Lesson 3.1: Equilibrium Concepts for Chemistry Olympiad
1. General form of equilibrium constant: law of mass action
2. Solids and pure liquids do not appear in the equilibrium constant expression (they are not
variables!).
3. Examples of equilibria:
a) weak acid dissociation (Ka)
b) vapor pressure
c) solubility product (Ksp)—used for salts with low solubility
d) Kw (equilibrium constant for water’s dissociation)
e) weak base equilibria (Kb)
Lesson 3.1: Equilibrium Concepts for Chemistry Olympiad
Heat of formation (Hf)
for ammonia = -46 kJ/mol
4. Le Chatelier’s Principle
3 H2 (g) + N2 (g)  2 NH3 (g)
5. G = -RT ln(Keq)
H =
H°rxn =
3 H2 (g) + N2 (g)  2 NH3 (g)
S°rxn =
G° at 300K =
G° at 600K =
Problem Set 3.21: Equilibrium Questions from 2001 Local Olympiad
Verify the answer choices marked below. Then refer to the video instruction!
Problem Set 3.22 Thermodynamics Questions from 2002 Local Olympiad
Try to answer these questions—then check the answer key!
Problem Set #3.3: Free Response AP Chem Equilibrium Question
1984 A
Sodium benzoate, C6H5COONa, is the salt of a the weak acid, benzoic acid, C6H5COOH. A 0.10
molar solution of sodium benzoate has a pH of 8.60 at room temperature.
(a) Calculate the [OH-] in the sodium benzoate solution described above.
(b) Calculate the value for the equilibrium constant for the reaction:
C6H5COO- + H2O <=> C6H5COOH + OH-
(c) Calculate the value of Ka, the acid dissociation constant for benzoic acid.
(d) A saturated solution of benzoic acid is prepared by adding excess solid benzoic acid to pure water at
room temperature. Since this saturated solution has a pH of 2.88, calculate the molar solubility of
benzoic acid at room temperature.