1. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Example Problem
Calculate the H for the following reaction using standard state enthalpy information:
CH4(g) O2(g)  CO2(g) H2O(g)
CH4(g) :
2H2(g) C(graphite)  CH4(g)
Hf= kJ
CH4(g)  H2(g) C(graphite)
Hf= kJ
CO2(g) :
C(graphite) O2(g)  CO2(g)
Hf= kJ
H2O(g) :
H2(g) ½ O2(g)  H2O(g)
Hf= kJ
2 H2(g) O2(g)  2 H2O(g)
Hf= kJ
CH4(g) O2(g)  CO2(g) H2O(g)
H= kJ
Hf= kJ

2. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
...an easier way...
Calculate the H for the following reaction using standard state enthalpy information:
CH4(g) O2(g)  CO2(g) H2O(g)
List all Hf values.
Standard State (0) kJ kJ kJ
Multiply by coefficients.
1(-74.85) kJ
1(-393.5) kJ
2(-241.8) kJ kJ kJ kJ
Calculate “Products – Reactants”.
Products
Reactants
Hf = Hfproducts – Hfreactants
= [(-393.5)+(-483.6)] – (– 74.85)
Hf = kJ

3. Example Problem 6.76 a) Calculate Hrxn for
SiO2(s) HF(g)  SiF4(g) H2O(l)
From Appendix B kJ
-273kJ kJ kJ
Hrxn= [2( ) ( )] – [1  (-910.9)  (-273)]
Hrxn= -184 kJ