同济大学普通化学 第一章、二章习题答案(详细).pdf

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1、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)1fHm/kJmol285.83241.821169.91188.83Sm/Jmol k1=44.01 kJmol1rHm(298k)=241.82(285.83)kJmol11=118.92 Jmol1k1r Sm(298k)=(188.8369.91)Jmol k(2)是等温等压变化1 Qp=rHm(298k)N=44.01 kJmol 2mol=88.02 kJW=PV=nRT=2 8.315 Jk1mol1 298k=4955.7 J=4.956 kJ(或 4.96kJ)

2、U=Qp+W=88.02 kJ 4.96kJ=83.06 kJ17 解（1）N2（g）+2O2（g）=2 NO2(g)10033.2fHm/kJmol11191.6205.14240.1Sm/Jmol k1 2=66.4 kJmol1 rHm(298k)=33.2 kJmol111111r Sm(298k)=(240.1 Jmol k )2(205.14 Jmol k )2 191.6 Jmol k=121.68 Jmol1k1 (2)3 Fe(s)+4H2O(l)=Fe3O4(s)+4 H2(g)10285.831118.40fHm/kJmol11Sm/Jmol k27.369.91146.

3、4130.6811rHm(298k)=1118.4(285.83 4)kJmol =24.92 kJmol11r Sm(298k)=(130.68 4+146.4)(27.3 3+69.91 4)Jmol k=(669.12 361.54)Jmol1k1=307.58 Jmol1k1118.解：2Fe2O3(s)+3C(s,石墨)=4 Fe(s)+3 CO2(g)1fHm(298k)/kJmol 824.211Sm(298k)/Jmol k87.45.7427.31742.2fGm(298k)/kJmol rGm=rHm T r Sm 301.32 kJmol1=467.87 kJmol1 2

4、98 kr Sm11r Sm=558.89 Jmol k111111 r S3 Sm=m(CO2(g)298k)+27.3 Jmol k 4 87.4 Jmol k 2 5.74 Jmol k 31111Sm(CO2(g)298k)=1/3(558.89+192.02 109.2)Jmol k =213.90 Jmol kfHm(298k,C(s,石墨)=0fGm(298k,C(s,石墨)=0fHm（298k,Fe(s)）=0fGm（298k,Fe(s)）=0rHm=3fHm(298k,CO2(g)2fHm(298k,Fe2O3(s)1467.87 kJmol1=3fHm(298k,CO2(g

5、)2 (824.2 kJmol)11fHm(298k,CO2(g)=1/3(467.871648.4)kJmol =393.51 kJmol同理 rGm=3fGm(298k,CO2(g)2fGm(298k,Fe2O3(s)1301.32 kJmol1=3fGm(298k,CO2(g)2 (742.2 kJmol )1=394.36 kJmol1fGm(298k,CO2(g)=1/3(301.32 1484.4)kJmol19.解6CO2(g)+6H2O（l）=C6H12O6(s)+6O2（g）1fGm(298k)kJmol394.36237.18902.90110 rGm(298k)=902.

6、9 (237.18 6)(394.36 6)kJmol =4692.14 kJmol所以这个反应不能自发进行。220解(1)4NH3(g)+5O2(g)=4NO(g)+6H2O(l)1f Gm(298k)/kJmol16.4086.57237.1811rGm(298k)=(237.18)6+86.57 4 (16.4)4 kJmol =1011.2 kJmol 0 此反应不能自发进行。21解（1）MgCO3(s)=MgO(s)+CO2(g)11111.88601.6393.51fHm(298k)/kJmol11Sm(298k)/Jmol k65.627.0213.81f Gm(298k)/kJ

7、mol1028.28569.3394.361 rHm(298k)=601.6+(393.51)(1111.88)=116.77 kJmol11rSm(298k)=213.8+27.0 65.6=175.2 Jmol k1rGm(298K)=(394.36)+(569.3)(1028.28)=64.62 kJmol111(2)rGm(1123K)=rHm(298k)T rSm(298k)=116.77 kJmol 1123k 175.2 Jmol k=116.77 kJmol1196.75 kJmol1=79.98 kJmol1又 RTlnK(1123k)=rGm(1123k)8.315 Jmo

8、l1k11123 kln K(1123k)=(79.98)kJmol13 K(1123k)=5.25 103(3)刚刚分解时rGm(T)=rHm(298k)T rSm(298k)=0116.77kJ mol1 分解温度 T 可求:T 666.5k11rSm(298k)175.2J molkrHm(298k)分解最低温度为 666.5 k22.解法一:K(298k)=5.0 101616111rGm(298k)=RTlnK(298k)=8.315 Jmol k 298kln(5.0 10)=95.26 kJmolrGm(298k)=rHm(298k)298k rSm(298k)95.26 kJm

9、ol1=92.31 kJmol1298krSm(298k)11rSm(298k)=9.90 Jmol krGm(500k)=rHm(298k)500k rSm(298k)=92.31 kJmol1500k9.90 Jmol1k1=97.26 kJmol111而 rGm(500k)=RTlnK(500k)=8.315 Jmol k 500kln K(500k)rGm(500k)97.26103J mol1=23.4011RT8.315J molk(500k)ln K(500k)=K(500k)=1.45 1010解法二：rHm(298k)1K(500k)1(ln=)R500298K(298k)(

10、92.31103J mol1)202()k=15.05=8.315J mol1k1500298K(500k)=2.9 10-7K(298k)K(500k)=2.9 10-7 K(298k)=2.9 10-7(5.0 1016)=1.45 1010423.解：N2(g)+3H2(g)=2NH3(g)1fHm(298k)/kJmol0045.911Sm(298k)/Jmol k191.6130.68192.811 rHm(298k)=2(45.9)kJmol =91.8 kJmol1111Sm(298k)=(2192.8 191.6 3130.68)Jmol k=198.04 Jmol krGm(

11、T)=rHm(298k)T rSm(298k)=0=91.8 kJmol1T(198.04 Jmol1k1)=091.8103J mol1 T=463.5 k198.04J mol1k1 T463.5 k 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)K(T)9730.61810730.90511731.2912731.66根据rGm(T)=RT ln K (T)=rHm(T)T rSm(T)rHmrSm)lnK(T)(RTRmmrH

12、mK1(T)11假设 rH(T),rS(T)不变，则ln()RT1T2K2(T)K1(T)K2(T)T1 T2即 T 则 K(T)，rHm 0该正反应为吸热反应。525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=1.615 molK(873k)(pSO2pSO3p2)P)2(pO20.772()1.615=29.430.2320.615)()(p1.6151.61532(pN2)(pp(NH312pH2p)2

13、6 解:(1)K(T)p)(2)K(T)(pCO2p)(pH2OpH2(3)K(T)(pp)4(pH2O4)pH2)427.解法一:根据阿仑尼乌斯公式:lgk1 AEa(T1=301k)2.303RT1Ea(T2=278k)2.303RT2lgk2 A结合上述两式，得:lgEak111()k22.303R T1T2T1T2k)lg1.(1)T2T1k2k1t2(2)k2t1改写此式，得Ea 2.303R(因为，反应速率与变酸时间成反比，因此，将(2)代入(1)，Ea 2.303R(T1T2t301k 278k48h)lg2 2.3038.315lgT2T1t1278k 301k4h=7.521

14、04 Jmol1=75.2kJmol127.解法二:k=AeEaRTEaEa EaT2T1k1RT1RT2()=e=ExpRT1T2k26又反应速率与变酸时间成反比 Eak1t2278k 301k48h=()=12=Exp11k2t1301k 278k4h8.315J molkEa=75.2kJmol128.解:I22 I (快)H2+2H 2HI（慢）C(I)2快反应，很快达到平衡，故有：K（T）=CC(I2)CI2=I2I2=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=kH2I2其中I2项可用 I2=K(T)I2 这一式子代入。v=kH2I2=kH2

15、（K(T)I2）=（kK(T)）H2I2=k H2I229解：v=kCONO2=0.50 dm3 mol-1s1 0.025 moldm3 0.040 moldm3=5.0104 mol dm-3s-130解：（1）（2）=（3）K(3)=7K(1)K(2)1.0103=2.0 101065.010第二章水基分散系习题(P68-69)1.19kg dm337.5%-311.解:(1)盐酸溶液的量浓度 C=12.23 moldm136.5g mol (2)溶液的质量摩尔浓度 m 为：12.23moldm31dm3331(1dm 1.19kgdm)(12.23mol36.5g mol)12.23m

16、ol12.23mol =16.44 molkg-1 (3)解:(1.19kg 0.446kg)0.744kg按照(2)计算结果,该盐酸溶液中,每 kg H2O 中含 16.44 molHCl,而 1kg H相当于1000g2O18g mol1=55.56 mol H2O 该溶液中溶剂 H2O 与溶质 HCl 的摩尔总数为:n总=(55.56+16.44)mol=72 molX16.44mol(HCl)=72mol=0.23X55.56mol(H2O)=72mol=0.771.19kg dm31dm3或37.5%36.5g mol1=12.23 mol1.19kg dm31dm3(137.5%)

17、18g mol1=41.32 molX12.23mol41.32mol(HCl)=(12.23 41.32)mol=0.23X(H2O)=(12.23 41.32)mol=0.771）60.6g60.0g 100.0g=37.5%60.0g2）46g mol11.30mol(60.0g 100.0g)=10.43 moldm31.28kg dm30.125dm33）乙醇的摩尔分数为：60.0g46g mol1(60.0g=0.6746g mol1100.0g154g mol160.0g4）46g mol1100.0g=13.04 mol kg18 12.（13.解：设应加入尿素的量为 m,P

18、P*x (x为溶液中尿素的摩尔分数)0.200 kPa=3.17 kPa xx=0.200kPa=0.06313.17kPam60g mol118g mol1m=0.063160g mol11000g(10.0631)m63.1g=60g mol118g mol1可得：m=224.5g14.解：P P*x设该溶质的摩尔质量为 M则 29.30 kPa 27.62 kPa=29.30 kPa M23g200gM46g mol123g23gM=7.308mol27.62 M=86.93 gmol115.解：(1)相同质量（10g）的情况下：M(NaCl)=58.5M(NH4NO3)=80M(NH4

19、)2SO4)=132 物质的质量摩尔浓度 m 分别为 0.171molkg-1 0.125 molkg-1 0.076 molkg-1考虑电离结合实际粒子浓度 m 0.342molkg-10.250 molkg-1 0.228 molkg-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)916.解:两溶液的凝固点下降相同:Tf1=Kfm1=Kfm2=Tf21.5g m1=m2即:60g mol1=0.2kg21.38g1kgM (M 为所求未知物的摩尔质量)M=171.04 gmol117.解:Tb=Kb

20、mTb=80.960C 80.150C=0.810C=0.81k0.324gM0.004kg 0.81k=2.53 kkgmol-12.53k kgmol10.324g得出溶液中溶解的单质硫的摩尔质量为 M=253 gmol-10.004kg0.81k253g mol1 该单质硫分子中含硫原子数为 X=8132g mol18解：Tf=Kfmm 8 k=1.86 kkgmol m=774.2 g19.解：Tf=Kfm-1180g mol11kg按题意可得：16k=6.85 kkgmol-1M0.20103kg0.014g (M 为该有机物的摩尔质量)0.014g6.85k kg mol1-1 M

21、=29.97 gmol30.2010kg16k100.804g20.解：Tb=Kbm 0.455 C=0.455 k=Kb0.455k 53.2103kg Kb=3.85 kkgmol-10.00628mol0128g mol1353.210kg21.解：=nRT=cRTV =0.20 moldm-3 8.315 Pam3k-1mol-1 298k=4.956105 Pa而 gh=式中=1000kgm-3为水柱密度，g=9.8 ms-2为重力加速度。4.956105Pa 树汁可被提升的高度：h=32 g(1000kg m)(9.8ms)4.956105m1kg s2 =50.57 m22980

22、0kg ms 22.解：（1）=cRT0.8g1g()11342g mol =180g mol1000g1000g dm3 8.315Jmol-1k-1 298k =(0.00444+0.00292)moldm-3 8.315 Jmol-1k-1 298k =0.00736 moldm-38.315 Pam3k-1mol-1 298k =18.24 kPa (2)18.24 kPa=M 8.315Jmol-1k-1 298k1103m32g2g 8.315Pam3mol1k1298k 化合物的摩尔质量为 M=333110m 18.2410 Pa =271.69 gmol-12g(3)溶液的凝固点下降：Tf=Kfm=1.86 kkgmol-1271.5g mol11kg =0.01358 k 0.014k溶液的实际冰点为 Tf=273 k 0.014 k=272.99 k1112