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 vapH° values …
http://en.wikipedia.org/wiki/Enthalpy_of_vaporizatio
n
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