大学化学(第二版)部分习题参考答案

第一章部分习题参考答案=0-2×(-16.45)=32.90(kJ·mol-1)[P33:6题]解题思路解:查附录4中有关数据如下：2NH3(g)=N2(g)+3H2(g)ΔfHmө(298.15K)/(kJ·mol-1)-46.1100Smө(298.15K)/(J·mol-1·K-1)192.45191.50130.684ΔfGmө(298.15K)/(kJ·mol-1)-16.4500△rGmө(298.15K)=∑vi△fGmө(生成物)-∑vi△fGmө(反应物)解:(1)T=298.15KΔrGmө(T)≈ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=92.22-398.15×198.65×10-3=13.13（kJ·mol-1）根据:ΔrGmө(T)≈ΔrHmө(298.15K)–T•ΔrSmө(298.15K)计算.(2)T=398.15K△rSmө(298.15K)＝∑viSmө(生成物)-∑viSmө(反应物)＝[191.50+(3×130.684)]-(2×192.45）=198.65(J·mol-1·K-1)△rHmө(298.15K)＝∑vi△fHmө(生成物)-∑vi△fHmө(反应物)＝0–2×(-46.11）=92.22(kJ·mol-1)2NH3(g)=N2(g)+3H2(g)(3)T=300℃=300+273.15K=573.15K根据ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ计算.ΔrGmө(T)=ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=92.22-573.15×198.65×10-3=-21.63(kJ·mol-1)Q===100(PNH3/Pө)2(PN2/Pө)1·(PH2/Pө)3(1000/100)1·(1000/100)3(1000/100)2ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ=-21.63+2.303×8.314×10-3×573.15×lg(100)=0.318(kJ·mol-1)p(O2)=101.325Kpa×21％=21.28KpaP(O2)Pө1P(O2)/Pө21.28Kpa100Kpa解：查表可知：Sn(s)+O2(g)=SnO2(s)ΔfGmө(298.15K)/(kJ·mol-1)00-519.7[P33:3题]解题思路Q====4.699当T=298.15K时，ΔrGmө(T)=△rGmө(298.15K)△rGmө(298.15K)=∑vi△fGmө(生成物)-∑vi△fGmө(反应物)=(-519.7)-0=-519.7(kJ·mol-1)(1)根据ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ计算.∴P(O2)＜10-89.04Kpa2.303RT-ΔrGmө(T)1P(O2)/PөP(O2)PөlgQ＞=91.04,则:Q＞1091.04即:ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ＞0,非自发。ΔrGmө(T)+2.303RTlgQ＞0,(2)使Sn在常温空气中不被氧化,应使反应非自发.Q==＞1091.04∴常温空气中使Sn不被氧化所需O2最高压力应低于10-89.04Kpa．=△rGmө(298.15K)+2.303×8.314×10-3×298.15×lg(4.699)=-515.86(kJ·mol-1)＜0,向正方向自发进行.∴金属锡制件在常温的空气中能被氧化。ΔrGm(T)=ΔrGmө(T)+2.303RTlgQSn(s)+O2(g)=SnO2(s)[P34:8题]设汽车内燃机内温度因燃料燃烧反应达到1300℃，试计算该反应:N2(g)+O2(g)=2NO(g)在1300℃时的标准摩尔吉布斯函数变和标准平衡常数Kө。解:查表知:N2(g)+O2(g)=2NO(g)ΔfHmө(298.15K)/(kJ•mol-1)0090.25Smө(298.15K)/(J•mol-1•K-1)191.50205.14210.76△rHmө(298.15K)＝∑vi△fHmө(生成物)-∑vi△fHmө(反应物）＝2×90.25＝180.50kJ•mol-1△rSmө(298.15K)＝∑viSmө(生成物)-∑viSmө(反应物）＝2×210.76–(191.5+205.14)＝24.88J•k-1•mol-1ΔrGmө(T)=ΔrHmө(T)-T·ΔrSmө(T)lgKө=2.303RT-ΔrGmө(T)lgKө=-△rGmө/2.303RT=-（141.36）/（2.303×8.314×10-3×1573.15）=-4.69T=1300℃=1300+273.15=1573.15KΔrGmө(T)≈△rHmө(298.15K)–T•△rSmө(298.15K)=180.50-1573.15×24.88×10-3=141.36kJ•mol-1Kө=10-4.69=2.0×10-5答：略。[P33:4题]解题思路查附录4可知：CaO(s)+SO2(g)+1/2O2(g)=CaSO4(s)ΔfHmө(298.15K)/(kJ·mol-1)-635.09-296.830-1433Smө(298.15K)/(J·mol-1·K-1)39.75248.22205.14106.7(1)在标准态下自发进行的温度T的计算：△rSmө(298.15K)＝∑viSmө(生成物)-∑viSmө(反应物)＝[1×106.7]–[1×39.75+1×248.22+1/2×205.14]＝-283.84J·mol-1·K-1△rHmө(298.15K)＝∑vi△fHmө(生成物)-∑vi△fHmө(反应物)＝[1×(-1433)]–[1×(-635.09)+1×(-296.83)+1×0]＝-500.08kJ·mol-1解:ΔrGmө(T)≈ΔrHmө(298.15K)-T·ΔrSmө(298.15K)＜0时自发进行。即:-500.08-T×(-283.84)×10-3＜0∴T＜1762K(2)T=1250K时：CaO(s)+SO2(g)+1/2O2(g)=CaSO4(s)根据ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ计算。ΔrGmө(T)≈ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=-500.08-1250×(-283.84)×10-3=-145.28kJ·mol-1ΔrGm(T)=ΔrGmө(T)+2.303RTlgQ=-145.28+2.303×8.314×10-3×1250×lg(22.37)=-112.98kJ·mol-1＜0p(SO2)=10kPa,p(O2)=20kPa,系统为非标准态。1[p(SO2)/pө][p(O2)/pө]1/2Q===22.371[10/100][20/100]1/2∴反应能向正方向自发进行。(3)T=1500K时：2.303RT-ΔrGmө(T)根据lgKө=进行计算。ΔrGmө(1500K)≈ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=-500.08-1500×(-283.84)×10-3=-74.32kJ·mol-12.303RT-ΔrGmө(1500K)lgKө===2.5882.303×8.314×10-3×1500-(-74.32)∴Kө=102.588=3.87×102查附录4可知：CuO(s)+CO(g)＝Cu(s)+CO2(g)【P33:第5题】解题思路ΔfHmө(298.15K)/(kJ·mol-1)-157.3-110.50-393.5Smө(298.15K)/(J·mol-1·K-1)42.6197.733.2213.7解:(1)T=298.15K△rHmө(298.15K)＝∑vi△fHmө(生成物)-∑vi△fHmө(反应物)＝[1×0+1×(-393.5)]–[1×(-157.3)+1×(-110.5)]＝-125.7kJ·mol-1△rSmө(298.15K)＝∑viSmө(生成物)-∑viSmө(反应物)＝[1×33.2+1×213.7]–[1×42.6+1×197.7]＝6.6J·mol-1·K-1ΔrGmө(298.15K)=ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=-125.7-298.15×6.6×10-3=-127.67kJ·mol-12.303RT-ΔrGmө(T)lgKө===22.362.303×8.314×10-3×298.15-(-127.67)∴Kө=1022.36=2.29×1022ΔrGmө(698.15K)≈ΔrHmө(298.15K)-T·ΔrSmө(298.15K)=-125.7-698.15×6.6×10-3=-130.31kJ·mol-1∴Kө=109.75=5.62×109(2)T=698.15K与(1)比较:表明温度不同,ΔrGmө(T)不同,平衡常数也不同。2.303RT-ΔrGmө(T)lgKө===9.752.303×8.314×10-3×698.15-(-130.31)第二章部分习题参考答案【P59:第4题】在烧杯中盛放20.0cm30.100mol·dm-3氨的水溶液,逐步加入0.100mol·dm-3HCl溶液,试计算:(1)当加入10.00cm3HCl后,混合液的pH;(2)当加入20.00cm3HCl后,混合液的pH;(3)当加入30.00cm3HCl后,混合液的pH。解:(1)当加入10.00cm3HCl后:NH3·H2O＋HCl=NH4Cl+H2O反应后（mol）0.00100.001反应后浓度:c(NH3·H2O)=0.001/0.03=0.033mol·dm-3pH=14–pOH=14-4.75=9.25(KbӨ=1.77×10-5)c(NH4+)=0.001/0.03=0.033mol·dm-3反应前（mol）0.1×0.020.1×0.010NH3·H2O与NH4Cl组成弱碱-弱碱盐缓冲体系：pOH=pKbӨ-lg=-lg(KbӨ)-lg=4.75c(NH3·H2O)c(NH4+)0.0330.033NH3·H2O＋HCl=NH4Cl+H2O反应后（mol）000.002反应后浓度:c(NH4+)=0.002/0.04=0.05mol·dm-3反应前（mol）0.1×0.020.1×0.020NH4+H++NH3，KaӨ=KwӨ/KbӨ=5.65×10-10∵c0/KaӨ=0.05/(5.65×10-10)＞500∴pH=-lg[ceq(H+)/cӨ]=5.27解:(2)当加入20.00cm3HCl后:∴ceq(H+)≈Kaθ·c0=5.32×10-6mol·dm-3【P59:第4题】在烧杯中盛放20.0cm30.100mol·dm-3氨的水溶液,逐步加入0.100mol·dm-3HCl溶液,试计算:(1)当加入10.00cm3HCl后,混合液的pH;(2)当加入20.00cm3HCl后,混合液的pH;(3)当加入30.00cm3HCl后,混合液的pH。(KbӨ=1.77×10-5)NH3·H2O＋HCl=NH4Cl+H2O反应后(mol)00.0010.002反应后浓度:c(HCl)=0.001/0.05=0.02mol·dm-3反应前(mol)0.1×0.020.1×0.030∴pH=-lg[ceq(H+)/cӨ]=1.70∵HCl为强酸,完全解离：HCl=H++Cl-∴ceq(H+)≈c(HCl)=0.02mol·dm-3答:(略)。NH4+为弱酸,解离常数小,解离出的H+可忽略。解:(3)当加入30.00cm3HCl后:c(NH4+)=0.002/0.05=0.04mol·dm-3【P59:第4题】在烧杯中盛放20.0cm30.100mol·dm-3氨的水溶液,逐步加入0.100mol·dm-3HCl溶液,试计算:(1)当加入10.00cm3HCl后,混合液的pH;(2)当加入20.00cm3HCl后,混合液的pH;(3)当加入30.00cm3HCl