Chapter 18 Chemical Equilibrium 18.1 The Nature of Chemical Equilibrium

Reversible Reactions A chemical reaction in which the products re-form the reactants Reactants ↔ Products Le Chatelier’s Principle When system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimizes the stress.

Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time All reactions carried out in a closed vessel will reach equilibrium

Chemical Equilibrium Sometimes, if little product is formed, equilibrium lies far to the left (towards reactants) Reactants Products Sometimes, if little reactant remains, equilibrium lies far to the right (towards products) Reactants Products

Dynamic Equilibrium Reactions continue to take place Reactant molecules continue to be converted to product Product continues to be converted to reactant (reverse reaction) Forward and reverse reactions take place at the same rate at equilibrium

Causes of Equilibrium Beginning of Reaction Only reactant molecules exist, so only reactant molecules may collide Middle As product concentration increases, collisions may take place that lead to the reverse reaction At Equilibrium Rates of forward and reverse reactions are identical

Suppose that compound A reacts with B to form C and D. The symbol            is used for system at equilibrium.

If we add more A what would happen to the amount of C at equilibrium? Adding A increases the amount of C and D at equilibrium.

If we add more C what would happen to the amount of A at equilibrium? adding more C increases the amount of A and B at equilibrium.

If we add A what would happen to the amount of B at equilibrium? A reacts with B to form C and D. Therefore, adding A decreases B.

Example - The Haber Process N2 (g) + 3H2 (g) ↔ 2NH3 (g) Hydrogen is consumed at 3x the rate of nitrogen Ammonia is formed at 2x the rate at which nitrogen is consumed

Points to Remember Rxns in a closed system always reach equilibrium When a rxn has reached equilibrium, that means the rate of the forward rxn equals the rate of the reverse rxn A system will remain at equilibrium until the system is disturbed in some manner

The Equilibrium Constant For the balanced forward equation: jA + kB → lC + mD (j, k, l, m) are coefficients K is a constant called the equilibrium constant K varies depending on temperature and upon the coefficients of the balanced equation K is determined experimentally [X] represents concentration of chemical species at equilibrium K = [C]l [D]m [A]j [B]k

The value of K will tell us the position of equilibrium Equilibrium Constant 4NH3(g) + 7O2(g)  4NO2(g) +6H2O(g) K = [NO2]4[H2O]6 [NH3]4[O2]7 The value of K will tell us the position of equilibrium

Practice N2(g) + 3H2(g)  2NH3(g) K = [NH3]2 [N2] [H2]3 Calculate K [equilibrium] [NH3] = 3.1 x 10-2 M [N2] = 8.5 x 10-1 M [H2] = 3.1 x 10-3 M Calculate K K = [3.1x 10-2]2 [8.5 x 10-1] [3.1 x 10-3]3 K = 3.8 x 104 no units

Practice N2(g) + 3H2(g)  2NH3(g) K = 3.8 x 104 [NH3] = 3.1 x 10-2 M [N2] = 8.5 x 10-1 M [H2] = 3.1 x 10-3 M The value of K will tell us the position of equilibrium If K >1, the rxn favors products If K <1, the rxn favors reactants If K = 1, equilibrium lies in the middle

The rxn was inversed so K is inversed Practice N2(g) + 3H2(g)  2NH3(g) Original rxn 2NH3(g)  N2(g) + 3H2(g) New rxn Original K Same [ ] K = 3.8 x 104 [NH3] = 3.1 x 10-2 M [N2] = 8.5 x 10-1 M [H2] = 3.1 x 10-3 M Calculate new K K = 1 3.8 x 104 The rxn was inversed so K is inversed K = 2.6 x 10-5

The coefficients were reduced by ½ Practice N2(g) + 3H2(g)  2NH3 (g) Original rxn ½N2(g) + 3/2H2(g)  NH3 (g) New rxn Original K Same [ ] K = 3.8 x 104 [NH3] = 3.1 x 10-2 M [N2] = 8.5 x 10-1 M [H2] = 3.1 x 10-3 M Calculate new K K = (3.8 x 104)1/2 The coefficients were reduced by ½ K = 2.0 x 102

Interpreting K If K = 1, then there are equal concentrations of reactants & products If K is small, the forward rxn occurs slightly before equilibrium is established, reactants are favored If K is large, reactants are mostly converted into products when equilibrium is established, products are favored

Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction Pure liquids are not the same as solutions, whose concentration can change

More Practice! At 25 °C, an equilibrium mixture of gases contains 6.4 x 10-3 M PCl3, 2.5 x 10-2 M Cl2, & 4.0 x 10-3 M PCl5. What is the equilibrium constant for the following reaction? PCl5 (g) ↔ PCl3 (g) + Cl2 (g) 2. At equilibrium, a 2.0 L vessel contains 0.36 mol of H2, 0.11 mol of Br2, & 37 mol of HBr. What is the equilibrium constant for the reaction at this temperature? H2 (g) + Br2 (g) ↔ HBr (g) K = 4.0 x 10-2 K = 3.5 x 104

Chapter 18 Chemical Equilibrium 18.2 Shifting Equilibrium

The Effect of Change in Pressure: Ways to Change Pressure Add or remove a gaseous reactant or product Add an inert gas (one not involved in the reaction) An inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products

The Effect of Change in Pressure Change the volume of the container When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system N2 (g) + 3H2 (g) → 2NH3 (g) shifts to the right to decrease the total molecules of gas present

The Effect of Change in Pressure When the container volume is increased, the system will shift so as to increase its volume N2 (g) + 3H2 (g) ← 2NH3 (g) shifts to the left to increase the total number of molecules of gas present

The Effect of Change in Concentration If a reactant or product is added to a system at equilibrium, the system will shift away from the added component (it will attempt to "use up" the added component) If a reactant or product is removed from a system at equilibrium, the system will shift toward the removed component (it will attempt to "replace" the removed component)

The Effect of a Change in Temperature An increase in temperature increases the energy of the system. Le Chatelier's principle predicts that the system will shift in the direction that consumes the energy For an exothermic rxn, energy is a product. The rxn will shift to the left to use up the excess energy For an endothermic rxn, energy is a reactant. The rxn will shift to the right to use up the energy A decrease in temperature will cause a system shift in the direction that "replaces" the lost energy

Le Chatelier’s Principle When a stress is applied to a system at equilibrium, the system will shift in a direction that will relieve the stress [changes] do not affect the value of K pressure changes do not affect the value of K temperature changes will affect the value of K

Which direction will the rxn shift to re-establish equilibrium if: Practice ([change]) As4O6(s) + 6C(s)  As4(g) + 6CO(g) Which direction will the rxn shift to re-establish equilibrium if: Add CO? left right no shift Add or remove C or As4O6? left right no shift Remove As4? left right no shift

Practice (change in pressure) Which direction will the rxn shift to reestablish equilibrium if the volume is reduced? P4(s) + 6Cl2(g)  4PCl3(l) left right no shift PCl3(g) + Cl2(g)  PCl5(g) left right no shift O2(g) + N2(g)  2NO(g) left right no shift

Practice (change in temperature) Which direction will the rxn shift to reestablish equilibrium if the temperature is increased? N2(g) + O2(g)  2NO(g) ΔHo = 181kJ left right no shift 2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ left right no shift How will the change affect K? N2(g) + O2(g)  2NO(g) ΔHo = 181kJ increase decrease 2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ increase decrease

Completed Reactions You do not need to consider equilibrium if you have a reaction that goes to completion. Examples: Formation of a gas in an open container Formation of a precipitate Formation of a slightly ionized product (polar) – neutralization

The Common-Ion Effect When the addition of an ion common to two solutes brings about precipitation or reduced ionization.

Solubility Product and the Common Ion Effect AgCl(s)  Ag+(aq) + Cl-(aq) What happens if a solution of NaCl is added to this system? Ksp = [Cl-][Ag+] = 1.8 x 10-10 solublility product NaCl(s)  Na+(aq) + Cl-(aq) The solubility of AgCl has actually decreased rxn shifts  AgCl(s)  Ag+(aq) + Cl-(aq)