2. A simpler problem: How much heat is given off when 1.6 g of CH 4 are burned in an excess of oxygen if  H comb = -802 kJ/mol? Step 1: Write the reaction equation. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) Step 2: Calculate molar amount involved Step 3: Calculate amount of heat given off  H rxn = (-802 kJ/mol)(0.100 mol CH 4 ) = -80.2 kJ Q: Is this an exothermic or endothermic reaction?

3. 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2 Energy 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2  H rxn = Heat content of products – heat content reactants  H rxn < 0 Reaction is exothermic But how do we determine the heat content in the first place?

4. Heat of formation,  H f The  H f of all elements in their standard state equals zero. The  H f of all compounds is the molar heat of reaction for synthesis of the compound from its elements  H f (AlBr 3 ): 2 Al + 3 Br 2 2 AlBr 3  H rxn = 2  H f (AlBr 3 )  H rxn 2  H f (AlBr 3 ) = Since the  H rxn can be used to find  H f, this means that  H f can be used to find  H rxn WITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!!

5. Hess’s Law:  H rxn =  H f (products) –  H f (reactants) 6 CO 2 (g) + 6 H 2 O (l)C 6 H 12 O 6 (s) + 6 O 2 (g)  H rxn = [  H f (C 6 H 12 O 6 ) + 6  H f (O 2 )] – [6  H f (CO 2 ) + 6  H f (H 2 O)] From  H f tables:  H f (C 6 H 12 O 6 ) = -1250 kJ/mol  H f (CO 2 ) = -393.5 kJ/mol  H f (H 2 O) = -285.8 kJ/mol  H rxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]  H rxn = +2825.8 kJ/mol

6. Using Hess’ Law with  H rxn What is the  H comb for ethane? C 2 H 4 (g) + 3 O 2 (g) → 2 CO 2 (g) + 2 H 2 O (g) -1323 kJ C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g) -137 kJ 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Target rxn: Given: Rxn 1: Rxn 2: 1) Rxn 1 doesn’t have enough oxygens, so multiply by 2 2(C 2 H 4 (g) + 3 O 2 (g) → 2 CO 2 (g) + 2 H 2 O (g))  H rxn 2(-1323 kJ) 2) Rxn 2 is going the wrong direction and doesn’t have enough C 6 H 6. Reverse reaction and multiply by 2. 2(C 2 H 6 (g) → C 2 H 4 (g) + H 2 (g)) 2(+137 kJ) Note: when you reverse a reaction, change the sign on the  H Rxn 3:2 H 2 (g) + O 2 (g) → 2 H 2 O (g)-242 kJ

7. 3) Combine the two reaction equations: 2 C 2 H 4 + 6 O 2 → 4 CO 2 + 4 H 2 O2(-1323 kJ) + 2 C 2 H 6 → 2 C 2 H 4 + 2 H 2 + 2(+137 kJ) 2 C 2 H 4 + 2 C 2 H 6 + 6 O 2 → 4 CO 2 + 4 H 2 O + 2 C 2 H 4 + 2 H 2 2 C 6 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Target rxn: 4) Still don’t have enough O 2 and need to get rid of H 2. Add in Rxn 3. + 2 H 2 + O 2 → 2 H 2 O + (-242 kJ) 2 C 2 H 6 + 2 H 2 + O 2 + 6 O 2 → 4 CO 2 + 2 H 2 O + 4 H 2 O + + 2 H 2 Final rxn: 2 C 2 H 6 + 7 O 2 → 4 CO 2 + 6 H 2 O Math: 2(-1323 kJ) + 2(+137) + (-242) = -2614 kJ =  H comb