Equilibrium

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. • rateforward = ratereverse

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. Chemical [Initial] [Equilibrium] SO2 2.00 1.50 O2 1.50 1.25 SO3 3.00 3.50 Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO2 + O2 2 SO3 • Determine the Equilibrium Expression • Plug the equilibrium concentrations into to Equilibrium Expression • Solve the Equation

9. Le Châtelier’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. Chemical [Initial] [Equilibrium] SO2 2.00 1.50 O2 1.50 1.25 SO3 3.00 ? Example – If the value of the Equilibrium Constant for the Reaction 2 SO2 + O2 2 SO3 is 4.36, Determine the Equilibrium Concentration of SO3 • Determine the Equilibrium Expression • Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression • Solve the Equation

14. Solubility & Solubility Product • Even “insoluble” salts dissolve somewhat in water • insoluble = less than 0.1 g per 100 g H2O • The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced AnXm(s)  n A+(aq) + m Y-(aq) • Equilibrium constant called solubility product Ksp = [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 Ksp = [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 Ksp [Ag+] = [I-] = x [Ag+] = 1.2 x 10-8 mol/L = [AgI] The solubility of AgI = 1.2 x 10-8 M