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Chapter 13: Chemical Equilibrium
The Equilibrium State 6/24/2018 Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. 2NO2(g) N2O4(g) Brown Colorless Equilibrium has been mentioned a few times prior to this chapter. Forward and reverse rates for reaction mechanisms in Ch 12 (Chemical Kinetics), evaporation of liquids in Ch 10 (Liquids, Solids, and Phase Changes), and dissociation of weak acids in Ch 4 (Reactions in Aqueous Solution) to name three. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Chapter 13: Chemical Equilibrium
The Equilibrium State 6/24/2018 2NO2(g) N2O4(g) At 25 °C. Chemical Equilibrium is dynamic. It’s not that all chemical reactions stop but that the rates of change become constant. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Chapter 13: Chemical Equilibrium
6/24/2018 This is a plot of how the rates of reactants and products change with time. Copyright © 2008 Pearson Prentice Hall, Inc.
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The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kc Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kc For a general reversible reaction: cC + dD aA + bB Double arrows show reversible reaction [A]a[B]b [C]c[D]d Products Equilibrium equation: Kc = Reactants Equilibrium constant “c” in Kc is for concentration. [A] still means molarity of A. No units for the equilibrium constant. Since the units would change depending on the coefficients of the reactants and products, they are left off. Equilibrium constant expression Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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The Equilibrium Constant Kc
2NO2(g) N2O4(g) For the following reaction: [N2O4] [NO2]2 Kc = = 4.64 x 10-3 (at 25 °C)
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Examples Write the equilibrium constant, Kc, for each of the following reactions: a. CH4(g) + H2O(g) CO(g) H2(g) b. 2 NO(g) N2(g) + O2(g)
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The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kc Copyright © 2008 Pearson Prentice Hall, Inc.
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Learning Check Experiment 2 Experiment 5 0.0429 (0.0141)2 [N2O4]
0.0337 (0.0125)2 = 4.63 x 10-3 Kc = = = 4.64 x 10-3
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The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kc The equilibrium constant and the equilibrium constant expression are for the chemical equation as written. [N2][H2]3 [NH3]2 2NH3(g) N2(g) + 3H2(g) Kc = [NH3]2 [N2][H2]3 1 Kc Kc = N2(g) + 3H2(g) 2NH3(g) = Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Equilibrium Constant, Kc
[N2]2[H2]6 [NH3]4 ´´ 4NH3(g) 2N2(g) + 6H2(g) K c = = Kc 2
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Examples Consider the following equilibria 2CO2(g) + 2 CF4(g) 4COF2(g)
Write the equilibrium constant K’c expresion
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The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kp 2NO2(g) N2O4(g) NO2 P 2 Kp = P is the partial pressure of that component N2O4 P “p” in Kp means pressure. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium 6/24/2018 The Equilibrium Constant Kp K mol L atm Kp = Kc(RT)Dn R is the gas constant, T is the absolute temperature (Kelvin). n is the number of moles of gaseous products minus the number of moles of gaseous reactants. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium 6/24/2018 Heterogeneous Equilibria CaO(s) + CO2(g) CaCO3(s) Limestone Lime [CaCO3] [CaO][CO2] (1) (1)[CO2] Kc = = = [CO2] Pure solids and pure liquids are not included. Heterogeneous refers to the reactants present in more than one phase. This reaction is the thermal decomposition of solid calcium carbonate (manufacturing of cement). Refer to section 13.4 for the mathematical reason for the simplification of the equilibrium expression. Kc = [CO2] Kp = P CO2 Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Examples In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is CO(g) + H2O(g) CO2(g) + H2(g) What is the value of Kp at 700K if the partial pressures in an equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm of H2?
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Examples Consider the following unbalanced reaction
(NH4)2S(s) NH3(g) + H2S(g) An equilibrium mixture of this mixture at a certain temperature was found to have [NH3] = M and [H2S] = M. What is the value of the equilibrium constant (Kc) at this temperature?
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Examples Consider the following reaction and corresponding value of Kc: H2(g) + I2(s) HI(g) Kc = 6.2 x 102 at 25oC What is the value of Kp at this temperature?
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Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium 6/24/2018 Heterogeneous Equilibria Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant 2H2(g) + O2(g) 2H2O(g) (at 500 K) Kc = 4.2 x 10-48 A very large value for K means that the numerator is much larger than the denominator. A very small value for K means that the denominator is much larger than the numerator. 2HI(g) H2(g) + I2(g) 2H2O(g) 2H2(g) + O2(g) (at 500 K) Kc = 57.0 (at 500 K) Kc = 2.4 x 1047 Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant cC + dD aA + bB [A]ta[B]tb [C]tc[D]td Reaction quotient: Qc = The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant If Qc < Kc net reaction goes from left to right (reactants to products). net reaction goes from right to left (products to reactants). If Qc > Kc no net reaction occurs. If Qc = Kc Emphasizing the relative magnitudes of the numerator and denominator may help students understand it. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Finding Equilibrium concentrations from Initial Concentrations
Chapter 13: Chemical Equilibrium 6/24/2018 Finding Equilibrium concentrations from Initial Concentrations Steps to follow in calculating equilibirium concentrations from initial concentration Write a balance equation for the reaction Make an ICE (Initial, Change, Equilibrium) table, involves The initial concentrations The change in concentration on going to equilibrium, defined as x The equilibrium concentration Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x Calculate the equilibrium concentrations form the calculated value of x Check your answers Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense Quadratic equation ax bx + c = 0
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Finding Equilibrium concentrations from Initial Concentrations
For Homogeneous Mixture aA(g) bB(g) cC(g) Initial concentration (M) [A]initial [B]initial [C]initial Change (M) ax bx cx Equilibrium (M) [A]initial – ax [B]initial – bx [C]initial - cx [products] Kc = [reactants]
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Finding Equilibrium concentrations from Initial Concentrations
For Heterogenous Mixture aA(g) bB(g) cC(s) Initial concentration (M) [A]initial [B]initial …….. Change (M) ax bx …….. Equilibrium (M) [A]initial – ax [B]initial – bx ………. [products] Kc = [reactants]
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant At 700 K, mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI . Kc is 57.0 at 700 K. 2HI(g) H2(g) + I2(g) Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant Set up a table: 2HI(g) H2(g) + I2(g) I 0.250 C +x -2x E x x I = initial concentrations. C = change in concentrations (use “x” for the change along with the stoichiometric coefficients). E = equilibrium concentrations. Substitute values into the equilibrium expression: Kc = [H2][I2] [HI]2 x2 ( x)2 57.0 = Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium 6/24/2018 Using the Equilibrium Constant Solve for “x”: x2 ( x)2 57.0 = x = Determine the equilibrium concentrations: H2: M There are a number of different problems that can be worked in addition to this one. Using the quadratic formula, successive approximations, etc. A way to check the equilibrium concentrations is to plug them back into the equilibrium expression and see if you get the equilibrium constant (approximately). I2: M HI: (0.0262) = M Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Examples Consider the following reaction
I2(g) + Cl2(g) ICl(g) Kp = 81.9 (at 25oC) A reaction mixture at 25oC intially contains PI2 = atm, PCl2 = atm, and PICl = atm. Find the equilibrium pressure of I2, Cl2 and ICl at this temperature. In which direction does the reaction favored?
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Examples The value of Kc for the reaction is 3.0 x Determine the equilibrium concentration if the initial concentration of water is 8.75 M C(s) + H2O(g) CO(g) + H2(g)
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Examples Consider the following reaction N2O4(g) 2 NO2(g)
A reaction mixture at 100oC initially contains [NO2] = 0.100M. Find the equilibrium concentration of NO2 and N2O4 at this temperature
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Le Châtelier’s Principle
Chapter 13: Chemical Equilibrium 6/24/2018 Le Châtelier’s Principle Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. The concentration of reactants or products can be changed. The pressure and volume can be changed. The temperature can be changed. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Concentration 2NH3(g) N2(g) + 3H2(g) Copyright © 2008 Pearson Prentice Hall, Inc.
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Chapter 13: Chemical Equilibrium
6/24/2018 Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Concentration In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance. the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance. The iron(III)-thiocyanate complex equilibrium is a nice demonstration since it has nice color changes. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Concentration
Add reactant – denominator in Qc expression becomes larger Qc < Kc To return to equilibrium, Qc must be increases More product must be made => reaction shifts to the right
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Altering an Equilibrium Mixture: Concentration
Remove reactant – denominator in Qc expression becomes smaller Qc > Kc To return to equilibrium, Qc must be decreases Less product must be made => reaction shifts to the left
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Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Concentration 2NH3(g) N2(g) + 3H2(g) at 700 K, Kc = 0.291 An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by increasing the N2 concentration to 1.50 M. [N2][H2]3 [NH3]2 (1.50)(3.00)3 (1.98)2 Qc = = = < Kc Since Qc < Kc, more reactants will be consumed and the net reaction will be from left to right. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Examples The reaction of iron (III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: Fe2O3(s) CO(g) Fe(l) + 3CO2(g) Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by: a. adding Fe2O3 b. Removing CO2 c. Removing CO; also account for the change using the reaction quotient Qc
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Examples Consider the following reaction at equilibrium
CO(g) Cl2(g) COCl2(g) Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following reaction mixture a. COCl2 is added to the reaction mixture b. Cl2 is added to the reaction mixture c. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc
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Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Pressure and Volume 2NH3(g) N2(g) + 3H2(g) Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Pressure and Volume In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that an increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas. a decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Pressure and Volume
If reactant side has more moles of gas Denominator will be larger Qc < Kc To return to equilibrium, Qc must be increased Reaction shifts toward fewer moles of gas (to the product) If product side has more moles of gas Numerator will be larger Qc > Kc To return to equilibrium, Qc must be decreases Reaction shifts toward fewer moles of gas (to the reactant)
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Altering an Equilibrium Mixture: Pressure and Volume
Reaction involves no change in the number moles of gas No effect on composition of equilibrium mixture For heterogenous equilibrium mixture Effect of pressure changes on solids and liquids can be ignored Volume is nearly independent of pressure Change in pressure due to addition of inert gas No change in the molar concentration of reactants or products No effect on composition
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Examples Consider the following reaction at chemical equilibrium
2 KClO3(s) KCl(s) O2(g) a. What is the effect of decreasing the volume of the reaction mixture? b. Increasing the volume of the reaction mixture? c. Adding inert gas at constant volume?
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Examples Does the number moles of products increases, decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume? PCl5(g) PCl3(g) Cl2(g) CaO(s) CO2(g) CaCO3(s) 3 Fe(s) H2O(g) Fe3O4(s) H2(g)
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Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Temperature 2NH3(g) N2(g) + 3H2(g) DH° = kJ Note the size of Kc as the temperature increases. Mathematically, the magnitude of the denominator increases with respect to the numerator as the temperature increases. As the temperature increases, the equilibrium shifts from products to reactants. Copyright © 2008 Pearson Prentice Hall, Inc.
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Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium 6/24/2018 Altering an Equilibrium Mixture: Temperature In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that the equilibrium constant for an exothermic reaction (negative DH°) decreases as the temperature increases. Contains more reactant than product Kc decreases with increasing temperature the equilibrium constant for an endothermic reaction (positive DH°) increases as the temperature increases. Contains more product than reactant Kc increases with increasing temperature A aqueous solution of cobalt(II) nitrate in a solution of hydrochloric acid will change colors from blue to pink as the temperature changes (it’s a reversible endothermic reaction) and makes for a nice demonstration. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.
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Examples The following reaction is endothermic
CaCO3(s) CaO(s) CO2(g) a. What is the effect of increasing the temperature of the reaction mixture? b. Decreasing the temperature?
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Examples In the first step of Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 2 NH3(g) O2(g) NO(g) H2O(l) ΔHo = -905 kJ How does the temperature amount of NO vary with an increases in temperature?
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The Effect of a Catalyst on Equilibrium
Chapter 13: Chemical Equilibrium 6/24/2018 The Effect of a Catalyst on Equilibrium Copyright © 2008 Pearson Prentice Hall, Inc.
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The Effect of a Catalyst on Equilibrium
Catalyst increases the rate of a chemical reaction Provide a new, lower energy pathway Forward and reverse reactions pass through the same transition state Rate for forward and reverse reactions increase by the same factor Does not affect the composition of the equilibrium mixture Does not appear in the balance chemical equation Can influence choice of optimum condition for a reaction
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