Answer to Question #87144 in General Chemistry for mike

Question

The standard enthalpy change for the combustion of 1 mole of propane is –2043.0 kJ.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Calculate ΔfH° for propane based on the following standard molar enthalpies of formation.


molecule ΔfH° (kJ/mol-rxn)
CO2(g) –393.5
H2O(g) –241.8


A. +104.7 kJ/mol
B. –4190.7 kJ/mol
C. –2043.0 kJ/mol
D. +1407.7 kJ/mol
E. –104.7 kJ/mol

Answer 1

$ΔfH°(C3H8)= 3(-393,5) + 4(-241,8)-5*0 - (- 2043)= –104.7 kJ/mol$

Answer:E. –104.7 kJ/mol