1. Enthalpy of Chemical Reactions CHEMISTRY 11 Feb. 16, 2010

2. Chemical Bonds A molecule is composed of atoms held together by chemical bonds For a chemical reaction to occur, bonds must be broken, and new chemical bonds must form To break a chemical bond, energy must be put into molecule. (Surroundings feel cold) When chemical bonds are formed, the molecule releases energy. (Surroundings feel hot)

3. Chemical Bonds Every reaction contains many chemical bonds breaking and new bonds forming The total energy of a chemical reaction is the net sum of the energy released and energy absorbed by molecules A reaction is either exothermic (heat releasing) or endothermic (heat absorbing)

4. Enthalpy If molecules absorb heat from their surroundings, the surroundings feel cold If molecules release heat into the surroundings, the surroundings feel hot Enthalpy = “heat energy” in molecules Can only measure the change in enthalpy ∆H = H PRODUCTS - H REACTANTS

5. ∆H > 0 Change in enthalpy is positive Products Reactants Reaction proceeding HCl H + Cl Add energy Enthalpy Changes (∆H) in Endothermic Reactions HCl + energy (kJ)  H + Cl HCl  H + Cl∆H = +energy (kJ) 431 kJ You can write change in enthalpy (∆H) in two ways:

6. Endothermic Chemical Reactions Endothermic reactions need more heat energy in order to start the reaction, so energy appears on the reactants side Molecules or atoms absorb energy from the surroundings, so ∆H is positive HCl + 431 kJ  H + Cl H + Cl  HCl∆H = +431 kJ

7. ∆H < 0 Change in enthalpy is negative Products Reactants H + Cl HCl Give off energy Reaction proceeding Enthalpy Changes (∆H) in Exothermic Reactions H + Cl  HCl + energy (kJ) H + Cl  HCl∆H = -energy (kJ) 431 kJ You can write change in enthalpy (∆H) in two ways:

8. Exothermic Chemical Reactions Exothermic reactions produce heat energy, so energy appears on the products side Molecules or atoms lose energy to the surroundings, so ∆H is negative H + Cl  HCl + 431 kJ H + Cl  HCl∆H = -431 kJ

9. Change in Enthalpy in Chemical Reactions H (g) + Cl (g)  HCl (g) ∆H = -431 kJ HCl (g)  H (g) + Cl (g) ∆H = +431 kJ H 2(g) + F 2(g)  2HF (g) ∆H = -542 kJ C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (l) ∆H = -1418 kJ

10. Exercises Energy & Enthalpy in Chemical Reactions Handout HEBDEN: pp. 120-122, q. 68-80 Test February 22, 2010

11. Enthalpy Changes (∆H) in Chemical Reactions