無(wú)機(jī)及分析化學(xué)答案第二版第二章

1、第二章第二章化學(xué)反應(yīng)一般原理化學(xué)反應(yīng)一般原理21苯和氧按下式反應(yīng)：C6H6(l)O2(g)?6CO2(g)3H2O(l)215在25℃100kPa下，0.25mol苯在氧氣中完全燃燒放出817kJ的熱量，求C6H6的標(biāo)準(zhǔn)摩爾燃燒焓?cH?m和該燃燒反應(yīng)的?rU?m。解：?=?B?1?nB=(?0.25mol)(?1)=0.25mol?cH?m=?rH?m==?817kJ0.25mol=?3268kJ?mol?1?rU?m=?rH?m??

2、ngRT=?3268kJ?mol?1?(6?152)?8.314?10?3?298.15kJ?mol?1=?3264kJ?mol?122利用附錄III的數(shù)據(jù)，計(jì)算下列反應(yīng)的?rH?m。(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)

3、解：(1)?rH?m=[4?(?241.818)?(?1118.4)]kJ?mol?1=151.1kJ?mol?1(2)?rH?m=[(?285.830)(?1130.68)?(?393.509)?2?(?425.609)]kJ?mol?1=?171.78kJ?mol?1(3)?rH?m=[6?(?241.818)4?90.25?4?(?46.11)]kJ?mol?1=?905.5kJ?mol?1(4)?rH?m=[2(?285.830

4、)2(?393.509)?(?484.5)]kJ?mol?1=?874.1kJ?mol?123已知下列化學(xué)反應(yīng)的標(biāo)準(zhǔn)摩爾反應(yīng)焓變，求乙炔(C2H2，g)的標(biāo)準(zhǔn)摩爾生成焓?fH?m。(1)C2H2(g)52O2(g)?2CO2(g)H2O(g)?rH?m=?1246.2kJ?mol?1(2)C(s)2H2O(g)?CO2(g)2H2(g)?rH?m=90.9kJ?mol?1(3)2H2O(g)?2H2(g)O2(g)?rH?m=483.6

5、kJ?mol?1解：反應(yīng)2?(2)?(1)?2.5?(3)為：2C(s)H2(g)?C2H2(g)?fH?m(C2H2)=2??rH?m(2)??rH?m(1)?2.5?rH?m(3)=[2?90.9?(?1246.2)?2.5?483.6]kJ?mol?1=219.0kJ?mol?124求下列反應(yīng)在298.15K的標(biāo)準(zhǔn)摩爾反應(yīng)焓變?rH?m。(1)Fe(s)Cu2?(aq)?Fe2?(aq)Cu(s)??Hr?=?141.71kJ?m

6、ol?10，反應(yīng)自發(fā)。26由軟錳礦二氧化錳制備金屬錳可采取下列兩種方法：(1)MnO2(s)2H2(g)?Mn(s)2H2O(g)(2)MnO2(s)2C(s)?Mn(s)2CO(g)上述兩個(gè)反應(yīng)在25℃，100kPa下是否能自發(fā)進(jìn)行？如果考慮工作溫度愈低愈好的話(huà)，則制備錳采用哪一種方法比較好？解：?rG?m(1)=[2?(?228.575)?(?466.14)]kJ?mol?1=8.99kJ?mol?1?rG?m(2)=[2?(?13

7、7.168)?(?466.14)]kJ?mol?1=191.80kJ?mol?1兩反應(yīng)在標(biāo)準(zhǔn)狀態(tài)、298.15K均不能自發(fā)進(jìn)行；計(jì)算欲使其自發(fā)進(jìn)行的溫度：?rH?m(1)=[2?(?241.818)?(?520.03)]kJ?mol?1=36.39kJ?mol?1?rS?m(1)=[2?188.82532.01?2?130.684?53.05]J?mol?1?K?1=95.24J?mol?1?K?1?rH?m(1)?T1?rS?m(1)

8、=0T1=36.39kJ?mol?1(95.24?10?3kJ?mol?1?K?1)=382.1K?rH?m(2)=[2?(?110.525)?(?520.03)]kJ?mol?1=298.98kJ?mol?1?rS?m(2)=[2?197.67432.01?2?5.740?53.05]J?mol?1?K?1=362.28J?mol?1?K?1?rH?m(2)?T1?rS?m(2)=0T2=298.98kJ?mol?1(362.28?1

9、0?3kJ?mol?1?K?1)=825.27KT1T2，反應(yīng)(1)可在較低溫度下使其自發(fā)進(jìn)行，能耗較低，所以反應(yīng)(1)更合適。27不用熱力學(xué)數(shù)據(jù)定性判斷下列反應(yīng)的?rS?m是大于零還是小于零。(1)Zn(s)2HCl(aq)?ZnCl2(aq)H2(g)(2)CaCO3(s)?CaO(s)CO2(g)(3)NH3(g)HCl(g)?NH4Cl(s)(4)CuO(s)H2(g)?Cu(s)H2O(l)解：反應(yīng)(1)、(2)均有氣體產(chǎn)生，

10、為氣體分子數(shù)增加的反應(yīng)，?rS?m?0；反應(yīng)(3)、(4)的氣體反應(yīng)后分別生成固體與液體，?rS?m?0。28計(jì)算25℃100kPa下反應(yīng)CaCO3(s)?CaO(s)CO2(g)的?rH?m和?rS?m，并判斷：(1)上述反應(yīng)能否自發(fā)進(jìn)行？(2)對(duì)上述反應(yīng)，是升高溫度有利？還是降低溫度有利？(3)計(jì)算使上述反應(yīng)自發(fā)進(jìn)行的溫度條件。解：?rH?m=[?393.509?635.091206.92]kJ?mol?1=178.32kJ?mol