1. Standard Enthalpy of Formation EQ: Why does the  Hfº for a free element equal zero?

2. A. Standard Enthalpy of formation (  H f º)  H f º : the energy change that occurs when 1 mole of a compound is formed (most are negative) A.Standard conditions = 25ºC & 1 atm (gases) & 1 Molar (aqueous solutions) B.values found in table or appendix C (pg 921) C.All free elements in their most stable form  H f º= zero Since you can not make elements This includes diatomics (Cl 2, H 2, O 2, etc.)

3. A. Standard Enthalpy of formation (  H f º)  Ex: reaction for the heat of formation for H 2 O (g)  Elements in water = H and O  Stable form = H 2 and O 2  H 2 (g) + ½ O 2 (g)  H 2 O (g)  H f º= -241.8 kJ/mol

4. B. Calculating  Hrxn The  Hºrxn can also be found using the  H f º values  use coefficients in the balanced reaction  pay attention to states (s, l, g, aq)  Hºrxn =  n  H f º(products) -  n  H f º (reactants) n= moles  (sigma) = sum of

5. Example 1 Use the standard enthalpies of formation on pg 510 to calculate  Hº rxn for: 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O (l)  Hº rxn = [ (4 mol x 33 kJ/mol) + (6 mol x -286 kJ/mol)] - [(4 mol x -46 kJ/mol)+ (7 mol x 0 kJ/mol)]  Hº rxn = [132 kJ + -1716 kJ] - [-184 kJ + 0kJ] = -1400 kJ. Round to the least number of decimal places out of all of the table values -46 0 (free element) 33 -286 From table

6. Example 2 Use the table on pg 921 and the reaction below to determine the standard enthalpy of formation for CCl 4 gas. CH 4 (g) + 4Cl 2 (g)  CCl 4 (g) + 4HCl(g) ∆H  = –397 kJ ∆H rxn =[(1 mol x ∆H f  CCl 4 (g)] + (4 mol x –92.307kJ/mol)] - [(1 mol x -74.81kJ/mol) + (4 mol x 0 kJ/mol)] [1 mol ∆H f  + -369.228 kJ ] - [-74.81 kJ] -397 kJ = -74.81 0? (gas)-92.307  From table

7. Example 2 (continued) [ 1 mol ∆Hf  + -369.228 kJ ] - [-74.81 kJ]-397 kJ = 1 mol ∆Hf  - 294.418 kJ -397 kJ = -102.6 kJ = 1 mol ∆Hf  1 mol ∆Hf  = -103 kJ/mol

8. Ex 3: Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  n  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x– 286 ] – [ 2x49.04 ] = -6536 kJ -6536 kJ 2 mol = - 3268 kJ/mol C 6 H 6