CHAP. 5: THERMOCHEMISTRY [CONTENTS] 133
Example
Calculate the standard reaction enthalpy of combustion of methane using the equation
CH 4 + 2O 2 =CO 2 + 2H 2 O(g).
at temperature 298.15 K andpst= 101.325 kPa. Is this reaction enthalpy the standard enthalpy
of combustion of methane? Data: AtT = 298.15 K andp= 101.325 kPa, ∆fH◦(H 2 O(g))
=−241.827 kJ/mol,∆fH◦(CO 2 )=−393.522 kJ/mol,∆fH◦(CH 4 )=−74.852 kJ/mol. The
enthalpy of formation of molecular oxygen is zero.
Solution
Based on equation (5.12) we have
∆rH◦ = ∆fH◦(CO 2 ) + 2∆fH◦(H 20 (g))−∆fH◦(CH 4 )−2∆fH◦(O 2 )
= − 393 .522 + 2×(− 242 .827)−(− 74 .852)− 2 ×0 =− 802. 324 kJ/mol
In this case the reaction enthalpy is not the standard enthalpy of combustion. In the above
chemical reaction water is in the gaseous state, but the most stable state of water at the given
temperature and pressure is liquid.
