Answer to Question #87144 in General Chemistry for mike

Question #87144
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
1
Expert's answer
2019-03-28T04:13:03-0400

"\u0394fH\u00b0(C3H8)= 3(-393,5) + 4(-241,8)-5*0 - (- 2043)= \u2013104.7 kJ\/mol"

Answer:E. –104.7 kJ/mol


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