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1、第1章化学热力学基础习题1.计算下列各体系的热力学能变化,已知:(1)体系从环境中吸热500J,同时,环境对体系做功600J。(2)体系向环境放热2000J,并对环境做功800J。解:(1)△U=Q+W二500+600=1100J(2)AU二Q+W二-2000-600=-2600J2.计算下行反应的标准反应焙变(1)2Al(s)+胆。3⑸A12O3(s)+2Fe(s)(2)C2H2(g)+H2(g)—C2H4(g)解:(1)2Al(s)+Fe2C)3(s)-*Al203(s)+2Fe(s)AfH^CkJ^mor1)
2、0-824.2・1675.70ArH9m=AfHem(Al203,s)+2AfHem(Fe,s)-2AfHem(Al,s)-2AfHem(Fe2O3,s)=-1675.7+2X0-2X0-(-824.2)=-851.5(kJ・mol“)(2)C2H2(g)+H2(g)-C2H4(g)△fHk(kJ・moP)226.73052.26ArH°m=AfHem(C2H4,g)-AfHem(C2H2,g)-AfHem(H2,g)=52.26-226.73一0=-174.47(kJ・moP)1):3・由下列化学方程式计算液体过
3、氧化氢在298K时的△fH0m(H2OArH°m=-241.82kJ・mol"ArHem=-926.92kJ-mol-1ArH0m=-1070.6kJ-mof1ArH°m=・498.34kJ^mof1ArHem=51.46kJ・mol"解:①由⑴式可toAfH0m(H2O,g)=・241.82kbmol20(g)=O2(g)ArHem二・49&34kJ・mol"ArH°m=AfH°m(O2,g)・2AfH0m(O,g)=0-2AfH°in(O,g)=-498.34kJ*mol_1AfH0m(O,g)=249.17k
4、J-mof1ArHem=・926.92kJ・mol③2H(g)+O(g)=H2O(g)△fHk(kJ・moP)249.17・241.82ArH°m=△lH°m(H2O,g)・2a(HkH,g)・△fH0m(O?g)=・241.82-2AfHem(H,g)-249.17=-926.92AfHem(H,g)=217.97kJ-mol4③2H(g)+20(g)=H2O2(g)ArH°m=・1070.6kJ・mol,AfH^XkJ^mor')217.97249.17ArH°m=AfHem(H2O2,g)-2AfHem(H,
5、g)-2AfHem(0,g)=AfHem(H2O2,g)-2X217.97-2X249.17=・1070.6AfHem(H2O2,g)=-136.32kJ-moF1④H2O2(1)=H2O2(g)ArH°m=51.46kJ-mof1△fH%kJ・mol」)-136.32ArHem=AfHem(H2O2,g)・AfH0m(H2O2,l)=-136.32-AfH0m(H2O2,1)=51.46AfHem(H2O2,1)=-187.78kJ-mol44•在373K,101.3kPa下,2.0mol出和l・0molO?反应
6、,生成2.0mol的水蒸气,总共放热484kJ的热量,求该反应的△『垣和公Uo解:2H2(g)4-O2(g)=2H2O(g)(1)定压下Qp=ArH=-484kJ即△rHni=-484kJ・mol"(2)AU=Q+W=Q・ZRT=_484_(2-2-1)x8.314x10*3x373=-480.9kJ5・用弹式量热计测得下列反应C7HI6(1)+11O2(g)->7CO2(g)+8H2O(1)在298K时Qv=-4804试求反应的等压热效应QpO解:Qp=Qv+△nRT=-4804+(7-11)x&314X10
7、'3x298=-4813.9kJ6.已知298K时:(1)C6H4(OH)2(aq)->C6H4O2(aq)+H2(g),ArHem,j=177.4kJ-mof1(2)O2(g)+H2(g)H2O2(aq),ArHem,2=-191.2kJ-mol*(3)l/2O2(g)4-H2(g)->H2O(g),ArH°m.3=-241.8kJ-mol1(4)H2O(g)->H2O(1),ArHem,4=・44.0kJ・mol"试计算下而反应的△rHem:C6H4(OH)2(aq)+H2O2(aq)->C6H4O2(aq)+
8、2H2O(1)解:C6H4(OH)2(aq)+H2O2(aq)->C6H4O2(aq)+2H2O(1)(5)*.*反应(5)=反应(1)■反应(2)+2x[反应(3)+反应(4)]AArHem.5=△血]-ArHem,2+2x(ArHem,3+ArHem,4)=177.4-(-191.2)+2x(-241.8・44・0)=-203kJ・mol"7.甘油三油酸