Lecture Presentation Chapter 15 Chemical Equilibrium 2012 Pearson Education, Inc. John D. Bookstaver St. Charles Community College Cottleville, MO The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
2012 Pearson Education, Inc. Equilibrium 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. Equilibrium 2012 Pearson Education, Inc.
A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. Equilibrium 2012 Pearson Education, Inc. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow: N2O4(g) 2NO2(g)
Equilibrium 2012 Pearson Education, Inc. The Equilibrium Constant Equilibrium 2012 Pearson Education, Inc. The Equilibrium Constant Forward reaction: N2O4(g) 2NO2(g) Rate law: Rate = kf[N2O4] Equilibrium
2012 Pearson Education, Inc. The Equilibrium Constant Reverse reaction: 2 NO2(g) N2O4(g) Rate law: Rate = kr[NO2]2 Equilibrium 2012 Pearson Education, Inc. The Equilibrium Constant Therefore, at equilibrium Ratef = Rater kf[N2O4] = kr[NO2]2 Rewriting this, it becomes
kf [NO2]2 = kr [N2O4] Equilibrium 2012 Pearson Education, Inc. The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes Keq = kf [NO2]2 =
kr [N2O4] Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions: (a) (b) (c) Solution Analyze We are given three equations and are asked to write an equilibrium-constant expression for each. Plan Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced chemical equation. Solve
(a) (b) (c) Practice Exercise Write the equilibrium-constant expression Kc for (a) (b) . Answer: (a) , (b) Equilibrium
The Equilibrium Constant Consider the generalized reaction aA + bB cC + dD The equilibrium expression for this reaction would be [C]c[D]d Kc = [A]a[B]b 2012 Pearson Education, Inc. Equilibrium The Equilibrium Constant Since pressure is proportional to
concentration for gases in a closed system, the equilibrium expression can also be written (PCc) (PDd) Kp = (PAa) (PBb) Equilibrium 2012 Pearson Education, Inc. Relationship Between Kc and Kp From the ideal-gas law we know that PV = nRT Rearranging it, we get n P=
RT V Equilibrium 2012 Pearson Education, Inc. 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)n where n = (moles of gaseous product) (moles of gaseous reactant) Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.2 Converting between Kc and Kp For the Haber process,
Kc = 9.60 at 300 C. Calculate Kp for this reaction at this temperature. Solution Analyze We are given Kc for a reaction and asked to calculate Kp. Plan The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must determine n by comparing the number of moles of product with the number of moles of reactants (Equation 15.15). Solve With 2 mol of gaseous products (2 NH3) and 4 mol of gaseous reactants, (1 N2 + 3 H2), n = 2 4 =2. (Remember that functions are always based on products minus reactants.) The temperature is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.08206 L-atm/mol-K. Using Kc = 9.60, we therefore have Practice Exercise For the equilibrium , Kc is 4.08 103 at 1000 K. Calculate the value for Kp. Answer: 0.335
Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Equilibrium Can Be Reached from Either Direction 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 2012 Pearson Education, Inc. Equilibrium Can Be Reached from Either Direction These are the data from the last two trials
from the table on the previous slide. Equilibrium 2012 Pearson Education, Inc. Equilibrium Can Be Reached from Either Direction It doesnt 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. Equilibrium 2012 Pearson Education, Inc. 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. Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.3 Interpreting the Magnitude of an Equilibrium Constant The following diagrams represent three systems at equilibrium, all in the same-size containers. (a) Without doing any calculations, rank the systems in order of increasing Kc. (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kc for each system. Solution
Analyze We are asked to judge the relative magnitudes of three equilibrium constants and then to calculate them. Plan (a) The more product present at equilibrium, relative to reactant, the larger the equilibrium constant. (b) The equilibrium constant is given by Equation 15.8. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.3 Interpreting the Magnitude of an Equilibrium Constant Continued Solve (a) Each box contains 10 spheres. The amount of product in each varies as follows: (i) 6, (ii) 1, (iii) 8. Therefore, the equilibrium constant varies in the order (ii) < (i) < (iii), from smallest (most reactant) to largest (most products). (b) In (i) we have 0.60 mol/L product and 0.40 mol/L reactant, Kc = 0.600/0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres/4 spheres = 1.5.) In (ii) we have 0.10 mol/L product and 0.90 mol/L reactant, giving Kc = 0.10/0.90 = 0.11 (or 1 sphere/9 spheres = 0.11). In (iii) we have 0.80
mol/L product and 0.20 mol/L reactant, giving Kc = 0.80/0.20 = 4.0 (or 8 spheres/2 spheres = 4.0). These calculations verify the order in (a). Comment Imagine a drawing that represents a reaction with a very small or very large value of Kc. For example, what would the drawing look like if Kc = 1 105? In that case there would need to be 100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw. Practice Exercise For the reaction , Kp = 794 at 298 K and Kp = 55 at 700 K. Is the formation of HI favored more at the higher or lower temperature? Answer: at the lower temperature because Kp is larger at the lower temperature Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction: N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) [NO2]2 Kc = [N2O4] [N2O4] Kc = [NO2]2 = 0.212 at 100 C
= 4.72 at 100 C Equilibrium 2012 Pearson Education, Inc. 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 Kc = [N2O4] N2O4(g) 2NO2(g)
2N2O4(g) [NO2]4 2 = (0.212) at 100 C 4NO2(g) Kc = 2 [N2O4] = 0.212 at 100 C Equilibrium 2012 Pearson Education, Inc. 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. Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed For the reaction that is run at 25 C, Kc = 1 1030.Use this information to write the equilibrium-constant expression and calculate the equilibrium constant for the reaction Solution Analyze We are asked to write the equilibriumconstant expression for a reaction and to determine the value of Kc given the chemical equation and equilibrium constant for the reverse reaction.
Plan The equilibrium-constant expression is a quotient of products over reactants, each raised to a power equal to its coefficient in the balanced equation. The value of the equilibrium constant is the reciprocal of that for the reverse reaction. Solve Writing products over reactants, we have Both the equilibrium-constant expression and the numerical value of the equilibrium constant are the reciprocals of those for the formation of NO from N2 and O2: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed Continued Comment Regardless of the way we express the equilibrium among NO, N 2, and O2, at it lies on the side that favors N2 and O2. Thus, the equilibrium mixture will contain mostly N 2 and O2 with very little NO present. Practice Exercise For , Kp = 4.34 103 at 300 C .What is the value of Kp for the reverse reaction? Answer: 2.30 102 Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Sample Exercise 15.5 Combining Equilibrium Expressions Given the reactions determine the value of Kc for the reaction Solution Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two. Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation. Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.5 Combining Equilibrium Expressions Continued Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives Now we have two equations that sum to give the net equation, and we can multiply the individual
Kc values to get the desired equilibrium constant. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.5 Combining Equilibrium Expressions Continued Practice Exercise Given that, at 700 K, Kp = 54.0 for the reaction and Kp = 1.04 104 for the reaction , determine the value of Kp for the reaction
6 HI(g) + N2(g) at 700 K. Answer: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Heterogeneous Equilibrium Equilibrium 2012 Pearson Education, Inc. The Concentrations of Solids and Liquids Are Essentially Constant Both the concentrations of solids and
liquids can be obtained by multiplying the density of the substance by its molar massand both of these are constants at constant temperature. Equilibrium 2012 Pearson Education, Inc. 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) Pb (aq) + 2Cl (aq)
2+ Kc = [Pb2+] [Cl]2 Equilibrium 2012 Pearson Education, Inc. 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. Equilibrium
2012 Pearson Education, Inc. Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression Kc for (a) (b) Solution Analyze We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions. Plan We use the law of mass action, remembering to omit any pure solids and pure liquids from the expressions. Solve (a) The equilibrium-constant expression is Because H2O appears in the reaction as a liquid, its concentration does not appear in the equilibrium-constant expression.
(b) The equilibrium-constant expression is Because SnO2 and Sn are pure solids, their concentrations do not appear in the equilibrium-constant expression. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Continued Practice Exercise Write the following equilibrium-constant expressions: (a) (b)
Answer: (a) (b) Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.7 Analyzing a Heterogeneous Equilibrium Each of these mixtures was placed in a closed container and allowed to stand: (a) CaCO3(s) (b) CaO(s) and CO2(g) at a pressure greater than the value of Kp (c) CaCO3(s) and CO2(g) at a pressure greater than the value of Kp (d) CaCO3(s) and CaO(s) Determine whether or not each mixture can attain the equilibrium Solution
Analyze We are asked which of several combinations of species can establish an equilibrium between calcium carbonate and its decomposition products, calcium oxide and carbon dioxide. Plan For equilibrium to be achieved, it must be possible for both the forward process and the reverse process to occur. For the forward process to occur, there must be some calcium carbonate present. For the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases, either the necessary compounds may be present initially or they may be formed by reaction of the other species. 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 cannot 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 equilibrium is attained. The presence of CaO initially makes no difference. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Sample Exercise 15.7 Analyzing a Heterogeneous Equilibrium Continued Practice Exercise When added to Fe3O4(s) in a closed container, which one of the following substances H 2(g), H2O(g), O2(g) allows equilibrium to be established in the reaction Fe3O4(s) + 4 H2(g)? Answer: H2(g) Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Equilibrium Calculations
Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.8 Calculating K When All Equilibrium Concentrations Are Known After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472 C, it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction Solution Analyze We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant. Plan Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for Kp. Solve Practice Exercise An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C:
[CH3COOH] = 1.65 102 M; [H+] = 5.44 104 M; and [CH3COO] = 5.44 104 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 C. The reaction is Answer: 1.79 105 Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. An Equilibrium Problem A closed system initially containing 1.000 x 103 M H2 and 2.000 x 103 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 103 M. Calculate Kc at 448 C for the reaction taking place, which is
H2(g) + I2(s) 2HI(g) Equilibrium 2012 Pearson Education, Inc. What Do We Know? Initially [H2], M [I2], M [HI], M 1.000 x 103
2.000 x 103 0 Change At equilibrium 1.87 x 103 Equilibrium 2012 Pearson Education, Inc. [HI] Increases by 1.87 x 103 M Initially [H2], M
[I2], M [HI], M 1.000 x 103 2.000 x 103 0 Change +1.87 x 103 At equilibrium 1.87 x 103
Equilibrium 2012 Pearson Education, Inc. Stoichiometry tells us [H2] and [I2] decrease by half as much. [H2], M [I2], M [HI], M Initially 1.000 x 103 2.000 x 103
0 Change 9.35 x 104 9.35 x 104 +1.87 x 103 At equilibrium 1.87 x 103 Equilibrium 2012 Pearson Education, Inc. We can now calculate the equilibrium
concentrations of all three compounds [H2], M [I2], M [HI], M Initially 1.000 x 103 2.000 x 103 0 Change 9.35 x 104
9.35 x 104 +1.87 x 103 6.5 x 105 1.065 x 103 1.87 x 103 At equilibrium Equilibrium 2012 Pearson Education, Inc. and, therefore, the equilibrium constant: [HI]2
Kc = [H2] [I2] = (1.87 x 103)2 (6.5 x 105)(1.065 x 103) = 51 Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations A closed system initially containing 1.000 103 M H2 and 2.000 103 M I2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 103 M. Calculate Kc at 448 C for the reaction taking place, which is
Solution Analyze We are given the initial concentrations of H2 and l2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant Kc for Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant. Solve First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table.
Second, we calculate the change in HI concentration, which is the difference Change inand [HI]initial = 1.87values: 103 M 0 = 1.87 103 M Between the equilibrium Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations Continued Third, we use the coefficients in the balanced equation to relate the change in
[HI] to the changes in [H2] and [I2]: Fourth, we calculate the equilibrium concentrations of H2 and I2, using initial concentrations and [H2changes ] = 1.000in 103 M 0.935 103 M = 0.065 103 M concentration. The equilibrium 3 [I2] =the 2.000 10 M 0.935 103 M = 1.065 103 M concentration equals initial concentration minus that consumed: Our table now looks like this (with equilibrium concentrations in blue for emphasis): Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed.
Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations Continued Finally, we use the equilibrium-constant expression to calculate the equilibrium constant: Comment The same method can be applied to gaseous equilibrium problems to calculate Kp, in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for Initial Change Equilibrium. 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 Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
Equilibrium 2012 Pearson Education, Inc. If Q = K, the system is at equilibrium. Equilibrium 2012 Pearson Education, Inc. If Q > K, there is too much product, and the equilibrium shifts to the left. Equilibrium 2012 Pearson Education, Inc. If Q < K, there is too much reactant, and the equilibrium shifts to the right.
Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.10 Predicting the Direction of Approach to Continued Equilibrium Practice Exercise At 1000 K the value of Kp for the reaction is 0.338. Calculate the value for Qp, And predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm. Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO 3. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc. Sample Exercise 15.11 Calculating Equilibrium Concentrations For the Haber process, , Kp = 1.45 105 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? Solution Analyze We are given an equilibrium constant, Kp, and the equilibrium partial pressures of two of the three substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the third substance (NH3). Plan We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in
the equation. Solve We tabulate the equilibrium pressures: Because we do not know the equilibrium pressure of NH3, we represent it with x. At equilibrium the pressures must satisfy the equilibrium-constant expression: We now rearrange the equation to solve for x: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.11 Calculating Equilibrium Concentrations Continued Check We can always check our answer by using it to
recalculate the value of the equilibrium constant: Practice Exercise At 500 K the reaction 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 Cl 2 in the equilibrium mixture? Answer: 1.22 atm Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.12 Calculating Equilibrium Concentration from Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 447 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? Solution Analyze We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species. Solve First, we note the initial concentrations of H2 and I2: Plan In this case we are not given any of the equilibrium concentrations. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample
Exercise 15.9, where we calculated an equilibrium constant using initial concentrations. [H2] = 1.000 M and [I2] = 2.000 M Second, we construct a table in which we tabulate the initial concentrations: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.12 Calculating Equilibrium Concentration Continued from Initial Concentrations Third, we use the stoichiometry of the
reaction to determine the changes in concentration that occur as the reaction proceeds to equilibrium. The H2 and I2 concentrations will decrease as equilibrium is established and that of HI will increase. Lets represent the change in concentration of H2 by x. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases. For each x mol of H2 that reacts, x mol of I2 are consumed and 2x mol of HI are produced: Fourth, we use initial concentrations and changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this:
Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.12 Calculating Equilibrium Concentration Continued from Initial Concentrations Fifth, we substitute the equilibrium concentrations into the equilibrium-constant expression and solve 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 to find the equilibrium concentrations: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.12 Calculating Equilibrium Concentration Continued from Initial Concentrations Check We can check our solution by putting these
numbers into the equilibrium constant expression to assure that we correctly calculate the equilibrium constant: Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation. Practice Exercise For the equilibrium , the equilibrium constant Kp is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl 5, PCl3, and Cl2 at this temperature? Answer: PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc. Le Chteliers Principle Equilibrium 2012 Pearson Education, Inc. Le Chteliers 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. Equilibrium 2012 Pearson Education, Inc.
The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. Equilibrium 2012 Pearson Education, Inc. The Haber Process If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.
Equilibrium 2012 Pearson Education, Inc. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Equilibrium 2012 Pearson Education, Inc. Changes in Temperature
Co(H2O)62+(aq) + 4Cl(aq) CoCl4(aq) + 6H2O(l) Equilibrium 2012 Pearson Education, Inc. Sample Exercise 15.13 Using Le Chteliers 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 pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased? Solution Analyze We are given a series of changes to be made to a system at equilibrium and are asked to predict what effect each change will have on the position of the equilibrium. Plan Le Chteliers principle can be used to determine the effects of each of these changes. Solve
(a) The system will adjust to decrease the concentration of the added N 2O4, so the equilibrium shifts to the right, in the direction of product. (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. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.13 Using Le Chteliers Principle to Predict Shifts in Equilibrium Continued (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. Practice Exercise For the reaction in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c) the volume of the reaction system is increased, (d) PCl3(g) is added? Answer: (a) right, (b) left, (c) right, (d) left Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.14 Predicting the Effect of Temperature on K
(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction (b) Determine how the equilibrium constant for this reaction should change with temperature. Solution Analyze We are asked to determine the standard enthalpy change of a reaction and how the equilibrium constant for the reaction varies with temperature. Plan (a) We can use standard enthalpies of formation to calculate H for the reaction. (b) We can then use Le Chteliers principle to determine what effect temperature will have on the equilibrium constant. Solve (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation, less the same quantities for the reactants. (See section 5.7). At 25 C, for NH3(g) is 46.19 kJ/mol. The values for H2(g) and N2(g) are zero by definition because the enthalpies of formation of the elements in their normal states at 25 C are defined as zero. (See section 5.7). Because 2 mol of NH3 is formed, the total enthalpy change is (2 mol)(46.19 kJ/mol) 0 = 92.38 kJ
Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Sample Exercise 15.14 Predicting the Effect of Temperature on K Continued (b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less NH3 and more N2 and H2. This effect is seen in the values for Kp presented in Table 15.2. Notice that Kp changes markedly with changes in temperature and that it is larger at lower temperatures. Comment The fact that Kp for the formation of NH3 from N2 and H2
decreases with increasing temperature is a matter of great practical importance. To form NH3 at a reasonable rate requires higher temperatures. At higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH3 is smaller. To compensate for this, higher pressures are needed because high pressure favors NH3 formation. Practice Exercise Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction Use this result to determine how the equilibrium constant for the reaction should change with temperature. Answer: H = 508.3 kJ; the equilibrium constant will increase with increasing temperature Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc. Catalysts Equilibrium 2012 Pearson Education, Inc. Catalysts Catalysts increase the rate of both the forward and reverse reactions. Equilibrium 2012 Pearson Education, Inc. Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.
Equilibrium 2012 Pearson Education, Inc.