2. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

3. Consider the combustion of methane, CH 4, to produce gaseous CO 2 and liquid H 2 O. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (l)∆H 1 = -890 kJ where ∆H 1 is the enthalpy change for the reaction. This reaction can be thought of as occurring in two steps. In the first step, methane is combusted to produce water vapor: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) In the second step, water vapor condenses from the gas phase to the liquid phase: 2H 2 O (g) → 2H 2 O (l)

4. Each of these reactions is associated with a specific enthalpy change: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g)∆H 1 = -802 kJ 2H 2 O (g) → 2H 2 O (l)∆H 2 = -88 kJ Combining these equations gives the following: CH 4 (g) + 2O 2 (g) + 2H 2 O (g) → CO 2 (g) + 2H 2 O (g) + 2H 2 O (l) We can cancel what is the same on both sides of the reaction and that gives the net reaction, or overall, reaction. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (l) ∆H 1 = ∆H 2 + ∆H 3 = -802 kJ + (-88 kJ) = -890 kJ

5. This energy relationship can be shown in an enthalpy diagram.