物理化學(xué)(復(fù)旦大學(xué)藥學(xué)院第一章習(xí)題答案-資料下載頁(yè)
2025-01-09 00:28本頁(yè)面
【正文】 – kJ mol 1 rUOm= rHOm– nRT=– +3 10 3= kJmol 1 由燃燒熱數(shù)據(jù)計(jì)算： rHOm=3 (– )– (– ) =– kJ mol 1 rUOm= rHOm– nRT=– +3 10 3= kJ mol 1 29. 反應(yīng) KCl(s) K+(aq, ) + Cl (aq, ) fHOm(kJ mol 1) – ? – rHOm= kJ mol 1 rHOm= (n fHOm)產(chǎn)物 – (n fHOm)反應(yīng)物 =[ fHOm (K+,aq, )– ]– (– ) fHOm (K+,aq, )=– kJmol 1 第一章習(xí)題解答 第 6 頁(yè) 共 6 頁(yè) 30. 生成反應(yīng) H2(g) + (g) = H2O(g) rHOm,298=– kJ mol 1 Cp,m(J K 1 mol 1) Cp=– (+ )= J K1 ? ????? TCHH p dOm ,29 8rOm ,37 3r =– + (373– 298) 10 3 =– kJmol 1 31. 反應(yīng) N2(g) + 3H2(g) = 2NH3(g) rHOm,298=– kJ mol 1 a b 103 c 107 N2(g) – H2(g) – NH3(g) – – – ? ????? TCHH p dOm ,29 8rOm ,39 8r ? ?? ???????? TcTbTaH d2Om ,2 9 8r 39829832Om ,29 8r 3121 ?????? ???????? cTbTaTH 3982983723Om ,29 8r )109117(31)10662(21)4162( ?????? ????????? ?? =– 92880+[– 6241+2178– 144]= –97086 J mol 1 32. 按圖示過(guò)程計(jì)算： HH2=nCp,m?T=(473–291)=5296 J HHI=nCp,m?T=2?(473–291)=10592 J HI2=?H1(s,291 ) + H2(s l) + ?H3(l, ) + H4(l g) + ?H5(g, 473K) = (– 291)+16736+ (– )+42677+ (473– ) =69632 J HH2+ HI2+?rHOm,473=?rHOm,291+ HHI 5296+69632+?rHOm,473=49455+10592 ?rHOm,473=– kJ mol 1 H2(g) + I2(s) H2(g) + I2(g) ?rHOm,291= kJ mol 1 2HI(g) 2HI(g) Cp,m= J?K–1?mol–1 T= ?熔 HOm=16736 J mol 1 Cp,m= J?K–1?mol–1 Cp,m=7R/2 J?K–1?mol–1 T= ?蒸 HOm=42677 J mol 1 s l l g ?rHOm， 473=？ HHI =7R/2 HI2 HH2 =7R/2
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