Chemical Equilibrium
2/26/2013
Unit 14: Chemical Equilibrium • Topics covered
• Reading Quiz
– The concept of equilibrium – The equilibrium constant – Heterogeneous equilibrium – Applications of equilibrium constants – Le Châtelier’s principle
– Read chapter 15 – Know definitions for all words in bold – Sample exercises 15.1, 15.2, 15.5
• The condition in which the concentrations of all reactants and products in a closed system cease to change with time is called chemical equilibrium • Chemical equilibrium occurs when opposing reactions are proceeding and equal rates, i.e. the rate of the forward reaction is equal to that of the reverse reaction – N2O4(g) 2NO2(g)
Chemical Equilibrium • As N2O4 is consumed, the rate of the forward reaction decreases • As NO2 is formed, the rate of the reverse reaction increases • When the two rates become equal, an equilibrium state is attained and there are no further changes in concentrations
The Equilibrium Constant • Given the following general equation: – aA + bB cC + dD
• The equilibrium condition is expressed by the equation: [C]c [D]d
K eq =
Chemical Equilibrium
[A]a [B]b
• We call this relationship the equilibrium expression, the constant Keq is called the equilibrium constant (no units)
Chemical Equilibrium • Regardless of starting concentrations, the equilibrium condition can be reached from either direction
The Equilibrium Constant K eq =
[C]c [D]d [A]a [B]b
• The equilibrium expression depends only on the stoichiometry of the reaction, not on its mechanism • The value of the equilibrium constant varies only with temperature • Concentration is measured in molarity or partial pressure for gaseous reactions
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Sample Problem • Write the equilibrium expression for the following reactions: – 2O3(g) 3O2(g) – 2NO(g) + Cl2(g) 2NOCl(g) – Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
Magnitude of Equilibrium Constants
• The larger the value of the equilibrium constant Keq, the farther the reaction proceeds to the right before reaching the equilibrium state
The Direction of the Chemical Equation and Keq • The equilibrium expression for a reaction in one direction is the reciprocal of the one written for the reaction in the reverse direction • Consequently, the numerical value of Keq for the forward reaction is the reciprocal of Keq for the reverse reaction
Magnitude of Equilibrium Constants • The magnitude of the equilibrium constant provides us with very important information about the composition of an equilibrium mixture • If the product is large, we say that the equilibrium lies to the right or to the product side • A very small equilibrium constant indicates that the equilibrium mixture contains mostly reactants or lies to the left
Sample Problems • Consider the following reaction: N2 + O2 2NO. At 25C, Keq = 1 x 10-30. Describe the feasibility of this reaction at room temperature. • The equilibrium constant for a reaction changes with temperature as follows: Keq = 794 at 298 K and Keq = 54 at 700 K. Is the formation of product favored at higher or lower temperatures?
Sample Problem • For the formation of NH3 from N2 and H2, Keq = 4.34 x 10-3 at 300C. What is the value of Keq for the reverse reaction?
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Other Manipulations of Keq • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number • The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps
Heterogeneous Equilibria • Equilibria that involve all substances in the same phase are called homogeneous equilibria • When substances in equilibrium are in different phases it is called heterogeneous equilibria
Sample Problem • Given the following: – HF H+ + F– H2C2O4 2H+ + C2O42-
Keq = 6.8 x 10-4 Keq = 3.8 x 10-6
• Determine the value of the equilibrium constant for the reaction: – 2HF + C2O42- 2F- + H2C2O4
Heterogeneous Equilibria • Consider the following reaction: – PbCl2(s) Pb2+(aq) + 2Cl-(aq)
• This presents a problem when writing the equilibrium expression, how do we express the concentration of a solid? • To solve this problem, pure solids, pure liquids, and solvents are not included in the equilibrium expression:
K eq = [Pb2+ ][Cl- ]2
Sample Problem • Write the equilibrium expression for the following reactions: – CO2(g) + H2(g) CO(g) + H2O(l) – SnO2(s) + 2CO(g) Sn(s) + 2CO2(g) – Sn(s) + 2H+(aq) Sn2+(aq) + H2(g)
KC and KP • Until now, we have used partial pressure to write equilibrium expressions for gases and molarity to write equilibrium expression for aqueous species • When Keq is calculated using molarity, it is often given the symbol KC • When Keq is calculated using partial pressure, it is given the symbol KP
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KC and KP • Occasionally, KC must be calculated for gaseous reactions • One way to do this is convert partial pressure of each species into molarity using the ideal gas law: – [A] = (PA/RT)
• Alternately, we can convert directly between the two constants using the following: – KP = KC(0.0821T)∆n – Temperature must be in K and pressure in atm – ∆n = moles of gas products - moles of gas reactants
Sample Problem
Sample Problem • Rewrite the following equilibrium expression using only units of concentration in terms of moles and temperature:
K eq =
[Sn 2+ ]PH 2 [H + ]2
Calculating Equilibrium Constants
• A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate KP and KC for this reaction: N2(g) + 3H2(g) 2NH3(g)
• We often don’t know the equilibrium concentrations of all chemical species in an equilibrium • If we know the equilibrium concentration of at least one species, however, we can generally use the stoichiometry of the reaction to deduce Keq • To do this, we can use the ICE box technique (Initial, Change, Equilibrium)
Sample Problem
Sample Problem
• Enough ammonia is dissolved in 5.00 L of water at 25C 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 4.64 x 10-4 M. Calculate Keq for the reaction NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
• Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g) 2SO2(g) + O2(g). Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate KP and KC at this temperature.
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Calculating Equilibrium Concentrations • Chemist frequently need to calculate the amounts of reactants and products present at equilibrium • If we know some of the equilibrium concentrations and Keq, we can solve for the remaining reactant concentrations
Sample Problem • For the Haber process, N2(g) + 3H2(g) 2NH3(g), KP = 1.45 x 10-5 at 500C. In an equilibrium mixture of the three gases, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of ammonia in the mixture?
Sample Problem
Calculating Equilibrium Concentrations
• At 500 K the reaction PCl5(g) PCl3(g) + Cl2(g) has KP = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2?
• In many situations, we know only the starting concentrations and the equilibrium constant • In these situations, we must solve for the equilibrium concentrations using the ICE box method
Sample Problem
Sample Problem
• A 1.000 L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448C. The value of KP for the reaction H2(g) + I2(g) 2HI(g) at 448C is 50.5. What are the partial pressures of all species in the flask at equilibrium?
• For the equilibrium PCl5(g) PCl3(g) + Cl2(g), KP = 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl 5 at an initial pressure of 1.66 atm. What are the partial pressures of all species at equilibrium?
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Predicting the Direction of a Reaction • When we substitute reactant and product concentrations of a system that is not necessarily at equilibrium into an equilibrium expression, the result is known as the reaction quotient (Q) – If Q = Keq, the sytem is at equilibrium – If Q < Keq, the system will proceed to the right to form more products – If Q > Keq, the system will proceed to the left to form more reactants
Le Châtelier’s Principle
Sample Problem • At 448°C Keq for the following reaction is 51. H2(g) + I2(g) 2HI(g) • Predict how the reaction will proceed if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2, and 3.0 x 10-2 mol if I2 in a 2.00 L container.
Change in Reactant or Product Concentrations
• If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance • This is known as Le Châtelier’s Principle
• If a chemical system is at equilibrium and we add a substance (either a reactant or product), the reaction will shift so as to reestablish equilibrium by consuming part of the added substance • Conversely, removing a substance will cause the reaction to move in the direction that forms more of that substance
Change in Reactant or Product Concentrations
Change in Reactant or Product Concentrations
• Consider the following system at equilibrium: N2(g) + 3H2(g) 2NH3(g) – Adding H2 would shift the equilibrium to reduce the newly increased concentration of H2 (i.e. to the right) – Removing N2 would cause the reaction to shift to produce more N2 (i.e. to the left) – If NH3 is added, which direction would the equilibrium shift?
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The Effects of Volume and Pressure Changes • Reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas • Conversely, increasing the volume causes a shift in the direction that produces more moles of gas
The Effects of Volume and Pressure Changes
• When the pressure is increased by decreasing the volume, the mixture is no longer at equilibrium • Reaction occurs from left to right, decreasing the total number of gaseous molecules until equilibrium is reestablished
Sample Problem • Consider the following equilibrium: – N2O4(g) 2NO2(g) ∆H = 58.0 kJ
• In what direction will the equilibrium shift when each of the following changes is made: (a) add N2O4; (b) remove NO2; (c) increase the total pressure by adding N2(g); (d) increase the volume; (e) decrease the temperature?
The Effects of Volume and Pressure Changes • Consider the equilibrium N2O4(g) 2NO2(g) – What happens as the total pressure increases by decreasing the volume of the container? – The equilibrium shifts to the left to reduce the pressure
• Keep in mind that changing either concentration or pressure does not change the value of Keq • Changes in pressure can change temperature, however, which may change the value of Keq
The Effect of Temperature Change • When the temperature is increased or decreased, it is as if we have added or removed a reactant or product • The equilibrium shifts in the direction that consumes the excess heat or replaces lost heat • Keq is affected by temperature in the following way: – Endothermic reaction: Increasing T results in an increase in Keq – Exothermic reaction: Increasing T results in a decrease in Keq
Sample Problems • For the reaction PCl5(g) PCl3(g) + Cl2(g), where ∆H = 87.9 kJ, in what direction will equilibrium shift when (a) Cl2 is removed; (b) the temperature is decreased; (c) the volume of the reaction system is increased; (d) PCl3 is added? • If the temperature is increased for the above equilibrium, what happens to the value of K P?
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The Effect of Catalysts • A catalyst increases the rate at which equilibrium is reached, but it does not change the composition of the equilibrium mixture • The value of Keq is not affected by the presence of a catalyst
Sample Problem • Use standard free energies of formation to calculate Keq at 25C for the following reaction: N2(g) + 3H2(g) 2NH3(g)
Free Energy and the Equilibrium Constant • At equilibrium, the change in standard free energy of a reaction is related to the equilibrium constant according to the following:
• The more negative ∆G is, the larger Keq is, and vice versa
Sample Problem • Calculate ∆G and Keq at 298 K for the following reaction: H2(g) + Br2(l) 2HBr(g)
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Unit 14: Chemical Equilibrium • Topics covered
• Reading Quiz
– The concept of equilibrium – The equilibrium constant – Heterogeneous equilibrium – Applications of equilibrium constants – Le Châtelier’s principle
– Read chapter 15 – Know definitions for all words in bold – Sample exercises 15.1, 15.2, 15.5
• The condition in which the concentrations of all reactants and products in a closed system cease to change with time is called chemical equilibrium • Chemical equilibrium occurs when opposing reactions are proceeding and equal rates, i.e. the rate of the forward reaction is equal to that of the reverse reaction – N2O4(g) 2NO2(g)
Chemical Equilibrium • As N2O4 is consumed, the rate of the forward reaction decreases • As NO2 is formed, the rate of the reverse reaction increases • When the two rates become equal, an equilibrium state is attained and there are no further changes in concentrations
The Equilibrium Constant • Given the following general equation: – aA + bB cC + dD
• The equilibrium condition is expressed by the equation: [C]c [D]d
K eq =
Chemical Equilibrium
[A]a [B]b
• We call this relationship the equilibrium expression, the constant Keq is called the equilibrium constant (no units)
Chemical Equilibrium • Regardless of starting concentrations, the equilibrium condition can be reached from either direction
The Equilibrium Constant K eq =
[C]c [D]d [A]a [B]b
• The equilibrium expression depends only on the stoichiometry of the reaction, not on its mechanism • The value of the equilibrium constant varies only with temperature • Concentration is measured in molarity or partial pressure for gaseous reactions
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2/26/2013
Sample Problem • Write the equilibrium expression for the following reactions: – 2O3(g) 3O2(g) – 2NO(g) + Cl2(g) 2NOCl(g) – Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
Magnitude of Equilibrium Constants
• The larger the value of the equilibrium constant Keq, the farther the reaction proceeds to the right before reaching the equilibrium state
The Direction of the Chemical Equation and Keq • The equilibrium expression for a reaction in one direction is the reciprocal of the one written for the reaction in the reverse direction • Consequently, the numerical value of Keq for the forward reaction is the reciprocal of Keq for the reverse reaction
Magnitude of Equilibrium Constants • The magnitude of the equilibrium constant provides us with very important information about the composition of an equilibrium mixture • If the product is large, we say that the equilibrium lies to the right or to the product side • A very small equilibrium constant indicates that the equilibrium mixture contains mostly reactants or lies to the left
Sample Problems • Consider the following reaction: N2 + O2 2NO. At 25C, Keq = 1 x 10-30. Describe the feasibility of this reaction at room temperature. • The equilibrium constant for a reaction changes with temperature as follows: Keq = 794 at 298 K and Keq = 54 at 700 K. Is the formation of product favored at higher or lower temperatures?
Sample Problem • For the formation of NH3 from N2 and H2, Keq = 4.34 x 10-3 at 300C. What is the value of Keq for the reverse reaction?
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Other Manipulations of Keq • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number • The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps
Heterogeneous Equilibria • Equilibria that involve all substances in the same phase are called homogeneous equilibria • When substances in equilibrium are in different phases it is called heterogeneous equilibria
Sample Problem • Given the following: – HF H+ + F– H2C2O4 2H+ + C2O42-
Keq = 6.8 x 10-4 Keq = 3.8 x 10-6
• Determine the value of the equilibrium constant for the reaction: – 2HF + C2O42- 2F- + H2C2O4
Heterogeneous Equilibria • Consider the following reaction: – PbCl2(s) Pb2+(aq) + 2Cl-(aq)
• This presents a problem when writing the equilibrium expression, how do we express the concentration of a solid? • To solve this problem, pure solids, pure liquids, and solvents are not included in the equilibrium expression:
K eq = [Pb2+ ][Cl- ]2
Sample Problem • Write the equilibrium expression for the following reactions: – CO2(g) + H2(g) CO(g) + H2O(l) – SnO2(s) + 2CO(g) Sn(s) + 2CO2(g) – Sn(s) + 2H+(aq) Sn2+(aq) + H2(g)
KC and KP • Until now, we have used partial pressure to write equilibrium expressions for gases and molarity to write equilibrium expression for aqueous species • When Keq is calculated using molarity, it is often given the symbol KC • When Keq is calculated using partial pressure, it is given the symbol KP
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KC and KP • Occasionally, KC must be calculated for gaseous reactions • One way to do this is convert partial pressure of each species into molarity using the ideal gas law: – [A] = (PA/RT)
• Alternately, we can convert directly between the two constants using the following: – KP = KC(0.0821T)∆n – Temperature must be in K and pressure in atm – ∆n = moles of gas products - moles of gas reactants
Sample Problem
Sample Problem • Rewrite the following equilibrium expression using only units of concentration in terms of moles and temperature:
K eq =
[Sn 2+ ]PH 2 [H + ]2
Calculating Equilibrium Constants
• A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate KP and KC for this reaction: N2(g) + 3H2(g) 2NH3(g)
• We often don’t know the equilibrium concentrations of all chemical species in an equilibrium • If we know the equilibrium concentration of at least one species, however, we can generally use the stoichiometry of the reaction to deduce Keq • To do this, we can use the ICE box technique (Initial, Change, Equilibrium)
Sample Problem
Sample Problem
• Enough ammonia is dissolved in 5.00 L of water at 25C 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 4.64 x 10-4 M. Calculate Keq for the reaction NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
• Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g) 2SO2(g) + O2(g). Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate KP and KC at this temperature.
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Calculating Equilibrium Concentrations • Chemist frequently need to calculate the amounts of reactants and products present at equilibrium • If we know some of the equilibrium concentrations and Keq, we can solve for the remaining reactant concentrations
Sample Problem • For the Haber process, N2(g) + 3H2(g) 2NH3(g), KP = 1.45 x 10-5 at 500C. In an equilibrium mixture of the three gases, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of ammonia in the mixture?
Sample Problem
Calculating Equilibrium Concentrations
• At 500 K the reaction PCl5(g) PCl3(g) + Cl2(g) has KP = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2?
• In many situations, we know only the starting concentrations and the equilibrium constant • In these situations, we must solve for the equilibrium concentrations using the ICE box method
Sample Problem
Sample Problem
• A 1.000 L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448C. The value of KP for the reaction H2(g) + I2(g) 2HI(g) at 448C is 50.5. What are the partial pressures of all species in the flask at equilibrium?
• For the equilibrium PCl5(g) PCl3(g) + Cl2(g), KP = 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl 5 at an initial pressure of 1.66 atm. What are the partial pressures of all species at equilibrium?
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Predicting the Direction of a Reaction • When we substitute reactant and product concentrations of a system that is not necessarily at equilibrium into an equilibrium expression, the result is known as the reaction quotient (Q) – If Q = Keq, the sytem is at equilibrium – If Q < Keq, the system will proceed to the right to form more products – If Q > Keq, the system will proceed to the left to form more reactants
Le Châtelier’s Principle
Sample Problem • At 448°C Keq for the following reaction is 51. H2(g) + I2(g) 2HI(g) • Predict how the reaction will proceed if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2, and 3.0 x 10-2 mol if I2 in a 2.00 L container.
Change in Reactant or Product Concentrations
• If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance • This is known as Le Châtelier’s Principle
• If a chemical system is at equilibrium and we add a substance (either a reactant or product), the reaction will shift so as to reestablish equilibrium by consuming part of the added substance • Conversely, removing a substance will cause the reaction to move in the direction that forms more of that substance
Change in Reactant or Product Concentrations
Change in Reactant or Product Concentrations
• Consider the following system at equilibrium: N2(g) + 3H2(g) 2NH3(g) – Adding H2 would shift the equilibrium to reduce the newly increased concentration of H2 (i.e. to the right) – Removing N2 would cause the reaction to shift to produce more N2 (i.e. to the left) – If NH3 is added, which direction would the equilibrium shift?
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The Effects of Volume and Pressure Changes • Reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas • Conversely, increasing the volume causes a shift in the direction that produces more moles of gas
The Effects of Volume and Pressure Changes
• When the pressure is increased by decreasing the volume, the mixture is no longer at equilibrium • Reaction occurs from left to right, decreasing the total number of gaseous molecules until equilibrium is reestablished
Sample Problem • Consider the following equilibrium: – N2O4(g) 2NO2(g) ∆H = 58.0 kJ
• In what direction will the equilibrium shift when each of the following changes is made: (a) add N2O4; (b) remove NO2; (c) increase the total pressure by adding N2(g); (d) increase the volume; (e) decrease the temperature?
The Effects of Volume and Pressure Changes • Consider the equilibrium N2O4(g) 2NO2(g) – What happens as the total pressure increases by decreasing the volume of the container? – The equilibrium shifts to the left to reduce the pressure
• Keep in mind that changing either concentration or pressure does not change the value of Keq • Changes in pressure can change temperature, however, which may change the value of Keq
The Effect of Temperature Change • When the temperature is increased or decreased, it is as if we have added or removed a reactant or product • The equilibrium shifts in the direction that consumes the excess heat or replaces lost heat • Keq is affected by temperature in the following way: – Endothermic reaction: Increasing T results in an increase in Keq – Exothermic reaction: Increasing T results in a decrease in Keq
Sample Problems • For the reaction PCl5(g) PCl3(g) + Cl2(g), where ∆H = 87.9 kJ, in what direction will equilibrium shift when (a) Cl2 is removed; (b) the temperature is decreased; (c) the volume of the reaction system is increased; (d) PCl3 is added? • If the temperature is increased for the above equilibrium, what happens to the value of K P?
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The Effect of Catalysts • A catalyst increases the rate at which equilibrium is reached, but it does not change the composition of the equilibrium mixture • The value of Keq is not affected by the presence of a catalyst
Sample Problem • Use standard free energies of formation to calculate Keq at 25C for the following reaction: N2(g) + 3H2(g) 2NH3(g)
Free Energy and the Equilibrium Constant • At equilibrium, the change in standard free energy of a reaction is related to the equilibrium constant according to the following:
• The more negative ∆G is, the larger Keq is, and vice versa
Sample Problem • Calculate ∆G and Keq at 298 K for the following reaction: H2(g) + Br2(l) 2HBr(g)
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