- “Heats” of formation fH° - DrU° ? CHEM 433 – 10/5/11 III. 1st Law: Thermochemistry (2.7-2.9) - H for chemical changes - Hess’ Law - “Heats” of formation fH° - DrU° ? READ: FINISH CHAPTER 2 HW #3 Due F. HW#4 out F.

ENTHALPY IS AN EXTENSIVE PROPERTY 1) How much heat is produced when 0.50 mol of H2O is produced via: 2 H2 (g) + O2 (g) —> 2 H2O (l) H = -572 kJ 2) What is H for: 4 H2 (g) + 2 O2 (g) —> 4 H2O (l) H = ?

Some DvapH° values … http://en.wikipedia.org/wiki/Enthalpy_of_vaporization Which value are highest? H2O vs NH3 ? Methane vs. propane vs. butane? Does PH3 have H-bonding?

So, how do you go about getting DH? CH4 (g) + 2 O2 (g) —> CO2 + 2 H2O (l) H = ? Experimentally: Measure qp & n => DrH° = qp / n (note ~ in kJ / mol) Theoretically: ? (CHEM 433 does not have a lab… or what if the reaciton does not go... or ….?)

Hess’ Law: In pictures…

Use Hess’ Law and the reaction below to calculate cH for C6H6(l) - which is rxnH for: __C6H6 (l) + __O2 (g) —> __CO2 (g) + __H2O (l) (balanced for one mol C6H6…) Data: C6H12 (l) + 9 O2 (g) —> 6 CO2 (g) + 6 H2O (l) H° = -3920 kJ C6H6 (l) + 3 H2 (g) —> C6H12 (l) H° = -205 kJ H2 (g) + 1/2 O2 (g) —> H2O(l) H° = -286 kJ

The Born-Haber Cycle - Hess’ Law in Action! Why are ionic compounds so stable ?

Standard Enthalpies of Formation: rH° to produce 1 mol of a given substance from elements in their reference states (most stable from, 1 bar, some specified T - often 298K )

Using “heats” of Formation : rH° for a given reaction can be obtained via: {The sum of rH°’s for the products} – {The sum of rH°’s for the reactants}