人卫版物理化学(第六版)课后习题答案详解 下载本文

即0.5=

??(??n)

1??n总=1+α+n α+n=2 代入上式,得: 0.5=

??2 1??解得:α=0.2=20% α下降

13.在 448~688K温度区间内,用 分光光度法研究了下面的气相反应:

2

I +环戊烯 = 2HI + 环戊二烯

得到KP=17.39-(1)

?51034

4.575T???计算在573K时,反应的ΔrGmΔrHm和ΔrSm

(2)始时用等量I2和环戊烯混合,温度为573K,起始总压为101.325Kpa,试求平衡后I2分压。 (3)若起始压力为1013.25Kpa,试求平衡后I2的分压。 解:(1) InK?p=17.39-

51034

4.575T51034

4.575?573015?T=573K时 InK?p=17.39-?In Kp=-2.073 Kp=0.125

ΔrGm=-RT InKp=-8.314×573×(-2.073)=9.88KJ

?InKp??=

??rHm1??C RT?51034?rHm? 解得:ΔrHm=92.74KJ ?R40575ΔrGm=ΔrHm—TΔrSm(恒温) 9.88=92.74-573ΔrSm 解得:ΔrSm=144.6J?K??1????

2)I2 + 环戊烯 = 2HI + 环戊二烯

开始P0 P0 0 0 平衡P0-X P0-X 2X X P0=

P总101.35??50.66KPa 222?PHI??P环戊二烯??2x??x?????????????x3101.325101.325?P??P???????22P25.33?(50.66?x)??50.66?xI??2???K? ??p=?P?101.325????x3即0.125?解得x?15.72kPa225.33?(50.66?x)2

14.CO2与H2S在高温下有如下反应

CO2(g)?H2S(g)?COS(s)?H2O(g)

今在610K时,将4.4×10Kg的CO2加入2.5dm体积的空瓶中然后再充入H2S使总压为1013.25Kpa。平衡后水的摩尔分数为0.02。同上实验,再620K,平衡后水的摩尔分数为0.03。(计算时可假定气体为理想气体)

(1)计算610K时的KP。 (2)求610K时的?rGm。 (3)计算反应的热效应?rHm。 解:(1)nCO2??-3

3

?4.4?10?3kg??0.1mol 44?10?3kg/mol开始时pV?nRT

PCO2?nRT0.1?8.314?610??202.86kPa ?3V2.5?10 CO2(g)?H2S(g)?COS(g)?H2O(g) 开始 202.86kPa 810.39kPa 0 0

平衡 202.86-p 810.39-p p p 平衡后:xH2O?pH2Op总?pH2O1013.25?0.02

解得:pH2O(g)?20.27kPa

pCOSpH2O???20.27?20.27?PP?KP??2.849?10?3 pCO2pH2S(202.86?20.27)?(810.39?20.27)?P?P?(2) ?rGm??RTlnKp??8.314?610?ln(2.849?10)?29.73kJ (3)620K时平衡后

???3 xHO?pH2Op总?pH2O1013.25?0.03

解得:pH2O(g)?30.40kPa

pCO2?202.86?30.40?172046kPa 则平衡后:pH2O?810.39?30.40?779.99kPa

pCOSpH2O????P?30.40?30.40?6.870?10?3 KP?PpCO2pH2S172.46?779.99???PP ln?KP2?KP1??rHm11???(?)

RT2T1??rHm6.870?10?311ln???(?) ?38.3146206102.849?10?解得:?rHm?276.7kJ

15.(1)在1120℃下用H2(g)还原FeO(s),平衡时混合气体中H2(g)的摩尔分数为0.54。求FeO(s)的分解压。已知同温度下,2H2O(g)?2H2(g)?O2(g)K??3.4?10?13;(2)在炼铁炉中,

氧化铁按如下反应还原: FeO(s)?CO(g)?Fe(s)?CO2(g) 求:1120℃下,还原1molFeO需要CO若干摩尔?

已知同温度下2CO2(g)?2CO(g)?O2(g) K??1.4?10?12 解:(1)FeO(s)?H2(g)?Fe(s)?H2O(g) ① 平衡气体中xH2?0.54,xH2O?1?0.54?0.46

???0 Kp,1?Kx??xH2OxH2?0.46?0.852 0.542H2O(g)?2H2(g)?O2(g) ②

?13 K?p,2?3.4?102FeO(S)?2Fe(s)?O2(g) ③

①×2+②=③

?2?13K??K??2.47?10?13 p,3?Kp,1p,2?0.852?3.4?10??2Kp,3??pO2p? pO2?Kp,3p?2.47?10???13?101.325?2.5?10?11KPa

(2)2CO2(g)?2CO(g)?O2(g) ④

FeO(s)?CO(g)?Fe(s)?CO2(g) ⑤

?KP,3????K?p,4?Kp,5??2.47?10?????1.4?10?12???12?13????0.42 ?12(③-④)÷2=⑤

设1120℃下,还原1molFeO需要nmolCO 2FeO(s)?CO(g)?Fe(s)?CO2(g) 1mol n 0

n-1 1mol ?V?0

1?0.42 n?1解得:n?3.38molk?p,5?kn?16.已知298.15K,CO(g)和CH3OH(g)标准摩尔生成焓?fH??m分别为-110.52

及-200.7KJ.mol?1.K?1,

?1CO(g).H2(g).CH3OH(I)的标准摩尔熵Sm分别为197.67,130.68及127J.mol?k.又知298.15K甲醇的饱和蒸气压为-16.59Kpa

摩尔汽化热?vapHm?38.0KJ.mol,蒸气可是为理想气体。

利用上述数据,求298.05K时,反应CO(g)+2 H2(g)= CH3OH(g)的?rGm及KP。

解: 101.325KPa CO(g)+2 H2(g)?CH3OH(g) ??S1

101.325Kpa CH3OH(I) ? ??S2

?S3 ??S4

16.59Kpa CH3OH(I)? CH3OH(g)

????S1?Sm,CH3OH(g)?2Sm,H2(g)?Sm,co(g)

???1? =127-2?130.68-197.67=-332.03J.mol?1 ?S2?0 ?S1??Hm,vapT38?103??127.45J.mol?1 298.15?S4?nRTlnp116.25?1?8.314?ln??15.04J.mol?1 p2101.35?rSm??S1??S2??S3??S4??219.62J.mol?1

????rHm??fHm,CH3OH(g)?2?fHm,H2(g)??fHm,CO(g)

=-200.7-(-110.52)=-90.18KJ.mol?1

????rGm??rHm?Tg?rSm??90.18?298.15?(?219.62)?103

=-24.73KJ.mol?1

???rGm=---RTlnKP

3?即—24.73?10??8.314?298.15?lnKP

KP=2.15?10

17求25℃时,下术反应的反应ka CH3COOH(I)+C2H5OH(I)=CH3COOC2H5(I)+H2O(I)已知个物质 的标准生成自由能?fGm

物质 ?fGm(kJ.mol)

??1?4??CH3COOH(I) —395.8 CH3COOC2H5(I) —341.1

H2O(I) —236.6 C2H5OH(I) —175.1

解:CH3COOH(l)+C2H5OH(l)=CH3COOC2H5(l)+H2O(l) ?rGm????vBB?fG?m?(-341.1)+(-236.6)-(175.1)-(-395.8)=-6.8KJmol-1

??? ?rGm??RTlnKa