同济大学普通化学第一章、二章习题内容答案(详细).doc

|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 / kJmol1 285.83 241.82mS / Jmol1k1 69.91 188.83 rH (298k) = 241.82(285.83) kJmol1 = 44.01 kJmol1mr S (298k) = (188.8369.91) Jmol1k1 = 118.92 Jmol1k1( 2 ) 是等温等压变化 Qp = rH (298k) N = 44.01 kJmol1 2mol = 88.02 kJmW = PV = nRT = 2 8.315 Jk1mol1 298k = 4955.7 J = 4.956 kJ (或 4.96kJ ) U = Qp + W = 88.02 kJ 4.96kJ = 83.06 kJ17 解（1） N2 （g）+ 2O2 （g） = 2 NO2 (g)fH / kJmol1 0 0 33.2mS / Jmol1k1 191.6 205.14 240.1 rH (298k) = 33.2 kJmol1 2 = 66.4 kJmol1mr S (298k) = ( 240.1 Jmol1k1 ) 2 (205.14 Jmol1k1 ) 2 191.6 Jmol1k1 = 121.68 Jmol1k1(2) 3 Fe(s) + 4H2O (l) = Fe3O4 (s ) + 4 H2 (g)fH / kJmol1 0 285.83 1118.4 0mS / Jmol1k1 27.3 69.91 146.4 130.68rH (298k) = 1118.4 (285.83 4 ) kJmol1 = 24.92 kJmol1mr S (298k) = (130.68 4 + 146.4 ) (27.3 3 + 69.91 4 ) Jmol1k1= ( 669.12 361.54 ) Jmol1k1 = 307.58 Jmol1k1|18. 解： 2Fe2O3 (s) + 3C (s ,石墨) = 4 Fe (s) + 3 CO2 (g)fH (298k)/ kJmol1 824.2mS (298k)/ Jmol1k1 87.4 5.74 27.3fG (298k)/ kJmol1 742.2m rG = rH T r Smm 301.32 kJmol1 = 467.87 kJmol1 298 kr Smr S = 558.89 Jmol1k1m r S = 3 S ( CO2(g) 298k) + 27.3 Jmol1k1 4 87.4 Jmol1k1 2 5.74 Jmol1k1 3S ( CO2(g) 298k) = 1/3 (558.89 +192.02 109.2 ) Jmol1k1 = 213.90 Jmol1k1mfH (298k, C (s ,石墨)=0 fG (298k, C (s ,石墨)=0mfH （298k, Fe (s)）=0 fG （298k, Fe (s) ）=0mrH =3fH (298k, CO2(g) ) 2fH (298k, Fe2O3 (s) ) m 467.87 kJmol1 =3fH (298k, CO2(g) ) 2 ( 824.2 kJmol1)m fH (298k, CO2(g) ) = 1/3 (467.871648.4) kJmol1 = 393.51 kJmol1m同理 rG =3fG (298k, CO2(g) ) 2fG (298k, Fe2O3 (s) )mm 301.32 kJmol1 = 3fG (298k, CO2(g) ) 2 (742.2 kJmol1 ) fG (298k, CO2(g) ) = 1/3 (301.32 1484.4 ) kJmol1 = 394.36 kJmol1m19.解 6CO2(g) + 6H2O（l） = C6H12O6 (s) + 6O2（g）fG (298k) kJmol1 394.36 237.18 902.9 0m rG (298k) = 902.9 (237.18 6 ) (394.36 6 ) kJmol1 = 4692.14 kJmol1 >0|所以这个反应不能自发进行。20解(1) 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(l)f G (298k) /kJmol1 16.4 0 86.57 237.18mrG (298k) = (237.18) 6 + 86.57 4 (16.4) 4 kJmol1 = 1011.2 kJmol1 0 此反应不能自发进行。21解 （1） MgCO3(s) = MgO(s) + CO2(g)fH (298k)/ kJmol1 1111.88 601.6 393.51mS (298k)/ Jmol1k1 65.6 27.0 213.8f G (298k) / kJmol1 1028.28 569.3 394.36m rH (298k) = 601.6 + (393.51) (1111.88) = 116.77 kJmol1rS (298k) = 213.8+ 27.0 65.6 = 175.2 Jmol1k1mrG (298K) = (394.36) +(569.3)(1028.28) = 64.62 kJmol1(2) rG (1123K) = rH (298k)T rS (298k) = 116.77 kJmol1 1123k 175.2 Jmol1k1mmm= 116.77 kJmol1 196.75 kJmol1 = 79.98 kJmol1又 RTlnK(1123k)= rG (1123k)m 8.315 Jmol1k11123 kln K(1123k) = (79.98) kJmol1 | K(1123k) = 5.25 103(3) 刚刚分解时 rG (T) =rH (298k)T rS (298k) =0mm 分解温度 T可求: kolJkkSmr 5.62.1756)298( 1 分解最低温度为 666.5 k22.解法一: K (298k)=5.0 1016rG (298k) = RTlnK(298k)= 8.315 Jmol1k1298kln(5.0 1016) = 95.26 kJmol1mrG (298k) = rH (298k)298k rS (298k)mm95.26 kJmol1 = 92.31 kJmol1298krS (298k)rS (298k) =9.90 Jmol1k1mrG (500k) = rH (298k)500k rS (298k)mm= 92.31 kJmol1500k9.90 Jmol1k1= 97.26 kJmol1而 rG (500k) = RTlnK(500k)= 8.315 Jmol1k1 500kln K(500k)m ln K(500k)= = = 23.40RTmr50)50(35.8269713moll K(500k) = 1.45 1010解法二： ln = )298(50k ()298RkHmr501)298= = 15.05(31.(13olJk0 = 2.9 10-7)298(50kK K(500k) =2.9 10-7 K(298k) = 2.9 10-7 ( 5.0 1016 ) = 1.45 1010|23.解： N2(g) + 3H2(g) = 2NH3(g)fH (298k)/ kJmol1 0 0 45.9mS (298k)/ Jmol1k1 191.6 130.68 192.8 rH (298k) = 2(45.9) kJmol1 = 91.8 kJmol1 mS (298k) = (2192.8 191.6 3130.68 ) Jmol1k1= 198.04 Jmol1k1rG (T) = rH (298k) T rS (298k) =0mmm= 91.8 kJmol1 T (198.04 Jmol1k1 ) =0 T = = 463.5 k304.198kolJ T>463.5 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) 973 1073 1173 1273K (T) 0.618 0.905 1.29 1.66根据 rG (T) = RT ln K (T) = rH (T) T rS (T)mmmlnRSrmr假设 rH (T), rS (T)不变， 则m )1()(ln221TRTmrK1 (T) K 2(T) T1 T 2 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/mol 1 1 0平衡时各物质的量/mol 1(10.615)2 (10.615)2=0.23 0.615 =0.77平衡时物质总量 = （0.23 +0.615 + 0.77 ）mol = 1.615 mol = 29.43)615.().23(7.)()()873( 2223 pPpkKOSO26 解: (1) (2) )()(32231pTKNH )()2pTKCO(3) 44)()()222 HOHOp27.解法一: 根据阿仑尼乌斯公式: (T1=301k) 1130.2lgRTEAka(T2=278k)2.l结合上述两式，得: )(30.2lg11Tka改写此式，得 .(1)22lg)(.kTREa因为，反应速率与变酸时间成反比，因此， (2)12t将(2)代入 (1)， hktTa 48lg30178235.0.lg)(30.1221 = 7.52104 Jmol1 = 75.2kJmol127.解法二: k= A RTEae = = Exp 21k21a)(21TREa|又反应速率与变酸时间成反比 = = Exp = 12 12tkh48 )278301(315.1kkmolJEa = 75.2 kJmol1aE28.解: I2 2 I (快) H2 + 2H 2HI （慢）快反应，很快达到平衡，故有： K（ T）= = CII)(2)(2I I2 = 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 H2I2 其中 I2 项可用 I2 = K(T) I2 这一式子代入。 v = kH2I2 = kH2（K (T) I2）= （kK (T)）H 2I2 = kH2I229解：v = kCONO2 = 0.50 dm3 mol-1s1 0.025 moldm3 0.040 moldm3= 5.0104 mol dm-3s-130解：（1）（2）=（3） = = = 2.0 1010)3(K)2(1630.5|第二章 水基分散系 习题 (P68-69)11.解: (1) 盐酸溶液的量浓度 C = =12.23 moldm-3135.6%.719molgdk(2) 溶液的质量摩尔浓度 m 为：).32.1()19.(3.2133 lldkgdmol= = = 16.44 molkg-146.0.lkg74.0(3) 按照(2)计算结果,该盐酸溶液中,每 kg H2O 中含 16.44 molHCl,而 1kg H2O 相当于 = 55.56 mol H2O18olg 该溶液中溶剂 H2O 与溶质 HCl 的摩尔总数为:=(55.56 + 16.44 ) mol =72 mol总n = = 0.23 = = 0.77)(HClXmol724.16)(2OHXmol756.或 = 12.23 mol135.%5.79.lgdk= 41.32 mol38).(1.ol = = 0.23 = = )(HClXml)2.41.)(2OHXmol)32.41.0.7712. 解: （1） = 37.5%g0.1.6（2） = = 10.43 moldm33128.).(46dmkgol 3125.0dol（3）乙醇的摩尔分数为：= 0.67111540.460.(. olgolgl（4） = 13.04 mol kg1ml0.|13.解：设应加入尿素的量为 m, ( 为溶液中尿素的摩尔分数)xP* 0.200 kPa = 3.17 kPa x = = 0.0631xkPa17.320 = 0.0631 1118060molgmolgl= 1)3.(l.63l可得： m = 224.5g14.解： xP*设该溶质的摩尔质量为 M则 29.30 kPa 27.62 kPa = 29.30 kPa × 146203molgMg= g2362.708.mol M = 86.93 gmol115. 解：(1) 相同质量（10g）的情况下：= 58.5 =80 )(NaClM)(34NOH=132)(42SONHM 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 -1 0.250 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:|（NH 4） 2SO4 > NH4NO3 > NaCl(3C) (2C) (2C) 16.解: 两溶液的凝固点下降相同: = m1 = m2 = 1fTfKf2fT m1 = m2 即: = (M 为所求未知物的摩尔质量)kgol2.0651kgM38.2 M = 171.04 gmol117.解: = Kbm T= 80.960C 80.150C = 0.810C = 0.81k 0.81k = 2.53 kkgmol-1 kgM4.32得出溶液中溶解的单质硫的摩尔质量为 M = = 253 gmol-1kgmolk81.04.32453 该单质硫分子中含硫原子数为 X = = 812lg18解： = m fTfK8 k = 1.86 kkgmol-1 kgmol180 m = 774.2 g19.解： = m fTfK按题意可得：16k = 6.85 kkgmol -1 kgM3102.4(M 为该有机物的摩尔质量)