Chapter 13 Chemical Equilibrium AP*

AP Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec )  LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K. (Sec )  LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions. (Sec 13.7)  LO 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached. (Sec 13.5)  LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K. (Sec 13.3, )

AP Learning Objectives  LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction. (Sec )  LO 6.7 The student is able, for a reversible reaction that has a large or small K to determine which chemical species will have very large versus very small concentrations at equilibrium. (Sec 13.2)  LO 6.8 The student is able to use Le Châtelier’s Principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium. (Sec 13.7)  LO 6.9 The student is able to use Le Châtelier’s Principle to design a set of conditions that will optimize a desired outcome, such as product yield. (Sec 13.7)  LO 6.10 The student is able to connect Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K. (Sec 13.7)

Section 13.1 The Equilibrium Condition AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 5 Chemical Equilibrium  The state where the concentrations of all reactants and products remain constant with time.  On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 6 Equilibrium Is:  Macroscopically static  Microscopically dynamic

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 7 Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 8 Chemical Equilibrium  Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 9 The Changes with Time in the Rates of Forward and Reverse Reactions

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 10 Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. CONCEPT CHECK!

Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 11 Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. CONCEPT CHECK!

Section 13.2 The Equilibrium Constant AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.7 The student is able, for a reversible reaction that has a large or small K to determine which chemical species will have very large versus very small concentrations at equilibrium.

Section 13.2 The Equilibrium Constant Consider the following reaction at equilibrium: jA + kB lC + mD  A, B, C, and D = chemical species.  Square brackets = concentrations of species at equilibrium.  j, k, l, and m = coefficients in the balanced equation.  K = equilibrium constant (given without units). Copyright © Cengage Learning. All rights reserved 13 j l k m [B][A] [D] [C] K =

Section 13.2 The Equilibrium Constant Conclusions About the Equilibrium Expression  Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse.  When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n.  K values are usually written without units. Copyright © Cengage Learning. All rights reserved 14

Section 13.2 The Equilibrium Constant  K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.  For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K.  Equilibrium position is a set of equilibrium concentrations. Copyright © Cengage Learning. All rights reserved 15

Section 13.3 Equilibrium Expressions Involving Pressures AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.

Section 13.3 Equilibrium Expressions Involving Pressures  K involves concentrations.  K p involves pressures. Copyright © Cengage Learning. All rights reserved 17

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Copyright © Cengage Learning. All rights reserved 18

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature: Copyright © Cengage Learning. All rights reserved 19

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Copyright © Cengage Learning. All rights reserved 20

Section 13.3 Equilibrium Expressions Involving Pressures The Relationship Between K and K p K p = K(RT) Δn  Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.  R = L·atm/mol·K  T = temperature (in Kelvin) Copyright © Cengage Learning. All rights reserved 21

Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p (3.9 × 10 4 ) from the previous example, calculate the value of K at 35°C. Copyright © Cengage Learning. All rights reserved 22

Section 13.4 Heterogeneous Equilibria AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.

Section 13.4 Heterogeneous Equilibria Homogeneous Equilibria  Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq) Copyright © Cengage Learning. All rights reserved 24

Section 13.4 Heterogeneous Equilibria  Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g) Copyright © Cengage Learning. All rights reserved 25

Section 13.4 Heterogeneous Equilibria  The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.  The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) Copyright © Cengage Learning. All rights reserved 26

Section 13.5 Applications of the Equilibrium Constant AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.  LO 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached.  LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.  LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.

Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction  A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.  Reaction goes essentially to completion. Copyright © Cengage Learning. All rights reserved 28

Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction  A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.  Reaction does not occur to any significant extent. Copyright © Cengage Learning. All rights reserved 29

Section 13.5 Applications of the Equilibrium Constant If the equilibrium lies to the right, the value for K is __________. large (or >1) If the equilibrium lies to the left, the value for K is ___________. small (or <1) Copyright © Cengage Learning. All rights reserved 30 CONCEPT CHECK!

Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q  Used when all of the initial concentrations are nonzero.  Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Copyright © Cengage Learning. All rights reserved 31

Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q  Q = K; The system is at equilibrium. No shift will occur.  Q > K; The system shifts to the left.  Consuming products and forming reactants, until equilibrium is achieved.  Q < K; The system shifts to the right.  Consuming reactants and forming products, to attain equilibrium. Copyright © Cengage Learning. All rights reserved 32

Section 13.5 Applications of the Equilibrium Constant Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)  Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? Copyright © Cengage Learning. All rights reserved 33 EXERCISE!

Section 13.5 Applications of the Equilibrium Constant Set up ICE Table Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial Change– 4.00– Equilibrium K = Copyright © Cengage Learning. All rights reserved 34

Section 13.5 Applications of the Equilibrium Constant Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)  Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN − (aq) (same temperature as Trial #1) Equilibrium: ? M FeSCN 2+ (aq) 5.00 M FeSCN 2+ Copyright © Cengage Learning. All rights reserved 35 EXERCISE!

Section 13.5 Applications of the Equilibrium Constant Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)  Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq) 3.00 M FeSCN 2+ Copyright © Cengage Learning. All rights reserved 36 EXERCISE!

Section 13.6 Solving Equilibrium Problems AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.  LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.  LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.

Section 13.6 Solving Equilibrium Problems 1)Write the balanced equation for the reaction. 2)Write the equilibrium expression using the law of mass action. 3)List the initial concentrations. 4)Calculate Q, and determine the direction of the shift to equilibrium. Copyright © Cengage Learning. All rights reserved 38

Section 13.6 Solving Equilibrium Problems 5)Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6)Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7)Check your calculated equilibrium concentrations by making sure they give the correct value of K. Copyright © Cengage Learning. All rights reserved 39

Section 13.6 Solving Equilibrium Problems Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #19.00 M5.00 M1.00 M Trial #23.00 M2.00 M5.00 M Trial #32.00 M9.00 M6.00 M Find the equilibrium concentrations for all species. Copyright © Cengage Learning. All rights reserved 40 EXERCISE!

Section 13.6 Solving Equilibrium Problems Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M Copyright © Cengage Learning. All rights reserved 41 EXERCISE!Answer

Section 13.6 Solving Equilibrium Problems A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH 3 (g) N 2 (g) + H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69 Copyright © Cengage Learning. All rights reserved 42 CONCEPT CHECK!

Section 13.6 Solving Equilibrium Problems A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = 4.00 × Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = M Concentration of NO 2 = 6.32 × M Copyright © Cengage Learning. All rights reserved 43 CONCEPT CHECK!

Section 13.7 Le Châtelier’s Principle AP Learning Objectives, Margin Notes and References  Learning Objectives  LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.  LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions.  LO 6.8 The student is able to use Le Châtelier’s Principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium.  LO 6.9 The student is able to use Le Châtelier’s Principle to design a set of conditions that will optimize a desired outcome, such as product yield.  LO 6.10 The student is able to connect Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K.  Additional AP References  LO 6.9 (see APEC #13, “Le Châtelier’s Principle”)

Section 13.7 Le Châtelier’s Principle  If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Copyright © Cengage Learning. All rights reserved 45

Section 13.7 Le Châtelier’s Principle Effects of Changes on the System 1.Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2.Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product). Copyright © Cengage Learning. All rights reserved 46

Section 13.7 Le Châtelier’s Principle Effects of Changes on the System 3.Pressure: a)The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b)Addition of inert gas does not affect the equilibrium position. c)Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas. Copyright © Cengage Learning. All rights reserved 47

Section 13.7 Le Châtelier’s Principle Copyright © Cengage Learning. All rights reserved 48 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

Section 13.7 Le Châtelier’s Principle Equilibrium Decomposition of N 2 O 4 Copyright © Cengage Learning. All rights reserved 49 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE