1. Section 7.4—Energy of a Chemical Reaction What’s happening in those hot/cold packs that contain chemical reactions?

2. Enthalpy of Reaction Enthalpy of Reaction (  H rxn ) – Net energy change during a chemical reaction +  H rxn means energy is being added to the system—endothermic -  H rxn means energy is being released from the system—exothermic

3. Enthalpy of Formation Enthalpy of Formation (H f ) – Energy change when 1 mole of a compound is formed from elemental states Heat of formation equations: H 2 (g) + ½ O 2 (g)  H 2 O (g) C (s) + O 2 (g)  CO 2 (g) A table with Enthalpy of Formation values can be found in the Appendix of your text Be sure to look up the correct state of matter: H 2 O (g) and H 2 O (l) have different H f values!

4. The overall enthalpy of reaction is the opposite of H f for the reactants and the H f for the products Reactants are broken apart and Products are formed. Breaking apart reactants is the opposite of Enthalpy of Formation. Forming products is the Enthalpy of Formation.  H rxn = sum of H f of all products – the sum of H f reactants Enthalpy of Formation & Enthalpy of Reaction This is not the way a reaction occurs—reactants break apart and then rearrange…remember Collision Theory from Chpt 2! But for when discussing overall energy changes, this manner of thinking is acceptable.

5. Example Example: Find the  Hrxn for: CH 4 (g) + 2 O 2 (g)  2 H 2 O (g) + CO 2 (g) H f (kJ/mole) CH 4 (g)-74.81 O 2 (g)0 H 2 O (g)-241.8 CO 2 (g)-393.5

6. Example Example: Find the  Hrxn for: CH 4 (g) + 2 O 2 (g)  2 H 2 O (g) + CO 2 (g) H f (kJ/mole) CH 4 (g)-74.81 O 2 (g)0 H 2 O (g)-241.8 CO 2 (g)-393.5  H rxn = -802.29 kJ

7. Let’s Practice #1 H f (kJ/mole) CH 3 OH (l)-238.7 O 2 (g)0 H 2 O (l)-285.8 CO 2 (g)-393.5 Example: Find the  Hrxn for: 2 CH 3 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O (l)

8. Let’s Practice #1 H f (kJ/mole) CH 3 OH (l)-238.7 O 2 (g)0 H 2 O (l)-285.8 CO 2 (g)-393.5 Example: Find the  Hrxn for: 2 CH 3 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O (l)  H rxn = -1453 kJ

9. Enthalpy & Stoichiometry The Enthalpy of Reaction can be used along with the molar ratio in the balanced chemical equation This allows Enthalpy of Reaction to be used in stoichiometry equalities

10. Example: If 1275 kJ is released, how many grams of B 2 O 3 is produced? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ Example

11. -1275 kJ kJ mole B 2 O 3 = ________ g B 2 O 3 -2035 1 43.62 mole B 2 O 3 g B 2 O 3 1 69.62  H = -1275 kJ (negative because it’s “released”) From balanced equation: -2035 kJ = 1 mole B 2 O 3 Molar mass: 1 mole B 2 O 3 = 69.62 g B 2 O 3 Example: If 1275 kJ is released, how many grams of B 2 O 3 is produced? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ

12. Let’s Practice #2 If you need to produce 47.8 g B 2 O 3, how many kilojoules will be released? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ

13. Let’s Practice #2 47.8 g B 2 O 3 g B 2 O 3 mole B 2 O 3 = ________ kJ 69.92 1 -1397 mole B 2 O 3 kJ 1 -2035 From balanced equation: -2035 kJ = 1 mole B 2 O 3 Molar mass: 1 mole B 2 O 3 = 69.62 g B 2 O 3 If you need to produce 47.8 g B 2 O 3, how many kilojoules will be released? B 2 H 6 (g) + 3 O 2 (g)  B 2 O 3 (s) + 2 H 2 O (g)  H = -2035 kJ