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

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

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!

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.

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) O 2 (g)0 H 2 O (g) CO 2 (g)-393.5

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) O 2 (g)0 H 2 O (g) CO 2 (g)  H rxn = kJ

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

Let’s Practice #1 H f (kJ/mole) CH 3 OH (l) O 2 (g)0 H 2 O (l) CO 2 (g) 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 = kJ

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

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 = kJ Example

-1275 kJ kJ mole B 2 O 3 = ________ g B 2 O mole B 2 O 3 g B 2 O  H = kJ (negative because it’s “released”) From balanced equation: kJ = 1 mole B 2 O 3 Molar mass: 1 mole B 2 O 3 = 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 = kJ

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 = kJ

Let’s Practice # g B 2 O 3 g B 2 O 3 mole B 2 O 3 = ________ kJ mole B 2 O 3 kJ From balanced equation: kJ = 1 mole B 2 O 3 Molar mass: 1 mole B 2 O 3 = 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 = kJ