1. Equilibrium

2. Reaction Dynamics  If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up.  However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction.

3. Reaction Dynamics  The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing  At the same time the reverse reaction speeds up as the concentration of the products increases.  Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium.  rate forward = rate reverse

4. Chemical Equilibrium  Equilibrium only occurs in a closed system!!  When a system reaches equilibrium, the amounts of reactants and products in the system stays constant  the forward and reverse reactions still continue, but because they go at the same rate the amounts of materials don't change.  There is a mathematical relationship between the amounts of reactants and products at equilibrium  no matter how much reactants or products you start with.

5. Equilibrium Constant   xA + yB ↔ nC + mD  Law Of Chemical Equilibrium  In this expression, K is a number called the equilibrium constant.  Do not include solids or liquids, only solutions and gases  For a reaction, the value of K for a reaction depends on the temperature  K is independent of the amounts of reactants and products you start with.

6. Position of Equilibrium  The relative concentrations of reactants and products when a reaction reaches equilibrium is called the position of equilibrium  Different initial amounts of reactants (and or products) will result in different equilibrium concentrations but the same equilibrium constant

7. Position of Equilibrium  If K is large then there will be a larger concentration of products at equilibrium than of reactants; we say the position of equilibrium favors the products.  If K is small then there will be a larger concentration of reactants at equilibrium than of products; we say the position of equilibrium favors the reactants.  The position of equilibrium is not affected by adding a catalyst.

8. Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2  2 SO 3 ¬ Determine the Equilibrium Expression ­ Plug the equilibrium concentrations into to Equilibrium Expression ® Solve the Equation 3.503.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical

9. Le Ch âtelier’s Principle  Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium  When a change is imposed on a system at equilibrium, the position of equilibrium will shift in the direction that will reduce the effect of that change

10. Concentration Changes and Le Châtelier’s Principle  The position of equilibrium can be affected without changing the equilibrium constant.  Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found  Position shifts toward the products  has the same K  Removing a reactant will increase the amounts of the other reactants and decrease the amounts of the products.  Position shifts toward the reactants  Removing a product can allow us to drive a reaction to completion!

11. Changing Pressure and Le Châtelier’s Principle  Changing the pressure of one gas is like changing its concentration  Has the same effect as changing the concentration on the position of equilibrium  Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules  Decreasing the volume of the system increases its pressure  Reduces the pressure by reducing the number of gas molecules  Opposite effect happens if the system pressure is decreased

12. Changing Temperature and Le Châtelier’s Principle  The equilibrium constant will change if the temperature changes  Exothermic reactions release heat, Endothermic reactions absorb heat.  For exothermic reactions, heating the system will decrease K  Think of heat as a product of the reaction  Therefore shift the position of equilibrium toward the reactant side  For endothermic reactions, heating the system will increase K  Think of heat as a reactant  The position of equilibrium will shift toward the products  Cooling an exothermic or endothermic reaction will have the opposite effects on K and equilibrium position

13. Example – If the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2  2 SO 3 is 4.36, Determine the Equilibrium Concentration of SO 3 ¬ Determine the Equilibrium Expression ­ Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression ® Solve the Equation ?3.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical

14. Solubility & Solubility Product  Even “insoluble” salts dissolve somewhat in water  insoluble = less than 0.1 g per 100 g H 2 O  The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced A n X m (s)  n A + (aq) + m Y - (aq)  Equilibrium constant called solubility product K sp = [A + ] n [Y - ] m  If undissolved solid in equilibrium with the solution, the solution is saturated  Larger K = More Soluble  for salts that produce same the number of ions

15. Example   Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ¬ Determine the balanced equation for the dissociation of the salt AgI(s)  Ag + (aq) + I - (aq) ­ Determine the expression for the solubility product  Same as the Equilibrium Constant Expression K sp = [Ag + ][I - ]

16. Example   Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ® Define the concentrations of dissolved ions in terms of x AgI(s)  Ag + (aq) + I - (aq) Stoichiometry tells us that we get 1 mole of Ag + and 1 mol I - for each mole of AgI dissolved Let x = [Ag + ], then [I - ] = x

17. Example   Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ¯ Plug the ion concentrations into the expression for the solubility product and solve for K sp [Ag + ] = [I - ] = x [Ag+] = 1.2 x 10 -8 mol/L = [AgI] The solubility of AgI = 1.2 x 10 -8 M