18 3 Reversible Reactions and Equilibrium Reversible Reactions

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions • You may have inferred that chemical reactions always progress in one direction. • This inference is not true. Some reactions are reversible. • A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur at the same time. 1 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Here is an example of a reversible reaction. 2 SO 2(g) + O 2(g) 2 SO 3(g) • The first reaction is called the forward reaction. • The second reaction is called the reverse reaction. 2 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions 2 SO 2(g) + O 2(g) 2 SO 3(g) The two equations can be combined into one using a double arrow. 2 SO 2(g) + O 2(g) 2 SO 3(g) The double arrow tells you that the reaction is reversible. 3 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Molecules of SO 2 and O 2 react to give SO 3. Molecules of SO 3 decompose to give SO 2 and O 2. 4 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Establishing Equilibrium When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. 5 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Interpret Graphs This graph shows the progress of a reaction that starts with concentrations of SO 2 and O 2, but with zero SO 3. This graph shows the progress of the reaction that begins with an initial concentration of SO 3, and zero concentrations for SO 2 and O 2. Notice that after a certain time, the concentrations remain constant. 6 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Conditions at Equilibrium Chemical equilibrium is a dynamic state. When the store opens, only the forward reaction occurs as shoppers head to the second floor. 7 Equilibrium is reached when the rate at which shoppers move from the first floor to the second is equal to the rate at which shoppers move from the second floor to the first. Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Conditions at Equilibrium At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reaction components. 8 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Concentrations at Equilibrium Although the rates of the forward and reverse reactions are equal at equilibrium, the concentrations of the components usually are not. • The relative concentrations of the reactants and products at equilibrium mark the equilibrium position of a reaction. 9 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Concentrations at Equilibrium The equilibrium position tells you whether the forward or reverse reaction is more likely to happen. • Suppose a single reactant, A, forms a single product, B. • If the equilibrium mixture contains 1% A and 99% B, then the formation of B is said to be favored. A B 1% 10 99% Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Reversible Reactions Concentrations at Equilibrium In principle, almost all reactions are reversible to some extent under the right conditions. • In practice, one set of components is often so favored at equilibrium that the other set cannot be detected. • When no reactants can be detected, you can say that the reaction has gone to completion, or is irreversible. • When no products can be detected, you can say that no reaction has taken place. 11 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

Affecting 18. 3 Reversible Reactions and Equilibrium > Factors Equilibrium: Le Châtelier’s Principle The French chemist Henri Le Châtelier (1850– 1936) 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. 12 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

Affecting 18. 3 Reversible Reactions and Equilibrium > Factors 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. 13 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Concentration Changing the amount, or concentration, of any reactant or product in a system at equilibrium disturbs the equilibrium. • The system will adjust to minimize the effects of the change. 14 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Concentration Consider the decomposition of carbonic acid (H 2 CO 3) in aqueous solution. H 2 CO 3(aq) < 1% CO 2(aq) + H 2 O(l) > 99% • The system has reached equilibrium. • The amount of carbonic acid is less than 1%. 15 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Suppose carbon dioxide is added to the system. Add CO 2 Direction of shift H 2 CO 3(aq) CO 2(aq) + H 2 O(l) • This increase in the concentration of CO 2 causes the rate of the reverse reaction to increase. • Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of the reactants. 16 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Suppose carbon dioxide is removed. H 2 CO 3(aq) Add CO 2 Direction of shift Remove CO 2 Direction of shift CO 2(aq) + H 2 O(l) • This decrease in the concentration of CO 2 causes the rate of the reverse reaction to decrease. • Removing a product always pulls a reversible reaction in the direction of the products. 17 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium Factors Affecting > Equilibrium: Le Châtelier’s Principle An equilibrium between carbonic acid, carbon dioxide, and water exists in your blood. • During exercise, the concentration of CO 2 in the blood increases. This shifts the equilibrium in the direction of carbonic acid. • The increase in the level of CO 2 also triggers an increase in the rate of breathing. With more breaths per minute, more CO 2 is removed through the lungs. • The removal of CO 2 causes the equilibrium to shift toward the products, which reduces the amount of H 2 CO 3. 18 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium Factors Affecting > Equilibrium: Le Châtelier’s Principle Temperature Increasing the temperature causes the equilibrium position of a reaction to shift in the direction that absorbs heat. • In other words, it will shift in the direction that reduces the stress. 19 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

Affecting 18. 3 Reversible Reactions and Equilibrium > Factors Equilibrium: Le Châtelier’s Principle Temperature N 2(g) + 3 H 2(g) Add heat Direction of shift Remove heat (cool) Direction of shift 2 NH 3(g) + heat Heat can be considered to be a product, just like NH 3. • Heating the reaction mixture at equilibrium pushes the equilibrium position to the left, which favors the reactants. • Cooling, or removing heat, pulls the equilibrium position to the right, and the product yield increases. 20 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

Affecting 18. 3 Reversible Reactions and Equilibrium > Factors Equilibrium: Le Châtelier’s Principle Pressure Equilibrium systems in which some reactants and products are gases can be affected by a change in pressure. • A shift will occur only if there an unequal number of moles of gas on each side of the equation. 21 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Pressure When the plunger is pushed down, the volume decreases and the pressure increases. Initial equilibrium 22 Equilibrium is disturbed by an increase in pressure. A new equilibrium position is established with fewer molecules. Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Pressure You can predict which way the equilibrium position will shift by comparing the number of molecules of reactants and products. N 2(g) + 3 H 2(g) Add pressure Direction of shift Reduce pressure Direction of shift 2 NH 3(g) • When two molecules of ammonia form, four molecules of reactants are used up. • A shift toward ammonia (the product) will reduce the number of molecules. 23 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > CHEMISTRY & YOU Fritz Haber and Karl Bosch figured out how to increase the yield of ammonia when nitrogen and hydrogen react. Their success came from controlling the temperature and pressure. In which direction did they adjust each factor and why? 24 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > CHEMISTRY & YOU Fritz Haber and Karl Bosch figured out how to increase the yield of ammonia when nitrogen and hydrogen react. Their success came from controlling the temperature and pressure. In which direction did they adjust each factor and why? An increase in pressure and a decrease in temperature would increase the yield of ammonia by shifting the equilibrium toward the production of ammonia. 25 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Factors Affecting Equilibrium: Le Châtelier’s Principle Catalysts and Equilibrium Catalysts decrease the time it takes to establish equilibrium. • However, they do not affect the amounts of reactants and products present at equilibrium. 26 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 2 Applying Le Châtelier’s Principle What effect will each of the following changes have on the equilibrium position for this reversible reaction? PCl 5(g) + heat a. b. c. d. 27 PCl 3(g) + Cl 2(g) Cl 2 is added. Pressure is increased. Heat is removed. PCl 3 is removed as it forms. Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 2 2 Solve Apply the concepts to this problem. Start with the addition of Cl 2. • Cl 2 is a product. • Increasing the concentration of a product shifts the equilibrium to the left. PCl 5(g) + heat 28 Add Cl 2 Direction of shift PCl 3(g) + Cl 2(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 2 2 Solve Apply the concepts to this problem. Analyze the effect of an increase in pressure. • Reducing the number of molecules that are gases decreases the pressure. • The equilibrium shifts to the left. PCl 5(g) + heat 29 Increase pressure Direction of shift For a change in pressure, compare the number of molecules of gas molecules on both sides of the equation. PCl 3(g) + Cl 2(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 2 2 Solve Apply the concepts to this problem. Analyze the effect of removing heat. • The reverse reaction produces heat. • The removal of heat causes the equilibrium to shift to the left. PCl 5(g) + heat 30 Remove heat Direction of shift PCl 3(g) + Cl 2(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 2 2 Solve Apply the concepts to this problem. Analyze the effect of removing PCl 3. • PCl 3 is a product. • Removal of a product as it forms causes the equilibrium to shift to the right. PCl 5(g) + heat 31 Remove PCl 3 Direction of shift PCl 3(g) + Cl 2(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? 4 HCl(g) + O 2(g) 32 2 Cl 2(g) +2 H 2 O(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? 4 HCl(g) + O 2(g) 2 Cl 2(g) +2 H 2 O(g) Reducing the number of molecules that are gases decreases the pressure. The equilibrium will shift to the right. 33 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants What does the size of an equilibrium constant indicate about a system at equilibrium? 34 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants Chemists express the equilibrium position as a numerical value. • This value relates the amounts of reactants to products at equilibrium. a. A + b. B c. C + d. D • In this general reaction, the coefficients a, b, c, and d represent the number of moles. 35 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants The equilibrium constant (Keq) is the ratio of product concentrations to reactant concentrations at equilibrium. a. A + b. B c. C + d. D • From the general equation, each concentration is raised to a power equal to the number of moles of that substance in the balanced chemical equation. [C]c x [D]d Keq = [A]a x [B]b 36 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants The value of Keq depends on the temperature of the reaction. • The flask on the left is in a dish of hot water. • The flask on the right is in ice. Dinitrogen tetroxide is a colorless gas. Nitrogen dioxide is a brown gas. N 2 O 4(g) 37 2 NO 2(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. 38 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Equilibrium Constants The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. • When Keq has a large value, such as 3. 1 x 1011, the reaction mixture at equilibrium will consist mainly of product. • When Keq has a small value, such as 3. 1 x 10– 11, the mixture at equilibrium will consist mainly of reactant. • When Keq has an intermediate value, such as 0. 15 or 50, the mixture will have significant amounts of both reactant and product. 39 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 3 Expressing and Calculating Keq The colorless gas dinitrogen tetroxide (N 2 O 4) and the brown gas nitrogen dioxide (NO 2) exist in equilibrium with each other. N 2 O 4(g) 2 NO 2(g) A liter of the gas mixture at equilibrium contains 0. 0045 mol of N 2 O 4 and 0. 030 mol of NO 2 at 10 o. C. Write the expression for the equilibrium constant (Keq) and calculate the value of the constant for the reaction. 40 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 3 1 Analyze List the knowns and the unknowns. Modify the general expression for the equilibrium constant and substitute the known concentrations to calculate Keq. 41 KNOWNS UNKNOWN [N 2 O 4] = 0. 0045 mol/L Keq (algebraic expression) = ? [NO 2] = 0. 030 mol/L Keq (numerical value) = ? Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 3 2 Calculate Solve for the unknowns. • Start with the general expression for the equilibrium constant. Place the concentration of the product in the [C]c x [D]d numerator and the Keq = [A]a x [B]b concentration of the • Write the equilibrium constant expression for this reaction. [NO 2]2 Keq = [N O ] 2 2 42 reactant in the denominator. Raise each concentration to the power equal to its coefficient in the chemical equation. Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 3 2 Calculate Solve for the unknowns. Substitute the concentrations that are known and calculate Keq. (0. 030 mol/L)2 (0. 030 mol/L x 0. 030 mol/L) Keq = (0. 0045 mol/L) Keq = 0. 20 mol/L = 0. 20 43 You can ignore the unit mol/L; chemists report equilibrium constants without a stated unit. Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 3 3 Evaluate Does the result make sense? • Each concentration is raised to the correct power. • The numerical value of the constant is correctly expressed to two significant figures. • The value for Keq is appropriate for an equilibrium mixture that contains significant amounts of both gases. 44 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 Finding the Equilibrium Constant One mole of colorless hydrogen gas and one mole of violet iodine vapor are sealed in a 1 -L flask and allowed to react at 450 o. C. At equilibrium, 1. 56 mol of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate Keq for the reaction. H 2(g) + I 2(g) 45 2 HI(g) Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 1 Analyze List the knowns and the unknown. Find the concentrations of the reactants at equilibrium. Then substitute the equilibrium concentrations in the expression for the equilibrium constant for this reaction. KNOWNS UNKNOWN [H 2] (initial) = 1. 00 mol/L Keq = ? [I 2] (initial) = 1. 00 mol/L [HI] (equilibrium) = 1. 56 mol/L 46 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 2 Calculate Solve for the unknown. First find out how much H 2 and I 2 are consumed in the reaction. x + x = 1. 56 mol 2 x = 1. 56 mol x = 0. 780 mol 47 Let mol H 2 used = mol I 2 used = x. The number of mol H 2 and mol I 2 used must equal the number of mol HI formed (1. 56 mol). Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 2 Calculate Solve for the unknown. • Calculate how much H 2 and I 2 remain in the flask at equilibrium. mol H 2 = mol I 2 = (1. 00 mol – 0. 780 mol) = 0. 22 mol • Write the expression for Keq. [HI]2 Keq = [H ] x [I ] 2 2 48 Use the general expression for Keq as a guide: [C]c x [D]d Keq = [A]a x [B]b Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 2 Calculate Solve for the unknown. Substitute the equilibrium concentrations of the reactants and products into the equation and solve for Keq = (1. 56 mol/L)2 0. 22 mol/L x 0. 22 mol/L Keq = 1. 56 mol/L x 1. 56 mol/L 0. 22 mol/L x 0. 22 mol/L Keq = 5. 0 x 101 49 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 4 3 Evaluate Does the result make sense? • Each concentration is raised to the correct power. • The value of the constant reflects the presence of significant amounts of the reactions and product in the equilibrium mixture. 50 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 Finding Concentrations at Equilibrium Bromine chloride (Br. Cl) decomposes to form bromine and chlorine. 2 Br. Cl(g) Br 2(g) + Cl 2(g) At a certain temperature, the equilibrium constant for the reaction is 11. 1. A sample of pure Br. Cl is placed in a 1 -L container and allowed to decompose. At equilibrium, the reaction mixture contains 4. 00 mol Cl 2. What are the equilibrium concentrations of Br 2 and Br. Cl? 51 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 1 Analyze List the knowns and the unknowns. Use the balanced equation, the equilibrium constant, and the equilibrium constant expression to find the unknown concentrations. According to the balanced equation, when Br. Cl decomposes, equal numbers of moles of Br 2 and Cl 2 are formed. 52 KNOWNS UNKNOWN [Cl 2] (equilibrium) = 4. 00 mol/L [Br 2] (equilibrium) = ? mol/L Keq = 11. 1 [Br. Cl] (equilibrium) = ? mol/L Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 2 Calculate Solve for the unknowns. • The volume of the container is 1 L, so calculate [Br 2] at equilibrium. 4. 00 mol [Br 2] = = 4. 00 mol/L 1 L • Write the equilibrium expression for the reaction. [Br 2] x [Cl 2] Keq = [Br. Cl]2 53 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 2 Calculate Solve for the unknowns. • Rearrange the equation to solve for [Br. Cl]2. [Br 2] x [Cl 2] [Br. Cl]2 = Keq • Substitute the known values for Keq, [Br 2], and [Cl 2]. 4. 00 mol/L x 4. 00 mol/L [Br. Cl]2 = 11. 1 = 1. 44 mol 2/L 2 54 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 2 Calculate Solve for the unknowns. • Calculate the square root. [Br. Cl] = 1. 44 mol 2/L 2 = 1. 20 mol/L Use your calculator to find the square root. 55 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Sample Problem 18. 5 3 Evaluate Does the result make sense? It makes sense that the equilibrium concentration of the reactant and the products are both present in significant amounts because Keq has an intermediate value. 56 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > HCl is formed when H 2 and Cl 2 react at high temperatures. H 2(g) + Cl 2(g) 2 HCl(g) At equilibrium, [HCl] = 1. 76 x 10– 2 mol/L, and [H 2] = [Cl 2] = 1. 60 x 10– 3 mol/L. What is the value of the equilibrium constant? 57 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > HCl is formed when H 2 and Cl 2 react at high temperatures. H 2(g) + Cl 2(g) 2 HCl(g) At equilibrium, [HCl] = 1. 76 x 10– 2 mol/L, and [H 2] = [Cl 2] = 1. 60 x 10– 3 mol/L. What is the value of the equilibrium constant? [HCl]2 (1. 76 x 10– 2 mol/L)2 Keq = = [H 2] x [Cl 2] (1. 60 x 10– 3 mol/L) x (1. 60 x 10– 3 mol/L) Keq = 121 58 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Key Concepts At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reactant components. Stresses that upset the equilibrium of a chemical system include changes in concentration of reactants or products, changes in temperature, and changes in pressure. 59 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Key Concept and Key Equation The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. Keq = 60 [C]c x [D]d [A]a x [B]b Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Glossary Terms • reversible reaction: a reaction in which the conversion of reactants into products and the conversion of products into reactants occur simultaneously • chemical equilibrium: a state of balance in which the rates of the forward and reverse reactions are equal; no net change in the amount of reactants and products occurs in the chemical system 61 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Glossary Terms • equilibrium position: the relative concentrations of reactants and products of a reaction that has reached equilibrium; indicates whether the reactants or products are favored in the reversible reaction • Le Châtelier’s principle: when a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress 62 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > Glossary Terms • equilibrium constant: 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 63 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.

18. 3 Reversible Reactions and Equilibrium > END OF 18. 3 64 Copyright © Pearson Education, Inc. , or its affiliates. All Rights Reserved.