已知298K时,丙烯加H2生成丙烷的反应焓变rHm=-123.9kJ•mol－1,丙烷定容燃烧热Qv=-2213.0kJ•mol－1,fHm(CO2,g)=-393.5kJ•mol－1,fHm(H2O,l)=-286.0kJ&

Question

已知 298K时,丙烯加 H2 生成丙烷的反应焓变  r Hm =-123.9 kJ•mol－
1,丙烷定容燃烧热Qv =-2213.0 kJ•mol－1 , f Hm (CO2,g)=-393.5 kJ•mol－1,
 f Hm (H2 O,l)= -286.0 kJ•mol－1 .
计算:(1) 丙烯的燃烧焓 ; (2) 丙烯的生成焓.

Answer 1

① C3H6(g) + 2 O2 (g) = 3 CO2 (g) + 3 H2O (l)； △rHmΘ1 = △cHmΘ
② H2 (g) + 1/2 O2 (g) = H2O (l)； △rHmΘ2 = -286.0 kJ/mol
③ C3H8 (g) + 5 O2(g) = 3 CO2(g) + 4 H2O (l)； Qv = -2213.0 kJ/mol
④ C3H6 (g) + H2 (g) = C3H8 (g)； △rHm4 = -123.9 kJ/mol
△rHmΘ3= Qv + △nRT= -2213.0 + (3-5-1)*8.314*298/1000 = -2 220.4 kJ/mol
由于 ① = ④ +③ - ②
所以
△cHmΘ(C3H6,g)
= △rHm1 = △rHm4 + △rHm3- △rHm2
= -123.9 – 2220.4 + 286.0
= -2 058.3 kJ/mol
即C3H6(g)的燃烧焓△cHmΘ = -2 058.3 kJ/mol
由反应①得,△rHmΘ1 = 3 △fHmΘ(H2O,l) + 3 △fHmΘ(CO2,g) - △fHmΘ(C3H6,g)
所以 △fHmΘ(C3H6,g) = 2 058.3 + 3 *(-286.0)+3*(-393.5)= 19.8 kJ/mol
即C3H6(g)的生成焓△fHmΘ(C3H6,g) = 19.8 kJ/mol