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Chapter 13 Chemical Equilibrium

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Chapter 13 Table of Contents Copyright © Cengage Learning. All rights reserved The Equilibrium Condition 13.2The Equilibrium Constant 13.3Equilibrium Expressions Involving Pressures 13.4Heterogeneous Equilibria 13.5Applications of the Equilibrium Constant 13.6Solving Equilibrium Problems 13.7Le Châtelier’s Principle

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 3 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.

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 4 Equilibrium Is: Macroscopically static Microscopically dynamic

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 5 Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 6 Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 7 The Changes with Time in the Rates of Forward and Reverse Reactions

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 8 Concept Check 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.

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Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 9 Concept Check 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.

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Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 10 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). j l k m [B][A] [D] [C] K =

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Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 11 Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When 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.

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Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 12 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 value for K. Equilibrium position is a set of equilibrium concentrations.

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 13 K involves concentrations. K p involves pressures.

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 14 Example N 2 (g) + 3H 2 (g) 2NH 3 (g)

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 15 Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature:

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 16 Example N 2 (g) + 3H 2 (g) 2NH 3 (g)

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 17 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)

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Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 18 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.

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Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 19 Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

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Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 20 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)

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Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 21 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)

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 22 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. The Extent of a Reaction

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 23 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. The Extent of a Reaction

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check 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)

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 25 Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Reaction Quotient, Q

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 26 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. Reaction Quotient, Q

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 27 Exercise 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?

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 28 Set up ICE Table Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial Change – 4.00 – Equilibrium K = 0.333

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 29 Exercise 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+

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Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 30 Exercise 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+

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 31 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. Solving Equilibrium Problems

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 32 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. Solving Equilibrium Problems

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 33 Exercise 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.

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 34 Exercise (Answer) 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

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 35 Concept Check 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

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Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 36 Concept Check 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 x 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 x M

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Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 37 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.

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Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 38 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).

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Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 39 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.

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Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 40

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Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 41 Equilibrium Decomposition of N 2 O 4

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