1. Chapter 18: Chemical Equilibria Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up. “Only” the forward reaction occurs. Reversible reactions: Both the forward and the reverse reaction occur to a significant degree. Au 2 O 3 (s) + 2 Fe (s) 2 Au (s) + Fe 2 O 3 (s)Ex: 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g) Forward: Reverse: Equilibrium: A reaction is at equilibrium when Rate forward rxn = Rate reverse rxn 3 H 2 (g) + N 2 (g) ⇋ 2 NH 3 (g)

2. General form of equilibrium constant, K eq : aA + bB ⇋ cC + dD If Keq > 1, then more products than reactants present at equilibrium If Keq < 1, then more reactants than products present at equilibrium

3. Time Concentration N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ⇋ H2H2 NH 3 N2N2 K eq = [NH 3 ] [N 2 ] [H 2 ] 2 3

4. Heterogeneous equilibria If the reactants and products are in more than one state, the equilibrium constant is for a heterogeneous equilibria. Because the concentration of a solid does not change significantly compared to gaseous or liquid reactants or products, solids NEVER appear in equilibrium constant calculations. Ex: BaCl 2 (s) ⇋ Ba 2+ (aq) + 2 Cl - (aq) Since this particular type of equilibrium involves the solubility of the product, it is given a special designation: K sp = solubility product constant

5. Heterogeneous equilibria (cont.) Since the concentration of a liquid solvent that the reaction takes place in also doesn’t change significantly, it also doesn’t appear in the equilibrium calculation. Ex: H 2 O (l) ⇋ H 2 O (g) What are the equilibrium constants for the following: 1.C 10 H 8 (s) ⇋ C 10 H 8 (g) 2.CaCO 3 (s) ⇋ CaO (s) + CO 2 (g) 3.C (s) + H 2 O (g) ⇋ CO (g) + H 2 (g) 4.FeO (s) + CO (g) ⇋ Fe (s) + CO 2 (g) K eq = [C 10 H 8 ] K eq = [CO 2 ] K eq = [CO][H 2 ] [H 2 O] K eq = [CO 2 ] [CO]