同济大学普通化学 第一章、二章习题答案(详细)

xxx公司同济大学普通化学第一章、二章习题答案(详细)文件编号：文件日期：修订次数：第1.0次更改批准审核制定方案设计，管理制度普通化学（新教材）习题参考答案第一章化学反应的基本规律(习题P50-52）16解（1）H2O(l)==H2O(g)fH/kJmol1285.83241.82S/Jmol1k169.91188.83rH(298k)=[241.82(285.83)]kJmol1=44.01kJmol1rS(298k)=(188.8369.91)Jmol1k1=118.92Jmol1k1(2)∵是等温等压变化∴Qp=rH(298k)N=44.01kJmol12mol=88.02kJW=PV=nRT=28.315Jk1mol1298k=4955.7J=4.956kJ(或4.96kJ)∴U=Qp+W=88.02kJ4.96kJ=83.06kJ17解（1）N2（g）+2O2（g）==2NO2(g)fH/kJmol10033.2S/Jmol1k1191.6205.14240.1∴rH(298k)=33.2kJmol12=66.4kJmol1rS(298k)=(240.1Jmol1k1)2(205.14Jmol1k1)2191.6Jmol1k1=121.68Jmol1k1(2)3Fe(s)+4H2O(l)==Fe3O4(s)+4H2(g)fH/kJmol10285.831118.40S/Jmol1k127.369.91146.4130.68∴rH(298k)=[1118.4(285.834)]kJmol1=24.92kJmol1rS(298k)=[(130.684+146.4)(27.33+69.914)]Jmol1k1=(669.12361.54)Jmol1k1=307.58Jmol1k118. 解：2Fe2O3(s)+3C(s,石墨)==4Fe(s)+3CO2(g)fH(298k)/kJmol1824.2S(298k)/Jmol1k187.45.7427.3fG(298k)/kJmol1742.2∵rG=rHT•rS∴301.32kJmol1=467.87kJmol1298k•rS∴rS=558.89Jmol1k1∴rS=3S(CO2(g)298k)+27.3Jmol1k1487.4Jmol1k125.74Jmol1k13∴S(CO2(g)298k)=1/3(558.89+192.02109.2)Jmol1k1=213.90Jmol1k1fH(298k,C(s,石墨))=0fG(298k,C(s,石墨))=0fH（298k,Fe(s)）=0fG（298k,Fe(s)）=0rH=3fH(298k,CO2(g))2fH(298k,Fe2O3(s))467.87kJmol1=3fH(298k,CO2(g))2(824.2kJmol1)∴fH(298k,CO2(g))=1/3(467.871648.4)kJmol1=393.51kJmol1同理rG=3fG(298k,CO2(g))2fG(298k,Fe2O3(s))301.32kJmol1=3fG(298k,CO2(g))2(742.2kJmol1)∴fG(298k,CO2(g))=1/3(301.321484.4)kJmol1=394.36kJmol119.解6CO2(g)+6H2O（l）==C6H12O6(s)+6O2（g）fG(298k)kJmol1394.36237.18902.90∴rG(298k)=[902.9(237.186)(394.366)]kJmol1=4692.14kJmol1>0所以这个反应不能自发进行。20．解(1)4NH3(g)+5O2(g)==4NO(g)+6H2O(l)fG(298k)/kJmol116.4086.57237.18∴rG(298k)=[(237.18)6+86.574(16.4)4]kJmol1=1011.2kJmol1<0∴此反应能自发进行。(2)2SO3(g)==2SO2(g)+O2(g)fG(298k)/kJmol1371.1300.190∴rG(298k)=[(300.19)2(371.1)2]kJmol1=141.82kJmol1>0∴此反应不能自发进行。21．解（1）MgCO3(s)==MgO(s)+CO2(g)fH(298k)/kJmol11111.88601.6393.51S(298k)/Jmol1k165.627.0213.8fG(298k)/kJmol11028.28569.3394.36∴rH(298k)=[601.6+(393.51)(1111.88)]=116.77kJmol1rS(298k)=[213.8+27.065.6]=175.2Jmol1k1rG(298K)=[(394.36)+(569.3)(1028.28)]=64.62kJmol1(2)rG(1123K)=rH(298k)TrS(298k)=116.77kJmol11123k175.2Jmol1k1=116.77kJmol1196.75kJmol1=79.98kJmol1又∵RTlnK(1123k)=rG(1123k)∴8.315Jmol1k11123klnK(1123k)=(79.98)kJmol1∴K(1123k)=5.25103(3)∵刚刚分解时rG(T)=rH(298k)TrS(298k)=0∴分解温度T可求:∴分解最低温度为666.5k22.解法一:K(298k)=5.01016∴rG(298k)=RTlnK(298k)=8.315Jmol1k1298kln(5.01016)=95.26kJmol1∵rG(298k)=rH(298k)298krS(298k)∴95.26kJmol1=92.31kJmol1298krS(298k)∴rS(298k)=9.90Jmol1k1∴rG(500k)=rH(298k)500krS(298k)=92.31kJmol1500k9.90Jmol1k1=97.26kJmol1而rG(500k)=RTlnK(500k)=8.315Jmol1k1500klnK(500k)∴lnK(500k)===23.40∴K(500k)=1.451010解法二：∵ln===15.05∴=2.910-7∴K(500k)=2.910-7K(298k)=2.910-7(5.01016)=1.45101023.解：N2(g)+3H2(g)==2NH3(g)fH(298k)/kJmol10045.9S(298k)/Jmol1k1191.6130.68192.8∴rH(298k)=2(45.9)kJmol1=91.8kJmol1S(298k)=(2192.8191.63130.68)Jmol1k1=198.04Jmol1k1rG(T)=rH(298k)TrS(298k)=0=91.8kJmol1T(198.04Jmol1k1)=0∴T==463.5k∴T>463.5k时反应能自发进行。24.解：（1）（2）得：CO2(g)+H2(g)⇌CO(g)+H2O(g)根据化学平衡得多重规则，此反应的K(T)为：K(T)=K1(T)K2(T)∴该反应各温度下的平衡常数为:T/(k)973107311731273K(T)0.6180.9051.291.66根据rG(T)=RTlnK(T)=rH(T)TrS(T) 假设rH(T),rS(T)不变，则∵K1(T)〈K2(T)T1〈T2即T↑则K(T)↑，rHm>0∴该正反应为吸热反应。25．解：平衡反应2SO2(g)+O2(g)⇌2SO3(g)初始时各物质的量/mol110平衡时各物质的量/mol1(10.615)2(10.615)2=0.230.615=0.77平衡时物质总量=（0.23+0.615+0.77）mol=1.615mol∴=29.4326解:(1)(2)(3)27.解法一:根据阿仑尼乌斯公式:(T1=301k)(T2=278k)结合上述两式，得:改写此式，得…….(1)因为，反应速率与变酸时间成反比，因此，……(2)将(2)代入(1)，=7.52104Jmol1=75.2kJmol127.解法二:∵k=A∴==Exp又∵反应速率与变酸时间成反比∴==Exp=12∴=75.2kJmol128.解:I2⇌2I(快)H2+2H⇌2HI（慢）快反应，很快达到平衡，故有：K（T）==∴[I]2=K(T)[I2]对于分步反应，总的反应速率由较慢的步骤决定，即由第二步反应决定，即：v=k[H2][I]2其中[I]2项可用[I]2=K(T)[I2]这一式子代入。∴v=k[H2][I]2=k[H2]（K(T)[I2]）=（kK(T)）[H2][I2]=k’[H2][I2]29．解：v=k[CO][NO2]=0.50dm3mol-1s10.025moldm30.040moldm3=5.0104moldm-3s-130．解：（1）（2）=（3）∴===2.01010第二章水基分散系习题(P68-69)11.解:(1)盐酸溶液的量浓度C==12.23moldm-3(2)溶液的质量摩尔浓度m为：===16.44molkg-1(3)按照(2)计算结果,该盐酸溶液中,每kgH2O中含16.44molHCl,而1kgH2O相当于=55.56molH2O∴该溶液中溶剂H2O与溶质HCl的摩尔总数为:=(55.56+16.44)mol=72mol∴==0.23==0.77或=12.23mol=41.32mol∴==0.23==0.7712.解:（1）=37.5%（2）==10.43moldm3（3）乙醇的摩尔分数为：=0.67（4）=13.04molkg113.解：设应加入尿素的量为m,∵(为溶液中尿素的摩尔分数)∴0.200kPa=3.17kPa∴==0.0631∴=0.0631=可得：m=224.5g14.解：∵设该溶质的摩尔质量为M则29.30kPa27.62kPa=29.30kPa×=∴M=86.93gmol115.解：(1)相同质量（10g）的情况下：=58.5=80=132∴物质的质量摩尔浓度m分别为0.171molkg-10.125molkg-10.076molkg-1考虑电离结合实际粒子浓度m’0.342molkg-10.250molkg-10.228molkg-1Tf最多次之最小（2）相同物质量（0.01mol）时的冰点下降顺序:（NH4）2SO4>NH4NO3>NaCl(3C)(2C)(2C)16.解:∵两溶液的凝固点下降相同:=m1=m2=∴m1=m2即:=(M为所求未知物的摩尔质量)∴M=171.04gmol117.解:=Kbm=80.960C80.150C=0.810C=0.81k∴0.81k=2.53kkgmol-1得出溶液中溶解的单质硫的摩尔质量为M==253gmol-1∴该单质硫分子中含硫原子数为X==818．解：=m8k=1.86kkgmol-1∴m=774.2g19.解：=m按题意可得：16k=6.85kkg