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物理化学习题解答(六)习题 p3893931、反应CO(g)+H2O(g)=H2(g)+CO2(g)的标准平衡常数与温度的关系为lgKp=2150K/T-2.216,当CO, H2O,H2,CO2的起初组成的质量分数分别为0.30,0.30,0.20和0.20时,总压为101.3kPa时,问在什么温度以下(或以上)反应才能向生成物的方向进行?解:xCO=0.30/28 /(0.30/28+0.30/18+0.20/2+0.20/44)=0.081214xH2O=0.30/18 /(0.30/28+0.30/18+0.20/2+0.20/44)=0.12633xH2=0.20/2 /(0.30/28+0.30/18+0.20/2+0.20/44)=0.758xCO2=0.20/44/(0.30/28+0.30/18+0.20/2+0.20/44)=0.034454Qp= G=-RTln Kp+RTlnQp=-RTln10(2150/T-2.216)+RTlnQp=-8.3142.302582150+8.3142.302582.216T+8.314Tln2.54549= -41158.8+50.19T0,T820K2、PCl5(g)的分解反应为PCl5(g)=PCl3(g)+Cl2(g),在523K和100kPa下达成平衡,测得平衡混合物的密度=2.695kgm-3,试计算:(1) PCl5(g)的解离度;(2) 该反应的Kp和rGm。解:(1) PCl5(g) =PCl3(g) + Cl2(g)起始时mol: x 0 0平衡时mol: x(1- ) x xn总= x(1+),m总= x(1- )208.2388+x137.3328+x70.906p总V总=n总RT,p总=m总RT/(V总M均)=RT/M均M均=RT/p总= (2.6951038.314523) /(100103)=117.1846 gmol-1推荐精选M均= m总/ n总= x(1- )208.2388+x137.3328+x70.906/ x(1+) =(1- )208.2388+137.3328+70.906/ (1+)= 117.1846(1-)208.2388+137.3328+70.906=117.1846 (1+)(208.2388+117.1846-137.3328-70.906)=208.2388-117.1846117.1846=91.0542,=0.777(2) x PCl5= x(1- )/ x(1+)= (1- )/ (1+)=(1-0.777)/(1+0.777)= 0.125483x PCl3= xCl2=x/ x(1+)= /(1+)=0.777/(1+0.777)= 0.437259Kp= = rGm=-RTln Kp=-8.314523ln1.5237=-1831.4 Jmol-13、373K时,2NaHCO3(s)=Na2CO3(s)+CO2(g)+H2O(g)反应的Kp=0.231。(1) 在10-2m3的抽真空容器中,放入0.1mol Na2CO3(s),并通入0.2mol H2O(g),问最少需通入物质的量为多少的CO2(g),才能使Na2CO3(s)全部变成NaHCO3(s)?(2) 在373K,总压为101325Pa时,要在CO2(g)及H2O(g)的混合气体中干燥潮湿的NaHCO3(s),问混合气体中H2O(g)的分压为多少才不致使NaHCO3(s)分解?解:(1) 2NaHCO3(s)=Na2CO3(s)+CO2(g)+H2O(g)起始时mol: 0 0.1 x 0.2平衡时mol: 0.2 0 x-0.1 0.1 pH2O=npH2ORT/V,pCO2=nCO2RT/VQp= pH2O pCO2/(p )2Kp0.1(x-0.1)(8.314373/10)2/10020.231x0.2311002102/0.1/(8.314373)2+0.1=0.34mol(2) 2NaHCO3(s)=Na2CO3(s) + CO2(g) + H2O(g)推荐精选平衡时kPa: x 0 101.325- pH2O pH2OQp= pH2O pCO2/(p )2Kp,(100- pH2O) pH2O/(p )2Kp(101.325- pH2O) pH2OKp(p )2=0.2311002( pH2O-50.625)250.6252-2310=15.903234.72 kPapH2O66.53kPa4、合成氨反应为3H2(g)+N2(g)=2NH3(g),所用反应物氢气和氮气的摩尔比为3:1,在673K和1000kPa压力下达成平衡,平衡产物中氨的摩尔分数为0.0385,试求:(1) 该反应在该条件下的标准平衡常数;(2) 在该温度下,若要使氨的摩尔分数为0.05,应控制总压为多少?解:(1) 3H2(g) + N2(g) = 2NH3(g)起始时kPa: 3x x 0平衡时kPa: 3x-3n0 x-n0 2n0 p总=4x-2n0,x氨=p氨/p总= 2n0/(4x-2n0)=0.0385,n0=0.074145x p总=4x-2n0=4x-20.074145x=1000,x=259.625 pH2=3x-3n0=3259.625 -30.074145259.625 =721.125kPa pNH3=2n0=20.074145259.625 =38.5kPa pN2=x-n0=259.625 -0.074145259.625 =240.375kPa (2) p总=4x-2n0,x氨=p氨/p总= 2n0/(4x-2n0)=0.05,n0=0.095238x x2=(20.095238)21002/27(1-0.095238)4/(1.64410-4)=121976.9x=349.25 推荐精选p总=4x-2n0=4349.25-20.095238349.25=1330.5kPa 5、反应C(s)+2H2(g)=CH4(g)的rGm(1000K)=19.29 kJmol-1。若参加反应气体的摩尔分数分别为xCH4=0.10,xH2=0.80,xN2=0.10,试问在1000K和100kPa压力下,能否有CH4(g)生成?解:Qp= pCH4 p /pH22= xCH4 p/xH22p=0.10100/(0.802100)=0.15625rGm=-RTln Kpln Kp=-rGm/RT=-19.29103/(8.3141000)=-2.32018Kp=0.098256QpKp,故不能生成CH4(g)。6、在723K时,将0.10mol H2(g)和0.20molCO2(g)通入抽空的瓶中,发生如下反应: (1) CO2(g) + H2(g)=H2O(g) + CO(g)平衡后瓶中的总压为50.66kPa,经分析知其中水蒸气的摩尔分数为0.10,今在容器中加入过量的氧化钴CoO(s)和金属钴Co(s),在容器中又增加了如下两个平衡:(2) CoO(s)+H2(g)=Co(s)+H2O(g)(3) CoO(s)+CO(g)=Co(s)+CO2(g)经分析知容器中的水蒸气的的摩尔分数为0.30,试分别计算这三个反应用摩尔分数表示的平衡常数。解:(1) CO2(g) + H2(g)= H2O(g) + CO(g)起始时mol: 0.20 0.10 0 0平衡时mol: 0.20-n1 0.10-n1 n1 n1n总=0.30,xH2O= n1/0.30=0.10,n1=0.03xCO2=(0.20-n1 )/0.30=(0.20-0.03)/0.30=17/30xH2=(0.10-n1 )/0.30=(0.10-0.03)/0.30=7/30xCO =n1/0.30=0.03/0.30=0.10推荐精选 Kx(1)= xCO xH2O/(xCO2 xH2)= 33/(177)=9/119 Kp (1)=Kp= Kx= 9/119 CO2(g) + H2(g)= H2O(g) + CO(g)平衡时mol:0.20-n1+ n3 0.10-n1-n2 n1+ n2 n1-n3 CoO(s) + H2(g) = Co(s) + H2O(g)平衡时mol: ? 0.10-n1-n2 ? n1+ n2CoO(s) + CO(g)=Co(s) + CO2(g)平衡时mol: ? n1-n3 ? 0.20-n1+ n3n总=nCO+nCO2+nH2O+nH2= n1-n3+ 0.20-n1+ n3+ n1+ n2+ 0.10-n1-n2= 0.30xH2O=(n1+n2 )/0.30=0.30,n1+n2 =0.09 = 9/119= 9/119, Kx(2)= xH2O/xH2=( n1+n2)/ (0.10-n1-n2 )=0.09/(0.1-0.09)=9Kx(3)=xCO2/xCO=(0.20-n1+ n3)/ (n1-n3 )=1197、有人尝试用甲烷和苯为原料来制备甲苯CH4(g)+C6H6(g)=C6H5CH3(g)+H2(g)通过不同的催化剂和选择不同的温度,但都以失败而告终。而在石化工业上,是利用该反应的逆反应,使甲苯加氢来获得苯,试通过如下两种情况,从理论上计算平衡转化率。(1) 在500K和100kPa的条件下,使用适当的催化剂,若原料甲烷和苯的摩尔比为1:1,用热力学数据估算一下,可能获得的甲苯所占的摩尔分数。(2) 若反应条件同上,使甲苯和氢气的摩尔比为1:1,请计算甲苯的平衡转化率。已知500K时这些物质的标准摩尔Gibbs自由能分别为:fGm(CH4,g)=-33.08 kJmol-1,fGm(C6H6,g)=162.0 kJmol-1fGm(C6H5CH3,g)=-172.4kJmol-1,fGm(H2,g)=0 推荐精选解:(1) CH4(g) + C6H6(g) = C6H5CH3(g) + H2(g)起始时mol: x x 0 0平衡时mol: x-n x-n n nrGm=-172.4-162.0+33.08=-301.32rGm=-RTln Kp,ln Kp=-rGm/RT=-301.32/8.314/500=-0.07248Kp=0.93008 Kp=n2/(x-n)2=0.93008,n/(x-n)=0.9644,n=0.49094xx甲苯= n/2x=0.49094x/2x=0.24547(2) C6H5CH3(g) + H2(g) =CH4(g) + C6H6(g) 起始时mol: x x 0 0平衡时mol: x-n x-n n nKp=1.075177Kp=n2/(x-n)2=1.075177,n/(x-n)=1.036907,n=0.50906x转化率=n/x=0.5090622、800K,100kPa时,C6H5C2H5(g)=C6H5C2H3(g)+H2(g)的Kp=0.05,试计算:(1) 平衡时乙苯的解离度;(2) 若在原料中添加水蒸气,使乙苯和水蒸气的摩尔比为1:9,总压仍为100kPa,求此时乙苯的解离度。解: (1) C6H5C2H5(g)=C6H5C2H3(g)+H2(g)起始时p/kPa: p 0 0平衡时p/kPa: p(1-) p pp总=p+ p=100,p=100/(1+)Kp=(p/100)(p/100)/p(1-)/100=0.05 p2/(1-)=5,2/(1-2)=0.052=0.05-0.052;5.05 2=0.05;2=9.42510-4;=0.0307推荐精选(2) C6H5C2H5(g)=C6H5C2H3(g)+H2(g) H2O(g)起始时p/kPa: p 0 0 9p平衡时p/kPa: p(1-) p p 9pp总=p+ p+9p=100,p=100/(10+)Kp= (p/100)(p/100)/p(1-)/100=0.05 p2/(1-)=5,2/(1-)(10+)=0.05,2=0.05(10-9-2) 1.052+0.45=0.5,2+0.428571=0.47619(+0.214286)2=0.47619+0.2142862=0.522108+0.214286=0.722571;=0.50828524、 CO2(g)与H2S(g)在高温的反应为:CO2(g)+H2S(g)=COS(g)+H2O(g),今在610K时将4.4g的CO2(g)加入体积为2.5dm3的空瓶中,然后再充入H2S(g)使总压为1000kPa.。达平衡后取样分析,得其中含H2O(g)的摩尔分数为0.02。将温度升至620K重复上述实验,达平衡后取样分析,得其中含H2O(g)的摩尔分数为0.03。视气体为理想气体,度计算:(1) 610K时的Kp;(2) 610K时的rGm;(3) 反应的标准摩尔焓变rHm (设其不随温度而变);(4) 在610K时,往该体积的瓶中充入不参与反应的气体,直至压力加倍,则COS(g)的产量有何变化,若充入不参与反应的气体,保持压力不变,而使体积加倍,则COS(g)的产量又有何变化?解: 起始时: pCO2=nCO2RT/V=4.4/448.314610/2.5=202.86kPa pH2S =1000-202.86=797.14kPa CO2(g) + H2S(g) = COS(g) + H2O(g) 起始时p/kPa: 202.86 797.14 0 0 平衡时p/kPa: 202.86-pH2O 797.14-pH2O pH2O pH2O(1) 610K时: p总=1000kPa,xH2O= pH2O/p总=0.02,pH2O=20kPa推荐精选 Kp= (20/100)(20/100)/(182.86/100)(777.14/100)=2.81510-3(2) rGm= -RTlnKp =-8.314610ln(2.81510-3)=29.78kJmol-1(3) 620K时:p总=1000kPa,xH2O= pH2O/p总=0.03,pH2O=30kPa Kp= (30/100)(30/100)/(172.86/100)(767.14/100)=6.78710-3根据公式: , rHm=RT1T2ln Kp,2 / Kp ,1/(T2-T1)=8.314610620ln(6.787/2.817)/(620-610)=276.49 kJmol-1(4) Kp=Kx(p) =Kx由于反应前后分子数不变,压力不影响化学平衡,故COS(g)的产量也无影响。25、 一个可能大规模制备氢气的方法是:将CH4(g) +H2O(g)的混合气通过灼热的催化床,若原料气体组成的摩尔比为nH2O:nCH4=5:1,温度为873K,压力为100kPa,并假设只发生如下两个反应:(1) CH4(g) +H2O(g)=CO(g)+H2(g) r1Gm=4.435 kJmol-1(2) CO(g) +H2O(g)=CO2(g)+H2(g) r1Gm=-6.633kJmol-1试计算达到平衡并除去H2O(g)后,平衡干气的组成,用摩尔分数表示。解:r1Gm= -RTlnKp1,lnKp1=-r1Gm /(RT)= -4.435103/(8.314873)=-0.61104Kp1=0.542786r2Gm= -RTlnKp2,lnKp2=-r2Gm /(RT)= 6.633103/(8.314873)=0.913873Kp2=2.493962(1) CH4(g) + H2O(g) = CO(g) +3H2(g)起始时kPa: x 5x 0 0平衡时kPa: x-n1 5x-n1-n2 n1-n2 3n1+n2推荐精选(2) CO(g) + H2O(g)=CO2(g)+H2(g)起始时kPa: 0 5x 0 0平衡时p/kPa:n1-n2 5x-n1-n2 n2 3n1+n2p总=nCO+nCO2+nH2O+nH2+nCH4 =n1-n2+ n2 +5x-n1-n2+3n1+ n2+x-n1= 6x+2n1=100n1=50-3xKp1=(3n1+n2)3(n1-n2)/( x-n1)( 5x-n1-n2)/(p)2=0.542786(3n1+n2)3(n1-n2)/( x-n1)( 5x-n1-n2)=5427.86Kp2=(3n1+n2)n2 /(n1-n2)( 5x-n1-n2)=2.493962p总(干)= nCO+nCO2+nH2+nCH4 = n1-n2+ n2 +3n1+ n2+x-n1=3n1+n2+xxCO (干)=(n1-n2)/(3n1+n2+x)=xCO2(干)=n2/(3n1+n2+x)=xH2(干)=(n1+n2)/(3n1+n2+x)=xCH4(干)=( x-n1)/(3n1+n2+x)=26、有如下两个反应在323K时达成平衡:(1) 2NaHCO3(s)=Na2CO3(s)+H2O(g)+CO2(g)(2) CuSO45H2O(s)= CuSO43H2O(s)+2H2O(g)已知反应(1)的解离压力为4.0kPa,反应(2)的水气压力为6.05kPa。试计算出NaHCO3(s),Na2CO3(s),CuSO45H2O(s),CuSO43H2O(s)所组成的系统,在达到同时平衡时CO2(g)的分压。解:(1) 2NaHCO3(s)=Na2CO3(s)+H2O(g)+CO2(g)Kp (1)=pCO2 pH2O/(p)2=22/1002=410-4(2) CuSO45H2O(s)= CuSO43H2O(s)+2H2O(g)Kp (2)=( pH2O/p)2=(6.05/100)2=3.6602510-3推荐精选(1) 2NaHCO3(s)=Na2CO3(s) + H2O(g) + CO2(g)平衡时p/kPa: 0 0 2pH2O + pCO2 pCO2 (2) CuSO45H2O(s)= CuSO43H2O(s) + 2H2O(g)平衡时p/kPa: 0 0 2pH2O + pCO2Kp (2)=( pH2O/p)2=(2pH2O + pCO2)2/(p )2=3.6602510-3(2pH2O + pCO2)2=36.6025(2pH2O + pCO2)=6.05Kp (1)=pCO2 pH2O/(p)2= pCO2(2pH2O + pCO2)/ (p)2=410-4pCO2(2pH2O + pCO2)=4pCO2=4/(2pH2O + pCO2)=4/6.05=0.661157kPa补充:N2O的热分解反应为2N2O(g)=2N2(g)+O2(g),在一定温度下,反应的半衰期与初始压力成反比。在970K时,N2O(g)的初始压力为39.2kPa,测得半衰期为1529s;在1030K时,N2O(g)的初始压力为48.0kPa,测得半衰期为212s。(1) 判断该反应的级数;(2) 计算两个温度下的常数;(3) 求反应的实验活化能;(4) 在1030K,当N2O(g)的初始压力为53.3kPa,计算总压达到64.0kPa所需的时间。解:(1) 反应的半衰期与初始压力成反比,故该反应为二级反应。 (2) -dA/dt=k2A2,1/A-1/A0= k2t,t1/2=1/pAk2 k2(970K)=1/39.2/1529=1.6684210-5kPa-1.s-1 k2(1030K)=1/48.0/212=9.8270410-5kPa-1.s-1(3) lnk2(T2)/k2(T1)= -Ea/R(1/T2-1/T1) ln9.82704/1.66842= - Ea/8.314(1/1030-1/970) Ea= 245.5kJ.mol-1(4) 2N2O(g) = 2N2(g) + O2(g)推荐精选 起始时kPa: 53.3 0 0 平衡时kPa: 53.3-2x 2x xp总= 53.3-x=64.0,x =10.7kPa1/(53.3-210.7)-1/53.3=9.8270410-5 t t=128s (注:可编辑下载,若有不当之处,请指正,谢谢!) 推荐精选
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