Question

跪求答案和过程！

Given the bond enthalpies in kJ mol-1: C-H 412, C-C 348 and H-H 436 (∆Hatom = 218 kJ mol-1).
Calculate the enthalpy of formation of propane if the enthalpy of atomisation of carbon (graphite) is 715kJ mol-1.

Answer 1

题目要求计算丙烷的标准摩尔生成焓。
告诉了一些键能数据/

丙烷的标准摩尔生成焓△fHmθ 是反应3 C(s) + 4 H2(g) = C3H8(g)的标准摩尔焓变△rHmθ
△fHmθ = △rHmθ = Σ(反应物) - Σ(生成物) = 3*715 + 4*436 - (8*412 + 2*348) = -103 kJ/mol

Answer 2

好难的问题啊，做不了