作业题解(第2章化学反应的方向和程度).ppt

1.不用查表，指出下列各组物质中熵值由大到小的顺序。解：（1）S(O3,g)S(O2,g)S(O2,l)（2）S(KCl,s)S(K,s)S(Na,s),第二章化学反应的方向和程度,2.判断下列反应，哪些是熵增加的过程，哪些是熵减少的过程，并说明理由。解：(1)(电解水)H2O(l)=H2(g)+1/2O2(g)反应后气体分子总数增加，反应是熵增的过程，即S0。(2)2CO(g)+O2(g)=2CO2(g)反应后气体分子总数减少，反应是熵减的过程，即S0。(4)2HI(g)=H2(g)+I2(s)反应后气体分子总数减少，反应是熵减的过程，即S0。,3.解：(1)4NH3(g)+5O2(g)=4NO(g)+6H2O(g)S298/JK-1mol-1192.34205.03210.65188.72fG298/kJmol-1-16.48086.57-228.59S298=(iS298)生成物-(iS298)反应物=4210.65+6188.72-4192.34-5205.03=180.41JK-1mol-1G298=(ifG298)生成物-(ifG298)反应物=486.57+6(-228.59)-0-4(-16.48)=-959.34kJmol-10反应能自发进行。,(2)CaO(s)+H2O(l)=Ca2+(aq)+2OH-(aq)S298/JK-1mol-139.7569.91-53.1-10.75fG298/kJmol-1-604.04-237.18-553.54-157.29S298=(iS298)生成物-(iS298)反应物=-53.1+2(-10.75)-39.75-69.91=-184.26JK-1mol-1G298=(ifG298)生成物-(ifG298)反应物=-553.54+2(-157.29)-(-604.04)-(-237.18)=-26.90kJmol-10反应能自发进行。,4.解：(1)SnO2(s)Sn(s)+O2(g)fH298/kJmol-1-580.700S298/JK-1mol-152.351.55205.03H298=(ifH298)生成物-(ifH298)反应物=0+0-(-580.7)=580.7kJmol-1S298=(iS298)生成物-(iS298)反应物=51.55+205.03-52.3=204.28JK-1mol-1GT=H298-TS2980,(2)SnO2(s)+C(s)Sn(s)+CO2(g)fH298/kJmol-1-580.700-393.50S298/JK-1mol-152.35.74051.55213.64H298=0-393.50-(-580.7)-0=187.2kJmol-1S298=51.55+213.64-52.3-5.740=207.15JK-1mol-1GT=H298-TS2980,(3)SnO2(s)+2H2(g)Sn(s)+2H2O(g)fH298/kJmol-1-580.700-241.82S298/JK-1mol-152.3130.57451.55188.72H298=0-2(-241.82)-(-580.7)-0=97.06kJmol-1S298=51.55+2188.72-52.3-2130.574=115.542JK-1mol-1GT=H298-TS2980所以选择第三种方法最合适。,5.解：C(石墨)+H2O(g)=CO(g)+H2(g)fH298/kJmol-10-241.82-110.520S298/JK-1mol-15.740188.72197.56130.574H298=(ifH298)生成物-(ifH298)反应物=(-110.52)+0-0-(-241.82)=131.30kJmol-1S298=(iS298)生成物-(iS298)反应物=197.56+130.574-5.740-188.72=133.674JK-1mol-1G=H-TS0,6.解：(1)SO2(g)+O2(g)=SO3(g)fH298/kJmol-1-296.830-395.72S298/JK-1mol-1248.11205.03256.65fG298/kJmol-1-300.190-371.08H298=(ifH298)生成物-(ifH298)反应物=(-395.72)-0-(-296.83)=-98.89kJmol-1S298=(iS298)生成物-(iS298)反应物=256.65-248.11-205.03=-93.975JK-1mol-1G298=(ifG298)生成物-(iGfH298)反应物=(-371.08)-0-(-300.19)=-70.89kJmol-1T=298K时，,T=298K时，G=H298-TS298=-98.89-500(-93.975)10-3=-51.90kJmol-1,(2)CaCO3(s)=CaO(s)+CO2(g)fH298/kJmol-1-1206.92-635.09-393.50S298/JK-1mol-192.939.75213.64H298=(ifH298)生成物-(ifH298)反应物=(-635.09)+(-393.50)-(-1206.92)=178.33kJmol-1S298=(iS298)生成物-(iS298)反应物=39.75+213.64-92.9=160.49JK-1mol-1T=298K时，G=H298-TS298=178.33-298160.4910-3=130.50kJmol-1,T=500K时，G500=H298-TS298=178.33-500160.4910-3=98.085kJmol-1,8.解：已知：(1)FeO(s)+CO(g)Fe(s)+CO2(g)，K1=0.403(2)FeO(s)+H2(g)Fe(s)+H2O(g)，K2=0.669(2)-(1)得：CO2(g)+H2(g)CO(g)+H2O(g)，K,9.解：已知：(1)H2(g)+S2(g)H2S(g)(2)3H2(g)+SO2(g)H2S(g)+2H2O(g)(2)2-(1)2，得：4H2(g)+2SO2(g)4H2O(g)+S2(g),10.解：设平衡时有xmolNO生成。N2+O22NO起始浓度/moldm-30平衡浓度/moldm-3,11.解：设有xmolH2O(g),设密闭容器体积为Vdm3。CO(g)+H2O(g)CO2(g)+H2(g)起始浓度/moldm-32/Vx/V00平衡浓度/moldm-31/V(x-1)/V1/V1/V,12.解：设开始时PCl5为1molPCl5(g)PCl3(g)+Cl2(g)起始时物质的量/mol100平衡时物质的量/mol1-180%0.80.8=0.2平衡时总物质的量为1.8mol,13.解：设最少需H2为xmol，设密闭容器总压力为P。Ag2S(s)+H2(g)2Ag(s)+H2S(g)起始量/mol1.0 x00平衡量/mol0 x-1.01.0则平衡时气体总物质的量为xmol，,14.解：（1）,（2）设有xmolFeO转化为铁。FeO(s)+CO(g)Fe(s)+CO2(g)起始浓度/moldm-30.50/100.10/10平衡浓度/moldm-3(0.50-x)/10(0.10+x)/10,15.解：,反应逆向进行。,反应正向进行。,16.解：已知：Fe(s)+CO2(g)FeO(s)+CO(g)(1)Fe(s)+H2O(g)FeO(s)+H2(g)(2)(1)-(2)得：CO2(g)+H2(g)CO(g)+H2O(g)K当T1=973K时，K1=1.47/2.38=0.618当T2=1073K时，K2=1.81/2.00=0.905当T3=1173K时，K3=2.15/1.67=1.29由于