2. Lecture 4: Standard Enthalpies Reading: Zumdahl 9.6 Outline –What is a standard enthalpy? –Defining standard states. –Using standard enthalpies to determine  H° rxn

3. Hess’ Law Enthalpy is a state function. As such,  H for going from some initial state to some final state is pathway independent.  H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate  H for a reaction.

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

5. Definition of  H° f (cont.) 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.

6. 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 (or 100,000 Pascal; 1 atm=101325) –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.

7. Definitions (cont.) Standard elemental states (cont.): –Hydrogen: H 2 (g) (not atomic H) –Oxygen: O 2 (g) –Carbon: C (gr) … graphite as opposed to diamond We will denote the standard state using the subscript “°”. –Example:  H° rxn

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

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

10. 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 (zero)

11. 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

12. 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)

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

14. 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)  H° f in Appendix 4: CH 3 OH(l) -238.6 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -286 kJ/mol

15. Example (cont.) 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!

16. 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/molHF(g) -271 kJ/mol

17. Another Example (cont.) Given the reaction of interest: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g)  H° rxn = -1196 kJ  H° rxn =  H° f (products) -  H° f (reactants)  H° rxn = 6  H° f (HF(g)) -  H° f (NH 3 (g)) -  H° f (ClF 3 (g))

18. Another Example (cont.)  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))

19. 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 0(l) = 75 J/mol.K C P H 2 0(g) = 36 J/mol.K

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)  H = ? nC P (H 2 O(l))  T nC P (H 2 O(g))  T

21. 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

22. 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