CHEM 433 – 10/4/11 III. 1st Law: Adiabatic changes (2.6)

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Presentation transcript:

CHEM 433 – 10/4/11 III. 1st Law: Adiabatic changes (2.6) Thermochemistry (2.7-2.9) - Standard enthalpy changes - H for physical changes - H for chemical changes READ: FINISH CHAPTER 2 —> HW Due Th – Maybe F (?)

Some basic stuff … • Thermochemistry: The study of heat produced or consumed by chemical reactions… • Most often at constant p —> qp=H - H < 0 —> exothermic (“heat out”) - H > 0 —> endothermic (“heat in”) • Standard enthalpy changes (H°, or H “Saturn”): H for a process in which the initial and final substances are in standard states: 1 bar, pure (s, l, g - not mixed, or 1M in solution), and some specified T (Data in Tables 2.6 & 2.7 at 298K)

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?

Hess’ Law: In pictures…

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 = ?

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