物化典型例題解答

1、 1.1mol1MPa0.2MPa12.1-imOK(12UQWAUH0AHT2aSag300.15K2 # #1211Wp(VV)e21()(5VV)0.8pV0.8nRT111110.81B8.314300.151996JQUW1996JSnRln(V/V)2118.314ln(5.1)13.38J1GHTS0300.1513.384016J # 1.298.15K1molO2101.325kPa.0101.325kPaQWs(10)U0H=02QWnRTrrlnP28.314298.15Inp11/4443J6S體系QrT體系4443J14.90JK298.15KS環(huán)境Q4443J14.

2、90JK1環(huán)境298.15KSSS0(1隔離體系環(huán)境 2.293K5933Pa(1)100gx；（2ppp(Pa)（4分）y（2分）11826Pa32和46(1)nA=100/32=3.125mol,n=100/46=2.17mol,x=n/(n+n)=2.17/(3.125+2.17)=0.41(2)理P甲=p*x1182610.41:p乙=p*x59330.41=2433Pappp6977Pa2433kPa(3)根yp/p=6977/9410=0.74142.80100kPa1分1分1分9410Pa1238.7kPa80y)=0.3003.(8分)應(yīng)NHHS(s)4NH(g)HS(g)32

3、空容器中，在298Km298K)93.72kJ.mol-1NHHS(s)4m(10)2分3分1分2分70KPam298K a3.298.15K9srmK(A)g(B)gfHmkmol-0.13110.16fGmkmol71.29150.67298.15KHrmGrmNHHS(s)NH(g)HS(g)432/KPa00/KPaPePe298.15K時，系統(tǒng)平衡總壓P=2Pe,1=70KPa(1Pe1=35Kpa1K1=(Pe,/P)2=0.1225(2rGm=RTIn0=5204.6jmol-i3rHmrGmrSmT393.721035204.62946.88j.molLK氧298.15CH3C

4、H2CHCH2(g)CH2=CHCH=CH2+H2g(1)GG(B)G(H)G(A)rmfmfm2fmTOC o 1-5 h z150.67071.2979.38kJmol12HH(B)H(H)H(A)rmfmfm2fm110.16O(0.13)110.29kJBnol2SHGtrmrmrmx97.2kPa2 #(11029079380)f298.15103.7Jnol3 x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #Gr=m-RTlnKK=1.2410-144.300K203.60min250K2分2018.4min(10 x97.2kPa2 #(11

5、029079380)f298.15103.7Jnol3 #1t-ln(1x)kA1k-ln(1x)tAx97.2kPa2 #(11029079380)f298.15103.7Jnol3 #x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #T1300KxA=0.20t=3.60minx97.2kPa2 #(11029079380)f298.15103.7Jnol3 #x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #k-ln(l0.20)min0.0620min1.033103S300K3.60T2250KxA=0.20t=

6、18.4min12.02104S1k5Lln(10.20)min0.0121min250K18.4x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #kln2k1Ea117訂x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #x97.2kPa2 (11029079380)f298.15103.7Jnol3 #8.31425013000.0121ln0.0620Ea20.38103jmol-i4.有(1)A2B+1/2Ck4.5,104S1(2

7、)500kPa8力為多少。(10分)1ln2t1540s12k21(2lna-kt(2ax500ln4.5,10酣4800.21650 xA2B+1/2C/kPa50000/kPa500-97.2297.20.597.2(2P402.8+194.4+48.6=645.8kPa2E/V=rGm、5.Zn(s)|ZnCI2(0.05molkg)|AgCI(s)|Ag(s)ET1.015-4.921%T/K-298)298K1molSHQ11rmrmrZn(s)Zn+2e-2AgCI(s)+2e-2Ag+2CJZn(s)+2AgCI(s)2Ag+ZnC=-4.9210VK2TT=298KE=1.01

8、5-4.92104(298298)V=1.015V2G=-mzEF=-11.01596500mo-97.95Km-1oI(1Srm=zF嚴=1TT96500(4.9210-4)=-47.45molK(1分Hrm=G+TSrmrm=-97.95+298(-47.4510-3)=-83.81KJmoI(1rQ=Ts=298rm(-47.4510-3)=-14.15KJmol(1px97.2kPa2 #(11029079380)f298.15103.7Jnol3 #x97.2kPa2 #(11029079380)f298.15103.7Jnol3 #5.Hg(s)!Hg2CI2(s)BCI-(aq)

9、AgCI(s)Ag(s)x97.2kPa2 (11029079380)f298.15103.7Jnol3 #12E/V=0.06804-3.12104(T/K-298)1 (11029079380)f298.15103.7Jnol3 #298KrGm298K、100kPa-Sm-Hm=2mol1 #(11029079380)f298.15103.7Jnol3 #1)負極反應(yīng)：Hg(l)+Cl-=1/2Hg2Cl2(s)+e-(1AgCI(s)+e-Ag(s)+CI-Hg(l)+AgCI(s)=1/2Hg2CI2(s)+Ag(s)22g=-zEF=-10.0680496500=6.565kJmol2rmrSm.=zF()196500-3.121C4)-30.11JKmol2rmTprHm=rGm+TrSm=-15.54kJmo12(3)W=-2g=13.13kJ2rm6.SnAg可形成一個不Ag3Sn統(tǒng)相圖如下圖所示。(9(1)指出1-7各(2)畫出ab1 #(11029079380)f298.15103.7Jnol3 #1 (11029079380)f298.15103.7Jnol3 #xAg(1)L,F=2,F=2L,F=1 (11029079380)f298.15103.7Jnol3 #Sn(s)+Ag3Sn(s),F=1LAg3Sn(s),F=1L+