Chapter 13 Equilibrium Dr. Walker DE Chemistry

Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time This occurs when the rate of the forward reaction equals the rate of the reverse reaction All reactions carried out in a closed vessel will reach equilibrium

Reversible Reactions A chemical reaction in which the products can react to re-form the reactants Usually represented by a double sided arrow Equilibrium can be to the left or the right depending on conditions 2 HgO(s)  2Hg(l) + O2(g)

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

How Equilibrium Occurs 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

2NO2(g)  2NO(g) + O2(g) Last chapter, we studied reaction rates by looking at changes in concentration over time Kinetics are irrelevant at equilibrium because the concentration is now constant! At this point, the concentrations are no longer changing This reaction is now at equilbirium

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

Law of Mass Action jA + kB  lC + mD For the reaction: Where K is the equilibrium constant, and is unitless

Product Favored Equilibrium Large values for K signify the reaction is “product favored” (“equilibrium to right”) When equilibrium is achieved, most reactant has been converted to product

Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored” (“equilibrium to the left”) When equilibrium is achieved, very little reactant has been converted to product

Writing an Equilibrium Expression Write the equilibrium expression for the reaction: 2NO2(g)  2NO(g) + O2(g) K = ???

Example For the reaction N2 + 3 H2  2 NH3, the concentrations at equilibrium are as follows: [N2] = 0.921 M [H2] = 0.763 M [NH3] = 0.157 M Calculate the equilibrium constant.

Example K = K = K = 6.03 x 10-2 [0.157]2 [0.763]3[0.921] For the reaction N2 + 3 H2  2 NH3, the concentrations at equilibrium are as follows: [N2] = 0.921 M [H2] = 0.763 M [NH3] = 0.157 M Calculate the equilibrium constant. [NH3]2 [H2]3[N2] K = [0.157]2 [0.763]3[0.921] K = K = 6.03 x 10-2

Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H2(g) + N2(g)  2NH3(g) KP is equilibrium constant in terms of partial pressure Dn = sum of gaseous product coefficients – sum of gaseous reactant coefficients

Examples Using Pressure

Examples Using Pressure [NOCl]2 [1.2]2 [NO]2[Cl2] [0.05]2[0.3] Kp = = = 1.9 x 103

Examples Using Pressure

Examples Using Pressure 1.9 x 103 = K [(0.08206 Latm/mol K) (298 K)]-1 K = 4.7 x 104 From coefficients 2 – (2-1) = -1

Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present Pure solids/liquids NOT included in equation No concentration involved Write the equilibrium expression for the reaction: PCl5(s)  PCl3(l) + Cl2(g) Pure solid Pure liquid

jA + kB  lC + mD The Reaction Quotient For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD

Significance of the Reaction Quotient If Q = K, the system is at equilibrium If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

Writing Heterogeneous Equilibrium Expressions

Writing Heterogeneous Equilibrium Expressions K = [Cl2] KP = PCl2 Other materials are pure liquids or solids K = [H2O]5 KP = (PH2O)5 All other materials are pure solids. Coefficient stays on the equation

Determining Equilibrium Concentrations This is a technique that will be used not just here, but in chapters 14-15 as well. As a result, it is critical that you can set up what we will go over without confusion….

Determining Equilibrium Concentrations In typical chemical reactions Reactants are broken down and their concentration decreases Products are formed and their concentration increases Higher [products] = higher K, higher [reactants] = lower K Concentrations of all reagents change until equilibrium (Q = K is reached)

Solving for Equilibrium Concentration Consider this reaction at some temperature: H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Assume you start with 8 molecules of H2O and 6 molecules of CO. How many molecules of H2O, CO, H2, and CO2 are present at equilibrium? Here, we learn about “ICE” – an important problem solving technique that we will use over the next few chapters -Determines concentration of unknown reagents given an equilibrium constant

Solving for Equilibrium Concentration H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Step #1: We write the law of mass action for the reaction:

Solving for Equilibrium Concentration Step #2: We “ICE” the problem, beginning with the Initial concentrations H2O(g) + CO(g)  H2(g) + CO2(g) Initial: Change: Equilibrium: 8 6 -x -x +x +x 8-x 6-x x x

Solving for Equilibrium Concentration Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H2O(g) + CO(g)  H2(g) + CO2(g) Equilibrium: 8-x 6-x x x = 4

Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H2O(g) + CO(g)  H2(g) + CO2(g) Equilibrium: 8-x 6-x x x = 4 Equilibrium: 8-4=4 6-4=2 4

ICE Example What will we do to solve this?

ICE Example What will we do to solve this? Set up an ICE table [I2] @ equilibrium is 6.61 x 10-4 M

ICE Example What will we do to solve this? Set up an ICE table [I2] @ equilibrium is 6.61 x 10-4 M This means [I-] also is 6.61 x 10-4 M at equilibrium Solve for x, which is [I3-], solve for K

ICE Example What will we do to solve this? Set up an ICE table [I2] @ equilibrium is 6.61 x 10-4 M This means [I-] also is 6.61 x 10-4 M at equilibrium Solve for x, which is [I3-], solve for K

Le Chatelier’s Principle

LeChatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away. When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water right The system temporarily shifts to the _______ to restore equilibrium.

LeChatelier Example #2 A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container. N2O4 (g) + Energy  2 NO2 (g) left The system temporarily shifts to the _______ to restore equilibrium.

LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor right The system temporarily shifts to the _______ to restore equilibrium.

LeChatelier Example #4 A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased. N2O4 (g) + Energy  2 NO2 (g) left The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.