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TR N HIP H

HABI TPm

HC I CNm H

H a h c l t h u y t c s )

(In ln th II)

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Thng thng gia l thuyt v bi tppj ca mt mn hc bao gi cng c gn kt cht ch vi nhau. lm c cc dng bi tp ngi hc phi hiu k thuyt v bit cch vn dng n vo tng trng hp c th, k c cc php chuyn i n v tnh ln th thut gii ton.

Cun B i tp ha hc a i cng (Ha hoc l thuytc s) nhm p ng cc yu cu ny.

Sch gm 17 chng gm hu ht cc vn l thuyt c

s ca ha hc v c trnh by di dng bi tp. mi

chng chng ti i phn lm 3 phn nh:

A. Tm tt l thuyt

B. Bi tp c li gii

c. Bi tp cha c li gii

Trong chng cui cng ca sch chng ti trch dn mt s thi tuyn sinh v p n ca mn hc ny nhm gip cho bn c hnh dung v mt thi tng hp v cch gii quyt n.

3



Ni dung cun bi tp c bin son theo ng chng trnh chun c hi ng chuyn ngnh i hc Quc gm H Ni thng qua.

Cc tc gi v Nh xut bn rt mong nhn c nhng kin ng gp ca c gi ln xut bn sau c hon thin hn.

Cc tc gi

4



MC LC

Trang

Khi nim v th nguyn, n v......................... 7

Chng I. Mt s'khi nim chung........... :.................... 13

Chng L ^ Nguyn l I ca nhit ng lc hc. Nhitha hc.-T......................................................... 25

Chng II. ^ Nguyn l II ca nhit ng lc hX-...45

Chng IV. 't Cn bng ha hc.-v........................................... 59

Chng V. y Dung dch....................................................... 83

Chng VI. J ng ha hc.................................................... 119Chng VII. V^ in ha hc. -y................ ............ ................... 139

y* Chng VIII. Ht nhn nguyn t ....................................... 161

Chng IX. Cu to nguyn t theo quan im c hclng t ............................................................. 171

Chng X. Nguyn t hidro............................................. 179

Chng XI. Nguyn t nhiu electron......................... 193

ChngXII. H thng tun hon cc nguyn t'ha hc.. 203

' Chng XZ7J/cc khi nim chung v lin kt thuyt VB. 215

Chng X iv^y Thuyt MO v lin kt............................... . 240

5



Chng XV. Lin kt gia cc phn t v trong phccht....................................... ............................ 263

Chng XVI. Lin kt ha hc trong tinh th...................... 279

Chng XVII. Mt s thi v hng n gii mn hahc l thuyt..................................................... 297

Phc 370

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KHI NIM V TH NGUYN, N V

I. T h nguyn. Cc i lng (vt l) cn o thng c

vit di dng mt biu thc ton hc v c biu din bng

mt phng trnh th nguyn.-Phng trnh th nguyn c th

xem nh mt biu thc ton v c biu din bng cc i

lng c s di dng mt tch s'

Tt c cc th nguyn ca nhng i lng cn o trong co'

hc u xut pht t 3 i lng c s l: Chiu di: L; khi

lng: M; thi gian: T. Cc i lng ny lp thnh h Li.M.T.

J 1 f _ on ng L !V d th nguyn cua tc [v] = v ~ = = L. I .

Th nguyn ca lc [F] = khi lng X gia tc = M.L.T2

Th nguyn ca cng (nng lng) [A] = lc X on ng

Nh vy th nguyn khng ch r cc i lng cn o mt n v c th no.

Mt i lng cn xc nh m cc th nguyn ca chng u b trit tiu s dn ti i lng khng th nguyn

II. n v. Khi ngi ta tin hnh o mt; i lng no

thi gian T

Th nguyn ca L tc [a] =thi gian T

= M.L.T' 2 X L = M.IT"2



tc l mun so snh i lng vi i lng cng loi ly lm chun so snh .gi l n v o .

Cc n v o c xc nh bi mu chun lu gi ti vin

cn o quc t. V d mt l n v o chiu di.

ln ca mt i lng vt l c th m theo qui c ly

gi tr bng s l 1 c gi l n v ca i lng vt l . V

d: mt, kilogam. Tp hp cc n v lm thanh mt h n v.

c mt s" h n v thng dng nh: h VKS (mt, kilogam,

giy); h CGS (xngtimt,'gam, giy)... /

Trong thc t, do thi quen, tng a phng, ung vng

'inh th, ngay c tng quc gia ngi ta s dng nhng n v

r t khc nhau cho cng mt i lng o.

V d n v chung cho chiu di l it, song ngi Anh

l dng Ins (Inch), pht (foot), trong khi ngi Vit li dng

trng, gang. tc...

R rng cch dng ny gy kh khn trong giao lu

quc t. V vy cn c mt n v quc t chun.

III. H n v SI. Nhn thy s bt li v vic s dng h

n v ty tin nn vo thng 10-1960 ti Hi ngh ln th XI v

cn o quc t hp Paris, cc nh khoa hc i n thng

nht cn xy dng mt h thng n v chung quc t. l n

v SI (Vit t t t ch Php - Systme International)

Di y chng ti lc ghi mt s ch dn quan trng nht

thuc h SI c lin quan n vic s dng cho cc bi tp ha i

cng.



1III.l. H n v c s

7 n v chnh thuc h SI

N Tn i lng n v K hiu

Ting Vit Ting Anh

1 Chiu di met metre m

2 Thi gian giy second s

3 Khi lng kilgam kilogram kg

4 Lng ch't mo mol mo

5 Nhit ken vin Kelvin K

6 Cng dng in Ampe Ampere A

7 Cng nh sng nn Cndela cd

IL2. Mt s" n v SI dn xut hay dng

T 7 n v c s nu trn ngi ta cn c th nh ngha mt s' n v dn xut thng dung trong h SI. V d:

- n v lc. chnh l lc tc dng ln mt vt c khi lng lkg gy ra mt gia tc bng lm/s2. n v dn xut thu c y gi l Newton (N)

IN = lkg.m.s"2

- n v p sut. Trong n v SI, p suai Pascal -(Pa).

p sut thu c l do c tc dng ln 1 n v din tch.

lPa = lc/din tch = = kg ms"2 /m 2 = kgm- 1s- 2m

Di y l mt s n v dn xut hay dng

N Tn i n v K Theo

lng Ting Vit

TingAnh

hiu nh ngha

1 Lc Niutn Newton N kgm"2

2 p sut Patean Pascal P kgm ^s^N/m 2)

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3 Nng lng Jun Joule J kgmV"

4 Cng sut Ot Watt w kgm"s_1(i /s)5 in tch Culng Coulomb c ' As6 in th Vn Volt V _J/As(j /s)

7 Tn s Hc Hertz Hz s ' 1

II.3. Mt s n v khc hay s dng cn chuyn v h SIHin'nay, bn cnh h SI l n v chnh thc, trong ha

hc ngi ta cn dng mt s" n v khc khng thuc h S gi l n v phi SI. d dng trong qu trnh gii cc bi tp hai cng chng ti ghi li bng di y mt s" n v ngoi h thng cng cc h s' chuyn i v h SI.

N Tn ai lng vt l

n vi Khiu

Theo nh nghaTing Vit Ting Anh

1 Chiu i micromtnnomtAngstrm

micrometrenanometreAngstrom

n m n m

0

10'9m 10 m

2 Th tch lt litre I 10W3 Nhit tu bch

phnCelsius c T(K) =

(C)+273,154 Thi gian Pht,

giminutehour

minh

60s > 3600s

5 p sut tmtphebartormiiimt thy ngn

Atmospherebartorrmillimetre Hg

atmbarTorr

mmHg

1,13.10bPa 105Pa * 1atm

133 322Pa 133,322Pa

6 Nng lng ec calo ot gi electron- vn

ergCalorie Watt hour electron Volt

ergcal

w.heV

10*0-4.184J3600J

1,602.10"19J

7 in tch n v tnh in

Unitelectrostatical

ues CGS^ 10_19c

2,99798 Lc yn dyne dyn 10'SN9

m men lng cc

bai Debye D ^ 10_29C m2,9979

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III.4. Quan h gia th nguyn v mt s n v thng dng

N i lng Phngtrnh

Thnguyn

n v

xc nh SI CGS

1 Din tch s = / 2 L22m cm2

2 Th tch

coII> m3 cra3

3 Vn tc lV = ' t

LT*1 m s ' 1 cm s ' 1

4 Gia tc Va = t

LT' 2.9

m s cm s ' 2

5 Lc F = ma MLT' 2 kg m s g cm s~2

6 p sutF

p = - s

ML'lT"2 kg m 'V 2 -] -2 g cm s

7 Khi lng ring D = V

M* kg m ' 3 g cm ' 3

8 Cng (nng lng)

A = FZ ML2T*'2 kg m2 s' 2 g cm2 s ' 2

9 Cng sut N = t

m l 2t -3 1 2 *3 kg m s li -3g cm s

1 0 Tn s' 1 . - 2

f = T

T 1 -1s -1s

II.5. Cc bc bi, bc c so vi n v c s

Khi s dng h SI ngi ta thng ly cc bc gin c l

n v bc bi lOn hay n v bc c 1 0 "n vi n l s nguyn.

Bng di y ghi li cch dng ny.

Tip ng u

K hiu quc t

Bc c Tip ng u

K hiu quc t

Bc bi

Deei d 1 0 '1 Deca da 1 0 1

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Centi c 1 0 '2 Hecto h 1 0 2

Mili m 1 0 " 3 Kilo k 1 0 3

Micro n 1 0 6 Mega M 1 0 e

Nano n 1 0 9 Giga G 1 0 '1

Pico P 1 0 '12 Tera T 1 0 12

Fern to f 1 0 - 15 Peta P 1 0 15

Atto a 10-18 Exa E 1 0 18

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Chng I

MT S KHI NIM CHUNG

A. TM TT L THUYT

1. Theo Rutherford (1911) th nguyn t c cu thnh

bi ht nhn gm prton p, v ntron (n); lp v gm cc electron

quay quanh ht nhn.

Vy ngun t gm:

- Ht nhn vi s" proton l z, in tch q=l,6.10'19c v N

l ntron. Hai i lng ny c lin h vi nhau bng s' khi

A theo h thc A = N + z

- Lp v electron c in tch ng bng in tch proton

nhng ngc du v khi lng electron ch bng 1/1836 kh

lng proton, ngha l khi lng tp trung ht nhn.

2 . l vi nguyn t, ngi ta t dng n v kg m dng

n v khi lng nguyn t (u). n v ny c nh ngha nh

sau:

1 1 2 . 1 0 3 , _-!>/? ir r27iTT : : kg = 1 ,6 6 . 1 0 kg12 Na

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3. Mol l lng cht cha cng mt s phn t cu trc

nh 1 mol nguyn t H} 1 mol phn t H2, 1 mol ion H+... T

suy ra:

- S" phn t cu trc c trong 1 mol ca cht chnh hng

s" Avogaro: NA = 6,022.10

4. Quan h gia khi lng tng i (klt) v kh"! lng

tuyt i (KLT):

kt:KLT =

NaXc nh khi lng phn t theo t khi ca cht kh.

M a5. T khi d ca kh A vi kh B l d =

Mb

Ma, M b - khi lng phn t ca A, B.

Kh B l khng kh th d =_ Ma29 \

6 . Xc nh khi lng phn t theo th tch moi.

PV PnVn Cng thc Boyle Mariotte:

T T0

p, V, T - p sut, th tch, nhit iu kin th nghim.

P0, v 0, T0 - ng vi iu kin tiu chun: 760mmHg; 22,4; 273K

7. Phng trnh trng thi ca cht kh l tng

PV = nRT = RT M

n - s' mol kh; m- khi lng khi tnh bng gam (g); M - khi lng ca mt moi kh.

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hoc p = RT = RT VM M

d - khi lng ring ca kh.

8 . Nu c mt hn hp kh l tng nhit T c th tchV th p sut ton phn PT ca h c xc nh theo nh lutDalton

PT = X p, hayi

P T = ^ Z n .V

P; - p sut ring ca kh th i i - S^ mol kh i trong hn hp

B- BI TP C LI GII

1.1. Hy in cc s" liu cn thit nhng trng trong bng sau v:

K hiu nguyn t l57N 1880 19 T7, 9* - -S" khi 15 - - 23 -S" in tch ht nhn 7 - - 1 1 -S" p ro ton - - - 14 vS" electron 7 - - 14s

1.2. 1) Trong mt th nghim in phn ngi ta thu c

27g nc. Hi: ,

a) C bao nhiu mol H20?

b) C bao nhiu nguyn t hiro?

2) Bit rng khi lng nguyn t tng i ca oxi l

15,99944. Tnh khi lng nguyn t tuyt i ca nguyn t ny.

Cho NA = 6,022.lO ^m or1.

BI GII

1) m = M.n suy ra n = = = l.mol H.;0M 18

Vy s phn t H20 l: 1,5.6,022.1023 = 9,0345.10~3 phn

t. Trong mt phn t HO th nguvn t H = 2 X s phn t

H30.

Vy s" nguyn t H l:

9,0345.1023.2 = 18,069.1023 nguyn t.

2) Khi lng nguyn t tuyt i ca oxi c tnh:

_____ 1 5 99944m(KLT) = - - 1 = 26,564.10 g.

6,022.1023

1.3. Trong nhiu php tnh ngi ta thng s dng hng s" kh R. Hy xc nh hng s cc h n v khe nhau.

a) Trong h n v SI.

b) Theo n v cal.K^.mol1.

c) Theo n v a tm i.K '1.moi'1.

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IBI GII

p dng phng trnh trng thi cho mt mol cht kh l

c) R = = 0,082 atm i.K^.m ol1273,15 \

1.4. Mt nguyn t X c bn knh l 1,44A, khi lng

ring thc tinh th l 19,36g/cm3. Nguyn t ny ch chim 74%

th tch ca tinh th, phn cn li l rng. Hy:

a) Xc nh khi lng ring trung bnh ca ton nguyn

t ri suy ra khi lng moi nguyn t.

b) Bit nguyn t X c 1 1 8 ntron v khi lng mol

nguyn t bng tng s* khi lng proton v ntron. Tnh s"

tng PV = RT. iu kin tiu chun

T0 = 273.15K; p0 = latm = 1,013.105 N /nr (Pa);

v 0 = 22,4/ = 2 2 ,4 . IC tV .

Theo n v SI: 1J - iN.m, vy:

R =

b) Do Ical = 4,184J nn gi tr R l:

R = i = 1,987 cal.K.m or4,184

proton.

B GII

a) Khi lng ring trung bin!



y 7 4 J J - 100 ,'_100.19,36 _ . 3d = d => d = -d =----- 1 = 26;16g/cm100 74 74

Mt khc: m = v.d = 7cr3.d 3

m = .3,14(1,44.10*8)3.26,16 = 32,704.10'23g3

Vy khi lng ca ml nguyn t l:

M = N.m =. 6,022.1023.32,704.102: = 196,976g/raol

hay M ~ l97g/mol.

b) Theo u bi ta c th vit:

M = mp + mn = mp + 118 = 197

T biu thc ny ta suy ra s ht/proton cn tm: mp = 79.

1.5. 1) Trong s" cc ht nhn nguyn t ca nguyn t" th

ch (^?7Pb) c t s" N/Z l cc i v heli ( H e ) c N/Z l cc

tiu. Hy thit lp t s" N/Z cho cc nguyn t vi 2 < z< 82.

2) Mt nguyn t X c tng s" cc ht l 58, s" khi ca n

nh hdn 40. Hy xc nh s proton, s' electron v s ntron ca

nguyn t .

BI GII

Theo h thc A = z + N ta c th suy ra N = A - z. Vy:

1) T s' ca ch (l?7 Pb) l:zt

N _ 207-82 _ , co

2 ) Nguyn t X c tng s' cc ht l 58 nn chc chn z nm trong gii hn 2 < z < 82.

Vy ta p dng t l phn 1 : 1 < < 1,524z z

Mt khc, ta li bit S = p + e + n = 2p + n=>n = S - 2 p

-HA - S - 2P s o ss r = -----suyra - 16,459 < p < 19,333 v 2p + n = 583,524 3

Da vo hai phng trnh ny ta lp bng bin lun:

p 17 18 19

n 24 2 2 2 0

A 41 40 39

Kt lun Loi Loi ng

1.6. Trong mt th nghim quang hp, kh oxi sinh ra, c

thu qua nc. Th tch kh thu c iu kin 22c v di p sut kh quyn 758mmHg l 186ml. Tnh khi lng oxi bit

rng p sut hi nc 22c l 19,8mmHg.



BI GII

Trc ht hy tnh p sut ring ca oxi. V p sut chung

bng tng p sut ring ca tng cht, nn

p0a =PT- P H , 0 = 758 - 19,8 = 738,2 mmHg = 0,971 atm

Khl ng oxi c tnh t phng trnh trng thi ca

kh l tng:

PV = RT M

___PVM _ 0,971.0,186.32m = = - = 0,239g.

RT 0,082.(273 + 22)

1.7. 7 kg oxi c cha trong mt bnh cu di p sut 35 atm. Sau mt thi gian s dng;, p sut o c l 12atm. Hi c bao nhiu kilgam oxi thot ra.

BI GII

H qu ca nh lut Boyle-Mariotte cho ta mi quan h gia t trng ca kh v p sut:

dL=pL d2 p2

rii.v P d2.v p3

y V l th tch ca bnh cu.

V d. V = m ; m l khi ng kh; nn

M = j j

m 2 P2

Thay mx = 7kg; P = 35atm; p.; = 1 2 atm vo phng trnh trn ta c

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_ _ IX11P2 _7.12_ 0 1112 = p T = 1 5 24k

Vy lng oxi thot ra trong qu trnh s dng bng: 7 - 2,4 = 4,6 kg.

1.8. Mt bnh dung teh 247,2 cm3, c khi lng 25,201 g cha khng kh. Mt lng benzen c a vo bnh ri un nng ti 100C. Benzen bay hi ko theo ton b khng kh ra khi bnh. Ngi ta bnh ngui tr' li nhit phng, trong trng thi m ri cn. Khi lng lc ny l 25,817g. p sut kh-quyn l 742mmHg. Tnh khi lng mol ca benzen v vit cng thc phn t benzen bit rng cht ny ch gm hai nguyn t' cacbon v hiro.

BI GII

T cc gi tr ca p, V v T c th tm c s mol be zen v t khi lng ca bnh trc v sau khi cha hi benzen c th tm c khi lng m ca benzen.

V = 247,2cm3 hay 0,2472/

T = 273 + 100 = 373K

p = 742mmHg hay 742/760 .= 0,976atm

S" moi n ca ben2 en:

n = PV/RT = 0,976.0,2472/0,082.373 = 7,88.10'3mol

Khi lng be zen = (khl lng bnh + khi lng khng kh + khi lng hi ngng t) - (khi lng bnh + khi lng

khng kh)

- =25',817 - 25,201 = 0,616g

Khi lng mol M = = 0,616/7,88.10'3 = 78,2 g/mol1

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Cng thc ca bezen (CH)X : X = xt 612

Vy benzen c cng thc C6H6

c-BI TP T GII

1.9. 1) Tnh khi lng mol nguyn t ca Mg; p nu bit khi lng tuyt i (KLT) ca chng l: mMt = 40,358.x.0"27kg; mp = 51,417.10'-7kg.

2) Xc nh-khi lng tuyt i ca N v AI nu bit khi lng tng i (klt) ca chng l: MN = 14,007u; MA1 = 26,982u.

p s: 1 ) MMg = 24,307 g/mol.

Mp = 30,986 g/mol.

2) mN = 23,255.10"24g.

~~~~~ mA1 = 44,798.10'24g.

1.10. Nguyn t bc (Ag) c khi lng moi nguyn t v khi lng ring trung bnh ln lt bng 107,87 g/mol v10,5 g/cm:i. Bit nguyn t ny ch chim 74% th tch ca tinh th. Hy xc nh bn knh nguyn t ca bc (Ag) theo A.

p s: rAg = 1,444A.

1.11. i vi nguyn t km (Zn) ngi ta bit bn knh nguyn t v khi lng mol nguyn t ln lt c cc gi tr l 1,38A; 65g/mol.

a) Xc nh khi ng ring trung bnh ca Zn (g/cm8)

b) Bit Zn khng phi l khi c m c khong rng nn trong thc t n ch chim 72,5% th tch ca tinh th. Hy cho bit khi lng ring thc ca Zn l bao nhiu?

p S: a) 9,81 g/cm3; b) 7,11 g/cm3

22



1.12. Kim loi M tc dng va vi 4,032 kh CL iu kin tiu chun thu c 16,02 g MC13 theo phng trnh:

2M + 3C12 = 2 MC3

a) Xc nh khi lng nguyn t ca kim loi M.

b) Tnh khi lng ring ca M; suy ra t l phn trm ca th tch thc vi th tch ca tinh th. Bit M c bn knh r = 1,43A; khi lng ring thc l: 2,7 g/cm3

- p s: a) M = 27b) d = 3,66 g/cm3; %: 73%

1.13. Mt cch gn ng gia bn knh ht nhn rn v s" khi A ca mt nguyn t c h thc: r n = 1,8.10"1;\ A!/:icm. Hy xc nh khi lng ring d(g/cm3) ca ht nhn nguyn t.

p s: = 6,80.1013 g/cm3

1.14. Da vo nh ngha hy xc nh khi lng nguyn t ra kg cho mt n V khi lng nguyn t (lu). T kt qu tnh c, hy suy ra khi ng nguyn t tuyt i ca oxi, bit oxi c khi lng nguyn t l 15,9974 u.

p s': lu = l , 6 6 .1 0 27kg; moxi = 26,567.10"24g1.15. Mt nguyn t X c tng s" cc loi ht l 193,

trong s" proton l 56.

a) Hy xc nh s" khi ca X

b) Tnh khi lng nguyn t v khi lng ht nhn ca nguyn t X va tm c. Cho bit t s" khi lng ny t nu nhn xt cn thit. Cc gi tr khi lng ca p, n, e xem bng ph lc (cui sch).

p s": a) Ax = 137 b) m = 229,3579.0 '2^ kg mh/nhn = 229,3070.10'"7kg

_E n/t_= 1 , 0 0 0 2 2^h /nhn

Khi lng nguyn t hu nh tp trung ht nhn

23



1.16. Hon thnh s liu ghi trong bng di y

Nguyn t A z N N/ZCa 40 2 0I 127 74

TI 204 1,52Pb 82 125Np 237 73Pb - 208 1,53

1.17. Mt qu bng c n hi cao, c th tch ban u1 ,2 lt 1 atm v 300K. Qu bng ny bay ln tng bnh lu c nhit v p sut tng ng bng 250K v 3.1CT3 atm. Tnh th tch ca qu bng trn tng bnh lu. Chp nhn kh l l tng.

p s": 3,3.10^ lt.1.18. Kh than t (CO + H;) c to ra khi t c vi hi

nc theo phng trnh phn ng: c + HoO = c o + H2. Khi t chy mt tn than cc trng hi nc c 1 0 0 0 c th to ra c mt th tch kh than t l bao nhu? Ti 2-0C di p sut lOOatm.

p s": 4,02.10104 lt1.19. Mt bnh dung tch 2 cha 3g C 02 v 0 ,1 0 g H2

17c. Tnh p sut ring ca tng kh v p sut ton phn cc kh tc dng vo thnh bnh (gi thit kh l l tng).

p s": PC0 2 = 0,812 atm; PH = 0,3; p = 1 ,1 1 -atm

1.20. i vi lmol kh N9 0c s ph thuc ca th tch vo p sut c cho di y:______________________

p/atm 1 3 5 v/cm a 22405 7461,4 4473,1

Xc nh hng s' kh R bng:a) Tnh tonb) th (V th PV/nT ph thuc v p ri ngoi suy

ti p = 0 )L21. Mt bong bng kh bn knh l,5cm y h c nhit

8,4c v c p sut 2 ,8 atm, ni ln mt nc p sut kh quyn latm, nhit 25c. Hi kh ti b mt ca h nc th bn knh bong bng l bao nhiu (th tch hnh cu bn knh R l 4/3ttR3)

p s": 2,2 cm.

24



Chong II

NGUYN L I CA NHIT NG Lc HC. NHIT HO HC

A- TM TT L THUYT

1, Ni dung ca nguyn l I

Mt h nhit ng khi trao i nng lng vi mi trng

xung quanh di dng nhit Q v cng A th tng i s" Q + A

lun lun l mt hng - S ch ph thuc vo trng thi u vcui ca h, hon ton khng ph thuc vo ng i:

\ 2 ' -(Trng thi u) 1 (Trng thi cui)

Q(l) + Al) = Q

Trong 3 i lng nhit ng , A, Q ch c i lng u l hm trng thi; 2 i lng cn li l nhng i ng c trng

cho qu trnh, tc l chng ph thuc vo ng i.

Quy c v du ca nhit v cng:

- Nhit Q v cng A c tnh l dng (Q > 0; A > 0) nu

h nhn nhit v nhn cng t bn'ngoi.

- Nhit Q v cng A c tnh l m (Q < 0 ; A < 0 ) nu h

nhng nhit v sinh cng cho bn ngoi.

T biu thc (1) ta thy rng i vi h c lp (khng trao

i g vi bn ngoi) Q = 0, A = 0, do AU = 0. Vy trong h c

lp ni nng c bo ton.

2. Cc biu thc v cng v nh >ng mt s qu trnh

- Cng do h thc hin cho b ngoi c xc nh bng phng trnh:

Vi Pc l p sut ngoi, 'dv l bin thin th tch. i vi

nhng bin i v cng chm, c th xem Pe = p vi p l p sut

ca h do cng dn n th tch s l:

a) Cng

SA = -pe. V

1

- Qu trnh ng tch dV = 0. Suy ra; Ay = 0

- Qu trnh ng p p = const

a ; = -P(V, - v o = -P.AV

26



I vi h ng th ca kh l tng, s bin thin th tch

2 trng thi 1 v 2 l do s bin thin s" mol 2 trng thi ,

nn

Ap = -AnRT (3)

Qu trnh ng nhit (T =const) i vi 1 mol kh l tng:

'> _2rdV v.>At = - P.dV = ~RT [ = -RTln

J J V v 71 1 1

T = const th tch ca kh l tng t l vi p sut, do :

At = - R T ln - = -R T ln-!- (4)^ V, p,

b) Nhit v nhit dung

Nhit dung c c xc nh nh sau:5Q = Q dt dT

QTrong iu kin ang tch ta c: Cy =

T

v trong iu kin dng p ta c: c =SQp

p CT

Trorig h SI, n v nhit ung l Jun/.

Vn dng nguyn l I cho kh l tng ta c:

d = Q + 5A = Q - P.dV.

Vi iu kin ng tch dV =. 0; d = Qv - Cv.T.

l vi 1 mol kh l tng:.. 67

A \

i V 27



Qv = AU = Jc vT = CV2-T ,)1

nu Cv khng i trong khng nhit t n T2.

Vi iu kin ng p:

d + PV = Q hay d(U + PV) = 5QP;

+ PV = H. H l Entanpi;

Vy: dH = 5QP = C.dT hay AH = Qp = jc pdT1

i vi mt mol kh l tng v Cp khng ph thuc nhit

th: AH = Qp = Cp (T2 - Ti).

Vi iu kin ng nhit T = const.

Ni nng u v entanpi H ca kh tng ch ph thuc nhit , khng ph thuc vo th tch cng nh p sut, do :

a u t a h t 0 .

- Quan h gia Qt) v Qv ca phn ng ho hc din ra trong pha kh:

Qp = Qv + AnRT (5)

Vi An l hiu gia s" mol kh v phi phng trnh phn

ng v s mol kh v tri ca phng trnh phn ng.

3. Nhit ca phn ng ha hc

T nh lut Hess v nhit ca phn ng ho hc c th

rt ra mt h qu ca s tnh nhit ca phn ng da vo nhit

hnh thnh v nhit t chy nh sau:

>

28



IAHPU = E(AHht) cui - I(AHkt) u (6 )

AHp = Z(AHe) u - L(AHdc) cui (7)

Nhit ca phn ng xc nh theo (6 ) v (7) l nhit nhit

khng i.

Khi nhit thay i th nhit ca phn ng cng thay oi theo. S ph thuc ny c biu th bng nh lut Kirchhoff

nh sau:

( ^ ) = A C P= 2 Cpcul- C pcr&

Sau khi ly tch phn ta c:

T

AHT -AHT] = jACpdT=ACp(T2 ) (8 )

T,

Vi ACp l hng s" trong khong T h> T2 .

T nhit ca phn ng ha hc c th tnh c nng lng

lin kt ca cc cht c mt trong phn ng.

AH = z nng lng lin kt cc cht u - nng lnglin kt cc sn phm (9).

Nng lng in kt y c nh ngha l nng lng

cn thit ph v lin kt ha hc to ra cc nguyn t t o

th kh.

B- BI TP C LI GII

y II .l . Gin n ng nhit 0,850 moi kh l tng t p sut

15 atm v nhit 300K ti p sut latm. Tnh cng gin n:

29



a) Trong chn khng;

b) Khi p sut ngoi khng i l 1 atm;

c) Khi qu trnh l thun nghch.

BI GII

a) V l chn khng nn P = 0 do A = -PC.AV = 0.

b) A = -P p(V2 -V 1).

-

2 ~~ p 7 _ ( 1 o2 _>. v9 - v x = nRT ------

V = 5 1 l p 2 p j ?

1 p, , ^ '

A = -n R T E ,(^ --^ -)2 1 , -

. 1 i r - - k - 0 , 8 5 0 ( ^ 3 0 0 . ! ^ - ^ ) ~ /Q g o

. ^

A = -19,5 a tm i hay A = -19,5. 101,34J = -1980J.

c) A = -PdV = dV = riRTdlnVy

A = -nRTn ^ = - is p T ln ^

= -0,850,8,314.300-ln 1

Vy A = -5740J.

IL2. Tnh Q, A, A trong qu trnh nn ng nhit, thun

nghch 3 mol kh He t 1 atm n atm 400K.

30



BI GII

Mt cch gn ng c th xem He l kh l tng, do

AUt = 0. T nguyn l I suy ra

5- Q = -A = nRTln = 3.8,314.400.1n - = 1,61.104J

P1 1

Vy: A = 1,61.104J; Q = -1,61.104J :

S nn ng nhit 3 mol He l mt qu trnh ta nhit.

II.3. So snh s khc nhau gia AH v A i vi cc bin

i vt l sau y:

a) 1 mol nc 1 mol ncc 273K v latm.

b) 1 mol nc -> 1 mol hi nc 373K v latm. Cho\bit 273K, th tch mol ca nc v nc lng bng 0,0196 \l/mol v 0,0180 1/mol v 373K th tch mol ca nc lng v

hi nc tng ng bng 0,0188 1/mol v 30,611/mol.

BI GII

Trong c hai trng hp qu trnh l ng p nn:

AH = AU + A(PV)

= A + P.AV AH - AU = P. AV

a) AV =V L-V R

= 0,0180-0,0196

= -0,16.10'2l/mol

AH - AU = p. AV = 1.(-0,16.102) = -Q,l6 .icr2 .atm

hoc: AH - A = - 0,16 J/mol

31



b) AV = VH - VL = 30,61 - 0,0188 = 30,59 1/mol

AH - A = P.AV = 1. 30,59 l.atm hoc 3100 J/mol

So snh (AH - A) (a) v (b) cho thy s khc bit gia

AH v A l qu nh, c th b qua i vi cc pha ngng kt,

song s khc bit gia hai i lng ny l rrg k nu l pha

kh. T (a) thy rng AU > AH do s gim th tch khi nc

nng chy, kt qu ca vic h nhn cng t bn ngoi.

X II.4. t chy mt mol Benzen lng 25JC, latm to ra

kh C 0 2 v nc (H20) (Z), ta ra mt nhit lng bng 3267kJ.

Xc nh nhit hnh thnh ca Benzen lng iu kin cho v

nhit v p sut, bit rng nhit hnh thnh chun ca co*,

H20 (Z) tng ng bng - 393,5 v -235,8 kJ/mol.

BI GII

S t chy mt mol benzen theo phng trnh phn ng

C6H6 + 1 - 0 2(k) = 6 CO(k) + 3H20 (1).

Gii phng ra mt nhit lng A Hp = -32,67kJ/mol\

p dng phng trnh (6 ) ta c:

AHp = 6AHtC0% + AH^tH2Q - AH^tCgH6 - 7 AH^t0i? r 3267 =

= ( - 6 X 3 9 3 , 5 ) + ( - 3 X 2 8 5 , 8 ) - AH(hlC;H( - 0

Vy AHjJtc H = 49kJ / mol

II.5. Trn 50ml dung dch HC1 0,20M vi dung ch NaOH

0,20M trong mt nhit lng k, nhit tng t 22,2c ln

23,5c.

32



Xc nh nhit trung ha (tnh ra kJ/mo) theo phn ng:

H30 + + OH' = 2H20.

Cho bit t trng ca hn hp dung dch long l Ig/m v

nhit dung ring ca nc l 4,18J/g.K

BI GII

Th tch ca hn hp khi pha trn bng 50 + 50 = 100ml, do c khi lng bng l.OOg. Bin thin nhit gn lin vi

phn ng trung ha l: 23,5 - 22,2 = l,3c = 1,3K.

Lng nhit Q = m.e.At = 100 X 4,18 X 1,3 = 540J-

S' mol HC c trong 50ml dung dch 0,20M.

^00 ,2 0 .-- = 0 , 0 1 moi.

1000

Tng t s moi NaOH bng 0,01 moi.

Vy 0.01 moi H;30 + .phn ng vi 0,01 moi 0H gii phng

ra 540J. Nhit trung ha ng vi 1 moi s l: 54000J hay

54kJ/mol

Vy vi phn ng: HC + NaOH = NaCl + KgO. AH = -

54 KJ/moI-

I.6 . I vi phn ng: N, + 0 9 = NO 25c v latm.

AH = 90,37 kJ. Xc nh nhit ca phn ng 558K, bit rng

nhit dung ng p vi lmol ca N2, 0 2, NO ln lt bng

29,12; 29,36 v 29,86 J/K.mol.

BI GII

p dng nh lut Kirchhoff ta c:

33



t 2

AH = AH^ + jACpdT

558

A H 558 = A H 298 + j A C p d T

298

V Cp ca cc cht khng ph thuc T nn:

AH58 = AH98 + ACp(558 - 298)

= 90,37 + (29,86 - . 29,12 - 29,36).icr3.(558 - 298)

AH? 58 = 90,53kJ

Ch :

- Trng hp Cp l mt hm ca T v ni chung s ph

thuc c dng Cp = a + bT + CT2 +...hoc Cp = a + bT + CT 2 thi

khi tnh, ACp khng c a ra ngoi du tch phn v c dng,

chng hn ACp = Aa + AbT + ACT2 + ...

II.7. Xc nh nng lng lin kt trung bnh ca mt lin

kt C-H trong metan bit nhit hnh thnh chun AHjCH = -

74,8kJ/mol; nhit thng hoa ca than ch bng 716,7 kJ/mo v

nng lng phn ly phn t H2 bng 436 kJ/mol.

BI GII

Nng lng lin kt trng bnh ca mt lin kt C-H trong phn t CH4 bng 1/4 nng lng. Theo nh ngha, nng lng

lin kt trong CH4 l AH(298 ca qu trinh:

CH4(k) C(k) + 4H(k).

34



Vn dng nh lut Hess, AHoy* ca qu trnh phn ly

phn t thnh cc nguyn t t do th kh c xc nh nh chu trnh sau:

Vi l nhit thng hoa ca C; AHpL l nang lng

phn ly phn t.

Vy:

A^ 2 9 8 ~ _ A ^ht.CH 4 + +2AHpL

= -(-74,8) + 716 + 2x436

~ 1663,5 kJ/mol

Phn t CH4 c 4 Hn kt C-H; do nng lng 1 lin kt

C-H bng = 4i6kJ/mol

Bi ton trn cp ti vic tnh nng lng in kt khi bit nhit hnh thnh. Ngc li, bit nng lng lin kt c th tnh c nhit ca phn ng, chng hn i vi phn ng:

AH&U = z nng lng lin k a C9 v HI - nng

lng lin kt ca I2 v HC1.

c r ' ' C(k)- + 4H(k)

Cgr + 2H2(k)

4

Ch :

Cl2 + 2 HI (k) = I2(k) + 2HC1 (k)

35



Nng lng lin kt ca Cl2> HI, I2 v HC1 tng ng bng

239, 297, 149 v 431 kJ/mo.

AHp = 239 + 2.297 - 149 - 2.431 = -178kJ

L 8 ^)l mol nc nng chy 0 c, 1 atm, hp th mt

nhit ng bng 6019,2J. Th tch mol ca nc v ca nc

lng bng 0,0196 v 0,0180 lt. Tnh AH v A i vi qu trnh

ny.

BI GII

V Qp = a h nn AH = 6019,2J

tnh A ta vn dng cng thc AH = A + A(P.V).

;i vi qu trnh ng p

A(P.V) = P.AV = P(V2 -Vj) =

=1(0,0180 - 0,0196) =

= -1,6.10' 3 atm.l = -0,1630J

A = AH - P.AV = 6019,2 - (-1,63.10-2) = 6019.2J

II.9. 25c v latm s hnh thnh 1 mol c o t graphit v oxi c AH = - 1I0,418J: Xc nh AU nu lmol graphit c th tch

bng 0,0053 lt.

BI GII

T phn ng hnh thnh CO:

Cgr+I0* 00

36



1S bin thin s moi kh l An = - = 2 2

Mt khc bin thin th tch ca h (AV) do s hnh thnh

kh l AV = 1/2. 24,4/ = 12,2Z, n hn rt nhiu so vi s gim

th tkh ca grapht do c th b qua th tch graphit.

. AH = A + AnRT

-110,418 = A + .8,314.298 2

AU = -1349, 2J

11.10. Tnh nng lng Hn kt O-H trong phn t nc

bit cc d kin:

K20 (Z) = H20 (k) AH = 40,6 kJ/mol (1)

2H (k) = H2 (k) AH = -435 kJ/mol (2)

0 2 (k) = 20 (k) AH =-489,6 kJ/mol (3)

2H2 (k) + 0 2 (k) = 2H2 (G) ) AH = -571,6 kJ/mol (4)

/B i g i i s\

Nng lng lin kt l Ting'lng trung bnh cn ph

v mt lin kt xc nh trong phn t v to ra cc nguyn t

hay cc gc. Phn t H2 c 2 lin kt o - H. Nng lng trang

bnh ca mt lin kt 0 - H s bng 1/2 hiu ng nhit ca phn

ng H20(k) = 2 H(k) + 0(k).

tnh hiu ng nhit ny ta vn dng nh lut Hess.

Ly phng trnh (4) nhn vi 1 /2 ri cng vo cc phng

trnh (1) v (2 ) ta c:

37



(-571,6. - ) + 40,6 + (-435) = -680,2 kJ 2

Ly phng trnh (3) nhn vi 1 /2 ri tr i tng'trn (-680,2):

(489,6. - ) - (-680,2) = 925 kJ 2

Vy hiu ng nhit ca phn ng-

H20 (k) = 2H (k) + o (k)

l 925 kJ suy ra nng lng to ra 2 lin kt 0 - H l -925 KJ,v

nng lng trung bnh ca 1 lin kt o - H l -925/2 = 462,5

kJ/mol

Ch : cc tnh ton trn u da vo s chuyn ha

c thit lp trn c s nh lut Hess:

H2(k) + | o 2(k) A l4) > H20 (l)

AH(1). 2

AH(3)

y

-AH(:

2H(k) + 0(k) H, ->H20 (k)

1 1 .1 1 . Tnh Hft|3 i vi phn ng c o + (>2 = CO bit

298K nhit hnh thnh chun ca c o v C02 l -110,5 v -

393,5 kJ/mol ;

Cp (C) = 26,53 + 7,7.10":iT J/K.moi

Cp (C02) = 26,78 + 42,26.10':iT J/K.mol

c p (02) = 25,52 + 13,60.10'3T J/K.mol

38



BI GII

AH9 8 (P.U) = AH298 ht(C02) - AH298.ht(CO)

= -393,5 - (-110,5) = -283 kj

Da vo nh lut Kirchhoff tnh AH 73 (P.U):

473

AH?73 = AH98+ ACpdT298

vi

ACp = Cp(C02) - [CCO) + c p 0 2] = -12,51 + 27,76.10'3T

473

Cui cng AH73 = -283000 + [(-12,51 + 27,76. lO3 T)dT298

= -283000 - 12,51(473 - 298) + 27,7610 3 (4732 - 2982)

- 283000 - 2190 + 1870

= -283320 J/mol

AH7 3 (P.) = -283,320 kJ/mol.

C- BI TP T GII

11.12. Xc nh cng dng nng mt vt khi lng 30

kg ln mt cao 2 m.p s: 588J.

11.13. Tnh cng thc hin bi phn ng gia km v axit sunfuric long khi thu c mt mol kh hydro iu kin 0 c v 1 atm.

p s: -2.27 kJ

39 .



11.14. Tnh AU v AH trong qu trnh un nng 55,4g Xe t 300 n 400K, bit rng i vi 1 mol kh Xe Cv= l2.47J/K.mol

p s: AU = 526J; AH = 877J

II. 15. Nhit dung ng p ca lmol ng dc cho bi

phng trnh Cp = 22,65 + 6,3.10: TJ/K.

Tnh AH khi t nng lmol ng t 300 n 400K

p s: AH == 24S5J.

11.16. Tnh nhit ca phn ng quy v kg nhm i vi

phn ng: 2AI + FesOfl = 2Fe + AI9O3.

Cho bit AI = 27; AHj[tA o = -1667-,82 kJ/mo v

^ h tP e o = -819,28 kJ/mol.

p s: 15712,6 kJ

11.17. Tnh nhit hnh thnh ca tan bit:

Cp. + 0 2 = CO2, AH = -393,5 kJ

H2 + 0 , = H20 (0 AH = -285,8kJ.2r

2C2H6 + 7 0 , = 4C 0, + 6HS0 ); AH = -3119,6kJ

p s': -84,6 kJ

11.18. Chic bt la gas cha butan lng (AHJtbutan =

127kJ/mol). Xc nh lng nhit ta ra khi Ig btan lng trong

bt la b t chy; gi thit rng sn phm ca s t chy l

C02 v hdi H20.

p s: -4.7kJ

40



% IL19. Nhit hnh thnh trong dung dch nc 2C ca

HFaq; OHaq; Faq, ln lt bng -320,lkJ/mol; -229,94 kJ/mol v -

329,1 lkJ/mol . Nhit hnh thnh 25c ca H20 lng bng - 285,84kJ/mol.

a) Tnh nhit trung ha ca HFaq theo phn ng:

H V + D H ^ - J - j+ H jO (1 )

b) Tnh nhit in ly ca HF trong dung dch

HFac=H:q +OH-q

Bit nhit trung ha ng vi phng trnh:

K q + 0H q = H2G 0) '55,83 kJ/mol.

p sp': a)-64,91; b) -9,08.

11.20. Tnh aH ^ 98 i vi cc phn ng sau:

a) 2H2S (k) + 302 = 2HaO () + 2S02.

. b) 2H2S (k) + 302 = 2H20 (k) + 2S02.

c) 2HNS (k) + 2 NO = H20 2 (l) + 4N2.

Bit nhit hnh thnh ca H2S, HoO(0, H20(k), S02. KNo,

NO v H5O2 ln lt bng -20,63; -285,83; -241,81; -298,83;

294,1; 90,25; -187,78 kJ/mol.

Cng i vi cc phn ng trn, tnh AH^, K bit:

Cht H2S . 0 2 H20 () H20 (k)1

S02

Cp [J/molK] 34,23 29,35 75,29 33,57 39.87

41



Cht h n 3 NO h 20 2 n 2

rp[J/mol.K] 4 3 , 6 8 29,84 89,1 29,12

p s: -1124,03; -1036; -956,5 kJ/mol; -1118,7; -1036,7.

IL21. vi phn ng: 2 C9 + Oo = 2C09; Nhit ung

ng p ca cc chttrong khong t 298 n 2000K c dng

chung Cp = a + bT + cT'2. Cc h s a,b,c ca cc cht c cho

di y:

Cht

0 2

CO

co*

a[cal/.molJ

7,16

6,79

10,55

b.1 0 3

1

0,89

2,61

-5C . 1 0

-0,40

-0,11

-2,04

Tnh AH^ooo

Bit AH,98 ca phn ng'bng -565,96kJ

p s: -564,41kJ

^ 11.22. Tnh nng lng lin kt trong phn t PC13, t

xc nh nng lng lin kt trng bnh ca mt lin kt p - C.

Cho bit:

- Nng lng lin kt c Cl = 239 kJ/mol.

- Nng lng thng hoa ca p = 316,2 kJ/mol

- Nhit hnh thnh ca PCI3 (k) = -287 kJ/mol

p s: 961,7 kJ/mol; 320,56.kJ/mol

42



11-23. Xc nh nhit t chy chun AHgg ca metan:

CH4 + 20 2 = C02 + 2H20 (k).

Bit nng lng lin kt trung bnh ca:

c - H bng 414 kJ/mol

0 = 0 498,8

c = o 724

0 - H 460

p s: -034,4 kJ

11.24. Mt kh l tng no c nhit dung mol ng tch

mi nhit Cv = 2,5R (R l hng s" kh).

Tnh Q; A; AU v AH khi lmol kh ny thc hin cc qu

trnh sau y:

a) Gin n thun nghch ng p t (latm; 20dm3) n

(1 atm; 40dm3).

b) Bin i thun nghch ng tch t trng thi (latm;

40m3) n (0,5atm; 40dm:).

c) Nn thun nghch ng nhit t (0,5 atm; 40n3) n

(latm; 2 0 dm3).

Hy phc ho mi qu trnh trn cng mt gin P-V, ri

tnh Q; A; A v H cho chu trnh ny.

' 11.25. Xc nh nhit hnh thnh Imol AlG da vo cc

phng-trnh nhit ha hc di y:

p s: a) Q = 7,09 kJ; A = 5,06kJ

b) A = 0; Q = -5,07 kJ; AU = -5,07; AH = 7,S.

43



Al20 3(r)+3 C0 Cl2(k)=3 C0 2(k)+2 AlC1.3(r) A! =-232,24kJ

CO(k) + Cl2(k) = COCl2(k) AH2 =-112,40kJ

2Al(r) + - 0 2(k) = Al20 3(r) AH3=-1668,20kJ2

bit rng nhit hnh thnh ca c o v co? tng ng bng

-110,40 v -393,13 kJ/mol.

p s: -694,71 kJ

11.26. Tnh lng nhit ta ra 25c trong s hnh thnh

32g Fe20 3 t cc nguyn t" iu kin ng tch, bit rng trong

s hnh thnh FeO AH = -268,77 kJ v s oxi ha FeO thnh

Fe20 3 ta ra 2027,30J, vi lg FeO iu kin ng p;

nbng nhit lng ny u c xc nh 25c.

p s: 165,15 kJ

44



Giing III

N6UYN L II CA NHIT NG LC HC

A- TM TT L THUYT

1. Entrpi v s tnh Enprpi trong mt s q u trnh

Nguyn l II thit lp c rng i vi qu trnh thun

nghch nhit ng t s" gia lng nhit v cng nh v nhit

tuyt i l mt vi phn ton phn ng ca mt hm s no ,

hm Entropi S:

V Entrpi l mt hm trng thi nn i vi qu trnh

thun nghch i t trng thi 1 n trng thi 2 , bin thin

Entrpi AS s bng:

(1)

2SQ

(2)

Nu qu trnh thun nghch l ng nhit (2) tr thnh:



Nu qu trnh thun nghch l on nhit (Q = .0) th dS = 0;

AS = 0. vi qu trnh khng thun nghch nhit ng th:

> (-^ t ) k t n _

Nu qu trnh khng thun nghch l ng nhit th:

> k t n

Nu qu trnh khng thun nghch l on nhit Q = 0 th

. dS > 0, AS > 0 (6 )

Mt cch tng qut biu thc ton ca nguyn l II l:

dS > ^ (7)T

Khi h nhit ng c gp vi mi trng xung quanh m thnh mt h c lp th:

c lp (A ^ h nhit ng ^ m i trng ) ^ ( 8 )

Nu trong h c lp ch din ra qu trnh thun nghch th:

SHcip = 0 s = const

Nu trong h c lp ch din ra trong qu trnh khng

thun nghch th ASHccip > 0 (S > St).

V ngha vt l, entrpi s c trng cho tnh hn lon ca h nhit ng.

AS trong mt s' qu trnh:

p = const

T

ASp = J^p

Vi Cp l hng s:

ASp = CoinT,

V = const:

ASV = Cvln Tj

(Cv l hng s' trong khong Tj, T2).

AS ca kh l tng:

i vi 1 mol kh l tng

AS = R ln ^ - + Cvln-Jjf (Cv = const)V1 I

hocp _ T

AS = -R In + Cp 111 (Cp = const) (12)

2. Th nhit dng G v F

T nguyn l II vi s xut hin hm entrpi s, ta bit rng trong h c lp ch nhng qu trnh no lm tng entrpi

(AS > 0) mi c th t xy ra.

vi nhng qu trnh xy ra khng trong iu kin c

lp, th c nhng hm s" khc vi nhng bin s" tng ng;

l nhng hm s" G v F vi cc bin s tng' ng l p, T v V, T. Cc hm s" G v F ny l cc th nhit ng v c nh ngha nh sau:

G = H - TS, 'F = u - TS.

Th nhit ng G cn c gi l nng lng t o Gibbs

(Gip-X): F cn c gi l nng lng t do Helmholtz (Hem Hon X). G v F u c gi l nng lng t do v l phn nng lng t o chuyn thnh cng.

(10)

(11)

(9)

47



T cc biu thc nh ngha i vi G v F ta c:

i vi G T, p = const; G = H - TS

hay vi qu trnh hu hn

AG = AH - TAS (13)

i vi F T, V = const; dF = d - TdS

hay i vi qu trnh hu hn

AF = A - TAS (14)

Cc hm G v F c dng lm tiu chun nh gi chiu hng ca qu trnh. Thc vy nu qu trnh t xy ra

T.P=const thi phi km theo s gim ca G tc l aG =G2-G1< 0.

Cn T,v = const qu trnh din bin theo chiu gim ca F tc l AF = F - Fi < 0.

vi phn ng ha hc, kh nng tham gia vo phn ng ca cc cht c c trng bng mt khi nim i lc ha

hc. o i c ha hc AG hoc AF tu theo iu kin din bin ca phn ng ng nhit, ng p hay ng nhit ng tch.

i vi phn ng ha hc din ra iu kin chun th:

y, AG^t . bin thin nng lng t do chun ca s

hnh thnh hp cht t cc n cht. i vi n cht AGJ{, =0 .

B - BI TP C LI GII

IIL l. Tnh AS trong qu trnh gin n ng nhit 2mol kh

l tng t 1.51 n 2,4L

AG=XA9ht.snphm (15)

48



BI GII

p dng phng trnh (1 1 ) trong iu kin T = const, i

vi 2 mol kh:

AS =.2. 8,314 X in ^ 4 = 7,8 J/K 1 ,0

IIL2. Tnh AS - trong qu trnh un nng 200 gam nc

t 10c n 20c p = const, bit Cp ca nc bng 75,3J/K.mo.

BI G

Ap dng cng thc

AS = nCpln -

1Ta c

AS = . 75,31n = 29J/K 18 283

( n i Tnh AS ca qu trnh khuch tn vo nhau ca

lmol kh N2 v lmol kh 0 2. trng thi nguyn cht mi cht

kh cng mt iu kin v nhit , p sut v th tch.

BI GII

S khuch tn ca hai kh lm tng gp 2 ln th tch

T = const do :

AS0 2 = R ln ^ - = Kln2 = 2,303Rlg2

= #2,303Rlg2

AS = AS0 + ASNs^2,303Rlg2 = 11,5J/K

49



IXL4. Tnh AS ca phn ng 4Fe + 30 = 2Fe90,3- Bit s ca Fe, 0 2 v Fe20 3 tng ng bng 27,3; 205 v 87,4 J/Kmol.

BI GII

T phng trnh (12) Ta c:

ASp = 2Spe2Q3 ~4Spe - 3 Sq 2

= (2. 87,4) - (4. 27,3) - (3.205) = -549,4 J/K.

II.5. Hy tin on du ca AS trong phn ng sau:

a) CaCOs = CaO + C02.

b) NH3 + H C l(k)-N H 4Cl(r)

c) BaO(r) + C02(k) = BaC03(r)

BIGI

a) AS > 0; b) AS < 0; c) AS < 0.

III.6 . Tnh AG9S trong s hnh thnh lmol nc lng bit

cc gi tr Entrpi chun ca H2 v 0 2 v H9O ln lt bng:

130,684; 205,133; v 69,91 J/K,moi v AH ca s hnh thnh

nc lng bng -285,83kJ/mo.

BI GII

AG = AH - TAS.a q 0 q O q O 1 gOAfc> ~ tH20 " aH2 2 2

= 69,901 - 130,684 - . = -163,34J/Kmo.2

AG!w = -285830 - 298.(-163,34) = -237,154kJ

50



1I7. Tnh AS? 98 , AHS98v AG98 i vi phn ng

phn huy nhit CaCO.3, bit:

CaCOs CaO C 02

S?9S /J . K_1.m o rl ... +92,9 . . .38,1 213,7

A H ^./kJ.m or1... - 1206,90 -635,10 -393,50

BI GII

CaCOs = CaO + C 02.

O qO , oO _ qO C aO + C02 _ CaC03

= 38,1 + 213,7 - 92,9 = 158,9 J/K

AH = AHtc. 0 + M t.C ^K ,C C 0 3

= -635,10 - 393,50 - (-1206,90) = 178,30 kJ

^ 2 9 8 = ^ 2 9 8 ~TAS29g

= 178,30 - (298.158,9.10:?) = 130,90kJ

G ? 98 > 0 chng t 25c v latm s nhit phn CaCO

khng xy ra c.

III.8 . Tnh AS trong s trn 10g nc 0c vi 50g nc lng 40c trong mt h c lp. Nhit nng chyr.ca nc

334,4 J/g; T nhit ca nc l 4,18 J/K.g.

BI GII

Gi t l nhit lc cn bng nhit sau khi trn, ta c:

(10.334,4) + (10.4,18xt) = 50.4,18(40-t)

51



T phng trnh ny xc nh c nhit cui cng sau

khi trn t = 20c

Gi ASj l bin thin entrpi khi chuyn lOg nc t

0c thnh nc lng 20c, ta c:

Gi bin thin entrpi khi chuyn 50g nc lng t 40c

xung 20c l AS->:

Bin thin entrpi trong s trn

AS = ASj + AS2 = 15,21 - 13,77 = 1,44 J/K

Vi AS trong h c lp bng 1,44J/K (AS > 0) chng t y

l mt qu trnh khng thun nghch nhit ng lc.

II.9. S g st din ra 25c. latm theo phng trnh

Vi nhit hinh thnh -824,2 kJ/mol.

Kt hp vi gi tr AS thu c bi tp (III.4) hy chng

t s g st l mt qu trnh t xy ra.

293

10 . 334,4 273

9QO+ (10-4,181x1 ) = 15,21 J/K.

9.73

29^= 50 X 418 ln = -13,77 J / K

' 313

phn ng

4Fe + 30 = 2 Fe93.

52



BI GII

S g ca st ta nng lng di, dng nhit ra mi

trng xung quanh mt lng bng -824,2.2 ='-1648,4 kJ do

lm tng entrpi ca mi trng mt lng bng

0 _ 1 6 4 8 4 0 0 = 5529 J/K.** 298

Mt khc t bi tp (.4) ta c AS = -549,4 J/K

Vy ASlp = ASft + As = 5529 - 549,4 = 4979,6 J/K

AS^ lp > 0 chng t s g l qu. trnh t ph hy ca kim

loi iu kin thng v nhit v p sut.

111.10. Ti nhit no s chuyn lmol nc lng thnh

hi nc p sut kh quyn latm l mt qu trnh t din bin,

bit nhit ha hi lmol nc lng bng 40587,80J v b ii thin

entrpi ca s chuyn trng thi ny bng 108,68 J/K.

BI GII

Tiu chun nh gi chiu hng t din bin ca cc qu

trnh xy ra iu kin ng nhit, ng p nng lng t do G.

G = H - TS

AG = AH - TAS i vi qu trnh ng nhit.

Khi bay hi ca nc p = latm:

H20(Z) = H20(h)

AH = 40587,80J

AS = 108,68J/K

53



Tac: AG = 40587,80 - T.108,68.

Hy tm nhit ! ti c cn bng lng - hi ca nc.

Mun vy hy cho AG = 0 do

40587,80 - 108,68T = 0 suy ra T = 373,46K

y l nhit si ca nc. iu kin nc lng

chuyn thnh hi ch xy ra theo mt chiu th AG < 0 . iu ny

- ^ V m 40587,80 0 ^0 ^s c thoa mn nu T > - > 373,46.108,68

IIL ll. i vi phn ng CO(k) +H20(k) = C02(k)+ H?(k)

Cho bit nhng gi tr ca bin thin entanpi v bin thin

entrpi chun 300K v 1200K nh sau:

AHgQQ = -41,16kJ /mol AH^qq = -32,93kJ / mol

a joq = -42,4W / mol Aj^QQ = -29,6kJ / mol .

Hi phn ng t din bin s theo chiu no 300K v

20K?

BI GIIA

Tnh A.G0 2 nhit da vo h thc AG = AH - TaS.

300K AG^ oo = -41160 - (300x -42,4) = -28440J.

1200K AGj200 = -32930 - (I200x - 29,6) = 2590J

Kt lun AGgoo < 0 vy phn ng cho t xy ra 300K

theo chiu t tri sang phi, song 1200K AJ9q0 > 0 phn ng

t din bin theo chiu ngc i.

54



c. BI TP T GII

I.12. Tnh AS ng vi s nng chy lmol nc ti 0c, bit nhit nng chy ca bng 6 kJ/mol.

p s: 22 J .k ']

111.13. Trn 35g nc 250C (A) vi 160g nc 86c (B).

a) Tnh nhit cui cng ca h vi,gi thit l s trn

c tin hnh mt cch on nhit.

b) Tnh bin thin Entrpi ca A, B v ton b h.

p s': a) 75,lc '

b) A: 22,7J/K; B: -20,6J/K; h: 2,lJ/K

III.14- Tnh AS trong qu trnh un nng 1 mol hidro t

300 n 400K.

Bit rng vi Imol hidro; Cj, = 1,554 + 2}2.10"3T.J/K.mol.

p s': 1.74J/K i t

111.15. Hy tin on du ca ASgg v AHyg vi cc

qu trnh sau:

a) (C2H5)20 (1) - (C2H5)20 (k)

b) Cl2(k) -> 2C1 (k)

c) C10H8(k) -> C10Hg(r)

d) t chy (COOH)s(r) thnh COo(k) v H20 (Z)

e) C2H4(k) + H2(k) C2H6(k)

55



III. 16. Tnh A; AH; AS vi qu trnh chuyn lmolH20

lng 25c v latm thnh lmol hi nc 1 0 0 c, latm , bit

Cp(H20 2) = 75,24J/K.ml v nhit ha hi i vi lmol nc bng

40629,6J/mol.

p SAH = 46272,6J/mol; AS = 112,95J/K AU = 4371,47.

111.17. Phn loi cc qu trnh cho di y thnh thun

nghch v khng thun nghch nhit ng:

a) S ng c ca nc 0 c v latm 'b) S ng c ca nc chm ng ti -10c v latm.

c) S t chy cacbon trong 0 2 c C02 ti 800K v latm.

d) S ln c ma st ca qu bng trn sn nh.

e) S gin n on nhit ca kh trong chn khng.

n i.1 8 . Tnh AS ca phn ng: N 2 + - H* = NH32 2

Bit s , = 191,489; = 130,586;

. Sj,H = 192,505J / K.mol

p S: l82,79J/k.mol

111.19. Mt s' vi khun trong t nhn c mt nng ^

lng cn cho s tng trng do s oxy ha nitrit thnh nitrat:

2NO aq + 0 9 = 2 NOg aq .

Bit rng s hnh thnh NO v NO3 , nng lng t o

chun tng ng bng -34,6 v -H0,5kJ/mol. Hy tnh nng

lng t do thot ra khi lmol NO b oxy ha thnh lmol NO3 .

p s: -75,9kJ

56



IIIL20. Tnh nng lng t do hnh thnh chun 4^298

ca lmol H20 lng: H + O2 = H20 () bit 6.9,91;2 fyo

=130,684 v Sq = 205,38J/K.mol. Nhit hnh thnh

chun 25c ca Imol H20() bng -285,830 kJ/mol.

p s: -237,129 kJ/mol.

II.21. 0,35 mol kh tng 15,6c c gin n ng

nhit thun nghich t 1,2 lt n 7,4 lt n^r 0 Vv , & L A-

_ . J L 2 Y ' . ' A - a v V 'Tnh A/Q, AU,'/v4>(i vi qu trnh ny. 7

^ p s: -1530J; 1530J; 0; 5, 3J; -1530J

I.22. Tnh AGg73 ca phn ng: '

CH4 + H20(k) = CO + 3H2

Bit nhit hnh thnh chun AK^oggca CH4, H20(k) v

CO ln lt bng -74,8; 241,8 v -liO,5KJ/mo.

Entrpi chun ca CH4, H9(k) v c o bng 186,2; 188,7 v

197,6J/K.mol (Trong tnh ton gir thit rng AH v AS khng

ph thuc T). y

a) T gi tr AG tm c c tK kt lun g v kh nng t

din bin ca phn ng 373K. /

b) Ti nhit no th phn ng cho t xy ra iu

kin chun.

p s' AG = l,26.105J/raol; T > 961K

57



111.23. Tnh AG ca s ng c -5c i vi nc lng

chm ng, bit rng AS ca qu trnh ny bng 21,3J/K.mol v

An0 = -5,8KJ/rrol ti nhit -5c. T kt qu tm -c, hy cho

bit v vic c hay khng trng thi cn bng gia nc lng v

nc -5c.

p s": AG = -92J/mol

111.24. a) Mt mol kh l tng gin 1 thun nghch nhit

ng lc t th tch 2 lt ti 20 lt- Tnh AS ca h v ca mi

trng xung quanh.

b) Cng 1 moi kh trn gin n ng nhit khng thun

nghch sao cho khng mt cng no c h thc hin. Tnh AS

ca. h v ca mi trng xung quanh.

c) Vi nhng kt qu nhn c t (a) v (b), hy chng t

, (bng con s" tnh ton) rng s nn t nhin kh l tng trong

h c lp s vi phm riguyn l II ca nhit ng lc hc.

p s a) ASh = 19,10J/K = -ASmt;

b) ASh = 19S10J/K, ASmt = 0

58



Chng IV

CN BNG HA HC

TM TT L THUYT

1. iu k i n cn b n g n h i t ng. nh lu t t c dng khilng

iu kin tng qut v cn bng ca mt h nhit ng l

hm G ^ Gmin, AG = 0 T, p = const. Nu h nhit ng l phn

ng ha hc din ra T.p = const, gia cc cht kh l tng th

i vi phn ng tng qut:

+ V22 + ... vnAn = vjAj + v 2A 2 +. . . + v mA m...

p V j p V

AG = AG + RTln - ~ (1)P VI p v 2r A, -r A 2 -

vi Pi l p sut ring phn c^cc cht i trong phn ng

trng thi no v nhit v p Sut.

Khi cn bng c thit lp AG = 0, t (1) ta c:

p v pV2

AG = - RTin 77 -P V1 p v 2r Ai A 2 "*



T.p = const; AG = const v t bng - RTlnKp. Vy:

p v l p v2

K = A' A;p P V1 p v2

r Ai A ,-(2)

Kp l hng s cn bng ng vi p sut l mt i lng hng nh mt nhit xc nh. Khc vi (1), (2rce gi tr p l p sut ring ng vi cn bng.

Biu thc (2 ) biu th thc cht ca nh lut tc dng khi lng do Guldberg v Waage thit lp.

Nu phn ng din ra th tch khng i th p dng phng trnh P = nv RT = CRT vo biu thc (2) ta s c biu thc hng s" cn bng ng vi nng K^ :

Cc i lng [ ] ng vi nng lc cn bng ca cc cht An v Am.

Cng nh Kp, Kc ph thuc nhit , khng ph tlmc nng .

Biu thc (3) cng c vn dng cho phn n^ ii n r:i trong dung dch lng tng.

vi dung dch khng l tng, nng Cj cc thay bng hot a. ' 1 ^

vi phn ng trong pha kh, hng s' cn bng cn c biu th qua nng phn mol Kx:

K = [ A i3 Vl. [ A 2 ]V2-(3)

(4)

60



V i X- = - ; n; l s m o i c h t i .. 1 ZRi

Gia Pj v X c mi lin h Pj = P.X.

p = , p - l p sut ton phn ca h:

Gia Kp, Kc, Kz c mi lin h:

Kp = Kc.(RT)n =K x.PAn (5)

Vi An = ^ V m - ] T v n

2. nh hng ca n h t , p su t n cn bng ho

hc. N guyn l chuyn d ich cn bng ha hoc

a) nh hng ca nhit

Hng s cn bng Kp l mt hm ca nhit , m t s

ph thc n ta c phng trnh Vari't Hoff sau:

^ A H dlnKp/dT = = - (6 )

RT

Trong khong nhit t T! n T2 nu xem AH = const th sau khi ly tch phn, (6 ) tr thnh

Kp.T AH 1 1

ln K ^ = E (7)

S ph thuc ca hng sp cn bng vo nhit , .trong trng hp n gin, cn c biu th bng phng trnh c lin h n nhit AH v bin thin entropi AS ca phn ng

. _ AH AS ...ln.Kp = (8 )

p RT R

Phng trnh Van't Hoff dng (8 ) ng nghim ng khi

AH v AS khng ph thuc vo nhit

61



b) nh hng ca p sut

T phng trnh lin h Kp v Kx ta c:

dlnK x /P= - = - ' (9)x p RT

T (9) thy rng hng s" cn bng Kx s thay i theo p sut T = const ty thuc vo s thay i th tch ca h \

c) S chuyn dch cn bng ha hc

Tho nguyn l Le Chatelier, khi lm thay i mt yu t"

no nh hng ti cn bng ha hc th v.tr ca cn bng s

chuyn dch theo chiu gim tc dng ca yu t" .

Xt yu t" nhit

T (6 ) thy rng Kp s ng bin theo T khi AH >0 (thu nhit). Vy i vi phn ng thu nhit, s tng nhit lm tng Kp tc to thm ra sn phm *

Xt yu t" p sut hay nng

T (1 ) -> aG = -RTlnK,, + RTln A'* p 1 p v 2

A ; r A -

t s" hng th hai v phi ca phng trnh bng i

lng RTlnQ. Khi : AG = R Tln-0-Kp

Du ca AQ c quyt nh bi t s: -Q -.

K P

Q < Kp AG < 0: Phn ng din ra theo chiu thun l chiu c tc dng lm Q tng ln cho ti khi Q = Kp th cn bang c thit lp.

62



* Q > Kp -> AG > 0: Phn ng din ra theo chiu nghch l

chiu c tc dng lm Q gim xung tin ti bng Kp, lc ny

t ti s cn bng.

S thm mt kh tr vo h cn bng trong pha kh.

S c mt ca mt kh tr trong h cn bng lm. tng p

sut chung ca h, song v tr cn bng c thay i hay khng

cn ty thuc vo iu kin p sut v th tch ca h c thay

i hay khng.

* S thm kh tr V = const khng lm thay i trng

thi cn bng v p sut ring cc kh trong phn ng khng

thay i.

* S thm kh tr p - const c th lm bin i trng thi cn bng do s gim p sut ring cc cht kh trong h

phn ng.

3. Cn bng pha

a) Quy tc p h a Gibbs

y l mt nh lut quan trng nht ca cn bng ha

hc trong h d th m trc ht cn bng pha. Quy tc pha

c pht biu nh sau:

Bc t do V ca mt h cn bng, bng s" cu t c tr i s"pha p cng 2 :

V = c - p + 2 * (10)

Bc t do V cho bit s" cc thng s' c th thay i mt

cch ty (d nhin trong mt gii hn xc nh) m khng xm

phm vo cn bng pha ca mt h cn bng.

63



Th d xt Gn bng pha lng

B - BI TP C GII

rv .l . 1000K hng s cn bng Kp ca phn ng:

2S02 + O = ^ S 0 3 bng 3,5G a tm '1.

Tnh p sut ring lc cn. bng ca S0 2 v SO3 nu p

sut "chung ca h bng latm v p sut cn bng ca 02 bng 0 ,1 atm.

BI GII

Gi X l p sut ring ca S0 2 th p sut ring ca S O 3

b n g : l-P o 2 - X = 0,90 - X.

Gii ra ta c X = tm; v P0 = 0,33 atm.O i b l .

'X' IV.2. Tnh hng s cn bng Kp i vi phn ng:

N2 + 3H2 ^ 2NH3 25c

Bit: AG", ca NH3 bng -16,64 kJ/mol.

Kp s thay nh th no khi phn ng cho c vit

di dng: N 2 + - H 2 ^ N H 3 2 2

BI GII

AG - -2x16,64 = -33,28 kJ

A p 0

- AG = -R TlnK p > lnK p =RT

65



-33,280 \ 8,314x298

InKp = ) = 13,43

Vy Kp = 6,80.10.

p 2NHs = K = 6,80.105

p N2 p h 2

t vi phn ng:

N 2 + -H 2^ NH3 25C2 2

Kp= f NH33 = # 7 =825

PN 'PHh 2

V.3. Mt bnh phn ng dung tch 1 0 lt cha 0,100 mo H2 v 0 , 1 0 0 moi I2,. 698K, bit hng s" cn bng Kc = 54,4.

Tnh nng cn bng ca H2, 12, v HI.

BI GII

H2 + I2 2HI

t = 0 0 , 1 0 0 0 , 1 0 0 0

t=toc (0 ,1 -x) (0 ,1 -x) , 2 x

(0,1 - x) (0,1-X ) 2s

10 10 10

K , . = - M - = 3 [[H2 ][I2]

= i M L -,0 ,1 - X ,0 ,1 -X . 0>'. '

10 ro

66



p '= 7,38-=>x = 0,0787

(0 ,1 - x)

[I2] = [H2] = (0?~x) = 0,00213 mol/1 10

n 0 7 X 7[HI] = -= 0,0157 mol/l.

1 0

_____ Cn bng ca phn ng kh c 2 bng C:

^ c + CO2 2 CO

Xy ra 1090K vi hng SQ cn bng Kp = 10.

a) Tm hm lng kh c o trong hn hp kh cn bng, bit

p sut chung ca h l 1,5 atm.

b) c hm lng c o bng 50% v th tch, th p sut

chung l bao nhiu?

BI GII

C02 + c ^ 2CO n

Lc u 1 mol 0 .moi 1 mol

Cn bng 0s1\ 2 x mol 1 + x

Phn moi1 - X1 + X

2 x1 + X

nh lut tc dng khi lng c vit:

Kp=.-5S2_ = ^ c o l l _a2lL = p Pco2 p -x c o ,

1 + X*

10

67



Suy r a x = 0,79. Vy hn hp lc cn bng cha 2 x = 2.0,79

= l,58mol CO v i - X = 1 - 0,79 = 0,21 mol C02 (8 8 % c o v 12%

C02 v s"mol cng nh v th tch).

tm p chung ti hn hp cn bng cha 50% v th

tch CO, ta vn dng nh lut tc dng khi lng:

T7- - Peo _ p-xo ^ Pco2 Xco

haylO = P ^ - p = 20 atm0,5

IV.5. Tnh AG v hng s cn bng K i vi phn ng

NO + O3 = n o 2 + o 2

Cho bit cc d kin sau:

n o 2 0 2 NO O3

AGh t W kJ/m olJ 51,79 0 86,52 163,02

AH^t 998 [kJ/mol] 33,81 0 90,25 142,12

ASht.298tJ/kmo1^ 240,35 240,82 210,25 237,42

ln ca hng s cn bng l h qu chu yu ca AH hay ca AS ca phn ng? Gii thch.

BI GII

Tnh AG ca phn ng:

AG u = Gt (N 02 ) + AGt (0 2 ) - AGt (NO) - AGt (O3 )

68



Thay s" vo ta c A Gpu = -197,71 kJ

AG0 =HRTlnK, do suy ra

_ 1Q-A0/2.303RT

- 1 0 '197710/2303'S'314'298

K = 5.1034

xt nh hng ca yu t" nhit hay entropi, ta tnh

ring r tng i lng:

AH.u = AHt (N 02 ) + AH, (0 2 ) - AHht (NO) -AHht(0 3)

Thay bng s' ta c

AHU = -198,55J/K

ASpU = S(N02) + s(02) - S(NO) - s(03)

Thay bng s ta c

A Spy = -2,5 J/K.

T mi lin h gia hng s" cn bng K vi cc i lng

AH0 v AS0 ta c h thc l X

K= 1 0 +AS/2,303R 1 0 "AH/2,303RT

K = 100 i3.1034S

Bin thin entrpi ca phn ng r t nh v cu trc hnh

hc phn t ca cht phn ng v sn phm rt gn nhau. V

vy ng lc thc y phn ng din ra mnh lit t tri sang

phi l yu 0 nhit nng, v mt nng lng th sn phm bn

hn cc cht u.

69



IV.6 . 25c phn ng

NO + - 0 2 = N022 2 2

C AG = -34,82 kJ v AH = 56,34 kJ. Xcjlnh hng s

cn bng 298K v 598K.

Kt qu tm c c ph hp vi nguyn l chuy dch cn

bng Le Chatelier khng?

BI GII

Tnh hng s" cn bng 25c bng cng thc

_^qAG/2.3RT

= 103482072,3.8,31.298

K = ,3.106

tnh hng s" cn bng K 598K, khng dng c cng

thc trn vi cha bit AG 598K. Tuy nhin nu chp thun

AH = -56,43kJ khng ph thuc nhit trong khong t 298 n 598K th c th vn dng cng thc

K2 = AH 1 1 . n K " R \ T2

trih hng s cn bng K? 598K. Thc vy

, K2 56430, i 1 ,in ~ ~ r = ----- (------------ ) suy ra

1,3.10 8,314 298 598

K2 = 1 2 . K < Kx: s tng nhit trong trng hp phn

ng ta nhit lm cn bng chuyn dch sang tri l ph.c tc

dng chng li s tng nhit l iu ph hp vi nguyn l Le

Chatelier.

70



rv .7 . Xc nh nhit ti p sut phn li ca NH4CI l

la tm bit 25c c cc d kin:

AHgt [kJ/mol] A Gt [kJ/mol] ,

r H 4Cl(r) -315,4 -203,9

HCI(k) -92,3 -95,3

NH3 (k) -46,2 -16,6

BI GII

i vi phn ng

NH4Cl(r) = HCl(k) + NH3(k)

Hng s' cn bng

Gi T l nhit phi tm th vi p sut phn li l latm

ta c Phci = Pnh3 = 0,5atm do

= 0,5 X 0,5 = 0,25 (atm)2

AGggs ca phn ng bng

AG98 =-95,3 - 16,6 + 203,9 92kJ

T cng thc AG = -RTlnK ta c

92000 = -8,31 X 298 X lnK298 suy ra

' . 'V. 'Mt khc vi phn ng cho, A298 bng:

h 98 = -92,3 - 46,2 + 315,4 = 176,9kJ

_ * ! . ( J _ . )K298 2,303R 298 T

71



lgKT-lgK 298 2,303x8,314 (298 t '

g0,25 + 16,12 = 78 0 ( - ) 2,303x8,314 298 T

T y tnh ra T = 597K.

IV.8 . Xc nh i lc ha hc ca st vi oxi ca khng kh

(Pq2 = 0.2atm) 1000K nu p sut ring ca oxi trn st (II)

oxit nhit ny bng 3.10' 18 mmHg.

BI GII

S oxi ha st l phn ng d th

2Fe + O2 = 2FeO

c hng s cn bng Kp = do lnKp = -ln P0p0 2

i lc ha hc c o bng cng cc i ca phn ng tc

l = -RTInPo- p sut ring ca oxi trong khng kh l "" - 2

0,2atm = 152mmHg.

3.1CT18p 2 152

Q 1 n~ 18A max = - 8,314 xlOOOx l n ---- = 376786J / mol

152

*TV.h 50c v di p sut l 0,334 atm phn ly a ca N20 4(k) thnh N 0 2 bng 63%.

Xc nh Kp, Kc, Kx.

72



Ban u

Cn bng

Phn mol

BI GII

N20 4 ^ 2N02 n1 0 1

1 - a 2a 1 + a

1- a 2a1 + a 1 + a

P 2 ^ p 2 X2 P-X2 p . [ ] 2 K 2 ^ r -a N 0 2 = NOg _ 1 + q

p _ K N204 ~ P-XN204' _ XN204 ~ L:._ai(1 + (x)

Thay p = 0,344 atm; a = 0,63 vo phng trnh Kp, tm c Kp = 0,867 atm.

Kc = Kp.(RT)'in ; An = 2 - 1 = 1 do

Kc = 0,867(0,082.323)'1 = 0,034.

K* = Kp.p-An= = =2,52p 0,344

IV.10. T v p xc nh mt hn hp kh cn bng gm

3mol No, lmol H2 v lmol NH3.

a) Xc nh hng s" cn bng K ca.phn ng:

; N2 + 3H2 ^ 2NH3.

b) Cn bng s chuyn dch theo chiu no, khi thm

0 ,lniol N2 vo hn hp phn ng T,p = const

BI GII

XvH 2Kx= V - , % =8,33

v k x 5 5

73



S thm N vo h T.p = const do Kx vn c gi tr

trc tc l 8,33.

Sau khi thm -0 , 1 mo N2> trc khi c s chuyn dch cn

bng th: - ' - ' '

Nh th Q > Kx v cn bng phi chuyn dch theo chiu no gim Q cho ti bng. Kx th cn bng mi s c lp

li. gim Q thi XN-t phi gm tc l cn bng chuyn dch

sang tri.

IV.11. 378K hng s' cn bng; Kp ca phn ng:

C2H6OH(k) ^ CH3CHO(k) + H2, bng 6.4.10'9. Nhit t

chy ca etanol v axetalehit l -1412 v -1196 kJ/mol.

Nhit hnh thnh ca nc bng -287 kJ/mol. Tm Kp ti

403K.

Chp nhn rng trong khong t 378 n 403K AH ca

phn ng l khng i, do vi dng phng trnh (7) ta c:

2

8,39

BI GII

lg K p .4 0 3 =

AH(403 - 378) + lgKp.378

2,303x403x378

A n 0 c xc nh theo nh lut Hess: /\

A AH .c.etanol - A H d.c.anehit AH ht.n

= -1412 - (-1196 - 287) = 71kJ.

74



lgKp.403 = ----------------- --------------------(403-378) +lg(6,4 X 109)2,303 X 8,31 X 10" 3 X 403 X 378

Suy ra Kp 403 = 2?6.108.

IV.12. i vi phn ng N20 4k:) 2N 02, Kp 25c bng

0,144 v 35c bng 0,321. Tm AH, AS v AG 25c vi

phn ng cho.

BI GII

: K p .3 0 8 AH 1____ _K p . 2 9 8 2 , 3 0 3 . 8 , 3 1 2 9 8 3 0 8

Thay s vo, tm c

AH = 66,619 kJ.

AG = -RTlnKp = -8,31.298.1n0,144 = 4,8 kJ

AS0= AHi_AG!_= 6 M li_ 4 8 0 0 = 207i45J/K T T 298 298

IV.13. Cn .bng ca phn ng NH4H S ( r ) ^ NH3 + H2S(k)

c thit lp 200c trong mt th tch V. Phn ng cho l

thu nhit. Cho bit p sut ring c NH3 s tng, gim hay

khng i khi cn bng c ti lp sau khi: j'$>

a) Thm NH3; ( e r i

b) Thm H2S; ^ ' ***

- c) Thm NH4HS

d) Tng nhit ; ^ ^ ^ ^, s U\7r^,

e) Ap sut ton phn s tng do thm Ar vo h,

f) Th tch bnh tng ti 2V. f ' (cc V %>. 7.

75



BI GII

a) Tng b) Gim

c) Khng i d) Tng

e) Khng i ) Tng.

IV.14. 0c v di p sut p = 1 atm phn y ca kh N20 4 thnh N 0 2 bng 1 1 %.

a) Xc nh Kp.

b) Cng ti 0c, khi p sut gii t latm xung 0 ,8 atm th phn li thay i th no?

c) Cn phi nn ng nhit hn hp kh ti p sut no

phn li bng 8 %.

BI GII

a) N20 4 ^ 2N02 Zn

t = 0 lmol 0 1

t = to f - a 2 a 1 + a

P2 f 2 a y\ l + aJ Tr 4a2 ^2 " I I ">Pmo \ 1 + a i 4(7- _

Kp Kp =P\T_n. ' _ ^ 1 tA 1 n'2*^4

1 + a

Thay a = 1 1 % v p = 1 atm vo phng trnh ny ta s c Kp = 0,049 atm.

4(X2 4 a 2b) 0,049 = r-.0j8 -^ = 0,0612

1 - a 1 - a

76



Suy ra a = 0,123. phn li tng do gim p sut chung

ca h

4a2c) 0,049 = .p

1 - a

Thay a = 8 % suy ra p = 1,9 atm.

IV. 15. Tnh tc bin thin p sut theo nhit chuyn

pha i vi cn bng L ca H20 nguyn cht ti 0c, bit

AHnc = 6,01 kJ/mol;

VL = 0,0180 l.mol'1;

VR = 0,0196 l.mol-1.

BI GII

Phng trnh (1 1 ) i vi cn bng L c vit:

d P _ _AHn,c

dT Tnc.AVnc

Chuyn n v t Jun sang l.atm, ta c:

1 J = 9,87.103 l.atm

P ----6010.9,87.10~3 _ _136at K -1. T 273.(0,0180-0,0196)

Kt qu ny cho thy h nhit nng chy ca nc

xung 1K thi phi tng p sut cn bng n 136atm. T y suy

ra p sut tng ln 1 atm, th im chy ca nc gim i

7,35.10'3K, iu ny gii thch ti sao li c th trt trn bng

c.

77



IV.16. Xc nh nhit bay hi ca Hg, bit rng ti 330c

p sut hi ca Hg bng 459,74mmHg v nhit ^si ca Hg

di p sut kh quyn l 357c.

BI GII

p dng cng thc (13) ta c:

In 76_0_= AH_ J _ 1_ A _ 5 8 1 5 kJ/moj459,74 8,314 603 630

IV. 17. Di p sut no, nc s si 97c?

Nhit ha hi ring ca nc bng 2254,757 kJ/kg.

BI GII

p dng phng trnh (13) ta c:

1 T> 1 T) A H , T g ~ T ^

g 1 " 2 " 2.303R T,.T2

lgp i = lg760_ i ^ 8 373-370

2,303x8,31 370 x373

= 2,8348 hay P = 683,6 mmHg.

c- BI TP T GII

IV. 18. 60K i vi phn ng:

H2 + C02. ^ , H 2(k) + C

Nng cn bng ca H, C9, H9O v c o ln lt bng

0,600; 0,459; 0,500 v 0,425 mol/1.



a) Tm Kc, Kp ca phn ng.

b) Nu lng ban u ca H2 v C 02 bng nhau v bng 1

mol c t vo bnh 51 th nng cn bng cc cht bao

nhiu?

p s: Kc = Kp = 0,772;[H2] = [C023 = 0,106 M

v [H20] = [GO] = 0,094;

IV. 19. Mt bnh 51 cha lmol H(k), c un nng ti

800c. Xc nh phn trm phn li ca HI 800c theo phn ng 2 HI = ^H 2 + l 2(k). Bit Ko - 6,34.104.

p s': 4,8%.

TVO. 25c hng s' cn bng Kp i vi phn ng N2 +

3H2 2 N H 3 bng 6 ,8 .1 0 5.

a) Tnh AG ca phn ng.

b) Nu cng nhit trn, p sut u ca N2, H, NH3

l 0,250; 0,550 v 0,950 atm. Tm AG ca phn ng.

p s: a) -33,28 kJ; b) -25,7kJ

.rv^:. Ngi ta tin hnh phn ng:

PC15 ^ P C 1 3 + C12

vi 0,3 moi PC15; p sut u l 1 atm. Khi cn bng c th i t-

lp, p sut o c bng 1,25 atm(V,T = const).

a) Tnh phn li v p sut ring ca tng cu t.

b) Thit lp biu thc lin h gia phn li v p sut chung ca h.

p s': a) 0,25; p = p 0(l + a)

79



IV.22. Xc nh nhit vi phn ng

' CaCO:^ Ca0 + C02

bit rng 800c p sut phn li bng 201,3mmHg v 900c

brig 992mmHg.

p s': -166,82 kJ/mol

IV.23. Trong mt th nghim ngi ta t mt mpun cha

N20 4 lng c m = 4,6g vo mt bnh phn ng ui ht khng

kh c dung tch 5,71. p v mpun ri a nhit ca bnh

phn ng ln 50C; Kt qu l N20 4 bay hi v b phn li, p

sut trong bnh o c l 0,4586^apn. Tnh phn li ca N20 4

v hng s" cn bng Kc vi phn ng N9O4 = ^ 2 N 0 2-

p s: 97,4%; Kc = 8,58

IV.24. Mt hn hp u gm 7% SO2, 11% 0 2, v 82% N*2

di p sut 1 atm, c un nng ti 1 0 0 0 K vi s c mt ca

mt cht xc tc. Sau khi cn bng c thit lp, trong hn hp

cn bng S0 2 chim 4,7%. Tm mc xy ha SO> thnh SO3 v

hng s" cn bng Kp v K ca phn ng 2S02 + 0 2;^ 2 SO3 (ghi

ch: mc oxi ha c o bng t s gia p sut cn bng v

p sut u).

p s': 32,9%; Kp = 2,44; K = 200

IV.25. Trong s tng hp NH3 400c tho phn ng N2

+ 3 H 2 ^ ^ 2 N H 3 , h n h p u g m N v H c l t h e o n g

t l hp thc ri a vo bnh phn ng dung tch 11. Trong hn

hp cn bng, ngi ta thy c 0,0385 mol NH3. Tnh K

IVJ26. Cho phn ng:

2N02 . ^ N 20 4(k); AH = -58,04kJ.

Hy tin on iu g xy ra cho h cn bng khi:

a) Tng nhit

b) p sut trong h tng.

c) Ar c cho vo h p = const v V = const

d) a cht xc tc vo h.

IV.27. 25c hng s'cn bng Kp ca phn ng thu

nhit 2NO + Br.?(k) ^ 2 NOBr(k) bng 116,6 atm*1.

a) Nu em trn NOBr c P=0,l08. atm vi NO c p =0,latm

v Br2 c p 0 , 0 1 atm to ra mt hn hp kh 0c th v tr '

cn bng s nh th no (cu tr li phi nh lng).

b) a NOBr c p = 5 atm vo bnh phn ng 50c th

thy trong hn hp cn bng c ^ p = 4,30atm. Tnh Kp

50c. So snh gi tr Kp ny vi Kp 25c. Gii thch?

p s: K p5()(,c = 179 atm"1.

TV28. Xc nh s" bc t do cc i ca h:

a) Mt cu t

b) Hai cu t.

c) Ba cu t.

IV.29. Xc nh s" pha cc i trong h:

a) Mt cu t.

b) Hai cu t.

81



IV.30. Xc nh h nhit nng chy ca Cd 100 atm

bit nhit nng chy ring ca d = 57,32 kJ/kg; nhit nng

chy ca Cd di p sut kh quyn l 320,9C; Khi4ng ring

ca Cd rn v Cd lng bng 8,366 v 7,989 g/cm3.

p s: AT : 0,59

IV.31. Xc nh nhit chuyn pha t lu hunh rombic

sang u hunh n t ti 95,5c nu

AV = v n t - Vrombic = 0,014 cm3/g v dP/A = 25,5 atxn/K.

IV.32. Xc nh nhit ti Bi s nng chy di p

sut lOatm. Bit rng ti nhit nng chy l 271c, t trng

ca Bi lng v rn tng ng bng 10,005 g/cm3 v 9,637 g/cm3.

Nhit nng chy l: 54,392 kJ/kg.

82



Chng V

DUNG DCH

A - TM TT L THUYT

1- Nhng tnh cht chung ca dung dch

a) Nng dung dch v cch biu th nng

Ngi ta gi ng cht tan trong mt n v khi lng

hoc trong n v th tch dung dch hay dung mi l nng .

C n h i u c c h b i u t h nng :

- Nng phn trm: c%

S" gam cht tan trong 1 0 0 gam dung dch.

Co = _________Sgam cht tan_________ 1 0 0S' gam cht tan + s gam dung mi

- S" gam cht tan 0 S" gam dung dch

-Nng moi:

S^ mo cht tan trong lt ung dch.

- Nng ng lng gam N:

S' ng lng gam cht tan trong mt lt dung dch.

- Nng molan m:

83



S" mol cht tan trong mt kg dung mi.

- Nng phn raol Xi:

S" mo cht i chia cho s" mol ca cc cht c mt trong

dung dch.

Hai loi nng m v Xi hay c dng trong tnh ton

ho l vi chng khng ph thuc vo nhit .

b) nh ut Henry v tan ca cht kh trong cht lng.

Nu c l nng cht kh trong cht lng v p l p sut

r i n g ca kh th: c = k.p

vi k l h s" Henry, ch ph thuc vo nhit .

c) Cc nh lut v dung dch long.

- nh lut Rault: gim tng i ca p sut hi bo

ho ca dung mi trn dung dch bng phn mol cht tan:

*A *A _ n B = V------------T----- A r >

p n B + n A

- H qu ca nh lut Rault: S tng im si v s h

im ng c ca dung dch so Yi dung mi:

tng im si ( h im ng c) ca dung dch t l

vi nng molan ca cht tan:

At = K.m

vi K l hng s' nghir si hay hng s' nghim lnh.

84



- nh lut Van Hoff v p sut thm thu 71 c,dung

dch: 7. V = nRT r '

Vi V l th tch dung dch

n l s" moi cht tan

R l hng s" kh

T l nhit tuyt i.

2. Cn bng ion trong dung dch

; - gii thch kh n n g dn in ca dung dch, Arrhenius

(1887) gi nh cht in ly (cht tan) trong dung mi thch

hp (nc chng han) b phn ly thnh cc ion dng v m.

Tu theo mc dn in ca dung dch m phn bit:

- Cht in ly mnh, phn ly hon ton thnh cc ion tri

du nhau mi nng .

- Cht in ly yu, phn ly mt phn thnh ion; kh nng

phn ly tng theo long.

Trng hp ny s phn ly l mt qu trnh thun nghch;

cn bng phn ly c thit lp trong dung dch gia phn t

khng phn ly v cc phn t phn ly (cc ion) l cn bng \ ion. Th d:

C H 3 C O O H ^ H+ + CHsCOCT

Fe(SCN)2+ ^ Fe3+ + SCN'

Kh nng phn ly c c trng bng in ly a.

_ S" mol b in lya = ----- : -

S mol ha tan

in ly a bao gi cung nh hn mt, ln hn khng:

0 < a < 1

85



cl) p dng nh lut tc dng khi lng vo cn bng

ion. Khi nim v ch s'pK

n gin, xt cn bng ion sau nhit T "v .nng

c xc nh.

AB ^ A+ + B'

Nng u c 0 0

N n g c n b n g C ( 1 - a ) C a C a

Cn bng ion c c trng bng hng s'cn bng K.

TJr _ [A+] EB]K " ! AB!

Thay cc nng [A+], [B-] v [AB] bng cc i lng Ca

v C(1 - a), ta c: K =( 1 - a )

Biu thc ny cho thy a gim vi s tng ca c v c

gi l nh lut pha long Ostwald.

Trong tnh ton v cn bng ion, mt i lng hay c s

dng l ch s" pK: pK = -lgK.

Th d i vi cn bng:

C H 3 C O O H ^ H + + C H 3 C O O .

[H+] [CHoCOO- ] _RK = ----- 3 = 1,8.10 pK = 4,74

[CH3C00H] v -

H2S ^ E T + HS:

K = y n | p = 9 ,u o -s pK = 7,05L1 19J

86



Cu(NH3)f- Cu2+ + 4NH3

[Cu2+] [NH,r _K = V = 4,6. Hr4 pK = 3,34

[Cu(NH3 ) f ]

b) Thuyt axit - baz.

Arrhenius

Axit l cht khi phn ly trong nc cho ion H+

Baz l cht khi phn ly trong

nc cho ion OH

Axit HA

Baz BOH

H+ + OH'

? B+ + OH'

Bronsted - Lowry

Axit l cht c kh nng cho proton trong dung dch

Baz l cht c kh nng nhn proton trong dung dch

A ^ B + H+

Axit Baz

A v B l axit v baz lin hp

c) Ap dng cc thuyt Arrhenius v Bronsted - Lowry vo

mt s'cn bng ion trong dung dch long nng ca nc

xem nh khng i bng 55,5 moi 1.

S phn ly ca nc: .

H20 ^ H+ + OH H20 + H20 ^ H30 + + OH\

Axit 1 Baz 2 Axit 2 Baz 1

K = m m = t H 3 ^ o = 10_16 2 5 0C

[H20] [H2 o ] 2

Tch s" ion ca nc:

K = [H+] [OH ] = [H30 +] [OH ] = 1 0 M pK = 14

(Arrhenius) (Bronsted)

87



Lc Axit:

Baz:

HA ^ H+ +A" HA + H20 ^ H30 + + A".

Axitl Baz 2 Axit 2 Bazl

_ [H30 +][A~3 Ka [HA][H20] 55,5

Hng s" Axt:

_ [H-] [A-] _ [nsa*] [A-] _T^ _K^ ~ H A ] - [ H A ] PK = - ^ a

' Th d:

H2S ^ H * + HS' e s s + H ,0 Hao* + HS'

HCN ^ H+ + CN" HCN + H20 ^ H30 + + CN'

Axit cng yu, pK cng ln.

v axit mnh, phn .ly hon ton, khng c cn bng

phn ly nh khng p dng nh lut tc ng khi lng-

HC1 = H+ + Cl" HC1 + h 20 = h 30 + + c r

Theo Bronsted, Cl- l mt baz cc k yu, khng c kh nng kt hp vi proton.

-Lcbaz:

BOH =5=: B+ + OH" B + H20 .5^ BIT + OH .

bazl axit2 axitl ba 2Hng s^ baz:

K _ [B+ ] [QH~3 _ [OH~3 [BH*] b [BOH] B]

88



Th d:

NH3 + H2.0 =5 ^ N H / + OH

baz axit axit baz

NH3 v NH4+ l mt cp baz axit' lin hp

OH' + HgO+ Ego + H20bazl axit2 axitl baz2

Lu rng vi cp axit, baz lin hp th:

Ka . Kb = Kw

3. Khi nim pH

pH ca mt dung dh l ogarit thp phn ca nng ion

H+ hay ion hironi H30*.

pH = -lg[H+] = -lg[H30 +]

i v i c n b n g i o n c a n c [ H + 3 [ O H ' ] = K W t a c :

pH = pOH = pKw = 14.

- Mi trng trung tnh 25c [H+] = [OH1 = 107.

pH = pOH = 7

- Mi trng axit [H+] > 1 0 7 ; pH < 7

- Mi trng baz [H+3 < 10 7 ; pH > 7

4. Gng thc tnh pH ca cc dung dch nc

- HA mt axit mnh nng C:

HA + H20 = H30 + + A

[HsO+] = [ A~ ] - c

pH = -lg[H30 +] = -lgC



Cng thc ny ch ng khi pH < 6,5 v b qua [H30 +]

do nc phn ly ra.

- HA l mt axit yu c nng C:

pH == (p K a - lgC)

- B l mt baz mnh: B + H20 = BH+ + OH\

pH - 14 + lgC

- B l mt baz yu c nng C:

pH = 14 - (p K b - IgC)

- BA l mui ca axit mnh v baz mnh,

BA = B+ + A".

Cn bng ion ca nc khng b vi phm, do : pH = 7.

- BA mui ca axit yu HA v ba.z mnh.

BA = B+ + A" ; A + H20 HA + OH .

pH = 7 + (pKa + IgC) c l nng ca mu BA.

- BA l mui ca axit mnh baz yu c nng c .

BA = B+ + A" H20 + B+ ^ H30 ++ B

pH = 7 - (p K b - IgC)

- AB l mui ca axit yu HA v baz yu B c nng C:

AB = A~ + B+.

A" + H20 ^ HA-+ OH~ k*

B H + + H 20 ^ B + H 30 + Kb

pH = 7 + ( p K a -pK b)



5. Hn hp m

Thng mt hn. hp m c to ra t mt dung dch

axit yu v mu ca axit yu hoc t mt dung dch baz yu v

mui baz yu.

Dung dch m c pH t thay i khi pha long hoc khi

thm mt lng va phi axit hoc baz vo dung dch. Mt hn

hp m gm axit yu c nng Ca v mt mui ca axit yu

nng Cm th:

pH = pKa + lg a -

Trng hp hn hp m ca baz yu nng C, v mui

ca n nng Cm th: pH = 14 - pKb - Ig-^53-c b

6. S thy phn ca mui

Thc nghim cho thy dung dch nc ca mt s" mui c

phn ng axit hoc phn ng baz.

Mui ca axit yu v baz mnh; c s thu phn ca anion.

Thuyt Arrhenius

A" + H 20 ^ H A + OH'

Hng s" thy phn Kh =

pH = 7 + ^(pKa + lgC)

K.

K.

Thuyt Bronsted

A" + H 20 ^ H A + OH'baz 1 axt 1

Kh = KVKa = Kh.

pH = 7+^(pKa+.lgC)

, c l nng mui (mol/I) Mu axt mnh v baz yu.

91



Co s tKu phn cation

B+ + 2H20 ^ BOH + B+ + 2H20 BOH + H3Oh^ac axit 1 baz 1

Kh = Kw/Kb K K UK h =Kw Kh = ^ p ^ =

Cation l mt axit

KbK

pH = 7 - < p K a + lgC)2 p H = ( p K a -lgC)

M ca axit yu v baz yu.

B++ A- ^ BOH + HA B++ A '+ H ,0 ^ B O H + HAbaz axit ax itl bazo' 2 bazl axit 2

A ? .'?

Kh = H/Ka. K Kh = Kw/ K A2 .KB]

= KAi/KA2

pH = 7 + I (pKa - pKb) pH = ( pK Ai + pK A )

7. Tch s' tan

Xt cn bng gia mui kh tan trong dung dch v cc ion

ca mui: AnBm(r) ^ nA"q +.mB+q

p dng nh lut tc dng khi lng vo cn bng ny ta

c biu thc nh ngha tch s' tanT.

T = [A]n . [B+]m

[ A- ] v [B+] biu th bng mol/1

Ngi ta gi tan s (tnh b n g mol/I hoc g/1) l nng

c mui tan to thnh dung dch bo ha.

92



1B- BI TP C LI GII

v . l . Ha tan 3,42g MgCl2; 2,63g NaCl vo 88,20g nc.

Tm nng % v khi lng ca NaC, MgCl2 v H20.

BI GII

Khi lng tng cng ca dung dch:

2,63 + 3,42 + 88,20 - 94,25g

Trong 94,25g dung dch c 2,63g NaCl vy nng %

VT l _ 2 >6 3 1 0 0NaC bng: ------ = 2,79%94,25

Nng % MgCl2: 3 ,4 2 10- = 3,63%94,25

Nng % HoO: 8 8 , 2 0 ' - 1 0 0 = 93,58%94,25

V.2. HN03 c nng 69% c khi lng ring l,41g.cm'3.

Tm th tch ca dung dch cha 14,2g HN03.

BI GII

Khi lng dung dich cha 14,2g HNO: = 20,6 g69

Th tch ca dung dieh s l: Q - = 14.6cm3 M1,41

V.3. Khi lng ring ca dung dch H2SO4 49% l

l,385kg/dm3.

Hi ly ba nhiu th tch ung dch H2S0 4 49% iu ch:

93



a) 1 lt dung dch nng 0 ,5 N-

b) 400cm3 dung dch IN.

c) 250cm3 dung dch 0 ,2 M.

BI GII

a) MH so = 98.1I10 H9SO4 cha 2 ng lng gam.

98 )0,5N tng ng vi - - 24jTg H2S0 4 nguyn cht.

4

Th tch H2SO4 49% cn dng pha lt 0,5N l:

^ =4 99. . 36,lem49 . 1,385

Pha ong th tch ny bng 1 lt bng cch thm nc.

b) 400cm3 dung dch axit IN, cha:

^ = 19,6g axit nguyn ch't.1000

Vy th tch axit 49% cn ly pha 400cm3 IN l:

19,6 . 100 OQ 3 s 29 cm49 . 1,385 ^

c) 250cm3 dung dch cha: = 4,9g H2S 0 4

49y l lng H9SO4 cha trong: = lOg dung dch 49%.

4,9

, 10 qVy th tch phi ly: = 7,2em

1,380 ,

94



QhxcQ-r/V.4. 45,20| ng tan vo 316g nc. Tnh im si, im

ho rn ca dung dch, bit cc hng s" nghim si, v nghim

lnh l 0,51 v 1 ,8 6

BI GII i '-

Nng moln ca dung dch l: = 0,418 . 342,3.316

ATs = K.m = 0,51 . 0,418 = 0,21

im si phi tm: > 373 + 0,21 = 373,2lK.

im ho rn: 2%* - (0,418. 1,86) = 272,2 2 K.C*c- c

V.5. S phn tch hemoglobin trong mu cho thy st

chim 0,328% khi lng,hemoglobin.

cOTm khi lng mol ti thiu ca hemoglobin.

^D ung dch nc cha 80g hemoglobin trong mt lt dung

dch c p sut thm thu bng 0,0260 atm 4c. Tm khi lng mol ca hemoglobin.

c) C bao nhiu nguyn t st c trong 1 phn t hemoglobin.

BI GII

a) Sc mol st c trong lOOg hemoglobin:

^ i = 5,87.1CT3 mol 56

Mt phn t hemoglobin phi cha t nht mt nguyn t

st; Vy trong lOOg hemoglobin s c 5,S7.103mol hemoglobin. Khi

vlng moi ti thiu ca hemoglobin:

95



b) S" mol hemoglobin trong dung dch:

= V _ 0,0260 1 n _ R T _ 0,082. 277

= 1,14 10 3 mol

Khi lng mol: ------ -1,14 . 10^

T = 70200 g

c) Khi lng mol hemoglobin y so vi khi lng mol

ti thiu, thy gp 4 ln do phn t hemoglobin cha 4

nguyn t st.

V.6 . Tnh p sut hi ca dung dch ng cha 24g ng

(Cl2H22On) trong 150g nc 20c nu nhit ny p sut

hi ca nc nguyn cht bng 17>54mmHg.

BI GI

(17 5 4 -P ) __ =0,00833 => p = 17,39mmHg

17,54 5

V.7.'Dung dch axit xianhiric HCN nng 0,2M c hng

s Ka = 4,9.10'10.

Xc nh nng H30 + v in ly a

96



BI GI

HCN + H20 ^ 0,2

0 ,2 - X

H3o + -+CN-0x_

0X

V Ka rt nh, HCN rt t in ly nn [H30 +] cng nh v

vy ly mt cch gn ng: X < 0,2; do 0,2 - X 0,2

- . 2

v = 4,9.1010 X = 9,9. 10-6 0 ,2 Vae;

c / -

_ 9 9 10-6 in ly a = 5 . cr0 hay 0,005%

v 0,2

V.8 . Cho 1 0 " 2 mol KSCN vo lOml dung dch mui Fe3+

nng 103M. Bit rng phc c to ra (mu sm) c cng

thc FeSCN2+, v nng ion Fe3+ t do, cha tham gia vo phc

l 8 -O^M. Tnh hng sbn ca phc.

BI GII

Xt cn bng to phc

Fe3+ +SN ^ FeSCNj+

lc u v 1 CT3 0

cn bng 8 .1 Q' 6 0,999. 0,001.

Hng s bn ca phc FeSCN2+ l:

97



K = : --------- = ------- - --------- = 125[Fe ] [SCN-] 8 . 10 . 0,999

Ch : - nng u ca KSCN l 10"2/10' 2 = IM^ *

- Cn bng chuyn dch mnh sang phi v SCN d.

[FeSCN2'1'] _ 10-3

.9) Xc inh nng Cu2+ t trong 500ml dung dch

c iu ch t 0 ,1 mol CuS04 v 2 mol amoniac NH3.

Cho bit:

[CuNHs)]/" c hng sbn l 2 .1 0 13

Cp N H //N H 3 c pK, = 9,2.

S to phc xy ra theo phng trnh phn ng

Cu2+ + 4NH3 [Cu(NH3)4]2+

V cn bng NH3 + H20 NH4+ + OH" chuyn dch

mnh sang tri nn c th b qua tc dng ca NH3 vi H20.

Da vo y ta c nng cc cht lc u v lc cn bng nh

BI GII

sau:

Hng s bn ca phc [Cu(NH 9 ) 4 ]2+ c vit:

98



V phc r t bn (K 1) nn c th coi X 1 . Mt cch gn

ng

ta c K * ----- ----- 5iL i------_ = 2.10 3.0,2(1 -x )(4 -0 ,8 x ) 4

Vy [Cu2*] = 0,2(1 - x) = 0,2/(3,2)4*.2.1013 = 9.5.10' 17

V.IO. Mt hn hp dung dch CH3COOH v NaOH c nng

(m ol/I) u t n g n g l a v X

Vit cc phng trnh cn bng axit - baz xy ra trong

hn hp.

T vic tnh hng s' cn bng, hy chng t rng phn

ng trong dung dch l hon ton. (Ka = 1 ,8 . 10'5X

Thit lp phng trnh pH ph thuc vo a, X, hng s" axit

Ka tch ion ca nc Kw trong 3 trng hp x < a , x = a v x > a

BI GII

a) CH3COOH + OH" ^ CH3C0 0 " + H20

H20 + H20 ^ HsO+ + OH~

CH3COOH + % 0 ^ CH3COO" + h 30 +.

b) Hng s" cn bng nhn c trong phn ng trung ho:

[CH3COCH [CH3COO-[H,Q+] Ka 1,8.10"

c [CH3COOH] [OH'] [CHgCOOH].! ^ icr14

Gi tr r t ln ca hng s" cn bng Kc chng t phn ng

l hon ton.

99



c) d dng tm phng trnh pH, vi phn ng trung

ho axit yu bng baz mnh ta lp bng sau

CH3COOH + OH~ = CHgCOO + H 20

X < a a - X b qua X const

X = a x(l - a) x(l - a) ax const

X > a bo qua X - a a const

Trng hp X < a:

[HgO+] = -> pH = pKa + Colg(a - x) + lgxX

Trng hp X = a: [CH3COOH] = [OH" ] = K^H sO 4]

[CHgCOO-]* x(ci 1)

[HsO+] = - i - > p H = ( p K a + pKw + lgx)X 2

Trng hp X > a: [ OH- ] = X - a -> [H3O I = K ^x-a) - 1

pH = pKw + lg(x - a)

v.ir. Dung dch axt benzoic (axit yu) IM c cng pH vi dung dch HC1 nng 8 . cr3]^.

a) Tm pH.

b) Tnh hng s" axit Ka ca axit benzoic

BI GII

HC1 l axit mnh phn ly hon ton nn:

100



[H30 1 = [ e r ] = 8.10'3 do pH = -lg8 .1 0 '3 = 2,1.

Axit benzoic C6H5COOH l axit yu, do trong dung dch

c cn bng.

C6H5COOH + H20 ^ H30 + + CgHgCOO".

[H30 +][C6H5C 0 0-] a [C6H5 COOH]

B qua s phn ly ca H2O nn [H3CT] = [C6H5COO"]

Vy: Ka = [H3OY = 8(.10r?) 2 = 6,4.10'5.

Y.I2 I Tnh pH ca dung dch NaCH3COO 0,1M, bit hng

s'baz Kb = 5,7.10'10.

BI GII

NaCHsCOO = Na+ + CH3COO"

CHgCOCr + H20 ^ CH3COOH + OH

0 ,1 0 0 V.

(0,1-x) X X

Kh = = 5,7JO10b (0,1-X) .

Mt cch gn ng 0,1 x.dp

X2 = 5,7.10"11 suy ra X = 7,6.10'6.

Gi tr nh tm c i vi.x p ng iu gi thit

X 0,1M.

101



Vy [0H~ ] = 7,6.10"6M

[H+] = = 1 ,3 .1 0 _9M pH = 8,9 7 ,6 . 1er0

V.13. Tnh pH ca h m gm 0,050mol axit axetic v

0,050mol natri axetat trong mt lt dung dictu pH s thay i th f

no khi thm vo h m ny 0 ,0 0 1 ,mol HC1? 1 V V'~- ^ ^

BI GII

NaCH3COO = Na+ + CH3CpO"

CH3COOH + H20 ^ CH3 COO" + HoO+

[CH3 C0CT] [H3O ] *

a [CH3COH]

CH3COOH + Hao ^ CH3COO" + H3Oh

Lc u: 0,05 0,05 0

cn bng: 0,05 - X 0,05 + X X

K = 1,8 .1 o- 5 =(0,05 + x)x(0,05 - x)

. yAxit axetic l axit yu t phn ly nn: X 0,05. ^

0,05 - X 0;05 + X 0,05, o 1,8 . 10" =0,050

Vy X 1,8.IO"5 pH = -lgl,8.l0 ' 5 = 4,47

Khi thm HC1 vo h m th HO* do HC1 phn ly ra khi

phn ng vi CHgCOCP chuyn n thnh CH3COOH:

102



H30 + + CHgCOO0^ CH3COOH + liso

Trc khi thmHCl-^OOC!)? 0,050 0,050

Sau khi thm: C/D o 0,05^0,001 0,05+0,001

Cn bng: X 0,049 + X 0,051 - X

Ka = l,8.icr5 = x(0,049 + ) => [H30 +] = Ka5-= 1,9;10-5 a (0,051 -x ) 3 a 0,051

vi X 0,049 X 0,051 .

pH = -

Bài Tập Hóa Học Đại Cương & Hóa Lý Cơ Sở - Lâm Ngọc Thiềm

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