Question

On the coldest day of 2022 winter, 287.6x106 m3 (measured at 1 atm, 0 °C) of natural gas (methane, CH4) was used in Istanbul. Calculate the total enthalpy change for the combustion of this quantity of methane. (AH°FCH4(E)=-74.8 kJ/mol, AH°FCO2(g)=-393.5 kJ/mol, AH°FH2O()=-285.83 kJ/mol)

Answer 1

287.6 m^3 = 287,600 liters @ 1 atm and zero C = 287,600 L x 1 mole/22.4 L = 12,839 moles CH4.
CH4 + 2O2 ==> CO2 + 2H2O
delta Hrxn = (n*dHo products) - (dHo reactants)
dHrxn = (1*dHo CO2 + 2*dHo H2O) - (1*dHo CH4 + 2*dHo O2)
Plug in your values for CO2, CH4, and H2O (note that dHo for O2 = 0) and solve for dHrxn in kJ/mol CH4.
Then dHrxn in kJ/mol x number moles from above = ?