200 kJ. Ni nng ca htng thm250 kJ. Vy trong bin i trn cng ca hthng cgitr:

a) 350 kJ, hsinh cng b) 50 kJ, hnhn cngc) 50 kJ, hsinh cng d) -50 kJ, hnhn cng

6.2 Hiu ng nhit ca cc qutrnh ha hc. nh lut Hess6.10Trong iu kin ng p, mt nhit xc nh, phn ng :A(r) + 2B(k) = C(k) + 2D(k) pht nhit. Vy:

a) U< H b) U= H c) U> H d) Cha dliu so snh6.11 Trong iu kin ng tch, phn ng pht nhit lphn ng c:

a) Cng A < 0 b) U < 0 c) H < 0 d) U > 06.12 Tnh hiu sgia hiu ng nhit phn ng ng p vng tch ca phn ng sau y 25oC:

C2H5OH (l) + 3O2 (k) = 2CO2(k) + 3H2O (l) (R = 8,314 J/mol.K)

a) 4539J b) 2270J c) 1085J d) 2478J

6.13 Phn ng Fe2O3(r) + 3CO(k) = 2Fe(r) + 3CO2(k) iu kin cho c298= -6,8 Kcal. Suy raU298(kcal) ca phn ng bng: (R 2.10-3kcal/mol.K)

a) +6,8 b)8,6 c)6,8 d)5,06.14 Chn pht biuchnh xc ca nh lut Hessa) Hiu ng nhit ca qutrnh ha hc chphthuc vo bn cht vtrng thi ca cc cht u vsn phm chkhng phthuc vo ng i ca qutrnh.b) Hiu ng nhit ng p hay ng tch ca qutrnh ha hc chphthuc vo bn cht ca cc chtu vsn phm chkhng phthuc vo ng i ca qutrnh.c) Hiu ng nhit ng p ca qutrnh ha hc chphthuc vo bn cht vtrng thi ca cc chtu vsn phm chkhng phthuc vo ng i ca qutrnh.


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d) Hiu ng nhit ng p hay ng tch ca qutrnh ha hc chphthuc vo bn cht vtrng thica cc cht u vsn phm chkhng phthuc vo ng i ca qutrnh.6.15 i lng no sau y khng phi lhm trng thi:

1. Ap sut (p) 2. Entanpi (H) 3. Cng (A) 4. Nhit (Q) 5. nhit (t)a) 1, 2 & 3 b) 2, 3 & 4 c) 3 & 4 d) 1, 2, 3 & 4

6.16 Chn p n ng:ca mt qutrnh ha hc khi hchuyn ttrng thi thnht (I) sang trng thi thhai (II)

bng nhng cch khc nhau cc im:a) Thay i theo cch tin hnh qutrnh.b) Khng thay i theo cch tin hnh qutrnh.c) Cthcho ta bit chiu tdin bin ca qutrnh cc nhit khc nhau.d) Chai c im b vc u ng.

6.17 Chn pht biu ng:a) Hiu ng nhit ca mt phn ng o iu kin ng pbng bin thin ca entanpi, hiu ng nhitca phn ng o iu kin ng tch bng bin thin ni nng ca h.b) Hphn ng> 0 khi phn ng pht nhitc) Uphn ng< 0 khi phn ng thu nhitd) Hiu ng nhit ca mt phn ng khng ty thuc iu kin (to, p sut), trng thi ca cc cht thamgia phn ng cng nhcc cht to thnh sau phn ng (sn phm).6.18 Mt phn ng cH = -200 kJ.mol-1. Da trn thng tin ny cthkt lun phn ng ti nhit ang xt nhsau:

a) ta nhit b) ctc nhanh c) txy ra c d) ca, b, c u ng6.19 Chn cu trli ng. Gitro298 ca mt phn ng ha hc

a) Ty thuc vo cch vitcc hstlng ca phng trnh phn ng.b) Ty thuc vo nhit lc din ra phn ng.c) Ty thuc vo cch tin hnh phn ng.d) Tt cu sai.

6.20Chn cu ng. Phn ng thu nhit mnh:a) Khng thxy ra tpht mi gitri nhit .b) Cthxy ra tpht nhit thp.c) Cthxyra tphtnhit caonu bin thin entropi ca ndng.d) Cthxy ra tpht nhit cao nu bin thin entropi ca nm.

6.21 Cho phn ng : N2(k) + O2(k) = 2NO (k) co298,p= +180,8 kJ.iu kin tiu chun 25oC , khi thu c 1 molkhNO tphn ng trn th:a) Lng nhit ta ra l180,8 kJ. b) Lng nhit thu vo l180,8 kJ.c) Lng nhit thu vo l90,4 kJ. d) Lng nhit ta ra l90,4 kJ.

6.22 Hiu ng nhit to thnh tiu chun ca CO2lbin thin entanpi ca phn ng:a) Ckim cng+ O2(k) = CO2(k) 0

oC, p sut ring ca O2vCO2u bng 1 atm

b) Cgraphit + O2 (k) = CO2(k) 25oC, p sut ring ca O2vCO2u bng 1 atm

c) Cgraphit + O2 (k) = CO2(k) 0oC, p sut chung bng 1atm

d) Cgraphit + O2 (k) = CO2(k) 25oC, p sut chung bng 1atm6.23 Hiu ng nhit ca mt phn ng bng:

a) Tng nhit to thnh sn phm trtng nhit to thnh cc cht u.b) Tng nhit t chy cc cht u trtng nhit t chy cc sn phm.c) Tng nng lng lin kt trong cc cht u trtng nng lng lin kt trong cc sn phm.d) Tt cu ng

6.24 Chn trng hp ng: iu kin tiu chun, phn ng: H2 (k) + 1/2O2 (k) = H2O (l) pht ramt lng nhit l245,17kJ. Ty suy ra:

a) Hiu ng nhit t chy tiu chun ca H2l245,17kJ/mol.


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b) Nhit to thnh tiu chun ca nc lng l245,17kJ/mol.c) Hiu ng nhit phn ng trn l245,17kJ.d) Cba cu trn u ng.

6.25 Bit rng nhit to thnh tiu chun ca B2O3(r), H2O (l) ,CH4(k) vC2H2(k) ln lt bng: -1273,5 ; -285,8; -74,7 ; +2,28 (kJ/mol). Trong 4 cht ny, cht dbphn hy thnh n cht nht l:

a) H2O b) CH4 c) C2H2 d) B2O3

6.26 Trong cc hiu ng nhit (H) ca cc phn ng cho di y, gitrno lhiu ng nhit tchy?

1) C (gr) + 1/2O2(k) = CO (k) o298= -110,55 kJ2) H2 (k) + 1/2O2 (k) = H2O (l) o273= - 571,20 kJ3) H2 (k) + 1/2O2 (k) = H2O (k) o298= -237,84 kJ4) C (gr) + O2 (k) = CO2(k) o298= -393,50 kJa) 4 b) 2,4 c) 1,2,3,4 d) 2

6.27 Lp cng thc tnh hiu ng nhit (0) ca phn ng B A , thng qua hiu ng nhit cacc phn ng sau :

A C 1C D 2

B D 3a) 0 = 3 - 1 - 2 b) 0 = 3 + 2 - 1c) 0 = 2 - 1 - 3 d) 0 = 1 + 2 + 3

6.28 Lp cng thc tnh hiu ng nhit (0) ca phn ng B A , thng qua hiu ng nhit cacc phn ng sau :

A C 1D C 2B D 3

a) 0 = 1 - 2 + 3 b) 0 = 3 + 2 - 1c) 0 = 2 - 1 - 3 d) 0 = 1 + 2 + 3

6.29 Thai phn ng:(1) A + B = C + D 1 (2) E + F = C + D 2

Thit lp c cng thc tnh 3ca phn ng A + B = E + F :a) 3= 1- 2 b) 3= 1+ 2 c) 3= 2- 1 d) 3= -1-2

6.30 Cho bit:2NH3 (k) + 5/2O2(k) 2NO (k) + 3H2O (k)

ott,298(kJ/mol) -46,3 0 +90,4 -241,8Hiu ng nhit ca phn ng trn l:

a)452 kJ b) 452 kJ c) +406,8 kJ d)406,8 kJ

6.31Chn gitrng. Khi t chy than chbng oxy ngi ta thu c 33g khcacbonic vc70,9kcal tht ra iu kin tiu chun, vy nhit to thnh tiu chun ca khcacbonic cgitr(kcal/mol).a) -70,9 b) -94,5 c) 94,5 d) 68,6

6.32 Tnh nhit to thnh tiu chun ca CH3OH lng, bit rng:C (r) + O2 (k) = CO2(k) Ho1= -94 kcal/molH2(k) + 1/2O2 (k) = H2O (l) Ho2= -68,5 kcal/molCH3OH (l) + 1 O2 (k) = CO2 (k) + 2H2O (l) Ho3= -171 kcal/mola) +60 kcal/mol b)402 kcal/mol c) +402 kcal/mol d)60 kcal/mol

6.33 Tcc gitrcng iu kin ca cc phn ng :


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(1) 2SO2(k) + O2(k) = 2SO3(k) H = -196 kJ(2) 2S(r) + 3O2(k) = 2SO3(k) H = -790 kJ

hy tnh gitrcng iu kin ca phn ng sau : S(r) + O2(k) = SO2(k)a) H = -594 kJ b) H = -297 kJ c) H = 594 kJ d) H = 297 kJ

6.34 Lng nhit ta ra khi t chy 3g kim loi Mg bng O2(k) to ra MgO(r) l76kJ iu kin tiuchun. Nhit to thnh tiu chun (kJ/mol) ca MgO(r) l: ( AMg= 24g)

a) +608kJ b) +304kJ c)608kJ d)304kJ6.35 Xc nh hiu ng nhit t chy 1 mol metan theo phn ng:

CH4(k) + 2O2 (k) = CO2 (k) + 2H2O (l)Nu bit hiu ng nhit to thnh tiu chun ca cc cht CH4(k), CO2(k) vH2O (l) ln lt bng:

-74,85; -393,51; -285,84 ( kJ/mol)

a)604,5 kJ b) 890.34 kJ c)890,34 kJ d) 604,5 kJ

CHNG 7: THNG P VCHIU CA CC QUTRNH HA HC7.1 Nguyn lthhai ca nhit ng hc. Entropi7.1Chn pht biu sai:

Nguyn lthhai nhit ng lc hc cthpht biu nhsau:a) Nhit khng thtng truyn tvt lnh hn sang vt nng hn.b) Cng cthchuyn ha hn tn thnh nhit vnhit cng cthchuynha hn tn thnh cng.c) Khng thcqutrnh trong nhit ly tmt vt c chuyn thnh thnh cng mkhng cbchnh.d) Khng thcng cvnh cu loi hai.7.2Chn pht biu ng:a) Bin thin entropi ca hphthuc ng i.b) Entropi cthuc tnh cng , gitrca nkhng phthuc lng cht.c) Trong qutrnh tnhin bt kta lun lun c: dS Q/T (du = ng vi qutrnh thunnghch,du > ng vi qutrnh bt thun nghch)d) Entropi c trng cho mc hn n ca cc tiu phn trong h. Mc hn n ca cc tiu phntrong hcng nh, gitrentropi cng ln.7.3 Chn pht biu ng:1) Entropi ca cht nguyn cht trng thi tinh thhn chnh, nhit khng tuyt i bng khng.2) khng tuyt i, bin thin entropi trong cc qutrnh bin i cc cht trng thi tinhthhn chnh u bng khng.3) Trong hhtt ccc qutrnh txy ra lnhng qutrnh ckm theo stng entropi.

4) Entropi ca cht trng thi lng cthnhhn entropi ca n trng thi rn.a) 1 b) 1,2 c) 1,2,3 d) 1,2,3,4

7.4 Mt cht trng thi nhit cng caoth:a) Entropi cng ln. b) Entropi cng b.c) Entropi khng thay i. d) Mt trong ba cu a, b ,c ng vi cht cth.

7.5 cng iu kin , trong scc cht sau, cht no c entropi ln nht?Cht (1) : O (k) Cht (2) : O2(k) Cht (3) : O3(k)a) Cht 1 b) Cht 2 c) Cht 3 d) Khng bit c

7.6 Bin i entropi khi i ttrng thi A sang trng thi B bng 5 con ng khc nhau (xem gin) cc tnh sau:


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P B4

5 3 2

A 1

V

a) S ging nhau cho c5 ng. b) Mi con ng cS khc nhau.c) Cc pht biu a, b, d u sai. d) S ca ng 3 nhnht vlcon ng ngn nht

7.7 Qutrnh chuyn pha rn thnh pha lng c:a) H < 0 , S > 0 b) H > 0, S > 0 c) H < 0, S < 0 d) H > 0, S < 0

7.8 Chn cu ng. Phn ng : 2A ( k) + B (l) = 3C (r) + D (k) c:a) S = 0 b) S > 0 c) S < 0 d) Khng dn c

7.9 Trong cc phn ng sau:N2 + O2 = 2NO (1)2CH4 = C2H2 + 2H2(2)

2SO2 + O2 = 2SO3 (3)Cc cht u thkh. Hy sp xp S ca cc phn ng theo thttng dn gitr:a) 1,2,3 b) 2,1,3 c) 3,1,2 d) 2,3,1

7.10Trong cc phn ng sauN2 (k) + O2(k) = 2NO (k) (1)

KClO4 (r) = KCl (r) + 2O2(k) (2)C2H2 (k) + 2H2 (k) = C2H6(k) (3)

Chn phn ng cS ln nht, S nhnht (cho kt qutheo thtva nu)a) 1 , 2 b) 2 , 3 c) 2 , 3 d) 3 , 1

7.11 Xc nh qutrnh no sau y cS < 0.a) O2(k) 2O (k) b) 2CH4 (k) + 3O2(k) 2CO (K) + 4H2O (k)

c) NH4Cl (r) NH3(k) + HCl (k) d)N2(k,25o

C,1atm) N2(k,0o

C,1atm)7.12 Cho 3 phn ng:

H2O (l) H2O (k) (1) S12Cl (k) Cl2(k) (2) S2C2H2 (k) + H2(k) C2H4(k) (3) S3Hy cho bit du ca S1 , S2, S3:

a) S1> 0 , S2< 0 , S3< 0 b) S1< 0 , S2< 0 , S3> 0c) Cba S u dng d) Cba S u m

7.13 Chn pht biu ng ventropi cc cht sau:1) S

oH 2O (l)> S

oH 2O (k) 2) S

oMgO(r)< S

oBaO(r) 3) S

oC3H8(k) > S

oCH4(k)

4) SoFe(r)< S

oH2 (k) 5) S

oCa(r) > S

oC3H 8(k) 6) S

oS(r)< S

os (l)

a) 1,2,3,4 b) 2,3,4,6 c) 2,3,6 d)1,2,3,5,67.14 Tnh So(J/mol.K) 25oC ca phn ng : SO2(k) + O2(k) = SO3 (k)

Cho bit entropi tiu chun 25oC ca cc cht SO2(k) , O2(k) vSO3(k) ln lt bng : 248 , 205v257 (J/mol.K)

a)93,5 b) 93,5 c) 196 d)196


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7.2 Thng p . Chiu ca cc qutrnh ha hc.7.15 Chn cu ng.1) Bin thin thng nhit ng p lphn nng lng ca hcthtdochuyn thnh cng cchkhi qutrnh xy ra trong iu kin ng nhit ng p.2) Cng cch ltt ccc cng sinh ra (kccng chng p sut ngi) khi hchuyn ttrng thi u

n trng thi cui.3) Thng p ca hgim dn trong qutrnh tdin bin ca h.4) Trong iu kin ng p, qutrnh txy ra gn lin vi stng thng p ca h.

a) 1 v3 b) 1 , 3 v4 c) c4 cu u ng d) 2 v37.16 i lng no sau y khng phi lhm trng thi:

1. Ap sut (p) 2.Thng p ng nhit (G) 3. Cng (A) 4. Nhit (Q) 5. Entropi (S)a) 1, 2 & 3 d) 1, 2, 3 & 4 c) 3 & 4 b) 2, 3 & 4

7.17 Chn phng n sai. Cc i lng di y u lhm trng thi:a) entanpi, entropi. b) nhit , p sut, thng tch, thng pc) Thng p , ni nng, cng. d) Thng p, entanpi, entropi, ni nng

7.18 Cc i lng no sau y lhm trng thi:

1. Ap sut (p) 2. Nhit dung ng p (c) 3. Nhit (Q) 4. Entropi (S)a) 1, 2 & 4 b) 2, 3 & 4 c) 3 & 4 d) 1, 2, 3 & 4

7.19 Chn cu trli ng.Mt phn ng iu kin ang xt cG < 0 th:

a) xy ra tpht trong thc t . b) ckhnng xy ra tpht trong thc t .c) trng thi cn bng. d) Khng xy ra.

7.20 Phn ng khng thxy ra bt cgitrnhit no nu ti nhit phn ng ny c:a) < 0 ; S > 0 b) > 0 ; S > 0 c) < 0 ; S < 0 d) > 0 ; S < 0

7.21 Chn p n y nht. Phn ng cthxy ra tpht trong cc trng hp sau:a) < 0 ; S > 0

> 0 ; S > 0> 0 ; S < 0

b) > 0 ; S > 0

< 0 ; S < 0< 0 ; S > 0

c) < 0 ; S < 0

> 0 ; S > 0> 0 ; S < 0

d) > 0 ; S < 0

< 0 ; S > 0< 0 ; S < 07.22 Chn cu ng. Phn ng thu nhit:

a) Khng thxy ra mi gitrnhit .b) Cthxy ra nhit thp.c) Cthxy ra nhit cao nu bin thin entropi ca ndng.d) Cthxy ra nhit cao nu bin thin entropi ca nm.

7.23 mt iu kin xc nh, phn ng A B thu nhit mnh cthtin hnh n cng. Cthrtra cc kt lun sau:

a) Sp> 0 vnhit tin hnh phn ng phi cao.b) Phn ng B A cng iu kin ca cu a cGp> 0.c) Phn ng B A cthtin hnh nhit thp vcSp< 0.d) Tt cu ng

7.24 Phn ng 3O2(k) 2O3(k) iu kin tiu chun cHo298= 284,4 kJ, So298= -139,8J/mol.K. Bit rng bin thin entanpi vbin thin entropi ca phn ng t bin i theo nhit . Vypht biu no di y lphhp vi qutrnh phn ng:

a) nhit cao, phn ng din ra tpht.b) nhit thp, phn ng din ra tpht.c) Phn ng xy ra tpht mi nhit .d) Phn ng khng xy ra tpht mi nhit .


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7.25 Chn cu chnh xc. Cho phn ng tng qut aA + bB cC + dD co298< 0.a) Phn ng lun xy ra bt knhit no.b) nhit cao, chiu ca phn ng cn phthuc vo S.c) Phn ng khng thxy ra nhit thng.d) Phn ng chxy ra nhit cao.

7.26 Chn trng hp ng: Bit rng 0oC qutrnh nng chy ca nc p sut kh

quyn cG = 0. Vy 383K qutrnh nng chy ca nc p sut ny cdu ca G l:a) G > 0 b) G = 0c) G < 0 d) Khng xc nh c vcn cc yu tkhc.

7.27 Chn trng hp ng:Bit rng -51oC qutrnh nng chy ca H2Te p sut khquyn cG = 0. Vy 200K

qutrnh nng chy ca hydro telurua p sut ny cdu ca G l:a) G > 0 b) G =0c) G < 0 d) Khng xc nh c vcn cc yu tkhc

7.28Tnh bin i S khi 1 mol hi nc ngng tthnh nc lng 1000C ,1 atm. Bit nhit bayhi ca nc nhit trn l549 cal/g.

a) S = -26,5 cal/mol.K b) S = 26,5 cal/mol.Kc) S = 1,44 cal/mol.K d) S = -1,44 cal/mol.K7.29 Chn cu phhp nht. Cho phn ng 2Mg (r) + CO2(k) = 2MgO (r) + Cgraphit. Phn ng nychiu ng nhit tiu chun Ho298=-822,7kJ. Vphng din nhit ng ha hc, phn ng ny cth: (cho bit So298(J/mol.K) ca Mg(r), CO2(k), MgO(r) vCgraphitln lt bng 33, 214, 27 v6)

a) Xy ra tpht nhit cao. b) Xy ra tpht mi nhit .c) Yu ttonh hng khng ng k d) Khng tpht xy ra nhit cao.

7.30 Chn p ny :Mt phn ng cthtxy ra khi:

1) H < 0 rt m , S < 0 , tothng. 2) H < 0 , S > 0.3) H > 0 rt ln , S > 0 , tothng. 4) H > 0 , S > 0 , tocao.

a) 1 v2 ng b) 1, 2, 3, 4 ng c) 1, 2 v4 ng d) 2 v4 ng7.31 a scc phn ng xy ra nhit cao c:

a) bi n thin entropi m. b) bi n thin entropi dng.c) bin thin entanpi m. d) bin thin entanpi dng.

7.32 Chn nhng cu ng: Vphng din nhit ng ha hc:1) a sphn ng cthxy ra tpht hn tn khi Gop< -40 kJ.2) Phn ng khng xy ra tpht trong thc tkhi Gop> 40 kJ.3) Phn ng khng xy ra tpht trong thc tkhi Gop> 0.4) a scc phn ng cthng p tiu chun nm trong khong -40 kJ < Gop< 40 kJ xy ra tpht thun nghch trong thc t.

a) 1,2,4 b) 3,4 c) 1,3,4 d) Tt ccc cu trn u ng7.33 Chn cu sai.

a) Phn ng cGo< 0 cthxy ra tpht.b) Phn ng c Go> 0khng thxy ra tpht.c) Phn ng ta nhit nhiu thng ckhnng xy ra nhit thng.d) Phn ng ccc bin thin entanpi ventropi u dng ckhnng xy ra nhit cao.

7.34Chn pht biusai:a) Mt phn ng ta nhit mnh cthxy ra tpht nhit thng.b) Mt phn ng thu nhit mnh chcthxy ra tpht nhit cao.


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c) Mt phn ng hu nhkhng thu hay pht nhit nhng lm tng entropi cthxy ra tpht nhit thng.d) Mt phn ng thu nhit mnh nhng lm tng entropi cthxy ra tphtnhit thng.

7.35 dn phn ng cthxy ra tpht hn tnnhit thng, ta cthda trn du cacc i lng sau:

1) Go< 0 2) Go< -40 kJ 3) Ho< 0 vi o lna) 2 b) 1,2 v3 c) 2,3 d) 1,3

7.36Chn trng hp sai:Tiu chun cthcho bit phn ng xy ra tpht c vmt nhit ng l:

a) Ho< 0, So> 0 b)Cng chng p sut ngi A > 0c) G0< 0 d) Hng scn bng K ln hn 1.

7.37 Phn ng CaCO3(r) CaO (r) + CO2(k) lphn ng thu nhit mnh. Xt du o, So, Goca phn ng ny 25oC :

a) Ho< 0; So< 0 ; Go< 0 b) Ho< 0; So> 0 ; Go> 0c) Ho> 0; So> 0 ; Go< 0 d) Ho> 0; So> 0 ; Go> 0

7.38 Chn trng hp ng.Cn ctrn du Go298ca 2 phn ng sau :

PbO2(r) + Pb (r) = 2PbO (r) Go

298< 0SnO2(r) + Sn (r) = 2SnO (r) Go298> 0

Trng thi oxy ha dng bn hn i vi cc kim loi chvthic l:a) Ch(+2), thic (+2) b) Ch(+4), thic (+2)c) Ch(+2), thic (+4) d) Ch(+4), thic (+4)

7.39Phn ng H2O2(l) H2O (l) + O2(k) ta nhit, vy phn ng ny c:a) H < 0; S > 0 ; G > 0 khng thxy ra tphtnhit thng.b) H < 0; S > 0 ; G < 0 cthxy ra tpht nhit thng.c) H > 0; S < 0 ; G < 0 cthxy ra tphtnhit thng.d) H > 0; S > 0 ; G > 0 khng thxy ra tphtnhit thng.

7.40 Chn p n y : Chn pht biu sai:

1) Cthkt lun ngay lphn ng khng xy ra tpht khi Goca phn ng ny ln hn 0.2) Cthkt lun ngay lphn ng khng txy ra khi G ca phn ng ny ln hn 0 ti iu kinang xt.3) Mt htxy ra lun lm tng entropi.4) Chcc phn ng cGop< 0 mi xy ra tpht trong thc t.

a) 1, 3 v4 b) 1 v3 c) 1 v4 d) 3CHNG 8 : CN BNG HA HC VMC DIN RA CA CC QUTRNH HAHC8.1 Cc khi nim cbn. nh lut tc dng khi lng trong cn bng. Hng scn bng.8.1 Chn pht biu ng:

i vi phn ng mt chiu, tc phn ng s:a) Khng i theo thi gian.b) Gim dn theo thi gian cho n khi bng mt hng skhc khng.c) Tng dn theo thi gian.d) Gim dntheo thi gian cho n khi bngkhng.

8.2 Phn ng thun nghch l:a) Phn ng cthxy ra theo chiu thun hay theo chiu nghch ty iu kin phnng.b) Phn ng xy rang thi theo hai chiu ngc nhau trongcng mt iu kin.c) Phn ng txy ra cho n khi ht cc cht phn ng.


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d) Cu a vb u ng.8.3 Chn pht biu ng vhcn bng:

a) Hcn bng lhtrong ctlthnh phn cc cht khng thay i khi ta thay i cc iukin khc.

b) Hcn bng lhcnhit vp sut xc nh.c) Hang trng thi cnbng lhccc gitrthng strng thi(to, P, C) khng thayi theo thi gian.d) Khng cpht biu no ng.

8.4 Kt lun no di y lngkhi mt phn ng thun nghch cGo< 0:a) Hng scn bng ca phn ng ln hn 0.b) Hng scn bng ca phn ng ln hn 1.c) Hng scn bng ca phn ng nhhn 1.d) Hng scn bng ca phn ng nhhn 0.

8.5 Cho phn ng aA (l) + bB (k) cC (k) + dD(l), chng scn bng Kc. Chn phtbiung:

a) G = Go+ RTlnKc, khi G = 0 thGo= -RTlnKcb) Hng scn bng Kctnh bng biu thc:

Vi CA, CB, CCvCDlnng cc cht ti lc ang xt.c) Phn ng lun cKP= KC(RT)

nvi n =nsp-ncca tt ccc cht khng phthuc vo

trng thi tn ti ca chng.d) Cba phtbiu u sai.

8.6 Gishang cn bng, phn ng no sau y c coi lxy ra hn tn:a) FeO (r) + CO (k) = Fe (r) + CO2(k) KCb= 0,403

b) 2C (r) + O2(k) = 2CO (k) KCb= 1.10

c) 2 Cl2(k) + 2 H2O (k) = 4 HCl (k) + O2(k) KCb= 1,88. 10- d) CH3CH2CH2CH3 (k) = CH3CH(CH3)2(k) KCb= 2,5

8.7Cho mt phn ng thun nghch trong dung dch lng A + B C + D. Hng scn bng Kciu kin cho trc bng 200. Mt hn hp cnng CA= CB= 10

-3M, CC= CD = 0,01M. Trng

thi ca hiu kin ny nhsau:a) Hang dch chuyn theo chiu thun.b) Hang dch chuy n theo chi u nghch.c) Hnm trng thi cn bng.d) Khng thdn c trng thi ca phn ng

8.8Phn ng CaCO3(r) CaO (r) + CO2(k) chng scn bng Kp= PCO2. p sut hi caCaCO3, CaO khng cmt trong biu thc Kpv:

a) Cthxem p sut hi ca CaCO3vCaO bng 1 atm.b) Ap sut hi ca cht rn khng ng kc) Ap sut hi cht rn khng phthuc vo nhit .d) p sut hi ca CaCO3vCaO lhng snhit xc nh.

8.9 Cho phn ng CO2(k) + H2(k) CO (k) + H2O (k). Khi phn ng ny t n trngthi cn bng, lng cc cht l 0,4 mol CO2, 0,4 mol H2, 0,8 mol CO v0,8 mol H2O trong bnh kncdung tch l1 lt. Kcca phn ng trn cgitr:

a) 8 b) 6 c) 4 d) 2

8.10 Chn pht biu ng : cho phn ng A (dd) + B (dd) C(dd) + D (dd)

b

B

a

A

d

D

c

C

C

CC

CCK


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Nng ban u ca mi cht A, B, C, D l1,5 mol/l. Sau khi cn bng c thit lp, nng ca C l2 mol/l. Hng scn bng Kcca hny l:

a) Kc= 1,5 b) Kc= 2,0 c) Kc= 0,25 d) Kc= 4

8.11 Chn pht biu ng:Phn ng H2(k) + O2(k) H2O (k) cG

o298= -54,64 kcal.

Tnh Kpiu kin tiu chun. Cho R = 1,987 cal/mol.Ka) K

p= 40,1 b) K

p= 10

, c) K

p= 10

- ,d) K

p= -40,1

8.12 mt nhit xc nh, phn ng: S (r) + O2(k) = SO2(k) chng scn bng KC=4,2.10

52. Tnh hng scn bng KCca phn ng SO2(k) = S (r) + O2(k) cng nhit .

a) 2,38.10 b) 2,38.10

- c) 4,2.10

- d) 4,2.10

-

8.2 Anh hng ca cc yu tn cn bng ha hc8.13 Chn pht biungtrong nhng pht biu sau y:

1) Vic thay i p sut ngi khng lm thay i trng thi cn bng ca phn ng ctng smolcht khca cc sn phm bng tng smol cht khca cc cht u.

2) Khi tng nhit , cn bng ca mt phn ng bt ksdch chuyn theo chiu thu nhit.3) Khi gim p sut, cn bng ca mt phn ng bt ksdch chuyn theo chiu tng sphn t

kh.4) Ht trng thi cn bng thlng cc cht thm vo khng lm nh hng n trng thicn bng.

a) 1, 2 v 3 b) 1 c) 2 v 3 1, 3 v48.14Chn y ng:1) Mt hang trng thi cn bng, nu ta thay i mt yu t(p sut, nhit , nng ) thcnbng schuyn dch theo chiu chng li sthay i .2) Khi tng nhit , cn bng schuyn dch theo chiu phn ng ta nhit; khi gim nhit , cnbng schuyn dch theo chiu phn ng thu nhit.3) Hng scn bng ca mt phn ng lmt i lng khng i nhit xc nh.4) Khi thm mt cht ( tc cht hay sn phm) vo hcn bng, cn bng sdch chuyn theo chiu lmgim lng cht .

a) 1 v3 b) 1 , 3 v4 c) 1 v4 d) 1 v28.15 Chn pht biu ng:

Phn ng A (k) B (k) + C (k) 300oC cKp= 11,5, 500oC cKp= 33

Vy phn ng trn lmt qutrnh:a) on nhit. b) thu nhit. c) ng nhit. d) ta nhit.

8.16 Mt phn ng txy ra cG0< 0. Githit rng bin thin entanpi vbin thin entropi khngphthuc nhit , khi tng nhit thhng scn bng Kps:

a) tng b) gim c) khng i d) cha thkt lun c8.17Cn bng trong phn ng H2(k) + Cl2(k) 2HCl (k) sdch chuyn theo chiu no nutng p sut ca hphn ng?

a) Thun b) Nghch c) Khng dch chuyn. d) Khng thdn.8.18 Cho cn bng CO2 (k) + H2(k) CO (k) + H2O (k)

Tnh hng scn bng Kcbit rng khi n cn bng ta c0,4 mol CO2; 0,4 mol H2; 0,8 mol COv0,8 mol H2O trong mt bnh cdung tch l1 lt. Nu nn hcho thtch ca hgim xung, cnbng schuyn dch nhthno?

a) Kc= 8 ; theo chiu thun b) Kc= 8 ; theo chiu nghchc) Kc= 4 ; theo chiu thun d) Kc= 4 ; khng i

8.19Xt phn ng: CH3COOH + C2H5OH CH3COOC2H5+ H2O Kc = 4


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Suy ra hng scn bng ca phn ng thy phn CH3COOC2H5l:a) KC= 1/4 b) KC= 1/2 c) KC= KC d) KC= -KC

8.20 Chngii php hp lnht:Cho phn ng :

N2 (k) + O2(k) 2NO (k) H 0.thu c nhiu NO ta cthdng cc bin php :a) Tng p sut vgim nhit b) Tng nhit .c) Tng p sut vtng nhit . d) Gim p sut.

8.21 Cho phn ng:2SO2(k) + O2(k) 2SO3(k) c < 0

c nhiu SO3hn , ta nn chn bin php no trong 3 bin php sau:1. Gim nhit . 2. Tng p sut. 3. Thm O2.

a) Chcbin php 1 b) Chc1 v2 c) C3 bin php. d) Chc1 v38.22Chn ng:

Tc ng no slm tng hiu sut phn ng :CaCO3(r) CaO (r) + CO2(k) , > Oa) Gim nhit b) Tng p sut c) Tng nhit d) Tng nng CO2

8.23 Phn ng N2(k) + O2(k) = 2NO(k) , > 0 ang nm trng thi cn bng.Hiusu

t ph

nngstng ln khi p dng cc bin php sau:

1) Dng xc tc . 2) Nn h.3) Tng nhit . 4) Gim p sut hphn ng.a) 1 & 2 b) 1 & 3 c) 1, 3 & 4 d) 3

8.24 Chn cu ng:Xt hcn bng CO (k) + Cl2(k) COCl2(k) , < OSthay i no di y dn n cn bng chuyn dch theo chiu thun:

a) Tng nhit b) Gim th tch phn ng b ng cch nn hc) Gim p sut d) Tng nng COCl2

8.25 Phn ng thy phn ca ester : ester + nc axit + rutng hiu sut phn ng (cn bng chuyn dch theo chiu thun) ta cthdng cc bin

php no trong 3 bin php sau:1. dng nhiu nc hn.2. bng cch tin hnh thy phn trong mi trng baz3. Loi ru

a) Chdng c bin php 1 b) Chdng c bin php 2c) Chdng c bin php 3 d) Dng c cba bin php

8.26Cho cc phn ng:(1) N2 (k) + O2(k) 2NO (k) o> 0(2) N2 (k) + 3H2(k) 2NH3(k) o< 0(3) MgCO3(r) MgO (r) + CO2(k) o> 0

Vi phn ng no ta nn dng nhit cao vp sut thp cn bng chuyn dch theo chiu thuna) Phn ng (1) b) Phn ng (2) c) Phn ng (3) d) Phn ng (1) v(2)

8.27 Cc phn ng di y ang trng thi cn bng 25OC.


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N2(k) + O2(k) 2 NO (k) H00. (1)N2(k) + 3H2(k) 2 NH3(k) H00. (2)MgCO3(r) CO2(k) + MgO (r) H00. (3)I2 (k) + H2(k) 2HI (k) H00 (4)

Cn bng ca phn ng no dch chuyn mnh nht theo chiu thun khi ng thi hnhit vtng p sut chung ca:

a) Phn ng 2 b) Phn ng 1 c) Phn ng 3 d) Phn ng 48.28 Chn trng hp ng:

Xt cn bng: 2NO2(k) N2O4(k) o298= -14kcal(nu) (khng mu)

Trong bn trng hp di, mu nu ca NO2sm nht khi:a) un nng n 373K. b) Lm lnh n 273K c) Tng p sut. d) Gi298K.

8.29 Chn bin php ng.Phn ng ta nhit di y t trng thi cn bng:

2 A(k) + B(k) 4D (k)dch chuyn cn bng ca phn ng theo chiu hng to thm sn phm, mt sbin php sau

y c sdng:

1) Tng nhit 2) Thm cht D 3) Gim thtch bnh phn ng4) Gim nhit 5) Thm cht A 6) Tng thtch bnh phn nga) 1, 3, 5 b) 4,5,6 c) 2,3 d) Gim thtch bnh

CHNG 10: TC VCCHPHN NG HA HC10.1 Cc khi nim cbn. nh lut tc dng khi lng trong ng ha hc. Lthuyt cchphn ng ha hc.10.1 Chn p n ng. Cho phn ng : 2A (k) + B(k) C (k)

Biu thc tc phn ng phi l:a) v = k.CA .CBb) v = k. Ccc) v = k.CA

m.CB

n,vi m vn lnhng gitrtm c tthc nghim.

d) v = k.CAm

.CBn

, vi m vn lnhng gitrtm c tphng trnh phn ng10.2 Phn ng phn hy oxit dinitcstng qut:2N2O (k) 2N2 (k) + O2(k) v = k[N2O]

Ngi ta cho rng phn ng tri qua hai bc scp:Bc 1: N2O N2 + O Bc 2: N2O + O N2 + O2

Vy, pht biu no di y phhp vi cc dliu trn:a) Phn ng phn hy dinitoxit cbc ng hc bng 2.b) Bc 1 cphn tsln phn t.c) Oxi nguyn tlxc tc ca phn ng.d) Bc 2 lbc quyt nh tc phn ng.

10.3 Chn sai: Cho phn ng aA + bB = cC + dD cv = kCAm

CBn.

Bc ca phn ng:1. bng (n + m) 2. t khi ln hn 3 3. Bng (c+d)(a+b)4. Cthlphn s 5. Bng a + ba) 2 v3 b) 3 v4 c) 3 v5 d) 2 , 3 v5

10.4 Chn pht biu ng :Phn ng 2A + B 2C cbiu thc tc phn ng l v = k.CA2.CB, nn :a) Phn ng bc 3. b) Phn ng trn lphn ng phc tp.c) Bc ca phn ng c tnh trc tip bng hstlng ca cc cht tham gia phn ng vbng 3.

d) Cu a vc u ng.


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10.5 Phn ng 2A + 2B + C D + E ccc c im sau :* [A], [B] khng i, [C] tng gp i, vn tc v khng i.* [A] , [C] khng i, [B] tng gp i, vn tc v tng gp i.* [A], [B] u tng gp i, vn tc V tng gp 8 ln.

Cba thnghim u cng mt nhit Biu thc ca vn tc V theo cc nng A, B, C l:

a) v = k[A][B][C] b) v = k[A][B] c) v = k[A] [B][C] d) v = k[A] [B]

10.6 Cho phn ng: CH3Br (dd) + OH-(dd) CH3OH (dd) + Br(dd). Bit rng:Tc phn ng tng ln 2 ln khi tng [OH-] ln 2 ln ([CH3Br] khng i).Tc phn ng tng ln 3 ln khi [CH3Br] tng ln 3 ln ([OH

-] khng i).

Vit biu thc tc phn ng.a) v = k [CH3Br] b) v = k [CH3Br] [OH

-] c) v = k [OH

-] d) v = k [CH3Br] [OH

-]

10.7 Mt phn ng A + 2B = C bc 1 i vi [A] vbc 1 i vi [B], c thc hin nhitkhng i.a) Nu [A], [B] v[C] u gp i, vn tc phn ng tng gp 8 ln vphn ng lphn ng n gin.b) Nu [A] v[B] u tng gp i, vn tc phn ng tng ln gp 4 ln vphn ng ny lphn ngn gin.

c) Nu [A] tng gp i, [B] tng gp ba, vn tc phnng tng ln gp 6 ln vphn ng ny lphnng phc tp.d) Nu [A] v[B] u tng gp ba, vn tc phn ng tng ln gp 6 ln vphn ng ny lphn ngn gin.

10.8 Cho phn ng 2NO (k) + O2 (k) = 2NO2(k)Biu thc thc nghim ca tc phn ng lv = d[NO2]/dt = k [NO]

2[O2].

Cthkt lun rng:1) Phn ng cbc mt i vi O2vbc 2 i vi NO.2) Bc ca phn ng c tnh trc tip tcc hstlng ca cc tc cht.3) Phn ng cbc chung bng 3.4) Vn tc phn ng trong biu thc trn lvn tc phn ng trung bnh.

Cc kt lun ng l:a) 1 ,2 v3 b) 1, 3 v4 c) 1 v3 d) 1, 2, 3 v4

10.9 Chn cu sai. Hng stc phn ng :a) khng phthuc cht xc tc. b) khng phthuc nng cht phn ng.c) phthuc nhit . d) phthuc nng lng hot ha ca phn ng

10.10 Chn cu Sai:Hng stc ca phn ng nA + mB = AnBm

a) phthuc vo nng CA vCB.b) cgitrkhng i trong sut qutrnh phn ng ng nhit.c) ltc ring ca phn ng khi CA= CB= 1 mol.d) bin i khi cmt cht xc tc.

10.11i vi phn ng thun nghch :a) Phn ng pht nhit cE t < E n b) Phn ng pht nhit cE*t E*nc) Phn ng thu nhit cE t < E n d) Phn ng thu nhit cE*t E*n

10.2 Cc yu tnh hng n tc phn ng10.12 Tc phn ng ng th khtng khi tng nng ldo:

a) Tng sva chm ca cc tiu phn hot ng. b) Tng entropi ca phn ng.d) Gim nng lng hot ha ca phn ng. c) Tng hng stc ca phn ng.

10.13 Chn pht biu ng:Nguyn nhn chnh lm cho tc phn ng tng ln khi tng nhit l:


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a) Tn sut va chm gia cc tiu phn tng.b) Lm gim nng lng hot ha ca phn ng.c) Lm tng entropi ca h.d) Lm tng sva chm ca cc tiu phn cnng lng ln hn nng lng hot ha.

10.14 Stng nhit ctc ng n mt phn ng thun nghch :a) Chlm tng vn tc chiu thu nhit.b) Chlm tng vn tc chiu ta nhit.c) Lm tng vn tc cchiu thu vta nhit,lm cho hmau t n trng thi cn bng mi.d) Tng ng u vn tc cchiu thu vta nhit nn cn bng khng thay i.10.15 Khi tng nhit to, vn tc phn ng tng vstng nhit :

a) lm cho G < 0.b) lm gim nng lng hot ha.c) chyu llm tng sln va chm gia cc phn t.d) lm tng nng lng ca cc tiu phn trong h.

10.16 Cht xc tc cnh hng nhthno n trng thi cn bng ca phn ng ta nhit?a) Lm cho phn ng nhanh t ti cn bng

b) Lm tng nng lng ca cc tiu phn.c) Lm cho phn ng nhanh xy ra hn tn.d) Lm cho hiu sut ca phn ng theo chiu thun tng ln..

10.17 Cht xc tc cnh hng nhthno n cn bng ha hc:a) Khng nh hng n cn bng.b) Lm cn bng dch chuyn theo chiu phn ng nghch.c) Lm cn bng dch chuyn theo chiu phn ng thun.d) Lm tng hng scn bng ca phn ng.

10.18 Hoca phn ng cphthuc vo cht xc tc khng?a) C, vcht xc tc tham gia vo qutrnh phn ng.b) Khng, vcht xc tc chtham gia vo giai on trung gian caphn ng vc phc hisauphn ng. Sn phm vtc cht vn ging nhkhi khng ccht xc tc.c) C, vcht xc tc lm gim nhit cn cphn ng xy ra.d) C, vcht xc tc lm gim nng lng hot ha ca phn ng.10.19 Chn cc c tnh ngca cht xc tc.Cht xc tc lm cho tc phn ng tng ln nhcc c tnh sau:1) Lm cho G ca phn ng m hn.2) Lm tng vn tc phn ng nhlm gim nng lng hot ha ca phn ng.3) Lm tng vn tc ca phn ng nhlm tng vn tc chuyn ng ca cc tiu phn.4) Lm cho G ca phn ng i du tdng sang m.

a) 1 , 2 v3 b) 1 v2 c) 2 v4 d) 210.20 Chn cu Sai. Cht xc tc:

a) Khng lm thay i cc c trng nhit ng ca phn ng.b) Chctc dng xc tc vi mt phn ng nht nh.c) Lm gim nng lng hot ha ca phn ng.d) Lm thay i hng scn bng caphn ng.

10.21 Chn sai:Tc phn ng cng ln khi:

a) nng lng hot ha ca phn ng cng ln.b) entropi hot ha cng ln.c) sva chm chiu qugia cc tiu phn cng ln.d) nhit cng cao.


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10.22 Chn cu ngTc ca phn ng dth:a) tngln khi tng bmt tip xc phab) ca bt kphn ng no cng tng ln khi khuy trnc) chc quyt nh bi tng tc ha hc ca bn thn cht phn ng.c) phthuc vo bmt tip xc pha mkhng phthuc vo nng cht phn ng.

10.23Chn cung. Tc ca phn ng ha tan kim loi rn trong dung dch axit s:1) Gim xung khi gim nhit phn ng2) Tng ln khi tng kch thc cc ht kim loi.3) Gim xung khi gim p sut phn ng.4)tng ln khi tng nng axt.

a) 1 , 2 v4 b) 1, 3 v4 c) 1, 2 v3 d) 1 v410.24Chn p n ng vy nht:

Cmt sphn ng tuy cG < 0 song trong thc tphn ng vn khng xy ra. Vy cthpdng nhng bin php no trong cc cch sau phn ng xy ra:

1. Dng xc tc 2. Tng nhit 3. Tng nng tc cht 4. Nghin nhcc tc cht rna) 1 v2 b) 1 v3 c) 1 , 2 v4 d) 1, 2, 3 v4

10.25 Chn cu trli y nht.tng tc ca phn ng dpha cstham gia ca cht rn ta cthdng nhng bin

php no trong cc bin php sau y :1/ Tng nhit . 2/ Dng xc tc.3/ Tng nng cc cht phn ng. 4/ Gim nng sn phm phn ng

trn bmt cht phn ng rn.5/ Nghin nhcc cht phn ng rn.a) Tt ccc bin php trn. b) Cc bin php 1, 2, 3, 5.c) Cc bin php 1 , 2 , 3. d) Cc bin php 1, 2 3, 4.

10.26 Phn ng CO (k) + Cl2(k) COCl

2(k) lphn ng n gin. Nu nng CO tng t0,1M

ln 0,4M; nng Cl2tng t0,3M ln 0,9M thtc phn ng thay i nhthno?a) Tng 3 l n b) Tng 4 l n c) tng 7 l n d) Tng 12 l n

Mt phn ng kt thc sau 3 gi20oC. nhit no phn ng skt thc sau 20 pht, bit hsnhit ca phn ng l3.

a) 30oC b) 40oC c) 50oC d) 60oC10.28 100oC, mt phn ng kt thc sau 3 gi. Hsnhit ca phn ng l3. Khi tng nhit phn ng ln 120oC ththi gian phn ng sl:

a) 20 pht. b) 60 pht. c) 9 gi. d) p skhc.10.29 Phn ng thun nghch A2(k) + B2(k) 2AB (k)

Chsnhit ca phn ng thun vphn ng nghch ln lt l2 v3. Hi khi tng nhit

cn bng dch chuyn theo chiu no vtsuy ra du ca Ho

ca phn ng thun.a) Nghch, H0 < 0 b) Nghch, H0 > 0 c) Thun, H0 < 0 d) Thun, H0 > 0


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P N PHN 2:LTHUYT DIN RA CC QUTRNH HA HC

Cu 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10p n b b d a b b b c d cCu 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20

p n b d c d c d a a a cCu 6.21 6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 6.30p n c b d d c b a b a aCu 6.31 6.32 6.33 6.34 6.35p n b d b c cCu 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10p n b c b a c a b c c cCu 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 7.20p n d a b a a c c a b dCu 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 7.30

p n b c d d b c a a b cCu 7.31 7.32 7.33 7.34 7.35 7.36 7.37 7.38 7.39 7.40p n b a b d c b d c b aCu 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10p n d b c b d b a d c dCu 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20p n b b a b b d c d a bCu 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 8.29p n c c d b d c a a bCu 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10

p n c b c a d b c c a acu 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 10.19 10.20p n a a d c d a a b d dCu 10.21 10.22 10.23 10.24 10.25 10.26 10.27 10.28 10.29p n a a d a a d b a a

.

PHN 3 : LTHUYT VDUNG DCH. PHN NG TRAO I IONVPHN NG OXY HA KHTRONG DUNG DCH

CHNG 11: DUNG DCH LNGnh ngha dung dch, nng , tan , ng lng11.1ng lng ca KMnO4(Phn tlng M) bng:

a) M/1 b) M/3 c) M/5 d) Ty thuc vo phn ng11.2. Cho phn ng: Al2(SO4)3 + 2NaOH = Na2SO4 + 2[Al(OH)2]2SO4


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ng lng gam ca Al2(SO4)3 vNaOH ln lt bng: (Cho bit phn tgam ca Al2(SO4)3bng342g vca NaOH bng 40g)

a) 342g ; 40g b) 171g ; 40g c) 57g ; 20g d) 114g; 40g

11.3 Cho phn ng: 2MnO2 + O2 + 4KOH = 2K2MnO4 + 2H2Ong lng gam ca MnO2vO2ln lt bng: (cho bit phn tgam ca MnO2bng 87g vca O2

bng 32g)a) 43,5g; 16g b) 87g ; 16g c) 43,5g; 8g d) 21,75g ; 8g

11.4 Cho phn ng:2KAl(SO4)2.24H2O + 4NaOH = 2Na2SO4 + [Al(OH)2]2SO4+ K2SO4+ 48H2O

ng lng gam ca KAl(SO4)2.24H2O vNaOH ln lt bng: (Cho bit phn tgam caKAl(SO4)2.24H2O bng 690g vca NaOH bng 40g)

a) 690g ; 40g b) 345g ; 40g c) 172,5g ; 20g d) 230g; 40g

11.5 Cho phn ng: MnO2 + 4HClc, nng= MnCl2 + Cl2 + 2H2Ong lng gam ca MnO2vHCl ln lt bng: (cho bit phn tgam ca MnO2bng 87g vca

HCl bng 36,5g)a) 43,5g; 36,5g b) 21,75g ; 18,25g c) 87g ; 35,5g d) 21,75g ; 35,5g

11.6 Chn cu ng. tan ca cc cht trong nc l:a) Sml kht tan tan ti a trong 100g nc iu kin cho.b) Sgam cht tan tan ti a trong 100ml nc iu kin cho.c) Smol cht in ly rn t tan tan ti a trong 1 lt nc iu kin cho.d) Ca, b, c u ng.

11.7 Chn pht biu ngtrong cc pht biu sau:a) tan ca a scht t tan gim khi nhit ca dung dch tng.b) tan ca cht t tan chphthuc vo bn cht cht t tan vnhit .c) tan cht t tan stng khi cho vo dung dch ion cng loi vi 1 trong cc ion ca chtt tan .d) Khngcpht biu no ng.

11.8 Chn cc pht biu sai:1) Dung dch lng ldung dch cha bo ha vnng cht tan nh.2) Dung dch lmt hng th.3) Thnh phn ca mt hp cht lxc nh cn thnh phn ca dung dch cththay i.4) Dung dch bo ha ldung dch m c.

a) 1, 3 b) 2, 4 c) 2, 3 d) 1, 4

11.9 Dung dch A cnng phn trm C%, nng mol CM, khi lng ring d (g/ml), phn tlngca A lM, s ltan tnh theo g/100g H2O. Biu thc sai l:

a) s = 100. C%/(100- C% b) CM= 10 C%.d/M

c) C%= CM. M/(10.d) d) C%= 100.s / (100-s)

11.10 Chn pht biu ng:1) Nng phn phn tgam lsphn khi lng (tnh theo n vgam) ca cht tan hoc ca dung

mi trong dung dch.2) Nng ng lng gam c biu din bng smol cht tan trong 1 lt dung dch.3) i vi mt dung dch, nng ng lng gam ca mt cht cthnhhn nng phn t

gam ca n.4) Nng molan cho bit smol cht tan trong mt lt dung dch.5) Cn bit khi lng ring ca dung dch khi chuyn nng phn trm C%thnh nng phn t

gam hoc nng ng lng gam.6) Khi lng ring ca mt cht lkhi lng (tnh bng gam) ca 1 cm3cht .

a) 1, 2, 5, 6 b) 1 , 5, 6 c) 5 , 6 d) 3, 5, 6


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11.11 Chn pht biu ng:pha ch100 ml dung dch H2SO410

-4N thsml dung dch H2SO42.10

-2N phi ly l:

a) 0,5 ml b) 1 ml c) 2 ml d) 0,25 ml

11.12 Tnh thtch dung dch (lt) HCl 4M cn thit cthpha thnh 1lit dung dch HCl 0,5M.a) 0,125 l b) 0,0125 l c) 0,875 l d) 12,5 l

Cc nh lut vdung dch lng11.13 Chn pht biu ng.

a) Mt cht lng si mt nhit ti psut hi bo ha ca cht lng bng p sut mitrng.

b) Khi ha tan mt cht A trong cht lng B, p sut hi bo ha ca B tng.c) Nc lun lun si 100oC.d) Nc mui si nhit thp hn nc nguyn cht.

11.14 Chn pht biu sai.a) cng nhit T, p sut hi bo ha ca dung mi trong dung dch nghch bin vi nng cht

tan.b) gim tng i p sut hi bo ha ca dung mi trong dung dch bng nng phn mol ca

cht tan.c) Nhit kt tinh ca dung mi trong dung dch nghch bin vi nng mol ca dung dch.

d) Ap sut hi bo ha ca dung dch lng phn t tlthun vi nng phn mol ca cht tan.11.15 Chn pht biu sai.

a)Nhit si ca cht lng lnhit p sut hi bo ha ca nbng vi p sutmi trng.b)Nhit si ca dung dch cha cht tan khng bay hi lun lun cao hn nhit sica dung mi nguyn cht cng iu kin p sut ngi.c)Nhit ng c ca dung mi nguyn cht lun thp hn nhit ngc ca dungmi trong dung dch.d)p sut hi bo ha ca dung mi trong dung dch lun nhhn p sut hi bo ha cadung mi tinh khit.

11.16 chn cu ng.a) gim tng i p sut hi bo ha ca dung mi trong dung dch bng phn mol cadung mi trong dung dch.b) p sut hi bo ha ca dung mi trong dung dch lun nhhn p sut hi bo ha cadung mi tinh khit cng gitrnhit .c) p sut hi bo ha ca dung mi trong dung dch tlthun vi phn mol ca cht tantrong dung dch.d) p sut hi bo ha ca dung dch lng phn tkhng phthuc vo bn cht ca chttan.

11.17 Chn cu trli chnh xc. Nhit si ca cht lng lnhit mti :a) p sut hi bo ha ca cht lng bng p sut bn ngi.b) p sut hi bo ha ca cht lng bng 760mmHg.

c) Apsut hi bo ha ca cht lng >760mmHg.d) Ap su t hi bo ha ca ch t lng < 760mmHg.

11.18 Vi i lng k trong cng thc nh lut Raoult 2: T = kCm, pht biu no sau y lchnhxc:

a) k lhng sphthuc vo nng cht tan, nhit vbn cht dung mi.b) k lhng sphthuc vo nhit vbn cht dung mi.c) k lhng schphthuc vo bn cht dung mi.d) k lhng sphthuc vo bn cht cht tan vdung mi.


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11.19 Chncu ng. p sut khng i, nng dung dch lng (ccht tan khng bay hi vkhng to dung dch rn vi dung mi) cng tng th:

a) Nhit si tng. b) Nhit si gim.c) Nhit ng c gim. d) Cc cu a vc u ng.

11.20 Ha tan 5 gam mi cht C6H12O6, C12H22O11vC3H5(OH)3trong 500 gam nc. Trong cc dysau, dy no xp cc cht trn theo nhit si ca dung dch tng dn: (cho bit nguyn tgam ca C=12, O = 16 vH = 1)(cc cht trn khng bay hi)

a) C12H22O11 < C6H12O6< C3H5(OH)3 b) Khng sp cc) C3H5(OH)3< C6H12O6< C12H22O11 d) C12H22O11< C3H5(OH)3< C6H12O6

11.21 Tnh p sut hi bo ha ca nc trong dung dch cha 5g cht tan trong 100g nc nhit 25

oC. Cho bit nhit ny nc tinh khit cp sut hi bo ha bng 23,76mmHg vkhi lng

phn tcht tan bng 62,5g.a) 23,4mmHg b) 0,34mmHg c) 22,6mmHg d) 19,0mmHg

11.22 25oC, p sut hi bo ha ca nc nguyn cht l23,76mmHg. Khi ha tan 2,7mol glyxerinvo 100mol H2O nhit trn thgim p sut hi bo ha ca dung dch bng:

a) 23,13mmHg b) 0,64mmHg c) 0,62mmHg d)23,10mmHg

11.23 So snh nhit si ca cc dung dch CH3OH (t1) , CH3CHO (t2) vC2H5OH (t3) cng cha Bgam cht tan trong 1000g nc c: (bit rng cc cht ny cng bay hi cng vi nc)

a) t3> t2> t1 b) t1> t2> t3 c) t2> t1> t3 d) khng dliu tnh.11.24 Dung dch nc ca mt cht tan khng in li si 373,52oK. Nng molan ca dung dchny l: ( Ks= 0,52):

a) 0,01 b) 1,0 c) 10 d) 0,1

11.25 Trong 200g dung mi cha A g ng glucckhi lng phn tM; hng snghim ng cadung mi lK. Hi biu thc no ng i vi T:

a) T=5k.(A/M) b) T= k.(A/M) c) T= 1/5k.(A/M) d) T= k.A11.26 Dung dch nc ca mt cht tan bay hi khng in li si 100,26oC. Nng molan ca dungdch ny l: (hng snghim si ca nc Ks= 0,52)

a) 0,75 b) 1 c) 0,5 d) khng dliu tnh11.27 Ha tan 0,4g mt hp cht hu cX vo 25g axit acetic thdung dch bt u ng c 16,15oC. Bit rng axit acetic nguyn cht ng c 16,60oC vhp cht X to dung dch rn vi axitacetic. Cho bit hng snghim lnh ca axit acetic l3,6. Tnh phn tgam ca X.

a) 228g b) 256g c) 128g d) khng dliu tnh11.28 Chn pht biu ng:

1) Ap sut thm thu ca dung dch cln bng p sut gy ra bi cht tan nu cht ny thkhltng, chim thtch bng thtch ca dung dch vcng nhit vi nhit ca dungdch.

2) Ap sut thm thu tlthun vi nhit ca dung dch.3) Ap sut thm thu ca mt dung dch in li vkhng in li cng nhit vcng nng mol

lkhc nhau.4) nh lut Vant Hoff ( vp sut thm thu) ng cho dung dch bt knng no.

5) Ap sut thm thu tnh theo nng ng lng gam ca dung dch.a) 1, 2 , 3 b) 1, 3 , 5 c) tt cu ng d) Chccu 4 sai

11.29 Biu thc tn hc ca nh lut Raoult II cdng:a) t = k.Cm (Cm nng molan). b) t = k.CM(CMnng mol)c) t = kCN(CNnng ng lng) d) t = kC(Cnng phn trm)

CHNG 12: DUNG DCH IN LiDung dch in li, in li (), hng sin li (K)12.1 Mt cht in ly trung bnh 25oC cin ly :

a) 0,03


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b) 0,03


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c) Nhit ng c cao hn, p sut hi bo ha cao hn.d) Ap sut hi bo ha thp hn, nhit ng c cao hn.

12.12 Htan 0.585 gam NaCl vo 1 lt H2O. Ap sut thm thu ca dung dch ny 25oC cgitr

l: (Cho bit MNaCl= 58,5 vR = 0,082 lit , atm / mol. K; coi = 1)a) 0.488 atm b) 0.244 atm c) 0.041 atm d) 0.0205 atm

12.13 Chn cc cu sai:1) Chcc cht in li mnh mi cn sdng khi nim hot (a) thay cho nng trong biu thc

nh lut tc dng khi lng.2) Khi pha lng dung dch thhshot (f) tng.3) Cc dung dch cht in li yu lun chshot (f) bng 1.

a) 1, 2, 3 b) 1, 2 c) 2, 3 d) 1, 3

12.14 Chn cusai:a) Hot ca cht lnng biu kin ca ntrong dung dch.b) Hot ca ion phthuc vo lc ion ca dung dch.c) Hng sin li thay i khi tnh theo nng ca ion .d) Nng ca ion trong dung dchthng nhhn hot ca n.

12.15 Chn cu ng. Cho cc dung dch nc lng ca C6H12O6, NaCl, MgCl2, Na3PO4. Bit chngccng nng molan vin li ca cc mui NaCl, MgCl2vNa3PO4u bng 1. gim psut hi bo ha ca nc trong chng thay i nhthno theo thtca dy trn:

a) Bng nhau b) Gim dn c) tng dn d) Khng cquy lut12.16 Trt tsp xp no ca cc dung dch 0,01M ca nhng cht cho di y lphhp vi sgim dn p sut thm thu (cc mui in li hn tn):

a) CaCl2NaClCH3COOHC6H12O6 b) CH3COOHNaClC6H12O6- CaCl2c) C6H12O6- CH3COOHNaCl - CaCl2 d) CaCl2- CH3COOHC6H12O6NaCl

12.17 Chncu ng.Qutrnh htan tinh thKOH trong nc xy ra km theo sthay i entropi chuyn pha (Scp) ventropi solvat ha (Ss) nhsau:

a) Scp> 0 , Ss< 0 b) Scp< 0 , Ss< 0c) Scp< 0 , Ss> 0 d) Scp> 0 , Ss > 0

CHNG 13: CN BNG ION CA AXITBAZCc lthuyt axit baz13.1 Chn pht biu ng vy nht. Cc cht lng tnh theo thuyt proton (thuyt bronsted)trong cc cht sau: NH4

+, CO3

2-, HCO3

-, H2O, HCl l:

a) HCO3-vCO3

- b) HCO3

-vH2O c) HCO3

-, H2O vHCl d) NH4

+vHCl

13.2 Chn p n ng:Cho cc cht sau: CH3COOH , H2PO4

-, NH4

+, theo thuyt proton, cc cp axit bazlin hp xut pht

tchng l:a) CH3COOH2

+/CH3COOH; CH3COOH/CH3COO

-; H3PO4/H2PO4

-; H2PO4

-/PO4

3-; NH4

+/NH3;

b) CH3COOH2+/CH3COO-; CH3COOH/CH3COO-; H3PO4/H2PO4-; H2PO4-/HPO42-; NH4+/NH3;

c) CH3COOH2+/CH3COOH; CH3COOH/CH3COO

-; H3PO4/H2PO4

-; H2PO4

-/HPO4

2-; NH5

2+/NH4

+

d) CH3COOH2+/CH3COOH; CH3COOH/CH3COO

-; H3PO4/H2PO4

-; H2PO4

-/HPO4

2-;NH4

+/NH3;

13.3 Chn cu trli ng. Cu no trong scc cu di y sai:1) Bazlin hp ca mt axit mnh lmt bazyu vngc li.2) i vi cp axit-bazlin hp NH4

+/NH3 trong dung dch nc ta c:KNH4+. KNH3 = Kn,

trong Kn ltch sion ca nc.3) Hng sin li ca NH3trong dung dch nc l1,8 x 10

-5, suy ra KNH4+= 5,62 x 10

-10.

a) 1 b) 2 c) 3 d) khng ccu sai


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13.4 Da vo i lc proton ca cc dung mi NH3vHCl cho bit CH3COOH thhin tnh cht gtrong dung mi :

a) Tnh baztrong HCl, tnh axit trong NH3. c) Tnh baztrong c2 dung mi.b) Tnh baztrong NH3, tnh axit trong HCl. d) Tnh axit trong c2 dung mi.

13.5 Cho cc phn ng:1) MgCl2+ 6H2O = MgCl2.6H2O

2) BH3 + NaH = Na[BH4]

3) Ni2+ + 4NH3 = [Ni(NH3)4]2+Cc cht axit vbaztrong cc phn ng trn theo thuyt electron (thuyt lewis) l:

a) axit: H2O , BH3vNiCl2; baz: MgCl2, NaH vNH3b) axit: MgCl2, BH3vNiCl2; baz: H2O, NaH vNH3c) axit: MgCl2, NaH vNiCl2; baz: H2O, BH3vNH3d) axit: MgCl2, BH3vNH3; bazH2O, NaH vNiCl2

Cc hng sin li axit, baz, tch stan, tch sion ca nc, pH ca dung dch, dung dch m,chthpH13.6 Chn cc pht biung.

1) Theo thuyt proton, cc hng sin ly ca axit, baz ccng bn cht vi hng sthyphn.

2) tan vin li ccng bn cht.3) Theo thuyt proton, in ly vthy phn ccng bn cht.

4) Hng sin li ca axit, hng sin li bazvtch stan u lhng scn bng tun theo nhlut GuldbergWaage.

a) 3 , 4 b) 1 , 2 , 4 c) 1, 3 , 4 d) Tt cu ng13.7 Chn pht biu ng:

1) Axit cng yu thpKacng ln.2) Dung dch mt bazyu cpH cng nhkhi pKbca ncng ln.3) Bazcng mnh khi pKbcng ln4) Gia pKavpKbca H2PO4

-cquan hpKa+ pKb= -lgKn.

a) 1, 3, 4 b) 2,3 c) 1, 2, 4 d) 1, 2

13.8 Khi htan H3PO4vo nc, trong dung dch stn ti cc ion H+; HPO42-; H2PO

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