Factors Affecting : Le Châtelier s Principle Pressure Factors Affecting : Le Châtelier s Principle Pressure When volume decreases, the pressure increases. systems in which some reactants and products are gases can be affected by a change in pressure. A shift will occur only if there are an unequal number of moles of gas on each side of the equation. Initial equilibrium is disturbed by an increase in pressure. A new equilibrium position is established to favor fewer molecules. 18 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 19 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Factors Affecting : 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 (g) + 3H (g) Add pressure Reduce pressure NH 3 (g) When molecules of ammonia (product) form, 4 molecules of reactants are used up. A shift toward ammonia (product) will reduce the number of molecules. 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? N (g) + 3H (g) CHEMISTRY & YOU NH 3 (g) + heat An increase in pressure and a decrease in temperature would shift the equilibrium toward the production of ammonia (product). 0 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 1 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 1
Factors Affecting : Le Châtelier s Principle Catalysts and Catalysts decrease the time it takes to establish equilibrium. However, they do not affect the amounts of reactants and products present at equilibrium. Sample Problem 18. Applying Le Châtelier s Principle What effect will each of the following changes have on the equilibrium position for this reversible reaction? PCl 3 (g) + Cl (g) a. Cl is added. b. Pressure is increased. c. Heat is removed. d. PCl 3 is removed as it forms. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 3 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18. Sample Problem 18. Solve Apply the concepts to this problem. Start with the addition of Cl. Cl is a product. Increasing the concentration of a product shifts the equilibrium to the left. Add Cl PCl 3 (g) + Cl (g) 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. Increase pressure For a change in pressure, compare the number of molecules of gas molecules on both sides of the equation. PCl 3 (g) + Cl (g) 4 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 5 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved.
Sample Problem 18. Sample Problem 18. Solve Apply the concepts to this problem. Solve Apply the concepts to this problem. Analyze the effect of removing heat. Analyze the effect of removing PCl 3. The reverse reaction produces heat. PCl 3 is a product. The removal of heat causes the equilibrium to shift to the left. Removal of a product as it forms causes the equilibrium to shift to the right. Remove heat PCl 3 (g) + Cl (g) Remove PCl 3 PCl 3 (g) + Cl (g) 6 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 7 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? 4HCl(g) + O (g) Cl (g) +H O(g) 4HCl(g) + O (g) Cl (g) +H O(g) Reducing the number of molecules that are gases decreases the pressure. The equilibrium will shift to the right. 8 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 9 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 3
What does the size of an equilibrium constant indicate about a system at equilibrium? 30 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 31 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Chemists express the equilibrium position as a numerical value. This value relates the amounts of reactants to products at equilibrium. aa + bb cc + dd In this general reaction, the coefficients a, b, c, and d represent the number of moles. K eq = [C]c [D] d [A] a [B] b The equilibrium constant (K eq ) is the ratio of product concentrations to reactant concentrations at equilibrium. aa + bb cc + dd 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. 3 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 33 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 4
The flask on the left is in a dish of hot water. The flask on the right is in ice. The value of K eq depends on the temperature of the reaction. The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. Dinitrogen tetroxide is a colorless gas. Nitrogen dioxide is a brown gas. N O 4 (g) NO (g) 34 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 35 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.3 The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. When K eq has a large value, such as 3.1 x 10 11, the reaction mixture at equilibrium will consist mainly of product. When K eq has a small value, such as 3.1 x 10 11, the mixture at equilibrium will consist mainly of reactant. When K eq has an intermediate value, such as 0.15 or 50, the mixture will have significant amounts of both reactant and product. Expressing and Calculating K eq The colorless gas dinitrogen tetroxide (N O 4 ) and the brown gas nitrogen dioxide (NO ) exist in equilibrium with each other. N O 4 (g) NO (g) A liter of the gas mixture at equilibrium contains 0.0045 mol of N O 4 and 0.030 mol of NO at 10 o C. Write the expression for the equilibrium constant (K eq ) and calculate the value of the constant for the reaction. 36 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 37 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 5
Sample Problem 18.3 Sample Problem 18.3 Start with the general expression for the equilibrium constant. K eq = [C]c x [D] d [A] a x [B] b Write the equilibrium constant expression for this reaction. K eq = [NO ] [N O ] Place the concentration of the product in the numerator and the concentration of the reactant in the denominator. Raise each concentration to the power equal to its coefficient in the chemical equation. 38 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Substitute the concentrations that are known and calculate K eq. K eq = (0.030 mol/l) (0.0045 mol/l) K eq = 0.0 mol/l = 0.0 You can ignore the unit mol/l; chemists report equilibrium constants without a stated unit. 39 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.5 Sample Problem 18.5 Finding Concentrations at Bromine chloride (BrCl) decomposes to form bromine and chlorine. BrCl(g) Br (g) + Cl (g) At a certain temperature, the equilibrium constant for the reaction is 11.1. A sample of pure BrCl is placed in a 1-L container and allowed to decompose. At equilibrium, the reaction mixture contains 4.00 mol Cl. What are the equilibrium concentrations of Br and BrCl? 40 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 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 BrCl decomposes, equal numbers of moles of Br and Cl are formed. KNOWNS [Cl ] (equilibrium) = 4.00 mol/l K eq = 11.1 UNKNOWN [Br ] (equilibrium) =? mol/l [BrCl] (equilibrium) =? mol/l 41 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 6
Sample Problem 18.5 Sample Problem 18.5 The volume of the container is 1 L, so calculate [Br ] at equilibrium. [Br ] = Write the equilibrium expression for the reaction. K eq = 4.00 mol 1 L [Br ] [Cl ] [BrCl] = 4.00 mol/l 4 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Rearrange the equation to solve for [BrCl]. [BrCl] = [Br ] [Cl ] K eq Substitute the known values for K eq, [Br ], and [Cl ]. [BrCl] = 4.00 mol/l x 4.00 mol/l 11.1 = 1.44 mol /L 43 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.5 Calculate the square root. [BrCl] = 1.44 mol /L = 1.0 mol/l Use your calculator to find the square root. HCl is formed when H and Cl react at high temperatures. H (g) + Cl (g) HCl(g) At equilibrium, [HCl] = 1.76 x 10 mol/l, and [H ] = [Cl ] = 1.60 x 10 3 mol/l. What is the value of the equilibrium constant? 44 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 45 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 7
HCl is formed when H and Cl react at high temperatures. At equilibrium, [HCl] = 1.76 x 10 mol/l, and [H ] = [Cl ] = 1.60 x 10 3 mol/l. What is the value of the equilibrium constant? [HCl] K eq = = [H ] x [Cl ] K eq = 11 H (g) + Cl (g) HCl(g) (1.76 x 10 mol/l) (1.60 x 10 3 mol/l) x (1.60 x 10 3 mol/l) 46 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. 8