Enthalpy of Formation

1. Enthalpy of Formation • DHrxn has been tabulated for many different reactions. • Often tabulated according to the type of chemical reaction or process • DHvap (heat of vaporization) • liquid to gas • DHfusion (heat of fusion) • solid to liquid • DHcombustion (heat of combustion)

2. Enthalpy of Formation • Enthalpy of formation (DHf) • also called heat of formation • enthalpy change associated with the formation of a compound from its constituent elements in their elemental forms • DHf for methane (CH4) would be the enthalpy change associated with the following reaction: C (graphite) + 2 H2 (g)  CH4 (g)

3. Enthalpy of Formation • The magnitude of any DH depends on factors such as • temperature • pressure • physical state (s, l, g) of reactants and products • To compare DHf for different reactions, we define a set of standard conditions called the standard state.

4. Enthalpy of Formation • The standard state of a substance is its pure form at 1 atm and 25oC (298 K). • The DHf measured under standard state conditions is known as the standard enthalpy of formation (DHof) • the enthalpy change for the reaction that forms 1 mole of a compound from its elements when all substances are in their standard states

5. Enthalpy of Formation • The DHof for ethanol (C2H5OH) is depicted in the following thermochemical equation. 2 C (graphite) + 3 H2 (g) + 1/2 O2(g)  C2H5OH (l) DHof = - 277.7 kJ/mol Notice: standard state of C = graphite H = H2 (g) O = O2 (g) can use fractional coefficients

6. Enthalpy of Formation The DHof for C2H5OH (or many other compounds prepared from their elements in the standard state) can be looked up in a table. • Table 5.3 • Appendix C of text

7. Enthalpy of Formation • By definition, the standard enthalpy of formation of the most stable form of any element is zero. • There is no formation reaction when the element is already in its standard state. • DHof for C (graphite) = 0 • DHof for H2 (g) = 0 • DHof for O2 (g) = 0

8. Enthalpy of Formation Example: Write a balanced chemical equation showing the formation of 1 mole of C2H5Cl (g) from its constituent elements in their standard states.

9. Enthalpy of Formation Example: Write a complete thermochemical equation showing the formation of 1 mole of solid sodium bicarbonate from its constituent elements in their standard states.

10. Enthalpy of Formation Remember: • A complete thermochemical equation must have two components: • balanced chemical equation • DHrxn (DHof)

11. Enthalpy of Formation • You will have problems similar to the previous two examples on your exam. • You must be able to write a balanced chemical equation for the formation of 1 mole of a product from its constituent elements in their standard states. • You must be able to look up the DHof for the product in the table that I give you.

12. Enthalpy of Formation • Who cares about DHof??? • Hess’s Law can be used to calculate the enthalpy change for any reaction for which we know DHof for all reactants and products. DHorxn = S n DHof (products) - S n DHof (reactants) where n indicates the number of moles of each reactant or product

13. Enthalpy of Formation Example: Calculate the DHorxn for the combustion of propane gas (C3H8). C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

14. Enthalpy of Formation

15. Enthalpy of Formation Example: Calculate the heat of formation of glucose using the following thermochemical equation and the standard heats of formation given below. C6H12O6 (s)  2 C2H5OH (l) + 2 CO2 (g) DHo = - 69 kJ

16. Enthalpy of Formation

17. Enthalpy of Formation • Your exam will include problems similar to the ones in the last two examples. • Given DHof for all reactants and products, you must be able to find DHorxn. • Given the DHorxn and DHof for all but one reactant or product, you will be able to find the DHof for that reactant or product.