Chapter 15 Lecture- Chemical Equilibrium
Upcoming SlideShare
Loading in...5
×
 

Chapter 15 Lecture- Chemical Equilibrium

on

  • 31,506 views

Chapter 15 Lecture for AP Chemistry on Chemical Equilibrium

Chapter 15 Lecture for AP Chemistry on Chemical Equilibrium

Statistics

Views

Total Views
31,506
Views on SlideShare
30,318
Embed Views
1,188

Actions

Likes
3
Downloads
513
Comments
0

7 Embeds 1,188

http://msjkim.weebly.com 887
http://james1971midgley.edublogs.org 180
http://www.slideshare.net 78
http://www.weebly.com 27
http://msjkim.weebly.com. 11
http://study.myllps.com 4
http://webcache.googleusercontent.com 1
More...

Accessibility

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Chapter 15 Lecture- Chemical Equilibrium Chapter 15 Lecture- Chemical Equilibrium Presentation Transcript

    • Chapter 15 - Chemical Equilibrium Sections 15.1 - 15.4 Equilibrium & the Equilibrium Constant (K)
    • The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
    • The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring.
    • The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate.
    • A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.
    • Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation N2O4 (g) 2 NO2 (g)
    • N2(g) + 3H2(g) ⇌ 2NH3(g) 1M N2 + 3M H2 2M NH3
    • The Equilibrium Constant
    • The Equilibrium Constant • Forward reaction: N2O4 (g) → 2 NO2 (g) ✤Rate law: Rate = kf [N2O4]
    • The Equilibrium Constant • Forward reaction: N2O4 (g) → 2 NO2 (g) ✤Rate law: Rate = kf [N2O4] • Reverse reaction: 2 NO2 (g) → N2O4 (g) ✤Rate law: Rate = kr [NO2]2
    • The Equilibrium Constant • Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2
    • The Equilibrium Constant • Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 • Rewriting this, it becomes kf [NO2]2 = kr [N2O4]
    • The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes kf [NO2]2 Keq = = kr [N2O4]
    • For the gas-phase reaction A ⇌ B the forward reaction rate is 3.0 × 10−4 s−1 and the reverse reaction rate is 1.5 × 10−2 s−1. What is the value of the equilibrium constant, Keq? 1. 0.02 2. 50 3. 0.0004 4. 2500
    • Correct Answer: 1. 0.02 2. 50 3. 0.0004 4. 2500
    • N2O4 (g) → 2 NO2 (g) Eq 15.1 kf [NO2]2 Eq 15.5 Keq = k = [N O ] r 2 4 1. moles of reactant always equals moles of product. 2. mass of reactant always equals mass of product. 3. the rates of opposing reactions are equal. 4. cannot be determined without temperature information.
    • N2O4 (g) → 2 NO2 (g) Eq 15.1 kf [NO2]2 Eq 15.5 Keq = k = [N O ] r 2 4 1. moles of reactant always equals moles of product. 2. mass of reactant always equals mass of product. 3. the rates of opposing reactions are equal. 4. cannot be determined without temperature information.
    • N2O4 (g) → 2 NO2 (g) Eq 15.1 kf [NO2]2 Keq = k = [N O ] Eq 15.5 r 2 4 b) the constant will be______ than 1
    • N2O4 (g) → 2 NO2 (g) Eq 15.1 kf [NO2]2 Keq = k = [N O ] Eq 15.1 r 2 4 b) the constant will be greater than 1
    • the concentrations of reactants and products no longer change.
    • The Equilibrium Constant • To generalize this expression, consider the reaction aA + bB cC + dD
    • The Equilibrium Constant • To generalize this expression, consider the reaction aA + bB cC + dD • The equilibrium expression for this reaction would be [C]c[D]d Kc = [A]a[B]b
    • The Equilibrium Constant • To generalize this expression, consider the reaction aA + bB cC + dD • The equilibrium expression for this reaction would be [C]c[D]d Kc = [A]a[B]b a.k.a. the LAW of MASS ACTION
    • products reactants =K
    • SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions:
    • SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions:
    • Kc is independent of the starting concentrations of reactants and products.
    • The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written in terms of partial pressures (PC)c (PD)d Kp = (PA)a (PB)b
    • Relationship between Kc and Kp • From the ideal gas law we know that PV = nRT • Rearranging it, we get n P= RT V
    • Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp = Kc (RT)Δn Where Δn = (moles of gaseous product) − (moles of gaseous reactant)
    • Kc represents the equil. constant when equilibrium concentrations are expressed in molarity and Kp represents the equilibrium constant when equilibrium concentrations are expressed in atm.
    • SAMPLE EXERCISE 15.2 Converting Between Kc and Kp In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.
    • SAMPLE EXERCISE 15.2 Converting Between Kc and Kp In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature. Solve: Δn = 2 – 4 = –2 (Remember that Δ functions are always products minus reactants) T = 273 + 300 = 573 K R = 0.0821 L-atm/mol-K Kc = 9.60
    • SAMPLE EXERCISE 15.2 Converting Between Kc and Kp In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature. Solve: Δn = 2 – 4 = –2 (Remember that Δ functions are always products minus reactants) T = 273 + 300 = 573 K R = 0.0821 L-atm/mol-K Kc = 9.60
    • Equilibrium Can Be Reached from Either Direction N2O4 (g) → 2 NO2 (g) As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.
    • Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide. N2O4 (g) → 2 NO2 (g)
    • Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.
    • What Does the Value of K Mean? • If K >> 1, the reaction is product-favored; product predominates at equilibrium.
    • What Does the Value of K Mean? • If K >> 1, the reaction is product-favored; product predominates at equilibrium. • If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.
    • SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen: The value for the equilibrium constant for this reaction at 25°C is Kc = 1 × 10–30. Describe the feasibility of fixing nitrogen by forming NO at 25°C.
    • SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen: The value for the equilibrium constant for this reaction at 25°C is Kc = 1 × 10–30. Describe the feasibility of fixing nitrogen by forming NO at 25°C. Solve: Because Kc is so small, very little NO will form at 25°C. The equilibrium lies to the left, favoring the reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25°C.
    • Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. [NO2]2 N2O4 (g) 2 NO2(g) Kc = = 0.212 at 100°C [N2O4]
    • Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. [NO2]2 N2O4 (g) 2 NO2(g) Kc = = 0.212 at 100°C [N2O4] [N2O4] 1 2 NO2 (g) N2O4(g) Kc = = [NO2]2 0.212 = 4.72 at 100°C
    • Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. [NO2]2 N2O4(g) 2 NO2(g) Kc = = 0.212 at 100°C [N2O4]
    • Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. [NO2]2 N2O4(g) 2 NO2(g) Kc = = 0.212 at 100°C [N2O4] [NO2]4 2 N2O4(g) 4 NO2(g) Kc = = (0.212)2 at 100°C [N2O4]2
    • Manipulating Equilibrium Constants 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.
    • the new equilibrium constant is equal to (Kp)3.
    • For the following hypothetical reaction: Y(g) + Z(g) 2 A(g) Keq = 4.0 × 10−2. Given that, what is the value of Keq for the reaction: 4 A(g) 2 Y(g) + 2 Z(g) 1. 25 3. 0.016 2. 5.0 4. 6.3 × 102
    • Correct Answer: 1. 25 2. 5.0 3. 0.016 4. 6.3 × 102 The second equation is twice the first and reversed. Thus, Keq of the second equation is related to the first as shown below: K2 = 630
    • SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the following information, determine the value of Kc for the reaction
    • SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the following information, determine the value of Kc for the reaction Solve: Multiplying the first equation by 2 and raising K to the power 2, we get
    • SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the following information, determine the value of Kc for the reaction Solve: Multiplying the first equation by 2 and raising K to the power 2, we get Reversing the second equation and taking the reciprocal of K gives
    • SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the following information, determine the value of Kc for the reaction Solve: Multiplying the first equation by 2 and raising K to the power 2, we get Reversing the second equation and taking the reciprocal of K gives With two equations that sum to give the net equation, we multiply the individual Kc values to get the desired equilibrium constant.
    • Heterogeneous Equilibrium
    • The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.
    • The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) Kc = [Pb2+] [Cl−]2
    • CaCO3 (s) CO2 (g) + CaO(s) As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.
    • Kp = PH2O
    • PRACTICE EXERCISE Write the following equilibrium-constant expressions:
    • PRACTICE EXERCISE Write the following equilibrium-constant expressions:
    • SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium (a) pure CaCO3, (b) CaO and a CO2 pressure greater than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?
    • SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium (a) pure CaCO3, (b) CaO and a CO2 pressure greater than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO? Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present. (a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained. There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) There is no CaO present, so equilibrium can’t be attained because there is no way the CO2 pressure can decrease to its equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essentially the same as in (a): CaCO3 decomposes until
    • Chapter 15 - Chemical Equilibrium Sections 15.5 & 15.6 Equilibrium calculations
    • Bellwork Write the equilibrium expression for the following equilibrated systems. 2 NO(g) + O2(g) ⇌ 2NO2(g) 6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(s) + 6 O2(g)
    • Equilibrium Calculations
    • For the following hypothetical reaction: 2 Y(aq) + 3 Z(aq) A(aq) Calculate Keq given the following equilibrium concentrations: [A] = 0.50 M, [Y] = 0.10 M, [Z] = 1.0 M [C ]c[D ]d 1. 2.5 K eq = 2. 5.0 [A ]a[ ]b 3. 0.40 B 4. 0.025 5. 50.
    • Correct Answer: [0.50] K eq = 2 3 1. 2.5 [0.10] [ ] 1.0 2. 5.0 3. 0.40 4. 0.025 5. 50.
    • Equilibrium Calculations A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448°C for the reaction H2 (g) + I2 (g) 2 HI (g)
    • What Do We Know? [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change At 1.87 x 10-3 equilibrium
    • [HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change +1.87 x 10-3 At 1.87 x 10-3 equilibrium
    • Stoichiometry tells us [H2] and [I2] decrease by half as much [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 At 1.87 x 10-3 equilibrium
    • We can now calculate the equilibrium concentrations of all three compounds… [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3 equilibrium
    • …and, therefore, the equilibrium constant [HI]2 Kc = [H2] [I2] (1.87 x 10-3)2 = (6.5 x 10-5)(1.065 x 10-3) = 51
    • PRACTICE EXERCISE Sulfur trioxide decomposes at high temperature in a sealed container: 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 the value of Kp at 1000 K.
    • PRACTICE EXERCISE Sulfur trioxide decomposes at high temperature in a sealed container: 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 the value of Kp at 1000 K. Answer: 0.338
    • The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
    • The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
    • If Q = K, the system is at equilibrium.
    • If Q > K, there is too much product and the equilibrium shifts to the left.
    • If Q < K, there is too much reactant, and the equilibrium shifts to the right.
    • For the following hypothetical reaction: 2 X(g) + Y(g) A(g) Keq = 250. At a point during the reaction, the concentrations are [A] = 1.0 M, [X] = 0.50 M, and [Y] = 0.10 M. How will the reaction proceed to achieve equilibrium? 1. From left to right (toward products) 2. From right to left (toward reactants) 3. Already at equilibrium
    • Correct Answer: 1. From left to right (toward products) 2. From right to left (toward reactants) 3. Already at equilibrium [ ] 1.0 Qc = 2 [0.50] [0.10] Qc = 40 Because Qc < Keq, the reaction will shift from reactants toward products to achieve equilibrium.
    • SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?
    • SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?
    • SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?
    • SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture? Comment: We can always check our answer by using it to recalculate the value of the equilibrium constant:
    • SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrationsfrom Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?
    • SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrationsfrom Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?
    • SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrationsfrom Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?
    • If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x: Solving the quadratic equation (Appendix A.3) leads to two solutions for x:
    • If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x: Solving the quadratic equation (Appendix A.3) leads to two solutions for x: When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations: Check: We can check our solution by putting these numbers into the equilibrium-constant expression:
    • Chapter 15 - Chemical Equilibrium Section 15.7 Le Chatelier’s Principle
    • BELLWORK Equilibrium is established between gases in the reaction 2A(g) + 3B(g) ⇌ 1C(g) + 2D(g) a) Write an expression for the equilibrium constant b) What happens to the forward reaction rate if more A is added? All reactants are involved in the forward reaction’s rate law. c) Would the equilibrium concentrations be the same as they were before more A was added? Explain. d) Would [D] increase, decrease, or stay the same?
    • a) [C][D]2 [A]2[B]3 b) Forward reaction rate increases if A is added c) No, to keep K constant when more A is added, new equilibrium concentrations will develop. If all added A stays as A the ratio is “bottom heavy”. If all added A reacts to form products the ratio is “top heavy” So SOME A will react to form products but not all. All of the concentrations will change, the amount of change will depend on K, stoichiometry, and the amount of A added. USE ICEBOX! d) D will increase as more products are formed.
    • Le Châtelier’s Principle
    • Le Châtelier’s Principle “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.”
    • An equilibrium position is a set of concentrations which fulfill an equilibrium expression.
    • An equilibrium position is a set of concentrations which fulfill an equilibrium expression. For a given reaction and temperature, there are infinite equilibrium positions, yet the value of K never changes.
    • An equilibrium position is a set of concentrations which fulfill an equilibrium expression. For a given reaction and temperature, there are infinite equilibrium positions, yet the value of K never changes. Some things can shift the equilibrium position, but not the value of K.
    • An equilibrium position is a set of concentrations which fulfill an equilibrium expression. For a given reaction and temperature, there are infinite equilibrium positions, yet the value of K never changes. Some things can shift the equilibrium position, but not the value of K. Only a change in temperature will affect the value of K.
    • Le Chatelier’s principle If a change is applied to a system at equilibrium, the position of the equilibrium will shift in a direction which counteracts the change. The system will shift to undo the change
    • Changing concentration A(g) + B(g) ⇌ C(g) + D(g) Adding reactant causes a shift to the right as the forward rate increases and reactants form products.
    • Changing concentration A(g) + B(g) ⇌ C(g) + D(g) Adding reactant causes a shift to the right as the forward rate increases and reactants form products. Removing reactant causes a shift to the left as the forward reaction rate slows, causing a shift towards reactants.
    • Changing concentration A(g) + B(g) ⇌ C(g) + D(g) Adding reactant causes a shift to the right as the forward rate increases and reactants form products. Removing reactant causes a shift to the left as the forward reaction rate slows, causing a shift towards reactants. Adding products causes a shift towards reactants as the reverse reaction rate increases.
    • Changing concentration A(g) + B(g) ⇌ C(g) + D(g) Adding reactant causes a shift to the right as the forward rate increases and reactants form products. Removing reactant causes a shift to the left as the forward reaction rate slows, causing a shift towards reactants. Adding products causes a shift towards reactants as the reverse reaction rate increases. Removing products causes a shift to the right and more product formation.
    • The system will shift to undo the change Changing concentration A(g) + B(g) ⇌ C(g) + D(g) Increase or decrease in concentration Equilibrium position shift to right or left
    • a. the equilibrium shifts to the right, using up some of the added O2.
    • b. the equilibrium shifts to the left, forming more NO.
    • The following reaction is at equilibrium: N2(g) + 3 H2(g) 2 NH3(g) If we remove NH3(g), in what direction will the reaction move to reestablish equilibrium? 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium
    • Correct Answer: 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium If [NH3] is decreased, to return to equilibrium, more NH3 must be produced from the reactants. Thus, the reaction shifts from the reactants to the products.
    • The system will shift to undo the change Changing volume N2(g) + 3H2(g) ⇌ 2NH3(g) 4moles of gas 2moles of gas
    • The system will shift to undo the change Changing volume N2(g) + 3H2(g) ⇌ 2NH3(g) 4moles of gas 2moles of gas Increasing pressure (decreasing volume) will shift equilibrium towards fewer moles of gas. Ex. Shifts right to decrease pressure
    • The system will shift to undo the change Changing volume N2(g) + 3H2(g) ⇌ 2NH3(g) 4moles of gas 2moles of gas Increasing pressure (decreasing volume) will shift equilibrium towards fewer moles of gas. Ex. Shifts right to decrease pressure Decreasing pressure shifts equilibrium towards more moles of gas. Ex. Shifts left to increase pressure
    • The system will shift to undo the change Changing volume N2(g) + 3H2(g) ⇌ 2NH3(g) Changing volume/pressure only affects gaseous equilibria Adding an inert gas does not change the equilibrium position
    • the equilibrium shifts to the left.
    • The following reaction is at equilibrium: N2(g) + 3 H2(g) 2 NH3(g) If we increase the volume while holding the temperature constant, in what direction will the reaction move to reestablish equilibrium? 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium
    • Correct Answer: 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium Increasing the volume causes a shift in the equilibrium in the direction that produces more gas molecules, in this case on the reactants side.
    • The following reaction is at equilibrium: N2(g) + O2(g) 2 NO(g) If we increase the pressure, in what direction will the reaction move to reestablish equilibrium? 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium
    • Correct Answer: 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium The number of gas molecules on both sides of the equation are equal; thus, changing the pressure will not change the equilibrium position.
    • The equilibrium constant (K) does not change with changing concentration or pressure.
    • The equilibrium constant (K) does not change with changing concentration or pressure. ONLY A TEMPERATURE CHANGE AFFECTS K
    • Changes that cause no shift • A shift does not occur when a catalyst is added because both forward and reverse rates are increased equally, but equilibrium will be established faster.
    • Changes that cause no shift • A shift does not occur when a catalyst is added because both forward and reverse rates are increased equally, but equilibrium will be established faster. • Adding pure liquid, solid or an inert gas to an equilibrium system will not cause a shift.
    • A temperature change will change K Rates for forward & reverse reactions are affected differently by temperature change because of activation energy differences.
    • A temperature change will change K Rates for forward & reverse reactions are affected differently by temperature change because of activation energy differences. Temperature change is the only thing that will affect the value of K for an ideal closed system.
    • The system will shift to undo the change Changing temperature N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ •Rewrite the equation with energy as a reactant for endothermic processes or as a product for exothermic processes.
    • The system will shift to undo the change Changing temperature N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ •Rewrite the equation with energy as a reactant for endothermic processes or as a product for exothermic processes. •Use Le Chatelier’s principle. N2(g) + 3H2(g) ⇌ 2NH3(g) + 92kJ Adding heat will shift the equilibrium to the left, in the direction that consumes energy
    • The system will shift to undo the change Changing temperature N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ 57kJ + N2O4(g) ⇌ 2NO2(g) Heating the system shifts the equilibrium right, consuming energy.
    • The system will shift to undo the change Changing temperature N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ 57kJ + N2O4(g) ⇌ 2NO2(g) Heating the system shifts the equilibrium right, consuming energy. Cooling the system shifts the equilibrium towards the reactants
    • evaporation is an endothermic process and increasing temperature shifts equilibrium to the right.
    • The following reaction is at equilibrium: N2(g) + O2(g) 2 NO(g) ΔH° = +180.8 kJ. In what direction will the reaction move to reestablish equilibrium if the temperature is decreased? 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium
    • Correct Answer: 1. From left to right (toward products) 2. From right to left (toward reactants) 3. No change in equilibrium As the temperature is lowered, equilibrium will shift to the side of the equation that produces heat. Thus, for an endothermic reaction, the equilibrium shifts from right to left toward the reactants.
    • Catalysts increase the rate of both the forward and reverse reactions.
    • Equilibrium is achieved faster, but the equilibrium composition remains unaltered.
    • No.
    • SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium Consider the equilibrium In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?
    • SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium Consider the equilibrium In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased? Solve: (a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to the right, in the direction of products. (b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the equilibrium shifts to the right. (c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium. (d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right (e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the equilibrium constant, K.