化學(xué)平衡答案

pK=p(H2O,g)10-2=pG=-2.30RTlgK=-2.300.00831293lg0.0230=9.17(kJ·mol-r 100℃，100kPa時(shí)，H2O(l)和H2O(g)處于平衡狀態(tài)，所以GrG=(-228.50)5+(-661.91)-(-1880.06)=75.65(kJ·mol-rlgK=

r ，K=5.2510-r 2.308.3110-3 rGm=75.65+2.308.3110-3298lg(3.1710-20.600)5=26.4(kJ·mol-CuSO4·5H2O不能風(fēng)化成CuSO4；CuSO4能潮解成CuSO4·5H2O

mH(1)<rHm G(2)應(yīng)增大，因?yàn)镾(2)r rK(2)應(yīng)減??；因?yàn)镚2.30RTlgKG(2)K r r

p rHTlg m( 2K2K 2.30R

K

873 p1.64

2.30

873pK=3.7110p

(1)+(2)-(3)得NH3+ NH4K=K1K2/Kw=4.510-41.810-5/1.010-14=8.1

反應(yīng)的吉布斯自由能變G=2(-5.5)=-11.0kJ·mol-rlgK=-G/2.303RT=-(-11.0103)/2.3038.314298rpK=p反應(yīng)的熵變S=2S(ICl,g)-(S(I2,s)+S(Cl2,r =2247.3-(223.0+116.8)=154.8(J·K-1·mol-H=G+TS=-11.0+298154.810-3=35.1(kJ·mol- r rH 所以H(ICl,g) m =17.6(kJ·mol-

1molNH4Cl1molHCl1mol所以pHCl= =p/2=0.501003pK=2.5010-pG=-RTlnK =-8.3110-35972.30lg(2.5010-1)=6.87(kJ·mol-r CO(g)+平衡時(shí)壓力 (

= =200))2(20)

=20011 191

100=8.0010-

(8.00

)1/2=6.3210-(1)200 1 100 G=-RTlnK=-8.31410-3373ln(8.0010-9)=57.8(kJ·mol-r H=G+TS=57.8+3730.1255=104.6(kJ·mol- r rn2SO2+ 0n n總0.240.620.76Kp=

(0.76)2 總 p總 )2p p總=100 K=p

開始時(shí)c/mol·dm- 2平衡時(shí)c/mol·dm- 1.0-3.310- 3.310- 13.310-2(NO)2(O (3.3105)2(13.3105Kc

)2

(1.03.310

=1.810-2 5

4HCl(g)+

初始濃度/mol·dm-00平衡濃度/mol·dm-Kc

0.0154

=8.9 100

2

2

pH Kp 22pCOpH 2

=9.7因?yàn)镚2.303RTlgK2.3038.3110-3373lg(9.7010435.6kJ·mol-r 而GHTHr H 41.2所以S rm =-15(J·mol-1·K-r

起始濃度/mol·dm- 平衡濃度/mol·dm- (2x)K(1.0x)

=2.310-x7.510- NO2000KN2O2NO的方

設(shè)PCl3

2.00-xpCl

1001.00x32.003

100 52.005

K x/(2.00

= (1.00x)2/(2.00x22

1.00x=32.0035pPCl=5n n n總Daltonpip 2.00x，

1.00x，

3.00

3.00

3.00 pPClxPClp，pPClxPClp，pClxCl p1.00x

3.00x 3.00xKp

p 2 2.00x 5(1.00 (3.00x)(2.00

3.00xpp=K

p=100解得xpPCl=7.5kPapPCl=117kPapCl75.5

Kp

CaO+ pCO800pCO1.16102 CaCO320/1000.20則由pV= n=pV1.16

=0.130 0.130molCO20.20molCaCO3分解得到的，未反應(yīng)的CaCO3[(0.20-0.13)/0.20] p=220.3 V=0.4720 T=1000n總=pV220.30.47200.01251 8.314初態(tài)nCOCl20.631098.90.00638xmol的COCl2CO n總=0.00638+x=0.01251(mol)所以x[Cl2]=0.0130mol·dm-[COCl2]=5.310-4mol·dm-Kc[CO][Cl2][COCl2 =0.40mol n 1.0 = 0.083 0.40-2x+x= x=0.160.40-2x=0.40-20.16=0.08K=(0.16/0.24)1.0= [(0.08/0.24)(2)溫度升高到111

初始?jí)毫?100 平衡壓力/100 p0- p0-(p01.00)0n=p0V0

2.36

= p0=2.36100=0.27 0.083N2+ pNH3

=K=

=3p3Hpp Hp

1.03 =1.0100

=pNH3

1.04.92=0.10 0.083nHCl1.0000.10 [H+]=pH=5.12

=7.510-6(mol·dm-的HCl存在，則溶液的pH值下降。 起始分壓力/100平衡分壓力/100

pNO2/p=121000.12pNO/p(100-12)/100pO

/p=(300-112)/100=2( /p) 而K =6.310- 2.94O2(NO)

K=pCl(g)/100

4.794rGm=-fGm(Cl2,l)=-RTlnK lgKp=- )=2.308.31K=6.95，K= /p， =6.95100 由G=2G(SO32G(SO22(-370.4)-2(-300.4)140.0(kJ·mol-r 及G=-r得lgK(298K) 140.0

= 2.3038.314p所以K(298K1024p又rH2H(SO3)-2fH(SO22(-395.2)-2(-296.9)196.6(kJ·mol- 所以

KppKpp

196.62.303

(298298

)=pKppKp

p1021 K(773K103342.2G=2G(CO,gG(C,石)-G(CO2r =2(-137.3)–0-(-394.4)=119.8(kJ·mol-因?yàn)镚0r 119.8因?yàn)閘gKp= rm 2.308.31

=-p所以K110-21910-p

(1.000x)= =2所以pNO2

=0.6681002

pNO=0.3321002轉(zhuǎn)化率0.3320.6682K

p2rHm1000(1 K 2.303 lg 57.5 1lg 2.3038.314 pKpG=-RTlnK=-8.3110-33102.30lg2.90=-2.7(kJ·mol-r p

(1)設(shè)原有PCl5nmoln總mol PCl3(g)+ n(1-) n總=n(1+)1.001.00=2.695(1+)0.0831 =

pKp =1G=-RTlnK=-1.90(kJ·mol-r

no(1- n總=no(1+2MNO=92g·mol- MNO=46g·mol2M921461

1

2=

(p K 1 4p

40.501.00=pppN2O

p11

14

1K p0.100100kPa時(shí)，所以=87％

1

p

= H=H(C5H10,g)-H(C

,l)=-77.2-(-105.9)=28.7(kJ·mol-

H

5 因?yàn)镾 m88(J·mol-1·K-TrTbH 所以Tb m =3.3mrS mG=H-TS=28.7-(2980.088)=2.5(kJ·mol-r rlgK= rm

2.5

=- =

2.308.31 K 2.308.31p=0.36100kPa=36

2NO(g)+ m2fH(NOF)=-312.96+90.372=-132.22(kJ·mol-m所以fH(NOF)66.11kJ·mol-m2NO(g)+ S=2248-203.3-2210.6=-128.5(J·mol-1·K-rG=H-TS=(-312.96)-298(-128.5)10-3=-274.7(kJ·mol-r rG=-RTlnK K=2r

lgKp

r T r T K 2.3 T 2K p

50029()(1.37 2.300.0083129850K p2=9K

△NH3(g)+ 2pp1p2p1p212Kpp1p2ppG=-12.9kJ·mol-rp459℃時(shí)：KpG=-20.9kJ·mol-r p

rH

T

因?yàn)镵

m

=1.6

(kJ·mol 2K 2.30 2

427GHTSr r(GSS r rm=0.25(kJ·mol-1·K-r

mrH=-235.4-(-277.6)=42.2(kJ·mol-mS=282-161=121(J·mol-1·K-rG=H-TS=42.2–298(0.121)=6.1(kJ·mol-r rlgK rm

2.308.31

=-pK=p在此Kp(C2H5OHg)/100pH 在沸點(diǎn)Tb，G= Tb m =Sr Sr

H2O(l,100 H2O(g,100G=-228.59-(-237.18)=8.59(kJ·mol-rH2O(l,飽和蒸氣壓 H2O(g,飽和蒸氣壓rGm=0(此時(shí)為平衡態(tài)H2O(l,1.33kPa)H2O(g,1.33G=G+2.30RTr r=8.59+2.300.00831298(-1.88)=-2.10(kJ·mol-過(guò)程(1)H2O(g)向H2O(l)轉(zhuǎn)變，水蒸氣凝聚為水。過(guò)程(2)：平衡態(tài)。過(guò)程(3)：H2O(l)向H2O(g)轉(zhuǎn)變，水蒸氣變成水G=3(-228.59)-(-741.0)=55.2(kJ·mol-r G lgKp rm=-

Kp=2.010-H在298K，水的飽和蒸氣壓為3.17kPa， =101.3-3.17=H2pHOQp H H2

=3.3710-pp

mrH=3(-241.83)-(-822.1)=96.6(kJ·mol-mS=3188.72+227.2-3130.59-90.0=138.8(J·mol-1·K-rG=H-TS=96.6-10000.139=-42.2(kJ·mol-r r lgK= K=1.6 pHOQp =3.3710- pp

C6H6(l)+ G=153.2+0-(118.5-16.63)=51.3(kJ·mol-rG=-2.30RTlgKr lgKp rm

51.3

=- 2.308.31pK=9.810-p

( /p 2.0起始分壓商Qp =1.010- H(pN/p)1/2( /p)3/ 4.01/ Hpp673K，QpK773K，QpKpp mrH=-235.4-(-277.6)=42.2(kJ·mol-mS=282-161=121(J·mol-1·K-rp12000m0.600100kPaK=p(C2H5OH,g)/p=0.600所以G=-2.30RTlg0.600prG=H-TSr r4.22104T1212.308.31TmTbrUm

=337

CaSO4·1H2O(s)+3 G=-1435.2+3(-228.6)-(-1795.7)=17.6(kJ·mol-r G lgKp= rm=-3.09，

Kp=8.110-K=( /100kPa)3/2， =8.710-1 8.7101n(H2O，g)= =0.35 8.31n(CaSO4·2H2O)=0.2320.23(136.2+182)40 2

H=9.2-233.2=-57.2(kJ·mol- S=304-2240=-176(J·K-1·mol-rG=H-TS=-57.2+298176/1000=-4.75(kJ·mol-r rG=-RTlnK lnK Kr / (1)K = (pNO2/p) 2=0.2424％

M=0,18 =95(g·mol-又設(shè)平衡時(shí)P2(g)為2x則

n總1x(431)1

1

(231)=x= 即P4(g)分解百分率pp940KG2.30RTlgK2.300.00831940lg0.500r p

rH

T2K m K 2.30RT2K H lg HH

2.30=180kJ·mol-

1040H- rH 180S rm r

PCl3+ n總= 1.00

= Kp=12p=1

p=0.0294 =0.02941 0.704=c(1+’)，c=0.570mol·dm-3[Cl2PCl30.5700.2360.135mol·dm-[PCl5]’=0.435(mol·dm- 減向右(NH3合成 增

pHg=2/35.16104=3.441042pO=1/35.16104=1.721042 pHg=2/31.08105=7.20104O =1/31.08105=3.60104O2 Kp(693K)=p2· =(3.44104)2(1.72 2Kp(723K)=p2· =(7.20104)2(3.60 O(7.20104)2(3.60104 H(723693)lg (3.44104)2(1.72104 2.3038.3169372mrH=307kJ·mol-m

p=93.3 T=293 Vtot=15.7pVtotntotpVtot93.315.7=0.602 8.310xntot=1.00–2x+x=1.00–x0.60x=0.40mol2所以nNO1.002x0.20mol)80NO2締合為N2O42

333[Fe(phen)2]=0.1mol·dm-3=33設(shè)平衡后，[Fephen)2x3 Ce4++Fe(phen) Ce3++Fe(phen) 平衡濃度/mol·dm- [Ce3][Fe(phen)3K 3[Ce4][Fe(phen)23

(0.1x)x

=13xFephen)2310-4mol·dm-3

p2HT

lgK

m( 1) KKpKp

2.30 C(s)+ (pH/p

p2 )

/ p1.00

1.001.00pV=(1.00+(2)代入 Kp2

1.00

8.31

1.00

= = p=2.4102

nNOBr=1.10/110=0.0100 2 平衡時(shí)總物質(zhì)的量=0.0100(1+)

n=p1V30.01.00=0.0134總 8.31總0.0100(1+)= = 2 p/p4 30.04K Br2 =

2 (12)2(1 0.3602總 n'=p2V35.01.00=0.0141總 8.310.0100(1+')= 30.04Kp2 =0.1742K

T p2

m( 1H2K 2.30R2

mrH=62kJ·mol-m p

T2K H K

T2H 1173 3.00 2.30mrH=166kJ·mol-m

(1173973

Kp=pCO=9.5110-31002所以VCO/V總=pCO/p 所以體積分?jǐn)?shù)為：(9.5110-1/100)1000.951

Fe2++ Fe3++ K1=1.1Cu++ Cu2++Co2+------------ K2=1.6②-①Cu++ Cu2++K=K

1.610

=1.5所以Fe3+可氧化Cu2+，而且反應(yīng)相當(dāng)完

Kp兩式對(duì)照可得rHm/2.303R=1.97104所以rHm377kJ·mol-

←←←→

2nCOC=0.631/98.91=0.006382p0=nRT/V=(0.006388.31900)/0.472=101 CO(g) 設(shè)分解率為初始分壓力/100 平衡分壓力/100 p0(1- ptp01-p0p0p01+所以ptp01187.2101)185.3pCOC=p0(1-)=0.997100(1-0.878)=12.22

CaO+

=3.9102=C+ Kp2=CaCO3+ CaO+ Kp3=pCO2= 2pCO=3.91022

=2.7102p總pCO+pCO6.61022不

減小

1.0x=6 x=[HCHO]=210-4mol·dm-x

lgx251000

x= 2.30 3030.42g/100g 694kPa

。CO2101.3kPaKp=p=101.3= lgKT1168

=5.60910-3=7.282-T3518因?yàn)镚RTr所以G=-8.3110-3(273250)2.30lgKp=-36.5(kJ·mol-1r因?yàn)镚=G2.30RTr r20.26=-36.5+2.308.3110-3523lg 1.10310=15.6(kJ·mol-1)>

Kp =1/K'Kc=Kp(RT)-n=0.535(0.08311000)1/K' M=wRT=1.208.311473/101=145(g·mol-1254(1)1272= =1 =p總H=101 =14.3 r pIp總

1

=10120.752=86.61Kp

(86.6/100)14.4/

=Kp='250.5= ’=1

=6.70=0.04962p'SO2

=0.04968.31357=147p'Cl2=101 SO2+pSO2Cl2=147–pSO2=pCl2=101+(101x)(x 100=147 2pSO=x=8322pSOCl=147-83=6422pCl=101+83=1842因?yàn)镚G+r r所以(1)G4.77+8.3110-3298

pN

/22r (pNO/p)22=-4.77+5.70lg1.071.00(105)(2.67104)=1.94(kJ·mol-(2)

=-4.77+5.70lg2.671041.00

=-8.37

r (1.07105)K1.5106，平衡常數(shù)很大，反應(yīng)完全；NO和O2NO25.610-(1)Zn置換Cu；(2)Zn置換Cu；(3)無(wú)反應(yīng)發(fā)生

[Cu(1) 3 [Cu2 [Fe2]2[I [Ca2 [Fe3]2[I

[H

2Cl2 (3)2p2p

[Zn2[H

pNH3

pH

[H2][CO2Kp=p2p

Kc[HH2O Kp

pCl222

Kc

[COCl2]p1/2 1/ [O]1/2[NKp=

Kc

因?yàn)榉磻?yīng)的n= 所以Kc=K= 起始濃度/mol·dm- V

V平衡濃度/mol·dm-x

0.50 V(0.50x)(0.5x)V2

=nCO=nHO=x=0.03022nCO=0.472H =0.020H2

4HSO-+ H3O++SO44 1.0–x x21.0

=1.010- x=[H+]=0.095mol·dm-=0.095=9.5

4HSO-+ H3O++SO44 0.10–x' 0.10

=1.010- x'=[H+]'=0.027mol·dm-'=0.027=27

K 1.29 1.6610K373=2.510-

2.300.0083137330

因?yàn)閞H(2)>H rH(2)>Hm 所以rH(2)最m

mmrH(1)=fH=-287.0kJ·mol-mmG=H-TSr r=-1614.9-298(282.4-18.8-2202.7)10-=-1572.6(kJ·mol-G=-2.30RTlgKrlgK 2.300.00831

K=1.3

K=1 K

(K3

(K 1.0(4.3107)(5.61011)(5.71010)

=1.3

2NO(g)+ 1.0- (/2)2(1.0)

=1.810- =3.310- 初始濃度/mol·dm- 99

=1.310- 1.310-2–x x20.013

= x=0.013mol·dm-n總0.0130.0130.026mol·dm-V總p=n總RT0.0268.3110002.2102總V

-2.30RTlgK(773K)=H-773S r-2.30RTlgK(298K)=H-298S r-49.8=H-773S r-139.8=H-298S r- -90.0=475SrS=-189J·mol-1·K-mrm代入

rH=-196kJ·mol-

(1)提 (2)降 (3)提降 (5)無(wú)變

平衡常數(shù)加大，因I2不變，因I2I(1)Ag2SO4(s)減少；(2)(3)(4)(5)

H2(g)+1S2(g)= Kp=22Br2(g)+H2S(g)=2HBr(g)+1 Kp=9.0022(1)+ H2(g)+Br2(g)=Kp=0.8009.00104=7.20n=0 Kc=Kp=7.20104(2)2

(1)

=7.20 =99.3(2)2的逆反應(yīng)-(1)式即得(3)Kc3

=2.410- (Kc2)2 0.551.4

I

I2+ [IKc=32]][I2

[I]=0.0060- =0.0060(mol·dm- [I]=0.2500-0.0060=0.2440(mol·dm-586586Kc0.0200

=7.2

Kc2.3所以GRTlnKc8.3110-3298ln(2.310319.2kJ·mol-rrGmRTlnKcRTlnQc=-19.2+8.3110-32.30298=-13.5kJ·mol-1<

10.101.0

2SO2(g)+O2(g)=mrH=-396.72-(-296.82)=-199.8(kJ·mol-mG=-371.12-(-300.22)=-141.8(kJ·mol-r因?yàn)镚RTlnKr 所以lnK=141.8 所以K=7.18 8.314 因?yàn)?p

T2

199.8

K H ) Kp

2K 2.30R2

2.303 673K

=3.10

Kp22.2

H2(g)I2(g) po-pKp H

pHIx

=pHI= x=0.0352xHI的解離度2x

1

0.03522100%=6.58

(1)設(shè)解離度為 ，開始時(shí)H2O(g)為1mol·dm-2 H2(g)+12開始物質(zhì)的量 2平衡物質(zhì)的量 1- 12

1/11][2(11所以Kp =(1)1/2=(1.10510-4)1/2(2.2110-4)=2.3210-21 211pK7.410О(3)pKKK pG=G2-G1=-RTln(10K)-(-RTlnK)=-RTln10=-5.7(kJ·mol-r r r

HCO H+(aq)+CO2- G=-528.10+587.06=58.96(kJ·mol-rlgK2

58.96 [H][CO2=- K2 32.3038.3143K2=4.810-rGmRTlnKcRTlnQ

[HCO CaCO3200.20Kp= /p=

pCO=1162pV= 116V=0.208.31 V=15初始濃度/mol·dm-0平衡濃度/mol·dm-0.0100(1-2Kc=K(RT)-n=0.133(0.08311558)-1=1.0310-(0.01002)0.0100(1

=1.0310-=2.610- =1

2pHO=9.110-32Kp=K=

9.1 )2=8.310-(1) = =p=34 K=pH2SpNHpp

(34)2=pH

p3p

=

pH 2= =13 H2p總=pNH+pHS=93+13=106 pp K 11.83=───= G=-2.30RTlgK=5.19kJ·mol-r G=r CO(g)+ 202(1-) (202)2K =8.010-202(1H 104[(2.308.31103373lg8.0109S rmr =0.124(kJ·mol-1·K-1)=124(J·mol-1·K-

O2(g)+ 2Cl2(g)+ 平衡時(shí)pHO=pCl =(207-187)=114 pHCl=(470-2187)=96 (187/100)2(187/100)Kp=

(114/100)(96/100)

=Kc=Kp(RT)-Kc=130.0831753=8.1 PCl3(g)+初始濃度/mol·dm- 平衡濃度/mol·dm- 0.70

Kc

[PCl3][Cl2][PCl5]

== 初始濃度/mol·dm- 平衡濃度/mol·dm- 0.20– 0.25+ 0.25+cK=[PCl3][Cl2](0.25x)c

= x=[PCl5

0.20[PCl5]=0.20-0.054=0.15(mol·dm-[PCl3]=[Cl2]=0.30mol·dm- 0.350.200.70

=67 PCl3(g)+ 2.00– 1.00+ x= x=1.82

總4.82( /p)( /p (1.82202)(2.82202所以Kp=K = 100=5(5Kp=K=704=G=0-G

/p

(

202r G(Cl2,l)=2.30RTlgKr

=2.300.00831298lg7.04=4.83(kJ·mol-G(700K)=161-7000.25=-14(kJ·mol-rG(732K)=161-7320.25=-22(kJ·mol-r Glg r-K(700K)= (p/p)2=p3.3102 p總6.6102K(732K)= (p'/p)2=p6.1102 p'總1.2103G(298K)=25.8kJ·mol-1=-2.30RTlgKr pK=2.9510-p(pH2O/100)3=2.9510-pH2O=3.09pV= 3.09kPa1.00=n8.31n=1.2510-3

=1.25 初始濃度/mol·dm- 6.0010-=6.0010-38.31318=15.9平衡時(shí)：p總25.915.9-15.9+215.9= =62.9(25.92)K= 1 = 25.91 1(2)G=H-TSr rS=72.81000(8.313182.30lg0.678)=232(J·mol-1·K-(3)

rK

72.82.30

(

1K= G=-RTlnKp=-11.4(kJ·mol- r

2未解離前：n'SO2p'SO

=6.7=0.050=0.0508.31357=14822 K 以pA'表示SO2Cl2的分壓 (pA')

p = 1=A(1=2pSOCl=148(1-)=148(1-0.70)=442 pSO=pCl=148=104

Qp

(71/100)(2.0/100)

=1.3103>K

nn總1.615( /p) (0.770/1.615)K = (pSO/p)(pO/p (0.230/1.615)(0.615 Kc=Kp(RT)-Δn=29.4(0.0831873)=2.13(2)Qc =10.0<K0.1002 9.20= 101平衡 n總 =0.120 8.310.100+x=0.120 x=0.020所以轉(zhuǎn)化率0.020100%20由(1)可知，平衡 =1010.080=67 =1010.040=34 (34/100)Kp=K =(3)G=-2.30RTlgKr G300K=-2.308.3110-3300lg0.17=4.4(kJ·mol-r由Gibbs可 -13.6=H-400S r4.4=H-300S r解得H=58.4kJ·mol- S=180J·mol-1·K- r(50.7/100)(4)Qp =0.51>(50.7(1)G=G(NiSO4,s)+6G(H2O,g)-Gr f0=2(1.000-=(67/100)6H2O,因?yàn)?72.96(-228.2)=77.5(kJ·mol-G=-RTlnK=-8.3110-32982.30lgKr lgK=- K=310- K=( /p pH

1lgK 2pHO=0.542MnO2(s)+ Mn(s)+mfG/kJ·mol- mG=2(-228.59)-(-465.18)=8.00(kJ·mol-rlgK 8.00

=- K= 2.308.31p (3.14Qp H2O =1.0310-2H [(1013.14)H2 pp2/1002 p所以 =4.90 即(146p

)2+

)-7.15= 22p=pNO=118 pNO=282

lg = rm 2.300.00831

=pHCl=pNH=3 ( )=

p=100 p=50

2 pCO=3.92( /p =2(pCO/p2pCO2/100=1.9 pCO=27

(21)(40Qp=100100=0.37>(50

( /p =6.810-(

/p)1/2(2O22O

/p)1/(pNO(10178%/100)1/2(10121%/100)1/pNO=2.810-14( /10 =6.610-(pNO2/10

=6.810-(10178%/100)1/2(2.81014/10)0=6.610-(pNO2/102pNO=3.810-82=-2.30RT Kp2=T1=450 G(450K)=9.32kJ·mol-rT2=550 G(550K)=0.880kJ·mol-rGHTS，假定HS不隨溫度變r(jià) r r9.32=H-450S r0.880=H-550S r由(1)(2)解得：S=84.4J·mol-1·K- H=47.3kJ·mol-1r

O298K時(shí)反應(yīng)器 =101.3O2823K時(shí)未反應(yīng)前反應(yīng)器中 =101.3823=280 MgCl2(s)+MgCl2(s)+1MgO(s)+ (2.80x)1/

2.80– = 4x2+3.06x-8.58= x=2pCl=21.13100=2262O =(2.80-1.13)100=167O2

0.150(1.01105)=1.521040.1501.01=

K

pHOpCOp

=

pHO 2 p=100p pCO+pHO=101 35kPa≤pHO≤ 0.35≤HO2 18.4= 0.40– n=(0.40( /p 100 0.40 =2(2

/p)

)0.40x n總0.24pV= 101V=0.248.31 V=5.9 H lg 2.300.00831300H所以 =-57kJ·mol-

pHO=3.1730%=0.9522pHO(1)=(9.3510-51002)1/2=0.9722pHO(2)=(5.4310-51002)1/2=0.7422pHO(1)>pHO>pHO2 故以CuSO4·3H2O形態(tài)存在

Qp

pHI I H

=1<

cHI 2.00

=1.9510-2(mol·dm-H2(g)+I2(g)= p'HI=1.9510-28.31553= pHI=89.6-89.62p

= p=

=

=8.0(p) pHI=89.6-16.0=73.6 (pHI)

71.0x 平衡時(shí)： = 2.0x/ =I2 ) x=11.9 pHI=71.0-11.9=59.1 =p=2.0+11.9/2=7.95

SO2(g)1 設(shè)解離度為2 2 nO(1- no ntot=(1- + +)no=no(1+ p=d MmixdRT0.9258.3190068.5g·mol- M(SO3)=80.0g·mol- M(SO2)=64.0g·mol- M(O2)=32.0g·mol-

1)1/

)1/

) /280.01+64.0 +32.0/2=68.51/=33.6

1/

1/G=G(NO2,g)+G(O2,g)-G(NO,g)-G(O3,r f =51.8+0-86.7-163.6=-198.5(kJ·mol-因?yàn)镚RTlnKr 198.5所以lgKp= rm 2.308.31

=Kp=7.1 因?yàn)閚= 所以Kc=7.1Kp 初始濃度/mol·dm- 210- 110- 210-平衡濃度/mol·dm- 210-9– 110-9– 210-n= Kp=KKc=[NO2][O2]

x(2103

=7.1[NO][O3

(2109x)(1109x2-310-9x+210-18= x=110-9mol·dm-所以平衡時(shí)[O3110-9110-90O3幾乎100%被破壞。

2①pHO/100=(6.8910-22pHO=1.6210-122②pHO/100=(5.2510-22pHO=3.5010-122③pHO/100=(4.0810-22pHO=3.6410-120.364=59.6 當(dāng)濕度>59.6 Na2SO4會(huì)吸潮K p空

N2N

)1/2( O2O

)1/p空=

=5.110-p空

791)1/2(

211)1/

H

26736.8

2.30 (2673mrH=89kJ·mol-m無(wú)影響起始濃度/mol·dm-+0平衡濃度/mol·dm-x0.250.50–(0.50x)x2(0.25x/

=5.0(0.50)0.25x

=5.0 [CO]=x=1.410-2mol·dm-pOH149.26pKb4cNH/cNH3H2O=1:44nNH3H2O=nNH=0.5000.100=0.05004故被吸收的nNH nNHHO+nNH=0.100 3600K =0.1008.31600=1.00 =503 600=1.01

=50.36001.01H [ /p

K=K =9.810- (p=100p2[2

/p]3[N2N

/pKc=Kp(RT)-n=9.810-4(0.0831600)2=

( /100) =(55pNO2)/2 1- 551=pNO=22 =2N2O4p=55 =441

N2O4

= (p總2 2 Kp=p

1

=6.6610- 1 1 '=n'(1 (99.720.139)Kp'= =7.8610-99.7 (1)正向移 (2)不 (3)逆向移 (4)影響不 (5)逆向移

2NH3(aq)+CO2(g)+ 2NH(aq)+CO2 3AgCl(s)+ Ag(NH)+(aq)+Cl3 CaO(s)+ CuSO4(s) 33HNO2(aq)=H+(aq)+NO(aq)+2NO(g)+3

1

0.183 0.408 0.408 pPCl 2(p55

=37.1

G=-RTlnK=-8.314523ln1.8=-2.63(kJ·mol-r

NH4HCO3(s)=NH3(g)+H2O(g)+ G=31.1kJ·mol-r pNH=pHO=pCO1.0 K

pH

則ΔG=G+RT =G+RTln 2 r r r =31.1+8.31410-3298ln0.0103=-3.1(kJ·mol-3pNH=pHCl1.03ΔrGm=G+RTlnK=91.22+8.31410-3298ln0.0102=68.4(kJ·mol-r H+(aq)+HS-mfH/kJ·mol- - -mm mmrH=22.1kJ·mol-mS=-58.2J·mol-1·K-rG=H-TS=39.4kJ·mol-r rpK1=6.91HS H+(aq)+S2mfH/kJ·mol- mm -mmrH=50.7kJ·mol-mS=-77.4J·mol-1·K-rG=73.8kJ·mol-rpK2=12.9

(1)增加B的濃 (2)增加A的濃(3)加壓、降溫、減小的C濃 (4)減

CO(g)+Cl2(g)=G=-2.30RTlgK K=2r2COCl2達(dá)到極限濃度：p2

/p=10- 根據(jù)假設(shè)：pCl/p= /p=10

106

=105<K可發(fā)生超過(guò)允許極限的COCl2，因此具有這樣的性

由數(shù)據(jù)看出，Kp1=Kp2時(shí)則皆達(dá)到平衡，此時(shí)T<873K， K=p /p lg

H m 2.30

973

H1=-17.7kJ·mol- H

973 m 9732.30

H2=30.4kJ·mol-lg0.871-lgKp1

-2.30

(873-T lg1.15-lg

=2.30

Kp1Kp2，lgKp1lgKp2，由(1)、(2)lg1.15

(873-T)=lg0.871 -

(873-T2.30

2.30

T=838代入(2)Kp1Kp22pCO/pCO2

(1)+ 1N2(g)+ 2K

2pNO/2

=K

·K p4( /p)( /p p 2得反應(yīng)(3)所以K=(K)2 (K·K)2=(1.31066.510-16)2=7.110- p p

反應(yīng) K=K=K =0.00354(0.0831=反應(yīng) G=-2.30RTr=-50.8(kJ·mol-

G=-228.59-16.5-394.36+666.1=26.7(kJ·mol-rlgK=-26.7/(2.300.00831 K=2.110-(p/p)32.110- p2.8kPa<3.17 G=H-TSr r=(-241.82-46.11-393.51+849.4)–313(0.189+0.192+0.214-=19.6(kJ·mol-K=5.310-p8.1kPa>7.38

K=[0.062/1.00]= [100/[0.030/K2

[0.062/

=4.410-K=KK=2.710- G(973K)=-2.30RTlg(1)2=8.19(kJ·mol-r G(1023K)=-2.30RTlg( )2=13.5(kJ·mol-r 8.19=H-973S r13.5=H-1023S r(1)- 50S=-rS=-1.1102(J·mol-1·K-r代入 H=-96kJ·mol- G(1000K)=-96+10000.11=14(kJ·mol-r

G=-2.30RTlgKr-418.4=-2.300.00831600lgKK=31

Q1.0104=1.0106<K1或ΔrGm418.42.30RTlg1.0104/100349.6(kJ·mol-1G=-2.30RTlg(0.68)2=9.32(kJ·mol-1rcK=c

0.68 -)2(0.831450)2=1.3510-n(Br2)

0.688.31

=1.810-4=1.810-42/0.10=0.36

G(873K)=H-TSr r=-157–873[0.4236=-76(kJ·mol-

1(0.20503)-2 - G=-76=-2.30RT 2)2=2.30RT

O2)r O =7.7710-5O2Sn2+(aq)+ Pb2+(aq)+mfG/kJ·mol- mm rG=2.0kJ·mol-1=-RTlnK=-m

c(Pb2+)/cc(Sn2+)/c(Pb2+c(Sn2+m逆反應(yīng)rG=-2.0kJ·mol-1RTlnKm

c(Sn2+)/cc(Pb2+)/c(Sn2+c(Pb2

NO(g)+1Br2(l)= K=3.610 Br2(l)= K=pBr/p=28.4/101.325= (1)–1 NO(g)+1Br2(g)= K 3.6

-

=6.8102 K (0.28)2

1 2CO(g)+ K=1.410-12 2Fe(s)+ K=2.4710-21(2)–1(1)：FeO(s)+ Fe(s)+ KKKK1.0molCOxmolFeOx1.0

1.0K x

x

1.0molFeO3.4molCO

開始NH3(1.48n=8.314(27320)=0.608開始時(shí)物質(zhì)的量平衡時(shí)物質(zhì)的量01開始時(shí)物質(zhì)的量平衡時(shí)物質(zhì)的量010322平衡時(shí)總和：n(1-1n+3nn 350℃時(shí)：pVn5.051031.00=0.608(1+)8.314=

1313xN

2(1

xH

2(1

H2O(l)G=-2.30RTlgK=-2.308.31291lg2.06=9.41103(J·mol-r G=1.48103J·mol-r(1)+ G=10.89103J·mol-rG=-2.30RTlgKr10.89103=-2.308.31291lgKHK= =1.11H2

(1)HI的解離度為 H2(g)+開始時(shí)物質(zhì)的量/ 平衡時(shí)物質(zhì)的量/ 1- 1 V[H2][I2

所以Kc K1.83K1.8310-+開始時(shí)物質(zhì)的量平衡時(shí)物質(zhì)的量0c/

3.0 2.0 2 4x Kc'=[H2][I2](3.0-x)(2.0-x)1.83x=所以生成HI2x3.6mol

2KCl(s)+3O2(g)=2 S=-247.2J·mol-1·K- rH/kJ·mol- - - H=44.7kJ·mol- G=-RTlnK=H-TS=44.7–298(-247.2)10-3=118.4(kJ·mol-r r-8.31410-3298lnK=K=1.7610-1KO( /p)3/O2O 6.91013kPaO2

C6H5C2H5(g)=C6H5C2H3(g)+ 1- 1 p p1

1

1K=

(p

K

2=K 1 1KP 1KPK1C6H5C2H5(g)+H2O(g)=++11-001 11

11

11 Kp=(11)(1-

(1+K)2+10K-11K= 1.0892+0.89-0.979==

1000K

1-2.010-

2.010-

12.010-21400K：開始時(shí)物質(zhì)的量

211.2710-221000K時(shí)分壓 2

= 2

12.0212.01072.010712.02=1.0102pCO=2.010-5O =1.010-5O221400K時(shí)分壓 pCO=1.010