Slide 1 of 39 Chemistry 18.2

© Copyright Pearson Prentice Hall Slide 2 of 39 Reversible Reactions and Equilibrium In the early 1900s, German chemists refined the process of making ammonia from elemental nitrogen and hydrogen.This process allows the manufacture of nitrogen fertilizers. You will learn how reaction conditions can influence the yield of a chemical reaction. 18.2

© Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Slide 3 of 39 Reversible Reactions How do the amounts of reactants and products change in a chemical system at equilibrium? 18.2

© Copyright Pearson Prentice Hall Slide 4 of 39 Reversible Reactions and Equilibrium > Reversible Reactions At chemical equilibrium, no net change occurs in the actual amounts of the components of the system. 18.2

Slide 5 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Reversible Reactions If the rate of the shoppers going up the escalator is equal to the rate of the shoppers going down, then the number of shoppers on each floor remains constant, and there is an equilibrium. 18.2

Slide 6 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Reversible Reactions A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. 18.2

© Copyright Pearson Prentice Hall Slide 7 of 39 Reversible Reactions and Equilibrium > Reversible Reactions 18.2 At equilibrium, all three types of molecules are present. SO 2 and O 2 react to give SO 3 SO 3 decomposes to SO 2 and O 2

Slide 8 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Reversible Reactions When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction. A + B 25% C + D Means that at equilibrium, only 25% of the reactants A & B have been converted to products C &D. 18.2

Slide 9 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Reversible Reactions 18.2

© Copyright Pearson Prentice Hall Slide 10 of 39 Reversible Reactions and Equilibrium > Reversible Reactions Animation 23 Take a close look at a generalized reversible reaction.

© Copyright Pearson Prentice Hall Slide 11 of 39 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle What three stresses can cause a change in the equilibrium position of a chemical system? 18.2

© Copyright Pearson Prentice Hall Slide 12 of 39 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Stresses that upset the equilibrium of a chemical system include changes in the concentration of reactants or products, changes in temperature, and changes in pressure. 18.2

Slide 13 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle The French chemist Le Châtelier proposed what has come to be called Le Châtelier’s principle: If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress. 18.2

© Copyright Pearson Prentice Hall Slide 14 of 39 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Simulation 24 Simulate Le Châtelier’s principle for the synthesis of ammonia.

Slide 15 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Concentration Rapid breathing during and after vigorous exercise helps reestablish the body’s correct CO 2 :H 2 CO 3 equilibrium, keeping the acid concentration in the blood within a safe range. 18.2

Slide 16 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > An increase in concentration of a particular entity will result in a shift to the opposite side of the equation. This is because an increase in concentration means a better chance of a reaction taking place A + B shifts right (more products) C + D

Slide 17 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > A decrease in concentration of a particular entity will result in a shift to the same side of the equation. This is because an decrease in concentration means less chance of a reaction taking place A + B shifts left (more reactants) C + D

Slide 18 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Temperature Dinitrogen tetroxide is a colorless gas; nitrogen dioxide is a brown gas. The flask on the left is in a dish of hot water; the flask on the right is in ice. 18.2

© Copyright Pearson Prentice Hall Slide 19 of 39 Reversible Reactions and Equilibrium > An increase in temperature results in a shift to the opposite side of the equation as the “heat”. This is because an increase in temperature means more collisions and a better chance of a reaction taking place A + B + HEAT shifts right (more products) C + D

© Copyright Pearson Prentice Hall Slide 20 of 39 Reversible Reactions and Equilibrium > A decrease in temperature will result in a shift to the same side of the equation as the “heat”. This is because a decrease in temperature means less collisions and less chance of a reaction taking place A + B + HEAT shifts left (more reactants) C + D

© Copyright Pearson Prentice Hall Slide 21 of 39 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Pressure Pressure affects a mixture of nitrogen, hydrogen, and ammonia at equilibrium 18.2

© Copyright Pearson Prentice Hall Slide 22 of 39 Reversible Reactions and Equilibrium > An increase in pressure (i.e. a decrease in volume) results in a shift to the side of the equation with less moles of gas entities. This is because an increase in pressure forces the smaller gas entities to come together to be more efficient in space (bigger molecules but less of them). 2 A (g) + B (g) shifts right (more products) A 2 B (g) (3 moles of gas) (1 mole of gas)

© Copyright Pearson Prentice Hall Slide 23 of 39 Reversible Reactions and Equilibrium > A decrease in pressure (i.e. an increase in volume) results in a shift to the side of the equation with more moles of gas entities. This is because an decrease in pressure forces the larger gas entities to break up to fill the void in space (smaller molecules but more of them). 2 A (g) + B (g) shifts left (more reactants) A 2 B (g) (3 moles of gas) (1 mole of gas)

© Copyright Pearson Prentice Hall Slide 24 of 39 Conceptual Problem 18.1

© Copyright Pearson Prentice Hall Slide 25 of 39 Conceptual Problem 18.1

© Copyright Pearson Prentice Hall Slide 26 of 39 Conceptual Problem 18.1

© Copyright Pearson Prentice Hall Slide 27 of 39 Practice Problems for Conceptual Problem 18.1 Problem Solving 18.6 Solve a similar problem with the help of an interactive guided tutorial.

© Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Slide 28 of 39 Equilibrium Constants What does the value of K eq indicate about the equilibrium position of a reaction? 18.2

Slide 29 of 39 © Copyright Pearson Prentice Hall Reversible Reactions and Equilibrium > Equilibrium Constants The equilibrium constant (K eq ) is the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation. 18.2

© Copyright Pearson Prentice Hall Slide 30 of 39 Reversible Reactions and Equilibrium > Equilibrium Constants A value of K eq greater than 1 means that products are favored over reactants; a value of K eq less than 1 means that reactants are favored over products. 18.2

© Copyright Pearson Prentice Hall SAMPLE PROBLEM Slide 31 of

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© Copyright Pearson Prentice Hall SAMPLE PROBLEM Slide 33 of

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© Copyright Pearson Prentice Hall Slide 35 of 39 Practice Problems for Sample Problem 18.1 Problem Solving 18.7 Solve Problem 7 with the help of an interactive guided tutorial.

© Copyright Pearson Prentice Hall SAMPLE PROBLEM Slide 36 of

© Copyright Pearson Prentice Hall SAMPLE PROBLEM Slide 37 of

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© Copyright Pearson Prentice Hall SAMPLE PROBLEM Slide 39 of

© Copyright Pearson Prentice Hall Slide 40 of 39 Practice Problems for Sample Problem 18.2 Problem Solving 18.9 Solve Problem 9 with the help of an interactive guided tutorial.

© Copyright Pearson Prentice Hall Slide 41 of 39 Section Quiz -or- Continue to: Launch: Assess students’ understanding of the concepts in Section 18.2 Section Quiz

© Copyright Pearson Prentice Hall Slide 42 of Section Quiz. 1. In a reaction at equilibrium, reactants and products a.decrease in concentration. b.form at equal rates. c.have equal concentrations. d.have stopped reacting.

© Copyright Pearson Prentice Hall Slide 43 of Section Quiz. 2. In the reaction 2NO 2 (g)  2NO(g) + O 2 (g), increasing the pressure on the reaction would cause a.the amount of NO to increase. b.the amount of NO 2 to increase. c.nothing to happen. d.the amount of O 2 to increase.

© Copyright Pearson Prentice Hall Slide 44 of Section Quiz. 3. Changing which of the following would NOT affect the equilibrium position of a chemical reaction? a.concentration of a reactant only b.concentration of a product only c.temperature only d.volume only

© Copyright Pearson Prentice Hall Slide 45 of For the following reaction, K eq = 1. A(g) + B(g)  C(g) + D(g) Therefore, at equilibrium a.[C] = [A]. b.[A][B] = 0. c.[AB] = [CD] = 1. d.[A][B] = [C][D] Section Quiz.

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