Chemical Equilibrium Concept of Equilibrium Equilibrium Constant Equilibrium expressions Applications of equilibrium constants Le Chatelier s Principle
The Concept of Equilibrium The decomposition of N 2 O 4 is a reversible process, meaning the products of the reaction can react to form reactants. N 2 O 4 (g) 2NO 2 (g) The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same. rate forward = k f [N 2 O 4 ] and rate reverse = k r [NO 2 ] 2
The Concept of Equilibrium Starting with N 2 O 4
Starting with NO 2
Reaction Rates Some important things to remember about equilibrium are: Equilibrium is a dynamic state both the forward and reverse reactions continue to occur, although there is no net change in reactant and product concentration over time. At equilibrium, the rates of the forward and reverse reactions are equal. Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products.
The Equilibrium Constant N 2 O 4 (g) 2NO 2 (g) rate forward = rate reverse k f [N 2 O 4 ] eq = k r [NO 2 ] 2 eq The subscript eq denotes a concentration at equilibrium. Rearranging k k = [ ] [ ] 2 The ratio of two constants (k f /k r ) is also a constant: equilibrium expression equilibrium constant K = [ ] [ ] 2
The Equilibrium Constant Note the relationship between the equilibrium constant and the balanced chemical equation: K = [ ] [ ] 2 N 2 O 4 (g) 2NO 2 (g)
Equilibrium Expressions When a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant.
Practice-Eq. Expression Write the equilibrium expression for K c, for the following reactions: (a) 2O 3 (g) 3O 2 (g) (b) 2NO (g) + Cl 2 (g) 2NOCl (g) (c) Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq)
Practice Problems Determine the vale of the equilibrium constant for the following reaction: Given: N 2 (g) + O 2 (g) 2NO 2 (g) N 2 (g) + O 2 (g) 2NO(g) K c1 = 4.3 x 10 25 2NO(g) + O 2 (g) 2NO 2 (g) K c2 = 6.4 x 10 9
Practice-Eq. Expression In an analysis of the following reaction at 100 C, Br 2 (g) + Cl 2 (g) 2BrCl(g) The equilibrium concentrations were found to be [Br 2 ] = 2.3 x 10 3 M, [Cl 2 ] = 1.2 x 10 2 M, [BrCl] = 1.4 x 10 2 M. Write the equilibrium expression and calculate the equilibrium constant for this reaction at 100 C.
Practice-Eq. Expression In an analysis of the following reaction at 100 C, Br 2 (g) + Cl 2 (g) 2BrCl(g) The equilibrium concentrations were found to be [Br 2 ] = 2.3 x 10 3 M, [Cl 2 ] = 1.2 x 10 2 M, [BrCl] = 1.4 x 10 2 M. Write the equilibrium expression and calculate the equilibrium constant for this reaction at 100 C. (a) calculate K for the reverse reaction. (b) calculate the equilibrium constant for the reaction: ½ Br 2 (g) + ½ Cl 2 (g) BrCl(g)
Equilibrium Expressions When the species in a reversible chemical reaction are not all in the same phase, the equilibrium is heterogeneous. Only gaseous species and aqueous species appear in equilibrium expressions, pure solids and pure liquids do not. CO 2 (g) + C(s) 2CO(g) K = 2 [ ] [ ] 2Fe(s) + 3H 2 O(l) Fe 2 O 3 (s) + 2H 2 (g) K = [ ] 2
The Equilibrium Constant The equilibrium constant gives the extent a reaction will proceed at a particular temperature. Three outcomes are possible: 1) The reaction will go essentially to completion and the equilibrium mixture will consist predominately of products. Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2+ (g) Kc = 1.5 x 10 7 (at 25 C) Large K c, product favored
The Equilibrium Constant The equilibrium constant gives the extent a reaction will proceed at a particular temperature. Three outcomes are possible: 2) The reaction will not occur to any significant degree, and the equilibrium mixture will consist predominantly of reactant. N 2 (g) + O 2 (g) 2NO(g) K c = 4.3 x 10 25 (at 25 C) Small K c, reactant favored 3) The reaction will proceed a significant degree but will not go to completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.
The Equilibrium Constant The reaction quotient (Q c ) is a fraction with product concentrations in the numerator and reactant concentrations in the denominator. Each concentration is raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation. The law of mass action: aa + bb cc + dd Q c [ ] [ ] a [ ] [ ] d = = b K (at equilibrium) applies to not only elementary reactions, but also to more complex reactions.
The Equilibrium Constant The value of the reaction quotient, Q, changes as the reaction progresses N 2 O 4 (g) 2NO 2 (g)
The Equilibrium Constant At any point during the progress of a reaction: aa + bb cc + dd c [ ] [ ] a [ ] [ ] Q = d b
Practice-Reaction Quotient CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) Write the reaction quotient for the following reaction:
Relationship between Q and K The equilibrium expression may be used to predict the direction of a reaction and to calculate equilibrium concentrations. Predictions are made based on comparisons between Q c and K c. There are three possibilities: 1) Q < K The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds in the forward direction. 2) Q = K The initial concentrations are equilibrium concentrations. The system is at equilibrium. 3) Q > K The ratio of initial concentrations of products to reactants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.
Practice- using K to solve problems The equilibrium constant, K c, for the formation of nitrosyl chloride from nitric oxide and chlorine, 2NO(g) + Cl 2 (g) 2NOCl(g) Is 6.5 x 10 4 at 35 C. In which direction will the reaction proceed to reach equilibrium if the starting concentrations of NO, Cl 2, and NOCl are 1.1 x 10 3 M, 3.5 x 10 4 M, and 1.9 M respectively?
Practice- Using K to solve problems Equilibrium concentrations can be calculated from initial concentrations if the equilibrium constant is known. K c = 24.0 (200 C) cis-stilbene trans-stilbene cis-stilbene trans-stilbene Initial concentration (M) 0.850 0
Practice- Using K to solve problems Consider the following reaction: H 2 (g) + F 2 (g) 2HF(g) 3.000 mol H 2 and 6.000 mol F 2 are mixed in a 3.000 L flask. If the equilibrium constant is 115 at room temperature, calculate the equilibrium concentrations of all three species.
Practice- Using K to solve problems K c for the reaction of hydrogen and iodine to produce hydrogen iodide, H 2 (g) + I 2 (g) 2HI(g) is 54.3 at 430 C. What will the concentrations be at equilibrium if the initial concentrations are: [H 2 ] = 0.00623 M, [I 2 ] = 0.00414 M, [HI] = 0.0424 M
Practice- Using K to solve problems Consider the following reaction: N 2 O 4 (g) 2NO 2 (g) K p = 0.25 A flask containing only N 2 O 4 (g) at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures.
Factors That Affect Chemical Equilibrium Le Châtelier s principle states that when a stress is applied to a system at equilibrium, the system, will respond by shifting in the direction that minimizes the effect of the stress. Stress refers to any of the following: The addition of a reactant or product The removal of a reactant or product A change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products A change in temperature
Changes in Concentration Consider the Haber process at 700 K: N 2 (g) + 3H 2 (g) 2NH 3 (g) At equilibrium: [N 2 ] = 2.05 M [H 2 ] = 1.56 M [NH 3 ] = 1.52 M K [ ] [ ][ ] ( 152. ) ( 205. )( 156. ) 2 2 = = = 2 3 3 0297. Applying stress by the addition of N 2 to give the following concentrations: [N 2 ] = 3.51 M [H 2 ] = 1.56 M [NH 3 ] = 1.52 M Q [ ] [ ][ ] ( 152. ) ( 351. )( 156. ) 2 2 = = = 0173. K 2 3 3 The reaction shifts to the right. N 2 (g) + 3H 2 (g) 2NH 3 (g)
Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)
Changes in Concentration Addition of a reactant or removal of a product will cause an equilibrium to shift to the right. Addition of a product or removal of a reactant will cause an equilibrium to shift to the left.
Changes in Volume and Pressure When volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas. N 2 O 4 (g) 2NO 2 (g) Equilibrium mixture: [N 2 O 4 ] = 0.643 M [NO 2 ] = 0.0547 M The reaction shifts to the left. N 2 O 4 (g) 2NO 2 (g) Volume decreases by half, concentrations are initially doubled: [N 2 O 4 ] = 1.286 M [NO 2 ] = 0.1094 M K [ ] [ ] ( 00547. ) ( 0643. ) 2 2 = = = 465. 10 3 Q [ ] [ ] ( 01094. ) ( 1286. ) 2 2 3 = = = 931. 10 K
Changes in Temperature Changes in volume and concentration do not change the value of the equilibrium constant. A change in temperature can alter the value of the equilibrium constant. Heat + N 2 O 4 (g) 2NO 2 (g) ΔH = 58.0 kj/mol Because the processes is endothermic, adding heat shifts the equilibrium toward products
Changes in Temperature For any endothermic reaction, heat is a reactant: heat + reactants products ΔH > 0 kj/mol Adding heat shifts the reaction towards products, Kc increases Removing heat shifts the reaction towards reactants, Kc decreases. For any exothermic reaction, heat is a product: reactants products + heat ΔH < 0 kj/mol Adding heat shifts the reaction towards reactants, Kc decreases Removing heat shifts the reaction towards products, Kc increases
Key Points The Concept of Equilibrium The Equilibrium Constant Calculating Equilibrium Constants Magnitude of the Equilibrium Constant Equilibrium Expressions Heterogeneous Equilibria Manipulating Equilibrium Expressions Using Equilibrium Expressions to Solve Problems Predicting the Direction of a Reaction Calculating Equilibrium Concentrations Factors that Affect Chemical Equilibrium Addition or Removal of a Substance Changes in Volume and Pressure Changes in Temperature