2002年1月16日期末考试答案（A）一填空题1．(1)2121/211ln22)b)c)kkktkkk+=a+(2)3.476×10-26.0×10-3mol·kg-1(3)某一电流密度下的电极电势与其平衡电极电势之差的绝对值；电化学极化；浓差极化（4）(5)江河中会有大量的SiO2溶胶，遇到海水相当于加入电解质聚沉，形成三角洲。σg-sσg-lσl-s(6)ij/i/jkTkTgegeεε−−(7)[(As2S3)mnHS-(n-x)H+]x-H+(8)a不影响化学平衡b有选择性(9)mm11bpppbpVbVV=++(10)增加减少二、计算和推证体题2．解：阴极反应：Zn(s)→Zn2++2e-阳极反应：2AgCl(s)+2e-→2Ag+2Cl-电池反应：Zn(s)+2AgCl(s)→2Ag+ZnCl2T=298K时，E=1.015-4.92×10-4×(298-298)V=1.015V4-4.9210VKpET−∂⎛⎞=−×⋅⎜⎟∂⎝⎠1-1-1rm21.015V96485Cmol195.19kJmolGzFE∆=−=××⋅=−⋅-14-1-1-1rm296485Cmol(4.9210)VK94.94JmolKpESzFT−∂⎛⎞∆==×⋅×−×⋅=⋅⋅⎜⎟∂⎝⎠-3-1-1rmrmrm{195.19+298(-94.9410)}kJmol224.2kJmolHGTS∆=∆+∆=−××⋅=−⋅-3-1-1rmrrm298(-94.9410)kJmol28.3kJmolHQTS∆==∆=××⋅=−⋅3．解：8220.07180.018ln0.0141.118.314298.15100010ppMpRTrpσρ−××====×××凸凸平平1.113.13kPa=3.47kPap=×凸8220.07180.018ln0.0141.118.314298.15100010ppMpRTrpσρ−××====×××平平凹凹3.13/1.11kPa=2.83kPap=凹ppp>>凸凹平4.(1)21122211222231lnln=ln1ln7.248101.3810SkSkSSSk−Ω=Ω=Ω∆−=ΩΩ==×Ω×(2)mlnlnTTRNTqVNkTnRTRTLpNkTNkTVVVVV⋅⋅∂∂⎛⎞⎛⎞=====⎜⎟⎜⎟∂∂⎝⎠⎝⎠V=3/23/2223/23/21/22222,2π2πlnlnlnln2π2π3322ln3322NVmkTmkTqVqhhqmkTmkTTThhTqUNkTNkTRTTT−⎛⎞⎛⎞==⎜⎟⎜⎟⎝⎠⎝⎠∂⎛⎞⎛⎞=⋅⋅=⎜⎟⎜⎟∂⎝⎠⎝⎠∂⎛⎞==⋅=⎜⎟∂⎝⎠V+5.解（1）111-1-1-1111-1-1-1a,-111-1-1lglglgsss200020004000lglglg4.0(4.0)8.0sss2.303lgln2.303lgsskkkkkkkkkTTTEkkCRT−−−−−==−=−=−+−−=−+==−+比较上两式得：-1-1a,-140002.3038.314Jmol76.59kJmolE=××⋅=⋅（2）T=400K时，-1-1111-1-120002000lg4.0=0.1slg4.0=0.1ssskkkkTT=−+=−ABt=00.50.05t=t0.5-x0.05+x1-1d(0.5-)-(0.05+)=0.1(0.5-)-0.01(0.05+)0.04950.11dxkxkxxxt==x−移项，积分：00d0.0495dln0.110.04950.110.0495-0.11xxxxtxx==−∫∫当t=10s时，-30.0495ln1.1=0.3moldm0.0495-0.11xx=⋅此时，-3-3AB0.2moldm0.35moldmcc=⋅=⋅三、分析能力考察题6.(1)首先说明动力学方程不能从计量方程得出，需...
