Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Enthalpy Enthalpy (H): heat flow for a chemical reaction. q constant P = H = H products H reactants exothermic:q < H H products < H reactants endothermic:q > H H products > H reactants Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Standard Enthalpy of Formation and Reaction Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law aA + bB cC + dD Standard Enthalpy of Formation and Reaction standard-state conditions = 25 o C & 1 atm Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.10 Standard Enthalpy of Reactions: Direct Method 2Al (s) + 1Fe 2 O 3 (s) 1Al 2 O 3 (g) + 2Fe (l) Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law 2Al (s) + 1Fe 2 O 3 (s) 1Al 2 O 3 (g) + 2Fe (l) Al (s) 0 kJ/mol Al 2 O 3(s) 1669.8kJ/mol Fe (l) 12.40kJ/molFe 2 O 3 (s)822.2kJ/mol Example 6.10 (cont) H = 0 kJ/mol for elements at 25 o C (e.g., Fe(s) = 0 kJ/mol; Fe(l) 0) Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.10 (cont) 2Al (s) + 1Fe 2 O 3 (s) 1Al 2 O 3 (g) + 2Fe (l) INDIRECT METHOD (Hesss Law) Hesss Law is used is convenient for obtaining values of H for reactions that are difficult to carry out in a calorimeter. E.g., C (s) + O 2 (g) CO (g) Nearly impossible to carry out in a calorimeter because when C is burned, CO 2 is almost always produced. Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law For H2 + Cl2 2HCl H = 185 kJ 185 kJ sign = exothermic coefficients represent moles (the -185 kJ refers to ONE MOLE of reaction) states of matter (phases) MUST be written standard temp = 25 o C (not STP: 0 o C) Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Rules: (For H2 + Cl2 2HCl H = 185 kJ 1.Magnitude of DH is directly proportional to the amount of reactant or product kJ1 mol H2-185 kJ 1 mol Cl kJ2 mol HCl 2.Reverse the sign for H for the reverse reaction. H for a reaction is the same, whether carried out in one step or several (State Function) Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (a) C (graphite) + O 2 (g) CO 2 (g) 393.5 (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) The values for H are determined by experiment, so they need to be given to you; either directly or in a table. H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (a) C (graphite) + O 2 (g) CO 2 (g) 393.5 Concentrate on aligning the substances in the step reactions to be on the same sides as the original equation. Notice: Theres 2 moles of C(graphite) in the original equation but only 1 mole in (a) (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 Double (a) to match the original equation. And H doubles, too (393.5 * 2 = 787.0 kJ/mol) (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 H2 is a reactant in both the original equation and in step (b). And, theres 1 mole H2 in the original equation and 1 mol H2 in step (b). (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 Align (c) C2H2 with original equation as a product by reversing step (c) H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (c) 4CO 2 (g) + 2H 2 O (l) 2C 2 H 2 (g) + 5O 2 (l) (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 Reverse (c) so it on the same side as the original equation. Include reversing the sign of H, too. Shown by (c). However, there 2 moles of C2H2 in step (c) but only 1 mole in the original equation. So, we need to halve step (c) H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (b) H 2 (g) + O 2 (g) H 2 O (l) 285.8 Divide step (c), including H, by 2 to match the original equations 1 mole C2H2. (shown by (c/2) (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (b) 1H 2 (g) + O 2 (g) 1H 2 O (l) 285.8 Determine the net equation for both substances & H by adding them together. (2a) 2C (graphite) + 2O 2 (g) 2CO 2 (g) 787.0 CO 2 : 2 2 O 2 : 2 + 5/2 H 2 O: 1 1 H = kJ/mol H kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) (-c/2) 1C 2 H 2 (g) (b) + 1H 2 (g) 285.8 Net equation of steps = original equation H = kJ/mol (2a) 2C (graphite) 787.0 H kJ/mol 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) kJ/mol Energy: Standard Enthalpy of Formation and Reaction Direct & Hesss Law Example 6.09 Standard Enthalpy of Reactions: Hesss Law 2C (graphite) + 1H 2 (g) 1C 2 H 2 (g) H = kJ/mol 330_06_05 Energy: Calorimetry Fin