Question

Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:
CH4+2O2→2H2O+CO2

ΔHrxn from a given table:
CH4 = 1656 kJ/mol
O2 = 498 kJ/mol
H2O = 928 kJ/mol
CO2 = 1598 kJ/mol
ΔHrxn = [(1656) + (2*498)] - [(2*928) + (1598)] = -802 kJ/mol ???

I've calculated the reaction's enthalpy change to a total of -802 kJ/mol but it's saying I'm wrong. Am I doing anything wrong?

Answer 1

Nevermind, I have found it the enthalpy change is -802 kJ. Not kJ per mole.