兰叶青 无机及分析化学课后习题答案(所有章节)
;(5)×;(6)×;
(7)×;(8)×;(9);(10)×;(11)×
2-11解:(1)敞开体系;(2)孤立体系;(3)敞开体系;
2-12解:(1) =100kJ W=?500 kJ △U = + W=?400 kJ
(2) =?100kJ W=500 kJ △U = + W=400 k
2-13解:因为此过程为可逆相变过程,所以
p =△H= 40.6kJ·mol?1
W=?p外△V≈?nRT =?8.314×373.15=?3.10 kJ·mol?1
△U = + W= 40.6+(?3.10)=37.5 kJ·mol?1
2-14解:(1)
?HΘΘ
rm???B?fHm(B)
B
?3?ΘΘ
fHm(CO2,g)+2?fHm(Fe,s)
?3?ΘΘ
fHm(CO,g)??fHm(Fe2O3,s)
?3?(?393.51)+2?0?3?(?110.52)?(?822.2)
??26.77kJ?mol?1
(2)
?Θ?Θ
rHm??B?fHm(B)
B
??HΘ,g)+?Θ
fm(CO2fHm(H2,g)
??HΘΘ
fm(CO,g)??fHm(H2O,g)
?(?393.51)+0?(?110.52)?(?241.82)
??41.17kJ?mol?1
(3)
?ΘHΘ
rHm???B?fm(B)
B
?6?Θl)+4?Θ
fHm(H2O,fHm(NO,g)
?5?(ΘOΘ
fHm2,g)?4?fHm(NH3,g)
?6?(?285.83)+4?(90.25)?5?0?4?(?46.11)
??1169.54kJ?mol?1
2-15解:乙醇的?HΘΘ
fm反应?rHm为:
(4) 2C(s,石墨)+3H2(g)+1/2O2(g)= C2H5OH(l),所以:
反应(4)=反应(2)×2+反应(3)×3?反应(1)
?ΘΘΘΘ
rHm(4)?2?rHm(2)?3?rHm(3)??rHm(1)
?