by Steven S. Zumdahl & Don J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6th Ed. Introductory Chemistry, 6th Ed. Basic Chemistry, 6th Ed. by Steven S. Zumdahl & Don J. DeCoste University of Illinois

Chapter 17 Equilibrium

Collision Theory of Kinetics Kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. In order for a reaction to take place, the reacting molecules must collide into each other. 2

Collision Theory of Kinetics (cont.) Once molecules collide they may react together or they may not, depending on two factors: Whether the reacting molecules collide in the proper orientation for new bonds to form Whether the collision has enough energy to "break the bonds holding reactant molecules together"

Effective Collisions Collisions in which these two conditions are met (and the reaction occurs) are called effective collisions. The higher the frequency of effective collisions the faster the reaction rate. 3

Effective Collisions (cont.) When two molecules have an effective collision, a temporary, high energy (unstable) chemical species called an activated complex (or transition state) is formed. Not a true molecule because its bonds aren’t complete At a higher energy level than the energy of reactants unstable and quickly dissociates as a result.

Activated Complex The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, Ea The larger the activation energy, the slower the reaction Activated complex video on you tube catalyst demo – with H2o2 and yeast 4

Activated Complex (cont.) The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat. Different reactions have different activated complexes and therefore different activation energies.

Reactions and Activated Complex

Factors Affecting Reaction Rate The kind of molecules and what condition the reactants are in Increasing temperature always increases reaction rate 5

Factors Affecting Reaction Rate (cont.)

Factors Affecting Reaction Rate (cont.) The larger the concentration of reactant molecules, the faster the reaction will go. Increases the frequency of reactant molecule collisions 6

Factors Affecting Reaction Rate (cont.)

Factors Affecting Reaction Rate (cont.) Catalysts are substances that affect the speed of a reaction without being consumed. Most catalysts are used to speed up a reaction, written above the arrow in the equation. Homogeneous = catalyst present in same phase as reactants (s, l, g) Heterogeneous = catalyst present in different phase as reactants 7

Factors Affecting Reaction Rate (cont.)

Factors Affecting Reaction Rate (cont.)

Catalysts Speed Up Reactions Catalysts work by providing an alternative pathway for the reaction with a lower activation energy. Catalysts are regenerated at the end. Enzymes are biochemical catalysts.

Catalysts Speed Up Reactions (cont.)

Molecular Interpretation of Factors Affecting Rate Increasing the temperature raises the kinetic energy of the reactant molecules, causing them to collide more frequently and to have more collisions with sufficient energy to form the activated complex. 8

Molecular Interpretation of Factors Affecting Rate (cont.) The larger the concentration of reactant molecules, the more frequently they collide and the larger the number of effective collisions will be. Therefore the speed will increase.

Molecular Interpretation of Factors Affecting Rate (cont.) Catalysts work by providing an alternative pathway for the reaction with a lower activation energy. Lowering the activation energy means more molecules have enough kinetic energy so that when they collide they can form the activated complex. The result is the reaction goes faster. 9

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. 10

Reaction Dynamics (cont.) The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing, while the reverse reaction speeds up as the concentration of the products increases. Eventually rateforward = ratereverse 11

Reaction Dynamics (cont.)

Chemical Equilibrium Equilibrium only occurs in a closed system. When a system reaches equilibrium, the amounts of reactants and products in the system stay constant. 12

Chemical Equilibrium (cont.)

Equilibrium Constant aA + bB  cC + dD = [C]c[D]d [A]a[B]b Keq Law Of Chemical Equilibrium In this expression, K is a number called the equilibrium constant. 13

Equilibrium Constant (cont.) aA + bB  cC + dD = [C]c[D]d [A]a[B]b Keq 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.

Self check p 524 Write the equilibrium expression for the following reaction 4NH3 (g)+7O2(g)-->4NO2(g)+6H2O(g)

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. 14

Position of Equilibrium (cont.) If K is large (>1), then there will be a larger concentration of products at equilibrium than of reactants, and we say the position of equilibrium favors the products. 15

Position of Equilibrium (cont.) If K is small (<1), then there will be a larger concentration of reactants at equilibrium than of products, and we say the position of equilibrium favors the reactants. The position of equilibrium is not affected by adding a catalyst because catalyst speeds up both sides of reaction equally.

Example #1: Determine the value of the equilibrium constant for the reaction 2 SO2 + O2  2 SO3 16

Example #1 (cont.) Determine the equilibrium expression Plug the equilibrium concentrations into the equilibrium expression Solve the equation 3.50 3.00 SO3 1.25 1.50 O2 2.00 SO2 [Equilibrium] [Initial] Chemical

http://www.sophia.org/tutorials/solving-equilibria-using-the-ice-table

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. 17

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. 18

Concentration Changes and Le Châtelier’s Principle (cont.)

Changing Volume and Le Châtelier’s Principle Changing the volume of a gas is like changing its concentration. Has the same effect as changing the concentration on the position of equilibrium Decreasing the volume of the system increases its pressure. Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules 19

Changing Volume and Le Châtelier’s Principle (cont.)

Changing Temperature and Le Châtelier’s Principle The equilibrium constant will change if the temperature changes. For exothermic reactions, heating the system decreases K. Think of heat as a product of the reaction Therefore shift the position of equilibrium toward the reactants 20

Changing Temperature and Le Châtelier’s Principle (cont.) For endothermic reactions, heating the system increases 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.

Self check p 531 Novelty devices for predicting rain contain cobalt (II) chloride and are based on the following equilibrium: CoCl2 (s)+6H2O(g)CoCl2.6H2O(s) Blue Pink What color will this indicator be when rain is likely due to increased water vapor in the air?

Self check p 535, 534 For each of the following reactions, predict the direction the equilibrium will shift when the volume of the container is increased. H2(g)+F2(g)2HF(g) CO(g)+2H2(g)CH3OH(g) 2SO3(g)2SO2(g)+O2(g)

For the exothermic reaction, 2SO2(g)+O2(g)2SO3(g) Predict the equilibrium shift caused by each of the following changes: SO2 is added SO2 is removed The volume is decreased The temp is decreased

Example #2: If the value of the equilibrium constant for the reaction 2 SO2 + O2  2 SO3 is 4.36, determine the equilibrium concentration of SO3 21

Example #2 (cont.) Determine the equilibrium expression Plug the equilibrium concentrations and equilibrium constant into the equilibrium expression Solve the equation ? 3.00 SO3 1.25 1.50 O2 2.00 SO2 [Equilibrium] [Initial] Chemical

Example #2 (cont.)

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) 22

Solubility and Solubility Product (cont.) Equilibrium constant called solubility product. Ksp = [A+]n[Y-]m If undissolved solid is in equilibrium with the solution, the solution is saturated. Larger K = more soluble For salts that produce same the number of ions

Example #3: Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10-16. 23

AgI(s)  Ag+(aq) + I-(aq) Example #3 (cont.) 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-]

Example #3 (cont.) 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 Plug the ion concentrations into the expression for the solubility product and solve for Ksp [Ag+] = [I-] = x 24

The solubility of AgI = 1.2 x 10-8 M Example #3 (cont.) [Ag+] = 1.2 x 10-8 mol/L = [AgI] The solubility of AgI = 1.2 x 10-8 M 25

Self check p 539 1.Write the balanced equation for the reaction describing the dissolving of each of the following solids in water. Also write the Ksp expression for each solid BaSO4 (s) B. Fe(OH)3(s) C. Ag3PO4(s) 2.Calculate the Ksp value for BaSO4, which has a solubility of 3.9X10-5 mol/L at 25ºC. 3. The Ksp value for PbCrO4 is 2X10-16 at 25ºC. Calculate its solubility at 25ºC.