新版无机化学习题解答【整理版】.doc

第一章习题解 1-97第一章习 题 解 答基本题1-1 有一混合气体，总压为150Pa，其中N2和H2的体积分数分别为0.25和0.75， 求H2和N2的分压。解：根据式(1-6)p(N2) = 0.25p = 0.25 150 Pa = 37.5 Pap(H2) = 0.75p = 0.75150 Pa =112.5 Pa1-2 液化气主要成分是甲烷。某10.0m3 贮罐能贮存 -164、100kPa下的密度为415kgm-3的液化气。计算此气罐容纳的液化气在20C、100kPa下的气体的体积。解：甲烷的物质的量为 n(4151000g.m-310m3/16.04g.mol-1) = 259103 mol 所以 1-3用作消毒剂的过氧化氢溶液中过氧化氢的质量分数为0.03，这种水溶液的密度为1.0gmL-1，请计算这种水溶液中过氧化氢的质量摩尔浓度、物质量的浓度和摩尔分数。解：1L溶液中，m( H2O2) = 1000mL1.0gmL-10.030 = 30g m( H2O) = 1000mL1.0gmL-1(1-0.030) = 9.7102g n( H2O2) = 30g/34gmoL-1=0.88mol n( H2O) = 970g/18g.mol-1=54mol b( H2O2)= 0.88mol /0.97kg = 0.91molkg-1 c( H2O2)= 0.88mol/1L = 0.88molL-1 x( H2O2) = 0.88/(0.88.+54) = 0.0161-4计算5.0%的蔗糖(C12H22O11)水溶液与5.0%的葡萄糖(C6H12O6)水溶液的沸点。解： b(C12H22O11)=5.0g/(342g.mol-10.095kg)=0.15molkg-1b(C6H12O6)=5.0g/(180g.mol-10.095kg)=0.29molkg-1蔗糖溶液沸点上升DTb=Kbb(C12H22O11)= 0.52Kkgmol-10.15molkg-1=0.078K蔗糖溶液沸点为：373.15K+0.078K=373.23K葡萄糖溶液沸点上升DTb=Kbb(C6H12O6)= 0.52Kkgmol-10.29molkg-1=0.15K葡萄糖溶液沸点为：373.15K + 0.15K = 373.30K1-5 比较下列各水溶液的指定性质的高低(或大小)次序。(l)凝固点：0.1molkg-1 C12H22O11溶液，0.1molkg-1 CH3COOH溶液，0.1molkg-1 KCl溶液；(2)渗透压：0.1molL-1 C6H12O6溶液，0.1molL-1 CaCl2溶液，0.1molL-1 KCl溶液，1molL-1 CaCl2溶液。（提示：从溶液中的粒子数考虑。）解：凝固点从高到低：0.1molkg-1 C12H22O11溶液0.1molkg-1 CH3COOH溶液0.1molkg-1 KCl溶液渗透压从小到大：0.1molL-1 C6H12O6溶液0.1molL-1 KCl溶液0.1molL-1 CaCl2 溶液1molL-1CaCl2溶液1-6 在20 时，将5.00g血红素溶于适量水中，然后稀释到 500毫升，测得渗透压为0.366 kPa，试计算血红素的相对分子质量。解：根据 P = cRT c =P/RT = 0.366/(8.314293.15) molL-1 = 1.5010-4 molL-150010-3L1.5010-4molL-1 = 5.00 g/MM = 6.67104gmol-1所以血红素的相对分子质量为6.671041-7 在严寒的季节里为了防止仪器中的水冰结，欲使其凝固点下降到-3.00，试问在500g水中应加甘油(C3H8O3)多少克? 解： Tf = Kf(H2O) b(C3H8O3) b(C3H8O3) =Tf / Kf(H2O) =3.00/1.86 molkg-1 =1.61 molkg-1 m(C3H8O3)=1.610.50092.09g =74.1g1-8 硫化砷溶胶是通过将硫化氢气体通到H3AsO3溶液中制备得到：2H3AsO3 + 3H2S As2S3 + 6H2O试写出该溶胶的胶团结构式。解： (As2S3)mnHS-(n-x)H+x-xH+1-9 将10.0 mL0.01 molL-1的KCl溶液和100 mL0.05 mo1L-1的AgNO3溶液混合以制备AgCl溶胶。试问该溶胶在电场中向哪极运动？并写出胶团结构。解: AgNO3溶液是过量的，胶核优先吸附Ag+，胶粒带正电，因此该溶胶在电场中向负极运动。胶团结构为： (AgCl)mnAg+(n-x)NO3-x+xNO3-1-10用0.1molL-1的KI和0.08molL-1的AgNO3两种溶液等体积混合，制成溶胶，在4种电解质NaCl、Na2SO4 、MgCl2、Na3PO4中，对溶胶的聚沉能力最强的是哪一种？解: KI溶液过量， 溶胶吸附I， 溶胶带负电， 阳离子对溶胶的聚沉起作用， 因此MgCl2对溶胶的聚沉能力最强。提高题1-11为了节约宇宙飞船中氧气的供应，有人建议用氢气来还原呼出的CO2，使其转变为水。每个宇航员每天呼出的CO2约1.00kg。气体转化器以600mLmin-1（标准状态下）的速率还原CO2。为了及时转化一个宇航员每天呼出的CO2，此转化器的工作时间百分比为多少？解：转化器一天还原的CO2为 600602410 -322.4441.70103 g此转化器的工作时间百分比为 1000g/1700 = 0.5881-12 人体肺泡气中N2， O2， CO2的体积百分数分别为80.5%、14.0%和5.50%，假如肺泡总压力为100kPa， 在人体正常温度下，水的饱和蒸气压为6.28 kPa， 计算人体肺泡中各组分气体的分压。解: p = 100kPa - 6.28 kPa = 93.7 kPap(N2) = 0.805p = 0.80593.7 kPa = 75.4 kPap(O2) = 0.140 p = 0.14093.7 kPa = 13.1 kPap(Ar) = 0.0550 p = 0.055093.7 kPa = 5.15 kPa1-13将某绿色植物经光合作用产生的干燥纯净气体收集在65.301g的容器中，在30及106.0kPa下称重为65.971 g， 若已知该容器的体积为0.500 L，计算此种气体的相对分子质量，并判断可能是何种气体。解:气体的质量为 m = 65.971g 65.301g 0.67g n PV/RT =(106.0103 Pa0.5 L) / (8.314103 PaLmol-1K-1303.15 K) 0.0210 mol M m/n 0.67g/0.0210 mol 31.9 gmol-1 气体的相对分子质量为31.9， 该气体可能为氧气。1-14医学上用的葡萄糖(C6H12O6)注射液是血液的等渗溶液，测得其凝固点下降为0.543，(l) 计算葡萄糖溶液的百分含量；(2) 如果血液的温度为37， 血液的渗透压是多少？解: (1) DTf= Kf(H2O)b(C6H12O6) b(C6H12O6) = DTf/Kf(H2O) =0.543K/1.86 Kkgmol-1=0.292 molkg-1w=0.292180/(0.292180+1000)= 0.0499(2) P = cRT = 0.292molL-18.314kPaLmol-1K-1(273.15+37)K =753kPa1-15孕甾酮是一种雌性激素，它含有9.5% H、10.2% O和80.3% C，在5.00g苯中含有0.100g的孕甾酮的溶液在5.18时凝固，孕甾酮的相对分子质量是多少？分子式是什么？解：DTf =Tf -Tf=278.66-(273.15+5.18)K=0.33KDTf = Kf(苯) b(孕甾酮)= Kf(苯)m(孕甾酮)/M(孕甾酮)m(苯)M(孕甾酮)= Kf(苯)m(孕甾酮)/Tfm(苯) =5.120.100/(0.330.00500)gmol-1=310 gmol-1 C:H:O =31080.3%/12.011 : 3109.5%/1.008 : 31010.2%/16.00 = 21 : 29 : 2 所以孕甾酮的相对分子质量是310，分子式是 C21H29O2。1-16海水中含有下列离子，它们的质量摩尔浓度分别为： b(Cl-)=0.57molkg-1；b(SO42-)=0.029molkg-1；b(HCO3-)=0.002molkg-1；b(Na+)=0.49molkg-1；b(Mg2+)=0.055molkg-1；b(K+)=0.011molkg-1；(Ca2+)=0.011 molkg-1；试计算海水的近似凝固点和沸点。解：DTf = Kf(H2O) b = 1.86 (0.57 + 0.029 + 0.002 + 0.49 + 0.055 + 0.011 +0.011)K = 2.17KTf = 273.15K 2.17K= 270.98K Tb=Kb(H2O)b = 0.52 (0.57 + 0.029 + 0.002 + 0.49 + 0.055 + 0.011 +0.011)K = 0.61K Tb = 373.15K + 0.61K = 373.76K所以海水的近似凝固点和沸点分别为270.98K和373.76K。1-17 取同一种溶胶各20.00mL分别置于三支试管中。欲使该溶胶聚沉，至少在第一支试管加入4.0 molL-1的KCl溶液0.53mL，在第二支试管中加入0.05molL-1的Na2SO4溶液1.25mL，在第三支试管中加入0.0033molL-1的Na3PO4溶液0.74mL，试计算每种电解质溶液的聚沉值，并确定该溶胶的电性。解：第一支试管聚沉值： 4.00.531000/(20.00+0.53) =1.0102 (m mo1L-1)第二支试管聚沉值： 0.0501.251000/(20+1.25) = 2.9(m mo1L-1)第三支试管聚沉值： 0.00330.741000/(20+0.74)=0.12(m mo1L-1)Na3PO4的聚沉值最小，负电荷起作用，说明溶胶带正电。1-18 The sugar fructose contains 40.0% C， 6.7% H， and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 C. The boiling point of ethanol is 78.35C， and Kb for ethanol is 1.20 Kkg.mol-1. What is the molecular formula of fructose? (C6H12O6)Solution: DTb=Tb-Tb=(78.59 -78.35)K=0.24K DTb=Kb(ethanol)b(fructose)= Kb(ethanol)m(fructose)/M(fructose)m(ethanol) M(fructose)= Kb(ethanol)m(fructose)/DTbm(ethanol)=1.2011.7/(0.240.325)gmol-1 =180gmol-1 C:H:O =18040%/12.011 : 1806.75%/1.008 : 18053.32%/16.00 = 6 : 12 : 6 Molecular formula of fructose is C6H12O6.1-19 A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol， C2H5OH. The boiling-point elevation of the solution is 1.27C. Is HgCl2 an electrolyte in ethanol? Show your calculations. (Kb=1.20K.kg.mol-1)Solution : If HgCl2 is not an electrolyte in ethanol b(HgCl2)=9.41/(271.50.03275) molkg-1=1.05molkg-1 now， DTb= Kb(ethanol)b(HgCl2)b(HgCl2)= DTb/Kb(ethanol)= 1.27/1.20 molkg-1=1.05 molkg-1 Therefor， HgCl2 is not an electrolyte in ethanol.1-20 Calculate the percent by mass and the molality in terms of CuSO4 for a solution prepared by dissolving 11.5g of CuSO45H2O in 0.1000 kg of water. Remember to consider the water released from the hydrate.Solution : m(CuSO4) = 11.5159.6/249.68g= 7.35gm(H2O) = 11.5 - 7.35g = 4.15g Percent by mass: 7.35/(100.0+4.15)= 0.071b = 7.35/(159.60.1042) molkg-1= 0.442 molkg-1 1-21 The cell walls of red and white blood cells are semipermeable membranes. The concentration of solute particles in the blood is about 0.6molL-1. What happens to blood cells that are place in pure water? In a 1molL-1 sodium chloride solution?Solution : In pure water: Water will entry the cell.In a 1 mo1L-1 sodium chloride solution: The cell will lose water.1-22 1946年，George Scatchard 用溶液的渗透压测定了牛血清蛋白的分子量。他将9.63g蛋白质配成1.00L水溶液，测得该溶液在25时的渗透压为0.353kPa，请你计算牛血清蛋白的分子量。如果该溶液的密度近似为1.00gml-1，能否用凝固点下降法测定蛋白质的分子量？为什么？ 解: = cRT c = /RT = 0.385kPa/(8.314kPaLmol-1K-1298.15K)= 1.4410-4molL-1 M = m/n = 9.63g/1.4410-4mol = 6.69104g.mol-1 如果用凝固点下降法测定蛋白质的分子量，将质量摩尔浓度近似看作1.4410-4molkg-1 DTf = Kf(H2O) b = 1.861.4410-4= 2.70 10-4 K因为下降值太小，温度计测不准，故不能用凝固点下降法测定。第二章 习 题 解 答基本题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。页：7 DrHym= -817 kJ/0.25mol= -3268 kJmol-1 DrUm=DrHm-DngRT =-3268kJmol-1-(6-15/2)8.31410-3298.15kJmol-1= -3264.3kJmol-1 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页：7 (1) D r Hym=4(-241.8) - (-1118.4)kJmol-1 = 151.2 kJmol-1(2) D r Hym=(-285.8)+( -1130.68) - (-393.509) - 2(- 425.609)kJmol-1 = -171.8kJmol-1(3) D r Hym=6(-241.8)+ 490.25 - 4(-46.11)kJmol-1 = -905.4kJmol-1(4) D r Hym=2(-285.8)+2(-393.509) -(-485.76)kJmol-1 = -872.9kJmol-1。(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。页：7 反应2(2)-(1)-2.5(3)为：2C(s)+H2(g)C2H2(g) D f Hym=2D r Hym(2)-D r Hym(1)- 2.5D r Hym(3)=290.9-(-1246.2) -2.5483.6 kJmol-1 =219.0 kJmol-1(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页：8 D r Hym(1)= -89.1-64.77=-153.9 kJmol-1 D r Hym(2)= -167.159-100.37-(-121.55) -(-127.068)= -18.91 kJmol-1D r Hym(3)= 2(-48.5)+3(-285.83) +824.2=-130.3 kJmol-1D r Hym(4)= (-153.89) -64.77=-218.66 kJmol-1。(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。页：8 (1) D r Hym= -566.0 kJmol-1D r Sym=-173.0 Jmol-1K-1D r Gym= -514.4kJmol-1(2) D r Hym= -905.3 kJmol-1D r Sym=181.3 Jmol-1K-1D r Gym= -959.4kJmol-1(3) D r Hym= -24.8 kJmol-1D r Sym=15.4 Jmol-1K-1D r Gym= -29.4kJmol-1 (4) D r Hym=-197.78kJmol-1D r Sym=-188.13Jmol-1K-1D r Gym=-141.732kJmol-1全部自发(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？页：9D r Gym(1)= 9.03 kJmol-1D r Gym(2)= 190.0kJmol-1 两反应均不能自发进行； 自发进行的温度为：T1=393.4K T2=813.4K 故反应（1）更合适。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) D r Sym 0；反应(3)、(4)的气体反应后分别生成固体与液体，D r Sym 0，不能自发进行。DrHym0，D r Sym0，升高温度有利。TDrHym/D r Sym=178.32/160.610-3=1110.3K解： DrHym = -393.509 - 635.09 + 1206.92 kJmol-1= 178.32 kJmol-1 DrSym = 213.74 + 39.75 - 92.9 Jmol-1K-1= 160.6 Jmol-1K-1(1) DrGym= 178.32 - 298.15 160.6 10-3 kJmol-1= 130.44 kJmol-1 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 页：11略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。页：11反应(1)-(2)为所求反应：Ky= Ky1/ Ky2=21015/51022=410-8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。页：11反应 (1) +2 (2)为所求反应：Ky= Ky1 (Ky2)2=210.0342=2.410-2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。页：11V,T不变，pn，各平衡分压为：p(NO) =150-25=125kPa；p(O2) =450-25/2=437.5kPa Ky=(p(NO2)/py)2(p(NO)/ py)-2(p(O2)/ py)-1=(25/100) 2(125/100) -2(437.5/100) -1=9.110-3 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。页：12 N2O4(g) 2NO2(g) x = 0.243 e (N2O4)= 24.3%起始n/mol 1 0 x = 0.174 e (N2O4)= 17.4%平衡n/mol 1-x 2x 增大压力，平衡向气体分子数减少的方向移动。平衡相对分压 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。页：13D r Gym=-197.15-228.572+394.359+216.45kJmol-1=1.537kJmol-1 lgKy=-1.537103/(2.3038.314298.15)= -0.269 Ky=0.542NH3(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？。页：13D r Gym=DrHym-TDrSym D r Gym,298.15K= -210.9 kJmol-1-298.15K131.810-3 kJmol-1K-1=-250.2 kJmol-1D r Gym,373.15K= -210.9 kJmo