The document discusses standard enthalpy of formation (ΔHf°), which is the amount of heat absorbed or released when one mole of a substance is formed from its elements in their standard states. Examples are provided of writing balanced chemical equations for formation reactions and using ΔHf° values to calculate the enthalpy change (ΔH°) of chemical reactions. The standard heat of combustion (ΔHc°) is also introduced, which is the enthalpy change when a substance undergoes complete combustion.

1. ENTHALPY OF FORMATION

2. ENTHALPY OF FORMATION Standard enthalpy of formation ( Δ H f °): Also called the standard heat of formation of a substance Standard heat of formation: The amount of heat absorbed or released when one mole of the substance is formed at 25°C and 100kPa (SATP) from its elements in their standard states Δ H f ° units are always in kJ/mol

3. ENTHALPY OF FORMATION Standard heat of formation: The amount of heat absorbed or released when one mole of the substance is formed at 25°C and 100kPa (SATP) from its elements in their standard states Example: What equation represents the formation of nitric acid? Nitric acid = HNO 3(l) hydrogen nitrogen oxygen Elements: What are these elements in their standard (most common) states? ½ H 2(g) + ½ N 2(g) + 3/2 O 2(g)  HNO 3(l)

4. ENTHALPY OF FORMATION The Δ H f ° for elements in their standard states are zero

5. ENTHALPY OF FORMATION Determine the equations for the formation of: a) KBrO 3 K (s) + ½ Br 2(l) + 3/2 O 2(g)  KBrO 3(s) b) C 2 H 5 OH 2C (s) + 3H 2(g) + 1/2O 2(g)  C 2 H 5 OH (l) c) NaHCO 3 Na (s) + ½H 2(g) + C (s) + 3/2O 2(g)  NaHCO 3(s) Don’t forget that the equation must result in the formation of one mole of the desired compound.

6. ENTHALPY OF FORMATION Now we can assign the Δ H° f of the product as the Δ H° of the whole reaction 2C (s) + 3H 2(g) + 1/2O 2(g)  C 2 H 5 OH (l) Δ H°= -235.2kJ/mol This additional equation may assist solving Hess’ Law questions.

7. ENTHALPY OF FORMATION How are these equations and Δ H° f values useful? CaO (s) + H 2 O (l)  Ca(OH) 2(s) By knowing the Δ H° f of each of the chemicals in the above reaction, you can calculate the Δ H° of the reaction without using thermochemical equations and Hess’ Law

8. ENTHALPY OF FORMATION  H° f Formation reactions and their Δ H° f values may be manipulated to determine balanced equation and the Δ H° value of chemical reactions. Δ H° = ∑ Δ H° f(products) - ∑ Δ H° f(reactants) ∑ = “sum of” Don’t forget that the equations should be multiplied by the appropriate factor, as necessary.

9. ENTHALPY OF FORMATION Using Δ H° f values, calculate the Δ H° of combustion of one mole of ethanol to produce carbon dioxide gas and liquid water. C 2 H 5 OH (l) + O 2(g)  CO 2(g) + H 2 O (l) 3 2 3 -235.2kJ/mol 0.0000kJ/mol -393.5kJ/mol -285.8kJ/mol Δ H° = ∑ Δ H° f(products) - ∑ Δ H° f(reactants) = [(2 mol x -393.5kJ/mol) + (3 mol x -285.8kJ/mol)] – [(1 mol x -235.2kJ/mol) + (3 mol x 0.0000kJ/mol)] = -1644.4kJ – (-235.2kJ) = -1409.2kJ Since this equation already combusts 1 mole of ethanol, then the final answer is: Therefore the heat of combustion is -1409kJ/mol of ethanol

10. ENTHALPY OF FORMATION Using Δ H° f values, calculate the Δ H° for the following reaction: NaOH (s) + HCl (g)  NaCl (s) + H 2 O (l) -425.6kJ/mol -92.30kJ/mol -411.2kJ/mol -285.8kJ/mol Δ H° = ∑ Δ H° f(products) - ∑ Δ H° f(reactants) = [(1 mol x –411.2kJ/mol) + (1 mol x -285.8kJ/mol)] – [(1 mol x -425.6kJ/mol) + (1 mol x -92.30kJ/mol)] = -697.0kJ – (-517.9kJ) = -179.1kJ Therefore the heat of the reaction is -179.1kJ

11. ENTHALPY OF FORMATION Enthalpy of Combustion: What is the reaction to produce C 6 H 12 O 6 ? 6 CO 2 + 6 H 2 O  6 O 2 + C 6 H 12 O 6 Where does this naturally occur? How do we measure it? In plant chloroplasts; it is not easy to measure.

12. ENTHALPY OF FORMATION Enthalpy of Combustion: C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O For complex organics, such as C 6 H 12 O 6 , it is difficult to directly measure its formation. Instead, the compound is combusted and the products analyzed to determine Δ H° f for the original compound.

13. ENTHALPY OF FORMATION Enthalpy of Combustion:  H° c standard heat of combustion - the Δ H° for the combustion of one mole of compound Ex. CH 3 OH (g) + O 2(g)  CO 2(g) + H 2 O (l) Δ H° c = -727kJ

14. ENTHALPY OF FORMATION Enthalpy of Combustion: C 2 H 4(g) + 3 O 2(g)  2 CO 2(g) + 2 H 2 O (l)  H° c = -1386 kJ Calculate the  H° f for C 2 H 4 . 2 CO 2(g) + 2 H 2 O (l)  C 2 H 4(g) + 3 O 2(g)  H° = +1386 kJ Can’t simply reverse this equation.

15. ENTHALPY OF FORMATION x 0.000kJ -393.5kJ -285.8kJ Δ H° = ∑ Δ H° f(products) - ∑ Δ H° f(reactants) -1386kJ= [(2 mol x -393.5kJ/mol) + (2 mol x -285.8kJ/mol)] – [(1 mol x x ) + (3 mol x 0.000kJ/mol)] -1386kJ= -1358.6kJ – x x = 27.4kJ Therefore the heat of formation is 27.4kJ/mol Calculate the  H° f for C 2 H 4 . C 2 H 4(g) + 3 O 2(g)  2 CO 2(g) + 2 H 2 O (l)  H° c = -1386 kJ

16. ENTHALPY OF FORMATION Enthalpy of Combustion: C 2 H 4(g) + 3 O 2(g)  2 CO 2(g) + 2 H 2 O (l)  H° c = -1386 kJ Calculate the  H° f for C 2 H 4 . 2 CO 2(g) + 2 H 2 O (l)  C 2 H 4(g) + 3 O 2(g)  H° = +1386 kJ C (s) + O 2(g)  CO 2(g)  H° = -393.5 kJ H 2(g) + 1/2O 2(g)  H 2 O (l)  H° = -285.8 kJ 2C (s) + 2H 2(g)  C 2 H 4(g) Δ H f ° = ? 1 2 3

17. ENTHALPY OF FORMATION Enthalpy of Combustion: C 2 H 4(g) + 3 O 2(g)  2 CO 2(g) + 2 H 2 O (l)  H° c = -1386 kJ Calculate the  H° f for C 2 H 4 . 2 CO 2(g) + 2 H 2 O (l)  C 2 H 4(g) + 3 O 2(g)  H° = +1386 kJ x 2 2 C (s) + 2 O 2(g)  2 CO 2(g)  H° = -787 kJ x 2 2H 2(g) + O 2(g)  2 H 2 O (l)  H° = -571.6 kJ 2C (s) + 2H 2(g)  C 2 H 4(g) Δ H f ° = ? 1 2 3 2C (s) + 2H 2(g)  C 2 H 4(g) Δ H f ° = 27.4kJ

18. ENTHALPY OF FORMATION