CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Enthalpy change for a given reaction is equal to = (sum of standard enthalpies of formation of products) − (sum of standard enthalpies of formation of reactants).
Given,
enthalpy of CO2=−393.5kJ/mol
enthalpy of H2O=−285.8kJ/mol
enthalpy of CH4=−74.9kJ/mol ∴ Enthalpy of combustion of methane =[−393.5+2×(−285.8)]−[−74.9+0] =−965.1+74.9 =−890.2kj/mol =16890.2kJ/g =−55.6kJ/g
The standard enthalpy of formation is taken as zero, because it is most stable form at 298K and 1 atm pressure.