mojo wrote:for eg. combustion of CH4:
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)
enthalpy of combustion: (-393.51) + 2(-241.82) - (-74.81) = -802.34 kj/mol
energy output per unit mass:
802.34 x 1000/16(molar mass) = -50146.25 kj/kg
energy output per unit volume:
-50146.25 x 0.645(mass density) = -32344.33 kj/m3
1 mole of CO2= 44g
1 mole of CH4= 16g
At STP the volume of an ideal gas occupies 0.0224 m3...
the CO2 energy output per mass....
-802.34 kj/mol divided by 12 g/mol equals 66.86 kj/g
am i doing the right thing, and how would you calculate the CO2 output per unit volume?
Many thanks.
I did not verify your enthalpy calculation. Taking it as correct, the enrgy output per unit mass is correct.
I don't know where your density number came from. I would use molar volume at STP. Note that you accidentally used molar mass of 12, not 16.
-802.34 kj/mol / 0.0224 m3/mol = -35818.75 kJ/m^3
As to CO2, 1 mol CH4 --> 1 mol CO2. Of course, it won't be 22.4 L, because it won't be at STP, it will be HOT! :) The obvious temperature change of combustion means it would be better to work in moles or in mass.
(or, alternatively, after it cools to STP, the mole of CO2 will occupy 22.4 L)
Note that methane is a stronger greenhouse gas than CO2 (by 21X) so it is better to burn methane than allow it to escape. Still better to keep it sequestered.