1. Standard Enthalpy Changes of Reaction Section 15.1

2. Introduction Enthalpy change of formation: specified by ΔH θ f is the heat change (at constant pressure) on production of one mole of the pure substance from its elements in their standard state under standard thermodynamic conditions (298 K and 1 atm pressure)

3. More Standard state: generally the most thermodynamically stable form of the pure element that exists under standard thermodynamic conditions For example, for carbon it is graphite

4. Example ΔH θ f of silver bromide is the enthalpy change for the reaction: Ag (s) + ½Br 2(l) → AgBr (s) ΔH θ f = -99.5kJ mol - 1 Notice the formation of one mole and the standard states 2Ag (s) + Br 2(l) → 2AgBr (s) makes 2 mol Ag (s) + ½Br 2(g) → AgBr (s) Br not in standard state

5. Calculations Enthalpy change of a reaction = the sum of enthalpies of formation of products – sum of enthalpies of formation of reactants ΔH = ΣΔH θ f [products] – ΣΔH θ f [reactants] reactantsproducts elements

6. Example Problem #1 Calculate the enthalpy change of the following reaction: 3CuO (s) + 2Al (s) → 3Cu (s) + Al 2 O 3(s) ΔH θ f [CuO] = -155 kJmol -1 ΔH θ f [Al 2 O 3 ] = -1669 kJmol -1 See the board for the working out

7. Example Problem #2 Calculate the enthalpy change of the following reaction: NH 4 NO 3(s) → N 2 O (g) + 2H 2 O (l) ΔH θ f [NH 4 NO 3 ] = -366 kJ mol -1 ΔH θ f [N 2 O] = +82 kJ mol -1 ΔH θ f [H 2 O] = -285 kJ mol -1

8. Continued ΔH θ reaction = ΔH θ f,products – ΔH θ f, reactants Have to consider 2 mol of water ΔH θ = (82 + (2) (-285)) - (-366) ΔH θ = (82 + (-570)) - (-366) ΔH θ = -488 + 366 ΔH θ = -122 kJ mol -1

9. Enthalpy Change of Combustion ΔH θ c is the enthalpy change (at constant pressure) when one mole of a pure substance undergoes complete combustion under standard thermodynamic conditions Products of complete combustion are CO 2 and H 2 O Can be used to calculate the total enthalpy change only if both sides can be burnt in oxygen

10. Using Hess's Law elements compounds + O 2 + O 2 combustion products ΔH θ =ΣΔH θ c [reactants] – ΣΔH θ c [products] Combustion is from the compound while formation is to the compound

11. Example Problem #3 Calculate the enthalpy change for the hydration of ethene according to the following equation: C 2 H 4(l) + H 2 O (l) → C 2 H 5 OH (l) ΔH θ c [C 2 H 4 ] = -1409 kJ mol -1 ΔH θ c [C 2 H 5 OH] = -1371kJ mol -1 ΔH θ = -1409 - (-1371) = - 38 kJ mol -1