无机及分析化学答案第二版第二章.doc

第二章 化学反应一般原理2-1 苯和氧按下式反应： C6H6(l) + O2(g) 6CO2(g) + 3H2O(l)在25100kPa下，0.25mol苯在氧气中完全燃烧放出817kJ的热量，求C6H6的标准摩尔燃烧焓DcHym和该燃烧反应的DrUym。 解： x = nB-1DnB = (-0.25 mol) / ( -1) = 0.25 molyDcHym = DrHym = = -817 kJ / 0.25 mol= -3268 kJmol-1 DrUym = DrHym - DngRT= -3268 kJmol-1 - (6 -15 / 2) 8.314 10-3 298.15 kJmol-1= -3264 kJmol-1 2-2 利用附录III的数据，计算下列反应的DrHym。(1) Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)(2) 2NaOH(s) + CO2(g) Na2CO3(s) + H2O(l)(3) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)(4) CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l)解： (1) DrHym = 4 (-241.818) - (-1118.4) kJmol-1 = 151.1 kJmol-1(2) DrHym = (-285.830) + (-1130.68) - (-393.509) - 2 (-425.609) kJmol-1= -171.78 kJmol-1(3) DrHym = 6 (-241.818) + 4 90.25 - 4 (-46.11) kJmol-1= -905.5 kJmol-1(4) DrHym = 2(-285.830) + 2(-393.509) - (-484.5) kJmol-1 = -874.1 kJmol-12-3 已知下列化学反应的标准摩尔反应焓变，求乙炔(C2H2，g)的标准摩尔生成焓D fHym。(1) C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(g) D r Hym= -1246.2 kJmol-1(2) C(s) + 2H2O(g) CO2(g) + 2H2(g) D r Hym = +90.9 kJmol-1(3) 2H2O(g) 2H2(g) + O2(g) D r Hym = +483.6 kJmol-1解：反应2 (2) - (1) - 2.5 (3)为：2C(s) + H2(g) C2H2(g)D f Hym (C2H2) = 2 D r Hym(2) - D r Hym(1) - 2.5D r Hym(3)= 2 90.9 - ( -1246.2) - 2.5 483.6 kJmol-1 = 219.0 kJmol-1 2-4 求下列反应在298.15 K的标准摩尔反应焓变D r Hym。(1) Fe(s)+Cu2+(aq)Fe2+(aq)+Cu(s)(2) AgCl(s)+Br-(aq)AgBr(s)+Cl-(aq)(3) Fe2O3(s)+6H+(aq)2Fe3+(aq)+3H2O(l)(4) Cu2+(aq)+Zn(s) Cu(s)+Zn2+(aq)解： D r Hym(1) = -89.1-64.77 kJmol-1= -153.9 kJmol-1 D r Hym(2) = -167.159 -100.37 - (-121.55) - (-127.068) kJmol-1= -18.91 kJmol-1D r Hym(3) = 2 (-48.5) + 3 (-285.830) + 824.2 kJmol-1= -130.3 kJmol-1D r Hym(4) = (-153.89) - 64.77 kJmol-1= -218.66 kJmol-12-5 计算下列反应在298.15K的DrHym，DrSym和DrGym，并判断哪些反应能自发向右进行。(1) 2CO(g) + O2(g) 2CO2(g)(2) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)(3) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) (4) 2SO2(g) + O2(g) 2SO3(g)解：(1) D r Hym = 2 (-393.509) - 2 (-110.525) kJmol-1= -565.968 kJmol-1D r Sym = 2 213.74 - 2 197.674 - 205.138 Jmol-1K-1= -173.01 Jmol-1K-1D r Gym = D r Hym - TD r Sym= -565.968 - 298.15 (-173.01 10-3) kJmol-1K-1= -514.385kJmol-1 0，反应自发。(2) D r Hym = 6 (-241.818) + 4 90.25 - 4 (-46.11) kJmol-1= -905.5 kJmol-1D r Sym = 6 188.825 + 4 210.761 - 5 205.138 - 4 192.45 Jmol-1K-1= 180.50 Jmol-1K-1D r Gym = D r Hym - TD r Sym = -905.5 - 298.15 180.50 10-3 kJmol-1= -959.3kJmol-1 0，反应自发。(3) D r Hym = 3 (-393.509) - 3 (-110.525) - (-824.2) kJmol-1= -24.8 kJmol-1D r Sym = 3 213.74 + 2 27.28 - 3 197.674 - 87.4 Jmol-1K-1= 15.4 Jmol-1K-1D r Gym = D r Hym - TD r Sym = -24.8 - 298.15 15.4 10-3 kJmol-1= -29.4kJmol-1 0，反应自发。(4) D r Hym = 2 (-395.72) - 2 (-296.830) kJmol-1= -197.78kJmol-1D r Sym = 2 256.76 - 205.138 - 2 248.22 Jmol-1K-1= -188.06Jmol-1K-1D r Gym = D r Hym - TD r Sym = -197.78 - 298.15 (-188.06 10-3) kJmol-1= -141.71kJmol-1 0，反应自发。2-6 由软锰矿二氧化锰制备金属锰可采取下列两种方法： (1) MnO2(s) + 2H2(g) Mn(s) + 2H2O(g)(2) MnO2(s) + 2C(s) Mn(s) + 2CO(g) 上述两个反应在25，100 kPa下是否能自发进行？如果考虑工作温度愈低愈好的话，则制备锰采用哪一种方法比较好？解： DrGym(1) = 2 (-228.575) - (-466.14) kJmol-1 = 8.99 kJmol-1 DrGym(2) = 2 (-137.168) - (-466.14) kJmol-1 = 191.80 kJmol-1 两反应在标准状态、298.15K均不能自发进行；计算欲使其自发进行的温度：DrHym(1) = 2 (-241.818) - (-520.03) kJmol-1 = 36.39 kJmol-1DrSym(1) = 2 188.825 + 32.01 - 2 130.684 - 53.05 Jmol-1K-1 = 95.24 Jmol-1K-1 DrHym(1) - T1DrSym(1) = 0 T1 = 36.39 kJmol-1 / (95.24 10-3 kJmol-1K-1)= 382.1KDrHym(2) = 2 (-110.525) - (-520.03) kJmol-1 = 298.98 kJmol-1DrSym(2) = 2 197.674 + 32.01 - 2 5.740 - 53.05 Jmol-1K-1 = 362.28 Jmol-1K-1 DrHym(2) - T1DrSym(2) = 0T2 = 298.98 kJmol-1 / (362.28 10-3 kJmol-1K-1) = 825.27 KT1 0；反应(3)、(4)的气体反应后分别生成固体与液体，D r Sym 0 反应不能自发进行；(2) DrHym 0，D r Sym 0，升高温度对反应有利，有利于DrGym DrHym /D r Sym= 178.32 / 160.610-3 K= 1110 K2-9 写出下列各化学反应的平衡常数Ky表达式。(1) CaCO3(s) CaO(s) + CO2(g) (2) 2SO2(g) + O2(g) 2SO3(g)(3) C(s) + H2O(g) CO(g) + H2(g) (4) AgCl(s) Ag+(aq) + Cl-(aq)(5) HAc(aq) H+(aq) + Ac-(aq) (6) SiO2(s) + 6HF(aq) H2SiF6(aq) + 2H2O(l)(7) Hb(aq)(血红蛋白) + O2(g)HbO2(aq)(氧合血红蛋白)(8) 2MnO4-(aq) + 5SO32-(aq) + 6H+(aq) 2Mn2+(aq) + 5SO42-(aq) + 3H2O(l)解：(1) Ky =/ py)= p(CO2)/py (2) Ky = (p(SO3)/py)2(p(O2)/py)-1(p(SO2) /py)-2 (3) Ky = (p(CO)/py)(p(H2)/py)(p(H2O)/py)-1 (4) Ky = (c(Ag+)/cy)(c(Cl-)/cy)(5) Ky = (c(H+)/cy)(c(Ac-)/cy)(c(HAc)/cy)-1(6) Ky = (c(H2SiF6)/cy)(c(HF)/cy)-6(7) Ky = (c(HbO2)/cy)(c(Hb)/cy)-1(p(O2)/py)-1(8) Ky=(c(Mn2+)/cy)2(c(SO42-)/cy)5(c(MnO4-)/cy)-2(c(SO32-)/cy)-5(c(H+)/cy)-6 2-10 已知下列化学反应在298.15K时的平衡常数：(1) CuO(s) + H2(g) Cu(s) + H2O (g) Ky1 = 21015(2) 1/2O2(g) + H2(g) H2O(g) Ky2 = 51022计算反应 CuO(s)Cu(s) + 1/2O2(g) 的平衡常数Ky。解：反应(1) - (2)为所求反应，根据多重平衡规则：Ky = Ky1 / Ky2 = 2 1015 / 5 1022 = 4 10-8 2-11 已知下列反应在298.15K的平衡常数：(1) SnO2(s) + 2H2(g) 2H2O(g) + Sn(s) Ky1 = 21(2) H2O(g) + CO (g) H2(g) + CO2(g)； Ky2 = 0.034计算反应 2CO(g) + SnO2(s)Sn(s) + 2CO2 (g)在298.15K时的平衡常数Ky。解：反应 (1) + 2 (2)为所求反应，所以Ky = Ky1 (Ky2)2= 21 0.0342 = 2.4 10-22-12 密闭容器中反应 2NO(g) + O2(g) 2NO2(g) 在1500K条件下达到平衡。若始态p(NO) = 150 kPa，p(O2) = 450 kPa，p(NO2) = 0；平衡时p(NO2) = 25 kPa。试计算平衡时p(NO)，p(O2)的分压及平衡常数Ky。 解：V、T不变，p n，各平衡分压为：p(NO) =150 kPa - 25 kPa = 125 kPap(O2) = 450 kPa - (25 / 2) kPa = 437.5 kPaKy = (p(NO2) / py)2(p(NO) / py)-2(p(O2) / py)-1= (25 / 100) 2(125 / 100) -2(437.5 / 100) -1= 9.1 10-3 2-13 密闭容器中的反应 CO(g) + H2O(g) CO2(g) + H2(g) 在750K时其Ky = 2.6，求：(1) 当原料气中H2O(g)和CO(g)的物质的量之比为1:1时，CO(g)的平衡转化率为多少？(2) 当原料气中H2O(g):CO(g)为4:1时，CO(g)的平衡转化率为多少？说明什么问题？解：(1) V、T不变 CO(g) + H2O(g) CO2(g) + H2(g) 起始n / mol 1 1 0 0 平衡n / mol 1-x 1-x x x Sn = 2(1 - x) + 2x = 2平衡分压 p总 p总 p总 p总Ky = (p(H2) / py)(p(CO2) / py)(p(H2O) / py)-1(p(CO) / py)-12.6 = ()2()-2x = 0.62a(CO) = 62%(2) V、T不变 CO(g) + H2O(g) CO2(g) + H2(g) 起始n / mol 1 4 0 0 平衡n / mol 1-x 4-x x x Sn = 5平衡分压 p总 p总 p总 p总 2.6 = (x / 5)2 (1 - x) / 5-1(4 - x) / 5-1 x = 0.90a(CO) = 90%H2O(g)浓度增大，CO(g)转化率增大，利用廉价的H2O(g)，使CO(g)反应完全。2-14 在317K，反应 N2O4(g) 2NO2(g) 的平衡常数Ky = 1.00。分别计算当系统总压为400 kPa和800 kPa时N2O4(g)的平衡转化率，并解释计算结果。解：总压为400 kPa时 N2O4(g) 2NO2(g) 起始n / mol 1 0 平衡n / mol 1-x 2x 平衡相对分压 x = 0.243a(N2O4) = 24.3%总压为800kPa时 x = 0.174a(N2O4) = 17.4%增大压力，平衡向气体分子数减少的方向移动，a(N2O4)下降。2-15 已知尿素CO(NH2)2的DfGym= -197.15 kJmol-1，求尿素的合成反应在298.15 K时的D r Gym和Ky。2NH3(g) + CO2(g) H2O(g) + CO(NH2)2(s) 解： DrGym = -197.15 - 228.575 + 394.359 + 2 16.45 kJmol-1= 1.53 kJmol-1lgKy = -DrGym / (2.303RT) = -1.53 103 / (2.303 8.314 298.15) = -0.268 Ky = 0.540 2-16 25时，反应2H2O2(g)2H2O(g) + O2(g)的DrHym为 -210.9 kJmol-1，DrSym为131.8 Jmol-1K-1。试计算该反应在25和100时的Ky，计算结果说明什么问题？。 解： DrGym = DrHym - TDrSym DrGym,298.15K = -210.9 kJmol-1 - 298.15 K 131.8 10-3 kJmol-1K-1= -250.2 kJmol-1lgKy = -DrGym / (2.303RT)= 250.2 103 / (2.303 8.314 298.15)= 43.83Ky298.15K = 6.7 1043DrGym,373.15K = -210.9 kJmol-1 -373.15 K 131.8 10-3 kJmol-1K-1= -260.1 kJmol-1lgKy = 260.1 103 / (2.303 8.314 373.15)= 36.40Ky373.15K = 2.5 1036该反应为放热反应，对放热反应，温度升高，Ky下降。2-17 在一定温度下Ag2O的分解反应为 Ag2O(s) 2Ag(s) + 1/2O2(g)。假定反应的DrHym，DrSym不随温度的变化而改变，估算Ag2O在标准状态的最低分解温度？ 解： DrHym = -DfHym(Ag2O) = 31.05 kJmol-1 DrSym = 2 42.5 + 205.138 / 2 - 121.3 Jmol-1K-1= 66.3 Jmol-1K-1T = DrHym /DrSym= 31.05 kJmol-1 /( 66.3 10-3 kJmol-1K-1)= 468 K2-18 已知反应 2SO2(g) + O2(g) 2SO3(g) 在427和527时的Ky值分别为1.0 105和1.1 102，求该反应的DrHym。 解： yyy yDrHym = -3.2 102 kJmol-1 2-19 已知反应 2H2(g) + 2NO(g) 2H2O(g) + N2(g) 的速率方程 v = k c(H2)c2(NO)，在一定温度下，若使容器体积缩小到原来的1/2时，问反应速率如何变化？ 解： 体积缩小为1/ 2，浓度增大2倍：v2 = k 2c1(H2)(2c1)2(NO) = 8 k c1(H2)(c1)2(NO)= 8v12-20 某基元反应 A + B C，在1.20 L溶液中，当A为4.0 mol，B为3.0 mol时，v为0.0042 molL-1s-1，计算该反应的速率常数，并写出该反应的速率方程式。 解：v = kcAcB k = 0.0042 molL-1s-1 / (4.0 mol / 1.20 L) (3.0 mol) / 1.20 L = 5.0 10-4 mol-1Ls-12-21 某一级反应，若反应物浓度从1.0 mol L -1降到0.20 mol L -1需30min，问：(1) 该反应的速率常数k是多少？ (2) 反应物浓度从0.20 mol L -1降到0.040 mol L -1需用多少分钟？解： (1) ln(0.20 /1.0) = -k30 mink = 0.054 min-1 ；(2) ln(0.040 / 0.20) = - 0.054 min-1t t = 30 n nmin2-22 From reactions(1)(5)below, select, without any thermodynamic calculations those reactions which have: (a) large negative standar entropy changes, (b) large positive standar entropy changes, (c) small entropy changes which might be either positive or negative.(1) Mg(s) + Cl2(g) = MgCl2(s)(2) Mg(s) + I2(s) = MgI2(s)(3) C(s) + O2(g) = CO2(g)(4 Al2O3(s) + 3C(s) + 3Cl2(g) = 2AlCl3(g) + 3CO(g)(5) 2NO(g) + Cl2(g) = 2NOCl(g)Solution: (1) large negative standar entropy changes:(1) ,(5).(2) large positive standar entropy changes:(4). (3) small entropy changes which might be either positive or negative (2),(3).2-23 Calculate the value of the thermodynamic decomposition temperature (Td) for the reaction NH4Cl(s).= NH3(g) + HCl(g) at the standard state. Solution: DrHym = - 46.11- 92.307 + 314.43 kJmol-1= 176.01 kJmol-1DrSym = 192.45 + 186.908 - 94.6 Jmol-1K-1= 284.8 Jmol-1K-1T = DrHym /DrSym=176.01 kJmol-1 / 284.758 10-3 kJmol-1K-1= 618.0 K2-24Calculate DrGym at 298.15K for the reaction 2NO2(g)N2O4(g).Is this reaction spontaneous? Solution: DrGym = 97.89 -2 51.31 kJmol-1= - 4.73 kJmol-1 0The reaction is spontaneous. 2-25 The following gas phase reaction follows first-order kinetics:FClO2(g) FClO(g) + O(g)The activation energy of this reaction is measured to be 186 kJmol-1. The value of k at 322 is determined to be 6.7610-4s-1.(1) What would be the value of k for this reaction at 25? (2) At what temperature would this reaction have a k value of 6.0010-2s-1?Solution: (1) k2 = 3.7010-20 s-1(2) T = 676 K2-26 某理想气体在恒定外压(101.3 kPa)下吸热膨胀，其体积从80 L变到160 L，同时吸收25 kJ的热量，试计算系统热力学能的变化。解： DU = Q + W = Q - pDV= 25 kJ - 101.3 kPa (160 - 80) 10-3 m3= 25 kJ - 8.104 kJ= 17 kJ2-27 蔗糖(C12H22O11)在人体内的代谢反应为：C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l)假设在标准状态时其反应热有30%可转化为有用功，试计算体重为70kg的人登上3000m高的山(按有效功计算)，若其能量完全由蔗糖转换，需消耗多少蔗糖？(DfHym(C12H22O11) = -2222kJmol-1) 解： W = -70 kg 3000 m = -2.1 105 kgm= -2.1 105 9.8 J = -2.1 103 kJ DrHy = -2.1 103 kJ / 30%= -7.0 103 kJD r Hym = 11 (-285.830 kJmol-1) + 12 (-393.509 kJmol-1) - (-2222 kJmol-1)= -5644 kJmol-1 x = DrH / DrHym= (-7.0 103) kJ / (-5644) kJmol-1= 1.2 mol m(C12H22O11) = n(C12H22O11) / M(C12H22O11)= 1.2 mol 342.3 gmol-1= 4.2 102 g 2-28 人体靠下列一系列反应去除体内酒精影响：CH3CH2OH(l)CH3CHO(l)CH3COOH(l)CO2(g)已知DfHym(CH3CHO, g) = -166.4 kJmol-1，计算人体去除1 mol C2H5OH(l)时各步反应的DrHym及总反应的DrHym(假设T = 298.15 K)。解： CH3CH2OH(l) + 1/2O2(g) CH3CHO(l) + H2O(l) DrHym(1) = -285.830 -166.4 + 277.69 kJmol-1 = -174.5 kJmol-1CH3CHO(l) + 1/2O2(g) CH3COOH(l)DrHym(2) = -484.5 + 166.4 kJmol-1 = -318.1 kJmol-1CH3COOH(l) + O2(g) 2CO2 + 2H2O(l)DrHym(3) = 2 (-285.830) + 2 (-393.509) + 484.5 kJmol-1= -874.2 kJmol-1DrHym(总) = DrHym(1) + DrHym(2) + DrHym(3)= -174.5 -318.1-874.2 kJmol-1= -1366.8 kJmol-12-29 Calculate the values of DrHym, DrSym, DrGym and Ky at 298.15K for the reaction NH4HCO3(s) = NH3 (g) + H2O(g) + CO2(g) 。Solution: DrHym=- 46.11 - 241.818 -393.509 + 849.4 kJmol-1= 168.0 kJmol-1DrSym =192.45 + 188.825 + 213.74 - 120.9 Jmol-1K-1= 474.1Jmol-1K-1DrGym=-16.45 - 228.572 - 394.359 + 665.9 kJmol-1= 26.5 kJmol-1lgKy = -DrGym / (2.303RT)= -26.5 kJmol-1 / (2.303 8.314 10-3 kJmol-1K-1 298.15K)= - 4.64Ky = 2.3 10-5.2-30 糖在人体中的新陈代谢过程如下：C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l)若反应的吉布斯函数变DrGym只有30%能转化为有用功，则一匙糖(3.8g)在体温37时进行新陈代谢，可得多少有用功？(已知C12H22O11的DfHym = -2222 kJmol-1， Sym = 360.2 Jmol-1K-1) 解： C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(l)DfHym / kJmol-1 -2222 0 -393.509 -285.830Sym / Jmol-1K-1 360.2 205.138 213.74 69.91DrHym = 11 (-285.830) +12 (-393.509) - ( -2222) kJmol-1= -5644 kJmol-1DrSym = 11 69.91 + 12 213.74 - 12 205.138 - 360.2 Jmol-1K-1= 512.0 Jmol-1K-1D r Gym = DrHym - TDrSym= -5644 kJmol-1 - 310.15K 512.0 10-3 kJmol-1K-1= -5803 kJmol-1x = DnB / nB = 3.8 g / 342 gmol-1 = 1.11 10-2 molW有用功 = 30%D r Gy = 30%D r Gymx = 30% (-5803 kJmol-1) 1.1110-2 mol= -19 kJ (负号表示系统对环境做功)2-31 在2033 K和3000 K的温度条件下混和等摩尔的N2和O2，发生如下反应：N2(g) + O2(g) 2NO(g)平衡混合物中NO的体积百分数分别是0.80%和4.5%。计算两种温度下反应的Ky，并判断该反应是吸热反应还是放热反应。 解： 体积分数等于摩尔分数 Vi / V = ni / n = xi pi = xi pKy = (p(NO) / py)2(p(O2) / py)-1(p(N2) / py)-1Ky2033K = 0.00802 (1 - 0.0080) / 2-2= 2.6 10-4Ky3000K = 0.0452 (1 - 0.045) / 2-2= 8.910-3T升高，Ky增大，该反应为吸热反应。2-32 14C的半衰期为5730y(y：年的时间单位)。考古测定某古墓木质样品的14C含量为原来的63.8%。问此古墓距今已有多少年？解：放射性同位素的衰变为一级反应 ln(cB / c0) = -kt ln50% = - k 5730 yk = 1.210 10-4 y-1ln63.8% = -1.210 10-4 y-1t t = 3714 y2-33 在301 K时鲜牛奶大约4.0 h变酸，但在278 K的冰箱中可保持48 h时。假定反应速率与变酸时间成反比，求牛奶变酸反应的活化能。 解： ln Ea = 7.5 104 Jmol-1 = 75 kJmol-12-34 已知青霉素G的分解反应为一级反应，37 时其活化能为84.8 kJmol-1，指前因子A为4.2 1012 h-1，求37时青霉素G分解反应的速率常数？ 解： = 4.2 1012 h-1 = 4.2 1012 h-1 5.2 10-15 = 2.2 10-2 h-12-35 某病人发烧至40时，使体内某一酶催化反应的速率常数增大为正常体温(37)的1.25倍，求该酶催化反应的活化能？ 解： Ea = 60.0 kJmol-12-36 某二级反应，其在不同温度下的反应速率常数如下：T / K 645 675 715 750k 103/mol-1 Lmin-1 6.15 22.0 77.5 250(1) 作lnk1/T图计算反应活化能Ea；(2) 计算700 K时的反应速率常数k。解：(1) 1 / T 1.5510-3 1.4810-3 1.4010-3 1.3310-3lnk -5.09 -3.82 -2.56 -1.37 用Excel 作图：从图中直线方程得： -Ea/R = -16.774 K Ea = 8.314Jmol-1K-1 16774K = 1.39105 Jmol-1 = 139 kJmol-1 (2) lnk = -16774K(1/T) + 20.944= -16774 / 700 + 20.944 = -3.02k = 4.89 10-22-37 It is difficult to prepare many compounds directly from the elements, so DfHym values for these compounds cannot be measured directly. For many organic compounds, it is easier to measure the standard enthalpy of combustion DcHym by reaction of the compounds with excess O2(g) to form CO2(g) and H2O(l). From the following standard enthalpies of combustion at 298.15K, determine DfHym for the compound.(1) cyclohexane, C6H12(l), a useful organic solvent: DcHym= -3920kJmol-1(2) phenol, C6H5OH(s), used as a disinfectant and in the production of thermo-setting plastics : DcHym= -3053kJmol-1 Solution: (1) C6H12(l) + 9O2(g) = 6CO2(g) + 6H2O(l) D r Hym = DcHym =nBDfHym(B) -3920 kJmol-1=6(-393.509) + 6 (-285.830) -DfHym(C6H12(l) kJmol-1DfHym(C6H12(l) =156 kJmol-1(2) C6H5OH(s) +O2(g) = 6CO2(g)+3H2O(l)D r Hym = DcHym =nBDfHym(B)-3053kJmol-1=6(- 393.509)+ 3(-285.830) -DfHym(C6H5OH(s)kJmol-1DfHym(C6H5OH(s) = -166 kJmol-1 2-38 Tb(铽)的同位素的半衰期= 6.9 d，求10 d后该同位素样品所剩百分数。解：同位素的衰变为一级反应t1/2 = 0.693 / kk = 0.693 / 6.9 d = 0.10 d-1 lncB / c0 = -k t = -0.10 d-1 10 d = 1.0cB / c0 = 0.37 即还剩37%