Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Standard Enthalpy of Formation and Reaction.

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Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Standard Enthalpy of Formation and Reaction

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 & Hess’s Law

Example 6.10 Standard Enthalpy of Reactions: Direct Method 2Al (s) + 1Fe 2 O 3 (s)  1Al 2 O 3 (g) + 2Fe (s) Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s 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 (con’t) Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law

Example 6.10 (con’t) 2Al (s) + 1Fe 2 O 3 (s)  1Al 2 O 3 (g) + 2Fe (l) Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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: There’s 2 moles of C(graphite) in the original equation but only 1 mole in (a) (b) H 2 (g) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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, there’s 1 mole H2 in the original equation and 1 mol H2 in step (b). (b) H 2 (g) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g)  1C 2 H 2 (g) (b) H 2 (g) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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) + ½O 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g)  1C 2 H 2 (g) (b) H 2 (g) + ½O 2 (g)  H 2 O (l) –285.8 Divide step (–c), including  H, by 2 to match the original equation’s 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 & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g)  1C 2 H 2 (g) (b) 1H 2 (g) + ½O 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 + ½  5/2 H 2 O: 1  1  H = kJ/mol  H kJ/mol Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law

Example 6.09 Standard Enthalpy of Reactions: Hess’s 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 Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law 2C (graphite) + 1H 2 (g)  1C 2 H 2 (g) kJ/mol

Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g)  1C 2 H 2 (g)  H = kJ/mol Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law

Fin 330_06_05 Energy: Calorimetry