2. Lecture 5: Standard Enthalpies Reading: Zumdahl 9.6 Outline: What is a standard enthalpy? Defining standard states. Using standard enthalpies to determine  H° rxn

3. Definition of  H° f We saw in the previous lecture the utility of having a series of reactions with known enthalpy changes. What if one could deal with combinations of compounds directly, rather than dealing with whole reactions, in order to determine  H rxn ?

4. Definition of  H° f Standard Enthalpy of Formation,  H° f : “The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances at standard state.” We envision taking elements at their standard state, and combining them to make compounds, also at standard state.

5. What is Standard State? Standard State: A precisely defined reference state. It is a common reference point that one can use to compare thermodynamic properties. Definitions of Standard State: –For a gas: P = 1 atm. –For solutions: 1 M (mol/L). –For liquids and solids: pure liquid or solid –For elements: The form in which the element exists under conditions of 1 atm and 298 K. Caution! Not STP (for gas law problems, 273 K)

6. Standard elemental states: –Hydrogen: H 2 (g) (not atomic H) –Oxygen: O 2 (g) –Carbon: C (gr) graphite, not diamond We will denote the standard state using the superscript ° For example, :  H° rxn

7. Importance of Elements We will use the elemental forms as a primary reference when examining compounds. Pictorally, for chemical reactions we envision taking the reactants apart to their standard elemental forms, then reforming the products.

8. Example: Combusion of Methane CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l)

9. Elements and  H° f For elements in their standard state:  H° f = 0 With respect to the graph, we are envisioning chemical reactions as proceeding through elemental forms. In this way we are comparing  H° f for reactants and products to a common reference state, namely  H° f = 0 for the elements.

10. Tabulated  H° f In Zumdahl, tables of  H° f are provided in Appendix 4. The tabulated values represent the  H° f for forming the compound from elements at standard conditions. C (gr) + O 2 (g) CO 2 (g)  H f ° = -394 kJ/mol

11. Using  H° f to determine  H° rxn Since we have connected the  H° f to a common reference point, we can now determine  H° rxn by looking at the difference between the  H° f for products versus reactants. Mathematically :  H° rxn =  H° f (products) -  H° f (reactants)

12. How to Calculate  H° rxn When a reaction is reversed,  H changes sign. If you multiply a reaction by an integer, then  H is also multiplied by the same factor (since it is an extensive variable) Elements in their standard states are not included in calculations since  H° f = 0.

13. Example Determine the  H° rxn for the combustion of methanol. CH 3 OH (l) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l) (From Appendix 4)  H° f CH 3 OH(l) -238.6 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -286 kJ/mol

14. CH 3 OH (l) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l)  H° rxn =  H° f (products) -  H° f (reactants) =  H° f (CO 2 (g)) + 2  H° f (H 2 O(l)) -  H° f (CH 3 OH(l)) = -393.5 kJ - (2 mol)(286 kJ/mol) - (-238.6 kJ) = -728.7 kJ Exothermic!

15. Another Example Using the following reaction: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g)  H° rxn = -1196 kJ Determine the  H° f for ClF 3 (g)   H° f NH 3 (g) -46 kJ/mol HF(g) -271 kJ/mol

16. Given the reaction of interest: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g)  H° rxn = -1196 kJ And Hess’ Law:  H° rxn =  H° f (products) -  H° f (reactants) Substituting, we have  H° rxn = 6  H° f (HF(g)) -  H° f (NH 3 (g)) -  H° f (ClF 3 (g))

17.  H° rxn = 6  H° f (HF(g)) -  H° f (NH 3 (g)) -  H° f (ClF 3 (g)) -1196 kJ = (6mol)(-271 kJ/mol) - (2mol)(-46 kJ/mol) - (2mol)  H° f (ClF 3 (g)) -1196 kJ = -1534 kJ - (2mol)  H° f (ClF 3 (g)) 338 kJ = - (2mol)  H° f (ClF 3 (g)) -169 kJ/mol =  H° f (ClF 3 (g))

18. Bonus Example Problem 9.81  H vap for water at the normal boiling point (373.2 K) is 40.66 kJ. Given the following heat capacity data, what is  H vap at 340.2 K for 1 mol? C P H 2 O(l) = 75 J/mol.K C P H 2 O(g) = 36 J/mol.K

19. Bonus Example (cont.) H 2 O(l) H 2 O(g)  H = 40.66 kJ/mol (373.2 K) H 2 O(l) H 2 O(g) (340.2 K)  H = ? nC P (H 2 O(l))  T nC P (H 2 O(g))  T

20. Bonus Example (cont.) H 2 O(l) H 2 O(g)  H = 40.66 kJ/mol (373.2 K) H 2 O(l) H 2 O(g) (340.2 K) nC P (H 2 O(l))  TnC P (H 2 O(g))  T  H rxn (340.2K) =  H (product) -  H (react) =  H water(g) (373.2K) + nC P (H 2 O(g))  T -  H water(l) (373.2K) - nC P (H 2 O(l))  T

21. Bonus Example (cont.)  H rxn (340.2K) =  kJ/mol + 36 J/mol.K(-33 K) - 75 J/mol.K(-33 K)  H rxn (340.2K) =  kJ/mol + 1.29 kJ/mol = 41.95 kJ/mol  H rxn (340.2K) =  H water(g) (373.2K) + nC P (H 2 O(g))  T -  H water(l) (373.2K) - nC P (H 2 O(l))  T