2.7 Calculation of Enthalpy Changes of a Class of Chemical Reactions 89
EXAMPLE2.29
Find the standard-state enthalpy change of the reaction of Eq. (2.7-6) at 298.15 K, using
values of enthalpy changes of formation from Table A.8.
Solution∆H◦ 2 ∆fH◦(NO 2 )+(−1)∆fH◦(N 2 O 4 )
2(33.095 kJ mol−^1 )−(9.079 kJ mol−^1 ) 57 .11 kJ mol−^1If the formation reaction for a substance cannot actually be carried out, the enthalpy
change of formation can be calculated from the enthalpy change of other reactions that
can be combined to give the reaction of interest, using Hess’s law.EXAMPLE2.30
The standard-state enthalpy change of combustion of methane at 298.15 K equals
−890.36 kJ mol−^1 , with liquid water as one of the products. Find the enthalpy change of
formation of methane at 298.15 K using the enthalpy changes of formation of H 2 O and CO 2.
Solution
The balanced reaction equation is0 CO 2 (g)+2H 2 O(l)−CH 4 (g)−2O 2 (g)so that− 890 .36 kJ mol−^1 ∆fH◦(CO 2 )+ 2 ∆fH◦(H 2 O)
+(−1)∆fH◦(CH 4 )+(−2)∆fH◦(O 2 )Because gaseous O 2 is the most stable form of oxygen at 298.15 K,∆fH◦(O 2 )0. Using
values of the enthalpy changes of formation of the other substances,∆fH◦(CH 4 ) 890 .36 kJ mol−^1 +(− 393 .522 kJ mol−^1 )
+2(− 285 .830 kJ mol−^1 )+(− 2 )(0)
− 74 .82 kJ mol−^1This value agrees fairly well with the value in Table A.8,− 74 .873 kJ mol−^1.Enthalpy Changes at Various Temperatures
An enthalpy change for a reaction at a temperature other than a temperature found in
a table can be calculated from heat capacity data. Consider the processes shown in
Figure 2.10. The reaction at temperatureT 2 is the process whose enthalpy change we
want to find. Call it process 2. An alternate pathway with the same initial and final
states consists of processes 3, 1, and 4. Process 3 is the change in temperature of the
reactants fromT 2 toT 1. Process 1 is the chemical reaction at temperatureT 1 , with a
known enthalpy change. Process 4 is the change in temperature of the products from