What does the magnitude of the equilibrium constant tell us? N2(g) + O2(g) N2 O2(g) N2(g) + 3H2(g) 2NH3(g) In Short 1
D. Altering Chemical Equations and the Effect on the Equilibrium Constant What is the expression for the equilibrium constant when we do the following? 1. Reverse a reaction? 2. Multiply or divide a balanced equation by a number? 3. Combine multiple reactions? 1. What happens to K when we reverse a rxn.? - when we reverse an equation, the equilibrium constant for the new reaction is equal to the inverse of the original K N2(g) + 3H2(g) 2NH3(g) K= [NH 3] 2 [N 2 ][H 2 ] 3 2NH3(g) N2(g) + 3H2(g) 2
2. What happens to K when we multiply a balanced equation by a factor n? - when an equation is multiplied by a factor n, the equilibrium constant for the new reaction is equal to the original K raised to the nth power N2(g) + 3H2(g) 2NH3(g) K= [NH 3] 2 [N 2 ][H 2 ] 3 3
3. How do we calculate the equilibrium constant for a reaction that is the combination of more than one reaction? - when we combine multiple equations, the equilibrium constant for the new reaction is equal to all of the equilibrium constants multiplied together A. 2NOBr(g) 2NO(g) + Br2(g) KA B. Br2(g) + Cl2(g) 2BrCl(g) KB 2NOBr(g) + Cl2(g) 2BrCl(g) + 2NO(g) (overall) A. K A = [NO]2 [Br 2 ] [NOBr] 2 B. K B = [BrCl]2 [Br 2 ][Cl 2 ] Overall 4
Example Problem Given the following information, HF(aq) +H2O(l) H3O + (aq) + F - (aq) K1=6.6x10-4 H2C2O4(aq) + H2O(l) H3O + (aq) + HC2O4 - (aq) K2=5.6x10-2 HC2O4 - (aq) + H2O(l) H3O + (aq) + C2O4 2- (aq) K3=5.4x10-5 What is the value of the equilibrium constant for the following reaction? 2HF(aq) + C2O4 2- (aq) 2F - (aq) + H2C2O4(aq) 5
E. Altering Equilibrium Conditions and the Effect on the Movement of Reactants and Products Le Chatelier s Principle Le Chatelier s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change Change and Chemical Reactions How can we use Le Chatelier s Principle to help us qualitatively predict the impact of the following changes? 1. Changes in product or reactant concentrations 2. Effects of volume and pressure changes 3. Effect of Temperature**** ****actually changes the equilibrium constant to favor either more reactants or products, but we can predict the direction 6
1. Changes in product or reactant concentrations O2(g) O2(aq) Hb(aq) + O2(aq) HbO2(aq) Hb(aq) + O2(g) HbO2(aq) What happens when you climb a mountain, and why is this potentially harmful to your health? Why do climbers stop at certain points before continuing their climbs? Summary 2HF(aq) + C2O4 2- (aq) 2F - (aq) + H2C2O4(aq) K = [F ] 2 [H 2 C 2 O 4 ] [HF] 2 [C 2 O 4 2 ] Removal of a reactant Removal of a product Addition of a reactant Addition of a product 7
What about the addition/removal of a solid or liquid reactant or product? CaCO3(s) CaO(s) + CO2(g) 8
2. Effects of volume and pressure changes How do pressure changes impact the equilibrium constant? A. Add or remove a gaseous reactant or product 2NOBr(g) + Cl2(g) 2BrCl(g) + 2NO(g) B. Add an inert gas (eg argon) (keep volume constant) 2NOBr(g) + Cl2(g) 2BrCl(g) + 2NO(g) K = [BrCl]2 [NO] 2 [NOBr] 2 [Cl 2 ] K= (P BrCl) 2 (P NO ) 2 (P NOBr ) 2 (P Cl2 ) 9
C. Changing the pressure by changing the volume of the system 2NOBr(g) + Cl2(g) 2BrCl(g) + 2NO(g) K = [BrCl]2 [NO] 2 [NOBr] 2 [Cl 2 ] = ( n BrCl V )2 ( n NO V )2 ( n NOBr V )2 ( n Cl 2 V ) or K = 1 V 4(n BrCl) 2 (n NO ) 2 1 = 1 (n BrCl ) 2 (n NO ) 2 V 3(n NOBr) 2 (n Cl 2 ) V (n NOBr ) 2 (n Cl 2 ) K= (P BrCl) 2 (P NO ) 2 (P NOBr ) 2 (P Cl2 ) Vi = 10.0 (equilibrium established) Vchange = 1.00 L (shift?) Vi = 1.00 (equilibrium established) Vchange = 5.00 L (shift?) Overall 10
3. Effect of Temperature***** Temperature or energy can be thought of as a reactant or product Endothermic: Reactants Products Exothermic: Reactants Products - all equilibrium constants change with Temperature N2(g) + O2(g) 2NO(g) Hrxn = 180.5 kj Temperature (K) 298 900 Equilibrium Constant In General Endothermic Rxn: Increasing T results in a new equilibrium Decreasing T results in a new equilibrium Exothermic Rxn: Increasing T results in a new equilibrium Decreasing T results in a new equilibrium 11
Example Problem Consider the following equilibrium, for which H < 0: 2SO2(g) + O2(g) 2SO3(g) How will each of the following changes affect an equilibrium mixture of the three gases? (a) O2(g) is added to the system; (b) the reaction mixture is cooled; (c) the volume of the vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) SO3(g) is removed from the system. 12
Example Problem For the reaction written below, ΔH = 2816 kj. 6CO2 (g) + 6H2O(l) C6H12O6(s) + 6O2(g) How will the following affect an equilibrium mixture of the reactants and products? (a) increasing PCO2, (b) increasing temperature, (c) removing CO2, (d) decreasing the total pressure, (e) removing some, but not all, of the C6H12O6, (f) adding a catalyst. 13
II. Using the Equilibrium Constant A. Predicting the Direction of a Reaction Two Types of Problems 1. Given equilibrium conditions calculate the equilibrium constant 2. Given the equilibrium constant and certain starting conditions, calculate the equilibrium concentrations or partial pressures of reactants and products Tool - Reaction Quotient, Q (crystal ball) j A + k B C + m D K l m [ C] [ D] j k [ A] [ B] 14
Tool Cont d If Q = K If Q < K If Q > K 15
B. Calculating Equilibrium Constants and Concentrations Procedure for solving equilibrium problems STEP 1 STEP 2 STEP 3 STEP 4 STEP 5 Write the balanced equation for the reaction Write the equilibrium expression List the initial concentrations Determine direction of the shift to equilibrium (if needed calculate Q) Define the change needed to reach equilibrium Define the equilibrium concentrations by applying the change to the initial concentrations STEP 6 STEP 7 Substitute the equilibrium concentrations determined in STEP 5 into the equilibrium expression. Solve for unknown. Check your result by comparison with any given numbers and check that any assumptions you made are valid. 16
Example Enough ammonia is dissolved in 5.00 liters of water at 25 C to produce a solution that is 0.0124 M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of OH - is 4.64x10-4 M. Calculate the equilibrium constant (Kc) at 25 C for this reaction. 17
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Example For the following reaction, Kc = 2.00 at 1000 C. 2CO2F2(g) CO2(g) + CF4(g) If a 5.00 L mixture contains 0.0290 M CO2F2, 0.0524 M CO2, and 0.0148 M CF4 at a temperature of 1000 C, what is the concentration of each reactant and product at equilibrium? 19
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Example A mixture of 1.00 mol NaHCO3(s) and 1.00 mol Na2CO3(s) is introduced into a 2.50 L flask in which the partial pressure of CO2(g) is 2.10 atm and that of H2O(g) is 0.941 atm. When equilibrium is established at 100 C, what will the total pressure be? 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) Kp = 0.23 at 100 C 21
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Example Using the information below, calculate the fraction of CH3COOH that reacts. CH3COOH(aq) + OH - (aq) H2O(l) + CH3COO - (aq) [CH3COOH]i = 0.100 M [OH - ]i = 0.100 M K = 1.8x10 9 23
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