物理化学总复习题答案

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1、答案：一、填空题1.0.5kB=0.5kC2.0.0693min3.[(AgI)mnI-·(n-x)K+]4.K3[Fe(CN)6]5.Eθ(Cl-/AgCl/Ag)－(RT/F)ln0.016.Eθ(Pb2+/Pb)＋(RT/2F)lna7.10mol8.0.6909.降压,通入惰性气体10.降温,增压11.2,112.2,1二、选择题1.C2.A3.C4.C5.A6.B7.A8.C9.B10.D11.C三、计算题1．解:(1)t1/2=[A]0/(2k)k=[A]0/(2t1/2)[A]=[A]0-kt=[A]0-[A]0t/(2t1/2)([A]/[A]0)

2、×100％=[1-t/(2t1/2)]×100％=0已反应完,反应掉100％.(2)k=0.6932/1000=6.932×10-4sln([A]/[A]0)=-kt=-6.932×10×2000=-1.386[A]/[A]0=0.25{([A]0-[A])/[A]0}×100％=(1-0.35)×100％=75％(3)x/(1-x)=[A]0kt=[A]0[1/([A]0t1/2)]t=t/t1/2=2x=2/3=66.7％2.解:(1)k=(1/t)ln[1/(1-x)]=(1/10)ln[1/(1-0.75)]=0.139min当t'=15min时,ln[

3、1/(1-x')]=kt'=0.139×15=2.085x'=0.88=88％(2)k=(1/t[A]o)[x/(1-x)]=(1/10[A]o)[0.75/(1-0.75)]=0.3/[A]ok=(1/t'[A]o)[x'/(1-x')]=0.3/[A]o[1/(15[A]o)][x'/(1-x')]=0.3/[A]ox'=0.82=82％(3)k=x[A]o/t=0.75[A]o/10=0.075[A]ox'=kt'/[A]o=0.075×15=1.13,A已反应完.3.解:R=0.5×10-3mγ=(1-0.8)γo=0.2γo=0.2×72.75=14.55m

4、N/m泡内压力p＝p0＋△p＝p0＋4γ/R＝101.325＋[4×14.55×10-3/(0.5×10-2)]×10-3＝101.337kPa4.解:2γM2×72.8×10-3×18×10-3r=────────=───────────────=54.3nmρRTln(pr/p0)1×103×8.314×293×ln1.02p=p0＋△p=p0＋(2γ/r)=101.3＋[2×72.8×10-3/(54.3×10-9)]×10-3=2.78×103kPa5.解:(1)(-)Fe(s)→Fe2+(a=1)+2e(+)2Fe3+(a=1)+2e=2Fe2+(a=1)电

5、池Fe(s)+2Fe3+(a=1)=3Fe2+(a=1)(2)Fe3+(a=1)+3e→Fe(s)(a)Fe2+(a=1)+2e→Fe(s)(b)Fe3+(a=1)+e→Fe2+(a=1)(c)∵(a)-(b)=(c),△G(c)=△G(a)-△G(b)Eθ(Fe3+/Fe2+)=3Eθ(Fe3+/Fe)-2Eθ(Fe2+/Fe)=3×(-0.04)-2×(-0.409)=0.698VEθ=Eθ(Fe3+/Fe2+)-Eθ(Fe2+/Fe)=0.698-(-0.409)=1.107V△Gθ=zEθF=-2×96485×1.107=-213.6kJ/mol△Sθ

6、=zF(əEθ/əT)p=2×96485×1.14×10-3=220J/(K.mol)△Hθ=△Gθ+T△Sθ=-213.6+298×220×10-3=-148kJ/mol6.解:(1)(-)Zn(s)→Zn2+[a(Zn2+)]+2e(+)PbSO4(s)+2e→Pb(s)+SO42-[a(SO42-)]电池Zn(s)+PbSO4(s)=ZnSO4(m=0.01mol/kg,γ±=0.38)+Pb(s)(2)E=Eθ-(RT/zF)ln[a(Zn2+)a(SO42-)]=Eθ-(RT/2F)ln(m±γ±)0.5477=Eθ(SO42-/PbSO4/Pb)-(-

7、0.763)-(8.314×298/2×96485)ln(0.01×0.38)Eθ(SO42-/PbSO4/Pb)=-0.359V(3)(-)Pb(s)=Pb2+(a1)+2e(+)PbSO4(s)+2e=Pb(s)+SO42-(a2)电池PbSO4(s)=Pb2+(a1)+SO42-(a2)故可设计电池:Pb(s)│Pb2+(a1)‖SO42-(a2)│PbSO4(s)│Pb(s)7.(1)V1=nRT/p=1×8.314×273/101.3=22.4dm3W=nRTln(V1/V2)=8.314×273ln(22.4/224)=-5226J∵恒