SGCh14 - Chapter Fourteen CHEMICAL EQUILIBRIUM • • •...

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Unformatted text preview: Chapter Fourteen CHEMICAL EQUILIBRIUM • • • • • Chemical Equilibrium and the Equilibrium Constant Writing Equilibrium Constant Expressions Calculating Equilibrium Constants What the Equilibrium Constant Tells Us Factors Affecting Equilibrium CHEMICAL EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT STUDY OBJECTIVES 1. 2. Describe the affect of a reverse reaction on the amount of product formed. Describe how equilibrium is established in a chemical reaction system. Reversible Chemical Reactions. In this chapter we will discuss how equilibrium is established in several types of chemical reaction systems. We will define the equilibrium constant and show how to calculate equilibrium concentrations of reactants and products. A state of c hemical equilibrium exists when the concentrations of reactants and products are observed to remain constant with time. When a mixture of SO2 and O2 , for instance, is introduced into a reaction vessel at a temperature of 700 K, a reaction that produces SO3 occurs: 2SO2 + O2 → 2SO 3 When a specific concentration of SO3 is reached, no additional SO3 is formed even though some SO 2 and O2 remain. From that time on, the concentrations of SO2 , O 2 , and SO3 stay constant, and we say the system has reached chemical equilibrium. The constant concentrations are the result of a r eversible chemical reaction. In the reverse reaction some SO 3 decomposes back into SO2 and O2 . 2SO3 → 2SO 2 + O2 When the rates of forward and reverse reactions are the same, no net chemical change occurs and a state of chemical equilibrium exists. The equilibrium state is referred to as d ynamic because of the continual conversions of reactants into products, and products into reactants at the molecular level. A reversible reaction is represented by the opposing arrows in the chemical equation: 2SO2 + O2 2SO 3 The three compounds are said to be in equilibrium. At 700 K all three species exist together in the reaction vessel. Equilibrium can be reached by the reverse reaction as well. That is, if only SO3 is added to the reaction vessel at 700 K, it will form an equilibrium amount of SO2 and O2 . 2 74 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 75 The Equilibrium Constant. In quantitative terms, the equilibrium state is described by the equilibrium constant expression which is also called the mass action expression. [SO3 ]2 Kc = [SO2 ]2 [O2 ] Here the brackets [ ] indicate concentration in moles per liter. In this expression, note that the concentration of the product is in the numerator, and the concentrations of the reactants appear in the denominator. Since there are two reactants their concentrations are multiplied . Also, the concentration of each component is raised to a power equal to its coefficient in the balanced equation. The value of this expression is called the equilibrium constant. The equilibrium constant for a reaction can be determined by measuring the concentrations of all components at equilibrium and substituting these values into the Kc expression. The equilibrium concentrations of SO 2 , O 2 , and SO3 are related to each other by the equilibrium constant expression, which at 700 K has the following value. [SO3 ]2 Kc = = 4.3 × 10 6 [SO2 ]2 [O2 ] WRITING EQUILIBRIUM CONSTANT EXPRESSIONS STUDY OBJECTIVES 1. 2. 3. Write the equilibrium constant expressions for homogeneous and heterogeneous chemical reactions, given balanced chemical equations. Given Kc for a reaction, calculate the value of Kc for the same reaction when it is balanced by using some multiple of the coefficients in the given equation. Also, find Kc for the reverse reaction. Convert Kc to K p for reactions involving gases. The Form of Kc and the Equilibrium Equation. In general the equilibrium constant expression has the form of a ratio of product concentrations over reactant concentrations at equilibrium. For the general equation aA + bB cC + dD where a, b, c, and d are the stoichiometric coefficients for the balanced equation. The equilibrium constant expression is [C]c [ D] d [A]a [B]b In an equilibrium constant expression each concentration is raised to a power equal to its stoichiometric coefficient. The numerator is found by multiplying together the equilibrium concentrations of the products, each raised to an exponent equal to its stoichiometric coefficient. The denominator is found by multiplying together the equilibrium concentrations of the reactants, each raised to an exponent equal to its stoichiometric coefficient. The equilibrium constant expression and its value depend on how the equation is balanced. Often an equation can be balanced with more than one set of coefficients, as shown below: Kc = 2SO2 + O2 1 SO2 + 2 O 2 2SO 3 SO 3 How does this affect the equilibrium constant? For the latter equation, the equilibrium constant expression is written: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 76 / Chemical Equilibrium Kc' = [SO3 ] [SO2 ][O2 ]1/2 where the prime (') is just to help us keep track of the constant to which we are referring. Note that Kc' is the square root of Kc from the previous section. [SO3 ]2 Kc = [SO2 ]2 [O2 ] [SO3 ] Kc' = [SO2 ][O2 ]1/2 Therefore, by inspection: Kc' = Kc The value of Kc' is: Kc' = 4 .3 × 1 0 6 = 2.1 × 10 3 The equilibrium constant expression for the reaction written in the reverse direction 2SO3 is Kc" = 2SO 2 + O2 [SO2 ]2 [O2 ] [SO3 ]2 Compare this to Kc for the original forward reaction: [SO3 ]2 Kc = [SO2 ]2 [O2 ] By inspection you can tell that Kc" is the reciprocal of Kc for the forward reaction. Kc" = [SO2 ]2 [O2 ] = [SO3 ]2 1 Kc Therefore the value of Kc" is 1 = 2.3 × 10 –7 4.3 × 1 0 6 Always use the Kc expression and value that are consistent with the way in which the equation is balanced. Kc" = Kp and Kc. For reactions involving gases, the equilibrium constant expression can be written in terms of the partial pressures, P i , of each gaseous component. For the equation: 2SO2 + O2 2SO 3 2 P SO3 Kp = 2 P SO2 P O2 The pressure, in atmospheres, of each gas at equilibrium is raised to a power corresponding to its coefficient in the balanced equation. The subscript p in Kp indicates that the equilibrium constant has a value that was calculated by using equilibrium partial pressures in atmospheres, rather than in units of moles per liter. The values of Kp and Kc are related. Recall that for an ideal gas PV = nRT, and so the pressure of an ideal gas is proportional to its concentration: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 77 P= n RT V where n/V = moles per liter. Substitution of (n/V)RT for the pressure of each gas gives the equation relating Kp and Kc that was derived in the text. Kp = Kc(RT)∆ n where R is the ideal gas constant (0.0821 L . atm/K. mol) and ∆ n is the change in the number of moles of gas when going from reactants to products. For the preceding reaction ∆ n = –1, and at 700 K, Kp is given by Kc 4 .3 × 1 0 6 Kp = Kc(RT)–1 = = = 7.5 × 10 4 RT (0.0821)(700) In general, Kp ≠ Kc except in the case when ∆ n = 0. Heterogeneous Equilibria. Whenever a reaction involves reactants and products that exist in different phases, it is called a heterogeneous reaction, and in a closed container a heterogeneous equilibrium will result. For example, when steam is brought into contact with charcoal the following equilibrium is established in a reaction vessel. C(s) + H 2 O(g) H2 (g) + CO(g) The equilibrium constant expression for this reaction will not be the same as for a homogeneous reaction. The usual expression for the constant would be [H2 ][CO] K= [C][H2 O] However, the concentration of a pure solid is itself a constant, and is not changed by the addition to, or removal of, some of that solid. Remember that solids have their own volume, and do not fill their containers as gases do. The concentration (mol/L) of a solid such as charcoal depends only on its density. This means that the equilibrium constant expression can be written [H2 ][CO] Kc = K[C] = [H2 O] where the constant concentration of the pure solid [C] has been combined with the equilibrium constant. Thus the equilibrium constant expression does not contain the solid. In general, concentrations of solids and pure liquids do not appear in equilibrium constant expressions. In this reaction, equilibrium will be maintained as long as some C(s) is present. The amount of solid carbon present does not affect the position of equilibrium. Multiple Equilibria. When the product molecules of one equilibrium reaction become reactants in a second equilibrium process, we have an example of multiple equilibria. In such a case the overall reaction is the sum of the two individual reactions. For instance, consider the reactions at 700°C: 1 NO2 NO + 2 O 2 K 1 = 0.012 followed by 1 SO2 + 2 O 2 SO 3 K 2 = 20 The overall reaction is the sum of the two steps: NO2 + S O 2 Back Forward Main Menu NO + SO3 TOC Kc = ? Study Guide TOC Textbook Website MHHE Website 2 78 / Chemical Equilibrium The equilibrium constant value for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Kc = K1 K2 = (0.012) (20) = 0.24 Guidelines for Writing Equilibrium Constant Expressions •The concentrations of the reacting species in the solution phase are expressed in mol/L. In the gaseous phase, the concentrations can be expressed in mol/L or in atm. The constant K c is related to Kp by a simple equation. • The concentrations of pure solids, pure liquids, and solvents are constants, and do not appear in equilibrium constant expressions. • The equilibrium constant (Kc or K p ) is a dimensionless quantity. • In stating a value for the equilibrium constant, we must specify the balanced equation and the temperature. • If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. _______________________________________________________________________________ EXAMPLE 14.1 The Equilibrium Constant Expression Write the equilibrium constant expressions for the following reversible reactions: a. b. 4NH3 (g) + 5O 2 (g) BaO(s) + CO2 (g) 4NO(g) + 6H 2 O(g) BaCO3 (s) •Method of Solution Remember that the equilibrium constant expression has the concentrations of the products in the numerator and those of the reactants in the denominator. Raise each concentration to a power equal to the coefficient of that substance in the balanced equation. Include concentration terms for gaseous components only; leave out the concentrations of pure liquids and solids as they are not included in the equilibrium constant expression. The expressions are: [NO]4 [H2 O]6 1 a. K c = b. Kc = 4 [O ]5 [CO2 ] [NH3 ] 2 _______________________________________________________________________________ EXAMPLE 14.2 The Equilibrium Constant for a Reverse Reaction The equilibrium constant for the reaction H2 (g) + I 2 (g) 2HI(g) at 400°C is [HI]2 = 64 [H2 ][I2 ] What is the equilibrium constant value for the reverse reaction? Kc = 2HI(g) H2 (g) + I 2 (g) •Method of Solution Look to see if the reactions are related in some way. Note that the equilibrium constant for the second reaction is the reciprocal of the equilibrium expression for the forward reaction. Kc' = Back Forward [H2 ][I2 ] [HI]2 = 1 Kc Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 79 •Calculation 1 1 = = 0.016 Kc 64 _______________________________________________________________________________ EXAMPLE 14.3 K p a nd K c V alues Kc' = What are the values of Kp and Kc at 1000°C for the reaction CaCO3 (s) CaO(s) + CO2 (g) if the pressure of CO2 in equilibrium with CaCO3 and CaO is 3.87 atm? •Method of Solution Enough information is given to find Kp first. Writing the Kp expression for this heterogeneous reaction: Kp = P CO2 = 3.87 Then to get K c, rearrange the equation, Kp = Kc(RT)∆ n where ∆ n, the change in the number of moles of gas in the reaction is +1. •Calculation Kc = Kp (RT)∆ n = 3.87 [(0.0821)(1273)]1 Kc = 0.0370 _______________________________________________________________________________ EXAMPLE 14.4 K p a nd Partial Pressures at Equilibrium At 400°C, K c = 64 for the reaction H2 (g) + I 2 (g) a. b. 2HI(g) What is the value of K p for this reaction? If, at equilibrium, the partial pressures of H 2 and I2 in a container are 0.20 atm and 0.50 atm, respectively, what is the partial pressure of HI in the mixture? •Method of Solution a. The equation relating Kp to K c is Kp = Kc(RT)∆ n Here the change in the number of moles of gas ∆ n is: ∆ n = 2 mol HI – 1 mol H2 – 1 mol I2 = 0 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 80 / Chemical Equilibrium Since, ∆ n = 0, Kp and Kc are the same. Kp = Kc(RT)0 = Kc Kp = Kc = 64 b. Writing the equilibrium constant expression 2 Kp = P HI P H2 P I2 = 64 and substituting the given pressures 2 P HI = 64 (0.20)(0.50) the partial pressure of HI is P HI = (0.20)(0.50)(64) P HI = 2.5 atm _______________________________________________________________________________ EXERCISES 1. Write the equilibrium constant expressions (Kc) for the following equations. a. 2HgO(s) 2Hg(l) + O2 (g) b. Ni(s) + 4CO(g) c. 2NO(g) + Br2 (g) 2. Ni(CO) 4 (g) 2NOBr(g) Kp for the decomposition of ammonium chloride at 427°C is NH4 Cl(s) NH3 (g) + HCl(g) Calculate Kc for this reaction. 3. Kp = 4.8 For the reaction H2 (g) + Br2 (g) 2HBr(g) Kp = 7.1 × 10 4 at 700 K. What is the value of Kp for the following reactions at the same temperature? a. 2HBr(g) H2 (g) + Br2 (g) 1 1 b. 2 H 2 (g) + 2 Br2 (g) HBr(g) 4. The following equilibrium constants were determined at 1123 K: C(s) + CO2 (g) CO(g) + Cl2 (g) Kc = 1.4 × 10 12 2CO(g) COCl2 (g) Kc = 5.5 × 10 –1 Write the equilibrium constant expression Kc and calculate the equilibrium constant at 1123 K for the following reaction: C(s) + CO2 (g) + 2Cl2 (g) Back Forward Main Menu 2COCl2 (g) TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 81 CALCULATING EQUILIBRIUM CONSTANTS STUDY OBJECTIVES 1. 2. Determine an equilibrium constant value, given equilibrium concentrations. Given initial concentrations and the equilibrium concentration of at least one component determine the equilibrium constant. Equilibrium Concentrations. The equilibrium constant for a reaction can be determined by measuring the concentrations of all components at equilibrium and substituting these values into the Kc expression. Two situations arise. In one all of the equilibrium concentrations are given, and a straight forward calculation yields the equilibrium constant. In the second, the initial concentrations of all reactants are given along with the equilibrium concentration of one product or one reactant. Examples 14.5 and 14.6 show these types of calculations. _______________________________________________________________________________ EXAMPLE 14.5 Calculating an Equilibrium Constant Ammonia is synthesized from hydrogen and nitrogen according to the equation: 3H2 (g) + N 2 (g) 2NH3 (g) An equilibrium mixture at a given temperature was analyzed and the following concentrations were found: 0.31 mol N 2 /L; 0.90 mol H2 /L; and 1.4 mol NH3 /L. What is the equilibrium constant value? •Method of Solution When the concentrations of each component of a chemical system at equilibrium are known, the value of Kc can be determined readily by substituting these concentrations into the Kc expression. [NH3 ]2 Kc = [H2 ]3 [N2 ] • Calculation (1.4)2 (0.90)3 (0.31) Kc = 8.7 Kc = _______________________________________________________________________________ EXAMPLE 14.6 Calculating an Equilibrium Constant When 3.0 mol of I2 and 4.0 mol of Br 2 are placed in a 2.0 L reaction chamber at 150°C, the following reaction occurs until equilibrium is reached: I2 (g) + Br2 (g) 2IBr(g) Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the equilibrium constant Kc for the reaction? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 82 / Chemical Equilibrium •Method of Solution We need to know the equilibrium concentrations of I2 , Br2 , and IBr. These, when substituted into the K c expression, will give the value of Kc. First find the number of moles of I2 and Br2 at equilibrium. Initially, the amounts of each component in the reactor are 3.0 moles of I 2 , 4.0 moles of Br2 , and zero moles of IBr. The number of moles of I 2 remaining at equilibrium is given by the initial number of moles of I2 minus the moles of I 2 reacted. This is also true for Br2 . The information that 3.2 mol IBr is formed tells us that 1.6 mol I2 and 1.6 mol Br 2 must have reacted. We know this because the balanced equation states that 2 mol IBr are formed whenever 1 mol I2 and 1 mol of Br2 react. •Calculation moles of I 2 at equilibrium = 3.0 mol – 1.6 mol = 1.4 mol I2 moles of Br2 at equilibrium = 4.0 mol – 1.6 mol = 2.4 mol Br2 Substituting the equilibrium concentrations into the equilibrium constant expression: Kc = [IBr]2 [I2 ] [Br 2 ] = 3.2 mol 2 2.0 L 1.4 mol 2.4 mol 2.0 L 2.0 L Kc = 3.0 •Comment The key to finding all the equilibrium concentrations was that the changes in I 2 and Br2 could be related to the change in IBr through the balanced chemical equation. Keep in mind that the equilibrium concentration equals the initial concentration plus the change in concentration due to reaction to reach equilibrium. The change in concentration is positive for a product, and negative for a reactant. _______________________________________________________________________________ EXERCISES 5. A sample of nitrosyl bromide was heated to 100°C in a 10.0 L container in order to partially decompose it. 2NOBr(g) 2NO(g) + Br2 (g) At equilibrium the container was found to contain 0.0585 mol of NOBr, 0.105 mol of NO, and 0.0524 mol of Br2 . Calculate the value of K c. 6. 1.25 mol NOCl was placed in a 2.50 L reaction chamber at 427°C. After equilibrium was reached, 1.10 mol NOCl remained. Calculate the equilibrium constant Kc for the reaction. 2NOCl(g) 7. 2NO(g) + Cl2 (g) The brown gas NO2 and the colorless gas N2 O4 exist in equilibrium. N2 O4 (g) 2NO 2 (g) 0.625 mol of N2 O4 was introduced into a 5.00 L vessel and was allowed to decompose until it reached equilibrium with NO2 . The concentration of N2 O4 at equilibrium was 0.0750 M. Calculate Kc for the reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 83 8. 5.0 mol of ammonia were introduced into a 5.0 L reaction chamber in which it partially dissociated at high temperatures. 2NH3 (g) 3H 2 (g) + N 2 (g) At equilibrium at a particular temperature, 80.0% of the ammonia had reacted. Calculate Kc for the reaction. WHAT THE EQUILIBRIUM CONSTANT TELLS US STUDY OBJECTIVES 1. 2. 3. Determine whether a reaction is at equilibrium or not, and if it is not, predict the direction in which a net reaction will occur. Predict the relative extents of several reactions given the values of their equilibrium constants. Given Kc and the initial concentrations, calculate the equilibrium concentrations of all components. Predicting the Direction of Reaction. Equilibrium constants provide useful information about chemical reaction systems. In this section, we will use them to predict the direction a reaction will proceed to establish equilibrium. We will also use them to determine the extent of reaction. The reaction quotient, Qc, is a useful aid in predicting whether or not a reaction system is at equilibrium. Take, for instance, the reaction 2SO2 + O2 2SO 3 The reaction quotient is: [SO3 ]2 0 Qc = [SO2 ]2 [ O 2 ]0 0 You will notice that Q has the same algebraic form of the concentrations terms as does K c. However, these concentrations are not necessarily equilibrium concentrations. We will call them initial concentrations, [ ]0 . When a set of initial concentrations is substituted into the reaction quotient, Qc takes on a certain value. In order to predict whether the system is at equilibrium or not, the magnitude of Qc must be compared with that of K c. • • When Qc = Kc, the reaction is at equilibrium, and no net reaction will occur. When Qc > Kc, the system is not at equilibrium, and a net reaction will occur in the reverse direction until Qc = Kc. • When Qc < Kc, the system is not at equilibrium, and a net reaction will occur in the forward direction until Qc = Kc. See Example 14.6 for an application of these criteria. The Magnitude of the Equilibrium Constant. The magnitude of the equilibrium constant is related to the degree of conversion of reactants to products before chemical equilibrium is reached. Since Kc is proportional to the concentrations of products divided by the concentrations of reactants present at equilibrium, the value of Kc tells us the relative quantities of reactants and products present at equilibrium. When Kc > > 1 more products are present than reactants at equilibrium. Conversely, when Kc << 1 more reactants are present at equilibrium than products. In this case equilibrium is reached before appreciable concentrations of products build Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 84 / Chemical Equilibrium up. The larger the value of Kc, the greater the extent of reaction before equilibrium is reached. See Example 14.7. In general, we can say: • When K c >> 1, the forward reaction will go nearly to completion. The equilibrium will lie to the right and favor the products. • When K c << 1, the reaction will not go forward to an appreciable extent. The equilibrium will lie to the left and favor the reactants. Calculation of Equilibrium Concentrations. Not only can we estimate the extent of reaction from the Kc value, but also the expected concentrations at equilibrium can be calculated from a knowledge of the initial concentrations and the Kc value. See Example 14.9 for this important type of calculation. In these types of problems it will be very helpful to use the following approach. 1. Express the equilibrium concentrations of all species in terms of the initial concentrations and an unknown x, which represents the change in concentration. 2. Substitute the equilibrium concentrations derived in part 1 into the equilibrium constant expression, and solve for x. The equilibrium concentration is given by: equilibrium concentration = initial concentration ± the change due to reaction where the + sign is used for a product, and the – sign for a reactant. 3. Use x to calculate the equilibrium concentrations of all species. _______________________________________________________________________________ EXAMPLE 14.7 Predicting Direction of Reaction At a certain temperature the reaction CO(g) + Cl2 (g) COCl2 (g) has an equilibrium constant Kc = 13.8. Is the following mixture an equilibrium mixture? If not, in which direction (forward or reverse) will reaction occur to reach equilibrium? [CO]0 = 2.5 M; [Cl2 ]0 = 1.2 M; and [COCl2 ]0 = 5.0 M •Method of Solution Recall that for the system to be at equilibrium Qc = Kc. Substitute the given concentrations into the reaction quotient for the reaction, and determine Qc. •Calculation Qc = [COCl 2 ]0 (5.0) = = 1.7 [CO]0 [Cl2 ]0 (2.5)(1.2) Compare Qc to K c. Since Qc < Kc the reaction mixture is not an equilibrium mixture, and a net forward reaction will bring the system to equilibrium. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 85 _______________________________________________________________________________ EXAMPLE 14.8 Extent of Reaction Arrange the following reactions in order of their increasing tendency to proceed toward completion (least extent → greatest extent). a. b. c. CO + Cl2 N2 O4 2NOCl COCl2 2NO2 2NO + Cl2 Kc = 13.8 Kc = 2.1 × 10 –4 Kc = 4.7 × 10 –4 •Method of Solution The larger the value of Kc, the more products there are at equilibrium compared to reactants, and the farther a reaction will proceed toward completion (the greater the extent of reaction). Therefore: b < c < a _______________________________________________________________________________ EXAMPLE 14.9 Calculating the Equilibrium Concentrations At 400°C, K c = 64 for the equilibrium H2 (g) + I 2 (g) 2HI(g) If 1.00 mol H2 and 2.00 mol I2 are introduced into an empty 0.50 L reaction vessel, find the equilibrium concentrations of all components at 400°C. •Method of Solution The equilibrium constant expression contains the desired concentrations, and we will use it to obtain the answer. Kc = [HI]2 = 64 [H2 ][I2 ] First we need expressions for the equilibrium concentrations of H2 , I 2 , and HI. Begin by tabulating the initial concentrations. _______________________________________________________ Concentration H2 + I2 2HI _______________________________________________________ Initial 1.00 mol/0.50 L 2.00 mol/0.50 L 0 Change — — — ________________________________________ Equilibrium — — — _______________________________________________________ Since the answer involves three unknowns, we will relate the concentrations to each other by introducing the variable x. Recall, that the equilibrium concentration = initial concentration ± change in concentration. Let x = the change in concentration of H2 . That is, let x = the number of moles of H2 reacting per liter. From the coefficients of the balanced equation we can tell that if the change in H2 is –x, then the change in I2 must also be –x, and the change in HI must be +2x. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 86 / Chemical Equilibrium The next step is to complete the table in units of molarity: __________________________________________________ Concentration H2 + I2 2HI ___________________________________________________ Initial (M) 2.0 4.0 0 Change (M) –x –x 2x _______________________________________ Equilibrium (M) (2.0 – x) (4.0 – x) 2x ___________________________________________________ Now substitute the equilibrium concentrations from the table into the Kc expression, (2x)2 = 64 (2.0 – x)(4.0 – x) Kc = and solve for x. (2x)2 x2 – 6 .0x + 8.0 = 64 Rearranging, we get: 4x2 = 64x 2 – 384x + 512 and grouping yields 60x 2 – 384x + 512 = 0 We will use the general method of solving a quadratic equation of the form ax2 + bx + c = 0 The root x is given by x= –b ± b2 – 4 ac 2a In this case, a = 60, b = –384, and c = 512. Therefore, x= x= –(–384) ± (384)2 – 4(60)(512) 3 84 ± 2 .5 × 1 0 4 = 2(60) 120 3 84 ± 158 = 1.9 and 4.5 mol/ L 120 Recall that x = the number of moles of H2 (or I2 ) reacting per liter. Of the two answers (roots), only 1.9 is reasonable, because the value 4.5 M would mean that more H2 (or I2 ) reacted than was present at the start. This would result in a negative equilibrium concentration, which is physically meaningless. We therefore use the root x = 1.9 M to calculate the equilibrium concentrations: [H2 ] = 2.0 – x = 2.0 M – 1.9 M = 0.1 M [I2 ] = 4.0 – x = 4.0 M – 1.9M = 2.1 M [HI] = 2x = 2(1.9 M) = 3.8 M Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 87 The results can be checked by plugging these concentrations back into the Kc expression to see if K c = 64: Kc = [HI]2 (3.8)2 = = 68 [H2 ][I2 ] (0.1)(2.1) Thus the concentrations we have calculated are correct. The difference between 64 and 68 results from rounding off to maintain the correct number of significant figures. Therefore, our result is correct only to the number of signficant figures given in the problem. _______________________________________________________________________________ EXERCISES 9. The reaction PCl 5 (g) P Cl 3 (g) + Cl2 (g) has the equilibrium constant value Kc = 0.24 at 300°C. a. Is the following reaction mixture at equilibrium? [PCl 5 ] = 5.0 mol/L, [PCl3 ] = 2.5 mol/L, [Cl2 ] = 1.9 mol/L b. Predict the direction in which the system will react to reach equilibrium 10. At 700 K, the reaction: 2SO2 (g) + O 2 (g) 2SO3 (g) has an equilibrium constant K c = 4.3 × 10 6 . a. Is a mixture with the following concentrations at equilibrium? [SO2 ] = 0.10 M; [SO3 ] = 10 M; [O 2 ] = 0.10 M. b. If not at equilibrium, predict the direction in which a net reaction will occur to reach a new equilibrium. 11. The decomposition of NOBr is represented by the equation 2NOBr(g) 2NO(g) + Br2 (g) K c = 0.0169 At equilibrium the concentrations of NO and Br2 are 1.05 × 10 –2 M and 5.24 × 10 –3 M, respectively. What is the concentration of NOBr? 12. At 400°C, the equilibrium constant for the reaction H2 (g) + I 2 (g) 2HI(g) is 64. A mixture of 0.250 mol H2 and 0.250 mol I2 was introduced into an empty 0.75 L reaction vessel at 400°C, find the equilibrium concentrations of all components. 13. For the equilibrium N2 O4 (g) 2NO2 (g) Kc = 0.36 at 100°C a sample of 0.25 mol N2 O4 is allowed to dissociate and come to equilibrium in a 1.5 L flask at 100°C. What are the equilibrium concentrations of NO2 and N2 O4 ? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 88 / Chemical Equilibrium FACTORS AFFECTING EQUILIBRIUM STUDY OBJECTIVE 1. Predict how the equilibrium concentrations of reactants and products are shifted by changes in concentrations of reactants and products, by pressure changes, and by temperature changes. Le Chatelier's Principle. When a reaction reaches a state of chemical equilibrium under a particular set of conditions, no further changes in the concentrations of reactants and products occur. If a change is made in the conditions under which the system is at equilibrium, chemical change will occur in such a way as to establish a new equilibrium. The factors that can influence equilibrium are change in concentration, change in pressure (or volume), and change in temperature. What effect does a change in one of these factors have on the extent of reaction? This question can be answered qualitatively by using Le Chatelier's principle: When a stress is applied to a system in a state of dynamic equilibrium, the system will, if possible, shift to a new position of equilibrium in which the stress is partially offset. To interpret this statement take, for example, the reaction 2NO2 (g) N2 O4 (g) First, the reaction must be at equilibrium. Let's add more N2 O4 as an illustration of a change in concentration. The concentration of N2 O4 increases, and the equilibrium is disturbed. The system will respond by using up part of the additional N2 O4 . In this case, a net reverse reaction will partially offset the increased N2 O4 concentration. The net reverse reaction brings the system to a new state of equilibrium. When equilibrium is reestablished, more NO2 will be present than there was before the N2 O4 was added. Thus, the position of equilibrium has shifted to the left. Le Chatelier's principle will predict the direction of the net reaction or "shift in equilibrium" that brings the system to a new equilibrium. In this case the stress of adding more N2 O4 was partially offset by a net reverse reaction that consumed some of the additional N2 O4 . The key to the use of Le Chatelier's principle is to recognize which net reaction, forward or reverse, will partially offset the change in conditions. Changes in Volume and Pressure. The pressure of a system of gases in chemical equilibrium can be increased by decreasing the available volume. This change causes the concentration of all components to increase. The stress will be partially offset by a net reaction that will lower the total concentration of gas molecules. Consider our previous example reaction 2NO2 (g) N2 O4 (g) When the molecules of both gases are compressed into a smaller volume their total concentration increases (this is the stress). A net forward reaction (shift to the right) will bring the system to a new state of equilibrium, in which the total concentration of molecules will be lowered somewhat. This partially offsets the initial stress on the system. Notice that when 2 moles of NO2 react only 1 mole of N2 O4 is formed. When equilibrium is reestablished, more moles of N2 O4 and fewer mole of NO2 will be present than before the pressure increase occurred. In this case the equilibrium has shifted to the right. In general, an increase in pressure by decreasing the volume will result in a net reaction that decreases the total concentration of gas molecules. A special case arises when the total number of moles of gaseous products and of gaseous reactants are equal in the balanced equation. In this case no shift in equilibrium will occur. Changes in Temperature. If the temperature of a system is changed, a change in the value of K c occurs. An increase in temperature always shifts the equilibrium in the direction of the endothermic reaction, while a temperature decrease shifts the equilibrium in the direction of the exothermic reaction. See Figure 14.8 in the text. Thus, for endothermic reactions the value of K c increases with increasing temperature, and for Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 89 exothermic reactions the value of Kc decreases with increasing temperature. In the case of our example reaction, Kc will decrease as the temperature is increased because the equilibrium will shift in the direction of the endothermic reaction, that is, in the reverse direction. 2NO2 (g) N2 O4 (g) o ∆ Hrxn = –58.0 kJ We can explain this in terms of Le Chatelier's principle. As heat is added to the system, it represents a stress on the equilibrium. The equilibrium will shift in the direction that will consume some of the added heat. This partially offsets the stress. In this reaction, the equilibrium shifts to the left, and Kc decreases. Remember, of these three types of changes; concentration, pressure, and temperature, only changes in temperature will actually alter the Kc value. There are two other factors related to chemical reactions that do not affect the position of equilibrium. The first of these is a catalyst. Catalysts speed up the rate at which equilibrium is reached by lowering the activation energy barrier. Of course, this speeds up the reverse reaction as well. The net result is that changes in the concentration of a catalyst will not affect the equilibrium concentrations of reactants and products, nor will they change the value of Kc. The second of the two factors is the addition of an inert gas to a system at equilibrium. This will cause an increase in the total pressure within the reactor. However, none of the partial pressures of reactants or products are changed, and so the equilibrium is not upset, and no shifting is needed to bring the system back to equilibrium. _______________________________________________________________________________ EXAMPLE 14.10 Changing Conditions Affecting Equilibrium For the reaction at equilibrium 2NaHCO3 (s) Na 2 CO3 (s) + H 2 O(g) + CO 2 (g) o ∆ Hrxn = 128 kJ state the effects (increase, decrease, no change) of the following stresses on the number of moles of sodium carbonate, Na2 CO3 , at equilibrium in a closed container. Note that Na 2 CO3 is a solid (this is a heterogeneous equation); its concentration will remain constant, but its amount can change. a. b. c. d. Removing CO 2 (g). Adding H2 O(g). Raising the temperature. Adding NaHCO3 (s). •Method of Solution Apply Le Chatelier's principle. a. b. c. d. If CO2 concentration is lowered, the system will react in such a way as to offset the change. That is, a shift to the right will replace some of the missing CO2 . The number of moles of Na2 CO3 increases. Addition of H2 O(g) exerts a stress on the equilibrium that is partially offset by a shift in the equilibrium to the left (net reverse reaction). This consumes Na2 CO3 as well as some of the extra H2 O. The number of moles of Na2 CO3 decreases. An increase in temperature will increase the Kc value of an endothermic reaction. There is a shift to the right, and more Na2 CO3 is formed. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The same equilibrium is reached whether the system contains 1 g of NaHCO3 (s) or 10 g of NaHCO3 . No shift in the equilibrium occurs. No change in the amount of Na2 CO3 occurs. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 90 / Chemical Equilibrium EXERCISES 14. Copper can be extracted from its ores by heating Cu2 S in air. Cu 2 S(s) + O2 (g) 2Cu(s) + SO2 (g) o ∆ Hrxn = –250 kJ Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions. a. Adding more O2 (g). b. Compressing the vessel volume in half. c. Raising the temperature. d. Adding more SO2 (g). e. Adding more Cu(s). f. Adding a catalyst. 15. Consider the following reaction at equilibrium: 2SO2 (g) + O 2 (g) 2SO3 (g) If the volume of the container is decreased at constant temperature, state which way the equilibrium will shift to reach a new equilibium: Left, right, or no change. 16. The following reaction is at equilibrium. Describe what change in the concentration of NO(g), if any, will occur if bromine is added to the reaction chamber. 2NOBr(g) 2NO(g) + Br2 (g) 17. The following reaction is exothermic: 2SO2 (g) + O 2 (g) 2SO3 (g) Describe what will happen to the concentration of SO2 when the temperature is increased. ________________________________________________________________________________ CONCEPTUAL QUESTIONS 1. Consider the following reaction at 400°C: H2 (g) + I 2 (g) 2HI(g) H2 and I2 were placed into a flask and allowed to react until equilibrium was reached. Then a small amount of H131 I was added. 1 31 I is an isotope of iodine that is radioactive. Will radioactive 1 31 I stay in the HI molecule or will some or all of it find its way into the I2 molecule forming I131 I? Explain your answer. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 91 2. Consider the following reaction at 400°C: H2 O(g) + CO(g) H2 (g)+ CO 2 (g) Amounts of H 2 O, CO(g), H2 (g), and CO2 (g) were put into a flask so that the composition corresponded to an equilibrium mixture. A lab technician added an iron catalyst to the mixture, but was surprised when no additional H2 (g) and CO2 (g) were formed even after waiting for many days. Explain why the technician should not have been surprised. PRACTICE TEST 1. Write the equilibrium constant expressions for the following reactions: a. 4NH3 (g) + 3O 2 (g) 2N2 (g) + 6H 2 O(g) b. 2N 2 O(g) + 3O2 (g) 2N2 O4 (g) c. 2ClO 2 (g) + F2 (g) 2FClO2 (g) d. H2 (g) + Br2 (l) 2HBr(g) e. C(s) + CO2 (g) 2CO(g) f. CuO(s) + H2 (g) 2. Cu(s) + H2 O(g) The equilibrium constant Kc for the reaction: Ni(s) + 4CO(g) Ni(CO) 4 (g) is 5.0 × 10 4 at 25°C. What is value of the equilibrium constant for the reaction? Ni(CO)4 (g) 3. Ni(s) + 4CO(g) The decomposition of HI(g) is represented by the equation 2HI(g) H2 (g) + I 2 (g) K c = 64 If the equilibrium concentrations of H2 and I2 at 400°C are found to be [H2 ] = 4.2 × 10 –4 M and [I2 ] = 1.9 × 10 –3 M, what is the equilibrium concentration of HI? 4. For the reaction CH4 (g) + 2H 2 S(g) C S 2 (g) + 4H 2 (g) Kp = 2.05 × 10 9 at 25°C. Calculate Kp and Kc, at this temperature, for 1 2H2 (g) + 2 C S 2 (g) 5. A 1.00 L vessel initially contains 0.777 moles of SO3 (g) at 1100 K. What is the value of Kc for the following reaction, if 0.520 moles of SO3 remain at equilibrium? 2SO3 (g) Back 1 H2 S(g) + 2 CH4 (g) Forward 2SO 2 (g) + O 2 (g) Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 92 / Chemical Equilibrium 6. Initially a 1.0 L vessel contains 10.0 moles of NO and 6.0 moles of O2 at a certain temperature. They react until equilibrium is established. 2NO(g) + O2 (g) 2NO2 (g) At equilibrium the vessel contains 8.8 moles of NO2 . Determine the value of K c at this temperature. 7. Given the reaction: N2 + O2 2NO K c = 2.5 × 10 –3 at 2130°C Decide whether the following mixture is at equilibrium, or if a net forward or reverse reaction will occur. [NO] = 0.005; [O2 ] = 0.25; [N 2 ] = 0.020 mol/L. 8. Hydrogen iodide decomposes according to the equation: 2HI(g) H 2 (g) + I 2 (g) Kc = 0.0156 at 400°C A 0.55 mol sample of HI was injected into a 2.0 L reaction vessel held at 400°C. Calculate the concentration of HI at equilibrium. 9. 2.00 mol of NOCl was placed in a 2.00 L reaction vessel at 400°C. After equilibrium was established, it was found that 24 percent of the NOCl had dissociated according to the equation: 2NOCl(g) 2NO(g) + Cl2 (g) Calculate the equilibrium constant Kc for the reaction. 10. For the reaction N2 (g) + O 2 (g) 2NO(g) Kp = 3.80 × 10 –4 at 2000°C what equilibrium pressures of N2 , O 2 , and NO will result when a 10 L reactor vessel is filled with 2.00 atm of N 2 and 0.400 atm of O2 and the reaction is allowed to come to equilibrium? 11. For the equilibrium N2 O4 (g) 2NO2 (g) Kc = 0.36 at 100°C Suppose 30.0 g N2 O4 is placed in a 2.5 L flask at 100°C. (a) Calculate the number of moles of NO2 present at equilibrium and (b) the percentage of the original N2 O4 that is dissociated. 12. What changes in the equilibrium composition of the reaction 2SO2 (g) + O 2 (g) 2SO 3 (g) o ∆ Hrxn = –197 kJ will occur if it experiences the following stresses? a. The partial pressure of SO3 (g) is increased. b. Inert Ar gas is added. c. The temperature of the system is decreased. d. The total pressure of the system is increased by reducing the available volume. e. The partial pressure of O2 (g) is decreased. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 93 13. For the chemical equilibrium PCl 5 (g) P Cl 3 (g) + Cl2 (g) o ∆ Hrxn = 92.9 kJ a. What is the effect on K c of lowering the temperature? b. What is the effect on the equilibrium concentration of PCl3 of adding Cl2 ? c. What is the effect on the equilibrium concentrations of compressing the mixture to a smaller volume? d. What is the effect on the equilibrium pressure of Cl2 of removing PCl3 ? 14. Consider the equilibrium SO2 (g) + NO2 (g) NO(g) + SO 3 (g) where Kc = 85.0 at 460°C. The following mixture was prepared in a reactor at 460°C. [SO2 ] = 0.0200 M; [NO2 ] = 0.0200 M; [NO] = 0.100 M; [SO3 ] = 0.100 M What will the concentrations of the four gases be when equilibrium is reached? 15. Arrange the following reactions in their increasing tendency to proceed toward completion: a. 2HF Kc = 1 × 10 –13 H2 F 2 b. 2H 2 + O2 Kc = 3 × 10 81 2H2 O c. 2NOCl Kc = 4.7 × 10 –4 2NO + Cl2 16. For the decomposition of calcium carbonate CaCO3 (s) CaO(s) + CO2 (g) o ∆ Hrxn = 175 kJ how will the amount (not concentration) of CaCO3 (s) change with the following stresses? a. CO 2 (g) is removed. b. CaO(s) is added. c. The temperature is raised. d. The volume of the container is decreased. 17. At 2000°C, 5.0 × 10 –3 mol CO2 is introduced into a 1.0 L container and the following reaction comes to equilibrium: 2CO2 (g) 2CO(g) + O2 (g) K c = 6.4 × 10 –7 a. Calculate the equilibrium concentrations of CO and O2 . b. What fraction of the CO 2 is decomposed at equilibrium? 18. In a closed container at 1900 K nitrogen reacts with oxygen to yield NO: N2 (g) + O 2 (g) 2NO(g) Kp = 2.3 × 10 –4 However, NO(g) quickly reacts with oxygen to produce NO2 (g): 2NO(g) + O2 (g) 2NO2 (g) Kp = 1.3 × 10 –4 Calculate Kp and Kc at 1900 K for the net reaction. N2 (g) + 2O 2 (g) Back Forward Main Menu 2NO2 (g) TOC Study Guide TOC Textbook Website MHHE Website 2 94 / Chemical Equilibrium ANSWERS Exercises 1. a. K c = [O2 ] b. K c = [Ni(CO)4 ] c. K c = [CO]4 [NOBr]2 [NO]2 [Br2 ] 2. Kc = 1.5 × 10 –3 3. a. K p = 1.4 × 10 –5 b . K p = 266 2 [COCl 2 ] 4. Kc = = 7.7 × 10 11 2 [CO2 ] [Cl2 ] 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. Kc = 0.0169 Kc = 5.6 × 10 –4 Kc = 0.13 Kc = 17.3 a. no b. reverse a. no b. forward [NOBr] = 5.85 × 10 –3 M [HI] = 0.532 M, [H2 ] = [I2 ] = 6.7 × 10 –2 M [NO2 ] = 0.17 M, [N2 O4 ] = 0.080 M a. right b. no shift c. left d. left e. no shift right decrease increase e. no shift Conceptual Questions 1. 2. Back Radioactive 1 31 I will find its way into the I2 molecule, but all of it will not form molecules of I131 I. Some will remain as H131 I. First of all 1 31 I will react just like any other iodine atom. The reaction system is at equilibrium which is a dynamic equilibrium. Therefore the forward and reverse reactions are occuring simultaneously and at the same rate. Any H131 I will eventually dissociate by the reverse reaction and form I131 I molecules. All of the 1 31 I will not end up in the I131 I molecules because the forward reaction converts I131 I back into H 131 I. Therefore the 1 31 I will be distributed between the two I-containing molecules. The technician should have known that a catalyst cannot change the position of equilibrium. A catalyst lowers the activation energy and so speeds up the rate of both the forward and reverse reactions and does not change the equilibrium constant. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Equilibrium / 2 95 Practice Test 1. a. K c = d. Kc = 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. [N2 ]2 [H2 O]6 [NH3 ]4 [O2 ]3 [HBr]2 [H2 ] b. Kc = e. Kc = [N2 O4 ]2 [N2 O]2 [O2 ]3 [CO]2 [CO2 ] c. Kc = f. Kc = [FClO 2 ]2 [ClO2 ]2 [F2 ] [H2 O] [H2 ] 2.0 × 10 –5 [HI] = 1.1 × 10 –4 M Kp = 2.2 × 10 –5 , K c = 5.4 × 10 –4 Kc = 0.031 Kc = 34 Net reverse reaction. [HI] = 0.22 M Kc = 0.012 P NO = 0.017 atm; PN2 = 1.99 atm; PO2 = 0.39 atm a. 0.36 mol NO2 b. 66 % N2 O4 dissociated a. The partial pressures of SO2 and O2 will increase. b. No changes in equilibrium partial pressures. c. The partial pressures of SO2 and O2 decrease while that of SO3 increases. d. The partial pressures of SO2 and O2 decrease while that of SO3 increases. e. The partial pressure of SO2 increases and that of SO3 decreases. a. K c will decrease b. [PCl3 ] decreases c. [PCl 5 ] increases; [PCl3 ] and [Cl2 ] decrease pressure increases [SO2 ] = [NO2 ] = 0.0118 M; [NO] = [SO3 ] = 0.108 M a. Least extent of reaction b. Greatest extent of reaction c. Intermediate of the three examples a. Decreases b. No change c. Decreases d. Increases a. [O2 ] = 1.5 × 10 –4 M, [CO] = 3.0 × 10 –4 M b. 6.1 percent Kp = 3.0 × 10 –8 Kc = 4.7 × 10 –6 d. Cl2 _________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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