Heat of Combustion, Enthalpy

Dear PF Forum,
But I don't understand what it means. Can someone help me?

https://en.wikipedia.org/wiki/Heat_of_combustion
A: Heat of combustion of CH4 is 50.09 MJ/kg
https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)
B: Enthalpy of water: -285.88 KJ/mol
C: Enthalpy of CO2: -393.509 KJ/mol

Then, I'll try to do a simple calculation.

A: CH4 heat of combustion is 50.09 MJ/kg.
What does it mean?
Does it mean that if we burn 1 kg of CH4 completely
CH4 + 2O2 -> CO2 + 2H2O
It will produce 50.09 mega joules?

B: Enthalpy of water is 285.88 kj/mol
What does it mean?
If we combine 2 moles of Hydrogen and 1 moles of Oxygen, it will explode and gives 285.88 kilo joules?

C: Then I did some calculation...
The heat of combustion of
1 moles CH4 + 2 moles O2 -> 1 moles CO2 + 2 moles H2O
16 gr CH4 + 32 gr O2 -> 44 gr CO2 + 36 gr H2O will gives 50.09 MJ/kg * 16 gr = 801.44 KJ

The enthalpy of
1 moles H2O: -285.88 KJ
1 moles CO2: -393.519 KJ

Combining those two:
2 moles H2O: -571.76 KJ/mol
1 moles CO2: -393.519 KJ/mol
= 965.269 KJ

801.44 KJ ≠ 965.269 KJ

Where did I go wrong?
Or my understanding of the concept of heat combustion and enthalpy is wrong.

Bystander
Homework Helper
Gold Member
Check the "phase" of water (liq/vap).

• Stephanus
Check the "phase" of water (liq/vap).
Of course. Thanks. The numbers are close now.

Chestermiller
Mentor
Of course. Thanks. The numbers are close now.
Liquid water vs water vapor accounts for only about half the difference. You also forgot to subtract the heat of formation of methane, which is -75 kJ/mole.

Chet

• Stephanus
James Pelezo
Gold Member
You might also read about Hess's Law Equation. That is, if you haven't already done so. Heat of Rxn (or, Heat of Combustion in this case) = (Sum of Enthalpies of Formation of Products) - (Sum of Enthalpies of formation of Reactants) Google 'Enthalpy of Formation Tables' You'll get many 'energy of formation' values to apply.

• Stephanus