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) anim

2. General form of equilibrium constant, K eq : aA + bB ⇋ cC + dD If K > 1, then more products than reactants present at equilibrium If K < 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 Occurs when the reactants and products are in more than one state Because the molar concentration of solids and liquids do not change, SOLIDS AND LIQUIDS NEVER appear in equilibrium constant calculations (they have a concentration value of 1). 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 K sp

5. Heterogeneous equilibria (cont.) 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]

6. The equilibrium constant for the reaction below at 700K is 0.44. What is the concentration of the carbon dioxide if [H 2 O] = 0.16 M, [CO] = 0.15 M and [H 2 ] = 0.14 M? H 2 O(g) + CO(g) ⇋ H 2 (g) + CO 2 (g) 1. Set up the equilibrium expression 2. Substitute in the known values and solve for the unknown. [CO 2 ] = 0.075 M

7. Molar solubility: the number of moles of the solute per liter in a saturated solution Ex 1: What is the molar solubility of lead(II) iodide if its K sp is 8.7 x 10 –9 ? Pb I 2 (s)  Pb 2+ + 2 I – x2x x = the molar solubility K sp = [Pb 2+ ][ I – ] 2 8.7 x 10 –9 = [x][2x] 2 8.7 x 10 –9 = 4x 3 x = 1.3 x 10 –3 M Ex 2: What is the molar solubility of Pb I 2 if 0.2 M K I is added? [ I – ] = 0.2 M 8.7 x 10 –9 = [x][0.2] 2 x = 2.2 x 10 –7 M

8. Ex 3: What is the K sp of Pb 3 (PO 4 ) 2 if it has a molar solubility of 1.5 x 10 -9 M? 1) Write out reaction equation: Pb 3 (PO 4 ) 2  3 Pb 2+ + 2 PO 4 3- 3x 2x 2) Write out K sp expression:K sp = [Pb 2+ ] 3 [PO 4 3- ] 2 K sp = (3x) 3 (2x) 2 K sp = (27x 3 )(4x 2 ) K sp = 108x 5 x = 1.5 x 10 -9 K sp = (108)(1.5 x 10 -9 ) 5 K sp = 8.2 x 10 -43

9. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe +3 (aq) + SCN -1 (aq) ⇋ FeSCN +2 (aq) Colorless ⇋ Dark red Initial color Left shift = more reactants (color is lighter) Right shift = more products (color is darker)

10. Other ways to cause a Le Châtelier Shift: 3 H 2 (g) + N 2 (g) + heat ⇋ 2 NH 3 (g) What kind of shift would you see if: *Pressure increased? [NH 3 ] increased? Heating temperature increased? Right shift   Left shift Right shift  *A gas–phase equilibrium will shift to the side of the reaction that takes up less space (smaller coefficient sum) when pressure is increased.

11. CH 4 (g) + 2 Cl 2 (g) ⇋ CCl 4 (g) + 2 H 2 (g) + heat What kind of shift would you see if: Pressure increased? Heating temperature increased? No Change  Left shift

12. 1324 2 drops 0.05 M NaSCN 2 drops 0.01 M Fe(NO 3 ) 3 3 drops H 2 O + 2 drops H 2 O + 2 drops 0.1 M Fe(NO 3 ) 3 + 2 drops 0.05 M NaSCN + 1 drop 1 M NaNO 3 0.1 M Fe(NO 3 ) 3 1 M NaNO 3 0.05 M NaSCN 0.01 M Fe(NO 3 ) 3 H2OH2O