5.7 Enthalpies of Formation

1. 5.7 Enthalpies of Formation Advanced Chemistry

2. Introduction • Using Hess’s Law, we can calculate enthalpy changes for many reactions from tabulated ∆H values • Enthalpies of vaporization (converting between liquids to gases) • Enthalpies of fusion (melting solids) • Enthalpies of combustion (combusting a substance in oxygen • ETC.

3. Enthalpies of Formation • Enthalpy of formation (∆Hf): the enthalpy change that accompanies the formation of a substance from the most stable forms of its component elements • Also called the heat of formation • The magnitude of any enthalpy change depends on temperature, pressure, and state of the reactants and products. • To compare enthalpies of different reactions, we must define a set of conditions, called a standard state.

4. Standard Enthalpy of Formation • The standard state of a substance in its pure form is 1 atm and 298 K. • Standard enthalpy of formation, ∆H⁰f, is the change in enthalpy for the reaction that forms one mole of the compounds from its elements with all substances in their standard states. • Reported in kJ/mol • The standard enthlapy of formation of the most stable form of any element is zero.

5. Standard Enthalpies of Formation

6. Equations Associated with Enthalpies of Formation • For which of these reactions at 25⁰C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose ∆H is an enthalpy of formation • 2Na(s) + ½ O2(g)  Na2O(s) • 2K(l) + Cl2(g)  2KCl(s) • C6H12O6  6C (diamond) + 6H2(g) + 3O2(g)

7. Practice • Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride CCl4.

8. Using Enthalpies of Formation to Calculate Enthalpies of Reaction • We can use Hess’s law and tabulations of ∆H⁰f values, such as those in Table 5.3, to calculate the standard enthalpy change for any reaction for which we known the ∆H⁰f values for all reactants and products. • For example, lets look at the combustion of propane gas, C3H8(g), to CO2(g) and H2O(l) under stander conditions. C3H8(g) + 5O2(g)  3CO2(g)+ 4H2O(l)

9. Example C3H8(g) 3C(graphite) + 4H2(g) ∆H1 = -∆H⁰f [C3H8(g)] 3C(graphite) + 3O2(g) 3CO2(g) ∆H2 = 3∆H⁰f [CO2(g)] 4H2(g) + 2O2(g)  4H2O(l) ∆H3 = 4∆H⁰f [H2O(l)] C3H8(g) + 5O2(g)  3CO2(g)+ 4H2O(l) ∆H⁰rxn = ∆H1 + ∆H2 + ∆H3

10. The example on the previous slide can be broken down to three reactions. • The example on the previous slide can be broken down to three reactions. • Decomposition: C3H8(g) 3C(graphite) + 4H2(g) is the reverse of the formation reaction for C3H8(g), so the enthalpy change for this decomposition reaction is a negative value of ∆H⁰f value for propane: -∆H⁰f [C3H8(g)]

11. The previous example can be broken down to three reactions. (2) Formation of CO2: The equation 3C(graphite) + 3O2(g) 3CO2(g) is the formation reaction for 3 mol of CO2(g). Because enthalpy is an extensive property, the enthalpy change is 3∆H⁰f [CO2(g)]

12. The previous example can be broken down to three reactions. (3) Formation of H2O: The enthalpy change for the equation 4H2(g) + 2O2(g)  4H2O(l) shows the formation of 4 mols. Therefore, the enthalpy change is 4∆H⁰f [H2O(l)]. • The reaction specifies that H2O(l) is produced, so be careful to use the value for 4H2O(l) and not the value for 4H2O(g)

13. Example C3H8(g) 3C(graphite) + 4H2(g) ∆H1 = -∆H⁰f [C3H8(g)] 3C(graphite) + 3O2(g) 3CO2(g) ∆H2 = 3∆H⁰f [CO2(g)] 4H2(g) + 2O2(g)  4H2O(l) ∆H3 = 4∆H⁰f [H2O(l)] C3H8(g) + 5O2(g)  3CO2(g)+ 4H2O(l) ∆H⁰rxn = ∆H1 + ∆H2 + ∆H3

14. Calculation of ∆H • Using the information that we have learned, we can use the equation: H⁰rxn = nH⁰f,products – mH⁰f,reactants • n and m are the stoichiometric coefficients (# of moles) of each compound. • You get this number from the chemical equation. • The symbol  (sigma) means “the sun of”

15. Example C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)] = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ

16. Calculating an Enthalpy of Reaction from Enthalpies of Formation • Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). C6H6(l) + O2  6CO2(g) + 3H2O(l).

17. More Practice • Calculate the enthalpy change for the combustion of 1 mol of ethanol C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)

18. Calculating the Enthalpy of Formation Using an Enthalpy of Reaction • The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178. 1 kJ. Calculate the standard enthalpy of formation of CaCO3(s).

19. More Practice • Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s). CuO(s) + H2(g)  Cu(s) + H2O(l) ∆H⁰ = -129.7 kJ