Final answer:
The heat of combustion of 1 mole of methane can be calculated using a thermochemical equation and the values of standard enthalpy of formation for CO2 and H2O, resulting in the release of 890.4 kJ of heat.
Step-by-step explanation:
To calculate the heat of combustion of methane, we use the standard enthalpy of formation values for the products and reactants involved in the combustion process. The balanced thermochemical equation for the combustion of methane is as follows:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + 890.4 kJ
This equation tells us that when 1 mole of methane is combusted, 890.4 kJ of heat is released. The enthalpy of combustion (ΔHcomb) can be calculated by the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:
ΔHcomb = [ΣΔHf,products] - [ΣΔHf,reactants]
Given the standard enthalpies of formation for CO2 and H2O are -394 kJ/mol and -286 kJ/mol respectively, and considering that the enthalpy of formation for O2 is zero (since it is in its standard state), the equation can be solved as:
ΔHcomb = [(-394 kJ/mol) + 2(-286 kJ/mol)] - (-74.6 kJ/mol) = -890.4 kJ/mol
Thus, the heat of combustion of 1 mole of methane under standard conditions is -890.4 kJ/mol, which means this amount of heat is released to the surroundings during the reaction.