2/24/2014. Chapter 14. Chemical Equilibrium. Hemoglobin. O 2 Transport. Hemoglobin Equilibrium System O 2. Chapter 14 Lecture Lecture Presentation

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1 Lecture Presentation Chapter 14 Chemical Equilibrium Hemoglobin Hemoglobin is a protein (Hb), found in red blood cells, that reacts with O 2. It enhances the amount of O 2 that can be carried through the bloodstream. Hb + O 2 HbO 2 The is used to describe a process that is in dynamic equilibrium. Sherril Soman Grand Valley State University Hemoglobin Equilibrium System Hb + O 2 HbO 2 The concentrations of Hb, O 2, and HbO 2 are all interdependent. The relative amounts of Hb, O 2, and HbO 2 at equilibrium are related to a constant called the equilibrium constant, K. A large value of K indicates a high concentration of products at equilibrium. Changing the concentration of any one of these necessitates changes the other concentrations to restore equilibrium. O 2 Transport In the lungs: High concentration of O 2 The equilibrium shifts to the right Hb and O 2 combine to make more HbO 2 Insert cartoon at top on page 650: shifting of reaction to the right O 2 Transport Fetal Hemoglobin, HbF HbF + O 2 HbFO 2 In the muscles: Low concentration of O 2, The equilibrium shifts to the right HbO 2 breaks down (dissociates) increasing the amount of free O 2. Insert cartoon at middle on page 650 : shifting of reaction to the left Fetal hemoglobin s equilibrium constant is larger than adult hemoglobin s constant. Fetal hemoglobin is more efficient at binding O 2. O 2 is transferred to the fetal hemoglobin from the mother s hemoglobin in the placenta. Hb + O 2 HbO 2 2 O 2 HbF + O 2 HbFO 2 O 2 1

2 Oxygen Exchange between Mother and Fetus Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates all the reactant molecules are converted to product molecules at the end. A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products. in these notes Reaction Dynamics When a reaction starts, the reactants are consumed and products are made. The reactant concentrations decrease and the product concentrations increase. As reactant concentration decreases, the forward reaction rate decreases. Eventually, the products can react to re-form some of the reactants, assuming the products are not allowed to escape. As product concentration increases, the reverse reaction rate increases. Processes that proceed in both the forward and reverse direction are said to be reversible. reactants products Dynamic Equilibrium As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate. Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal. Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant because the chemicals are being consumed and made at the same rate. H 2 (g) + I 2 (g) 2 HI(g) H 2 (g) + I 2 (g) 2 HI(g) At time 0, there are only reactants in the mixture, so only the forward reaction can take place. [H 2 ] = 8, [I 2 ] = 8, [HI] = 0 At time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place. [H 2 ] = 6, [I 2 ] = 6, [HI] = 4 2

3 H 2 (g) + I 2 (g) 2 HI(g) H 2 (g) + I 2 (g) 2 HI(g) At time 32, there are now more products than reactants in the mixture, the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated. [H 2 ] = 4, [I 2 ] = 4, [HI] = 8 At time 48, the amounts of products and reactants in the mixture haven t changed; the forward and reverse reactions are proceeding at the same rate. It has reached equilibrium. H 2 (g) + I 2 (g) 2 HI(g) H 2 (g) + I 2 (g) 2 HI(g) As the concentration of product increases and the concentrations of reactants decrease, the rate of the forward reaction slows down, and the rate of the reverse reaction speeds up. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of reactants and products no longer change. Equilibrium Equal An Analogy: Population Changes The rates of the forward and reverse reactions are equal at equilibrium. But that does not mean the concentrations of reactants and products are equal. Some reactions reach equilibrium only after almost all the reactant molecules are consumed; we say the position of equilibrium favors the products. Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed; we say the position of equilibrium favors the reactants. When Country A citizens feel overcrowded, some will emigrate to Country B. 3

4 An Analogy: Population Changes Equilibrium Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them. The relationship between the chemical equation and the concentrations of reactants and products is called the law of mass action. However, after a time, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, but not necessarily equal. Equilibrium For the general equation aa + bb cc + dd, the law of mass action gives the relationship below. The lowercase letters represent the coefficients of the balanced chemical equation. Always products over reactants K is called the equilibrium constant. Unitless Writing Equilibrium Expressions So, for the reaction 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) the equilibrium constant expression is as follows: What Does the Value of K eq Imply? What Does the Value of K eq Imply? When the value of K eq >> 1, when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules. When the value of K eq << 1, when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules. The position of equilibrium favors products. The position of equilibrium favors reactants. 4

5 A Large Equilibrium A Small Equilibrium Disturbing a Chemical Equilibrium Disturbing and Restoring Equilibrium The equilibrium between reactants and products may be disturbed in three ways: (1) by changing the temperature (2) by changing the concentration of a reactant (3) by changing the volume (for systems involving gases) A change in any of these factors will cause a system at equilibrium to shift back towards a state of equilibrium. This statement is often referred to as Le Chatelier s principle. Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same. However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored. The new concentrations will be different, but the equilibrium constant will be the same, unless you change the temperature. Le Châtelier s Principle An Analogy: Population Changes Le Châtelier's principle guides us in predicting the effect various changes in conditions have on the position of equilibrium. It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance. Disturbances all involve making the system open. When the populations of Country A and Country B are in equilibrium, the emigration rates between the two countries are equal so the populations stay constant. 5

6 An Analogy: Population Changes An Analogy: Population Changes When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established; the new populations will have different numbers of people than the old ones. Disturbing Equilibrium: Adding or Removing Reactants After equilibrium is established, a reactant is added, as long as the added reactant is included in the equilibrium constant expression. That is, not a solid or liquid How will this affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? How will it affect the value of K? Le Châtelier s Principle Changes in Concentration continued Remove Add Remove Add Change aa + bb Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) cc + dd Shifts the Equilibrium left right right left 34 The Effect of Concentration Changes on Equilibrium The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4. When N 2 O 4 is added, some of it decomposes to make more NO 2. 6

7 The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, some of it decomposes to make more NO 2. The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right. Increasing its partial pressure increases its concentration. It does not increase the partial pressure of the other gases in the mixture. Adding an inert gas to the mixture has no effect on the position of equilibrium. It does not affect the partial pressures of the gases in the reaction. Le Châtelier s Principle Effect of Volume Change on Equilibrium Changes in Volume and Pressure Change Increase pressure Decrease pressure Increase volume Decrease volume A (g) + B (g) C (g) Shifts the Equilibrium Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas When the number of gas moles on either side is the same, there is no effect. 39 Decreasing the volume of the container increases the concentration of all the gases in the container. It increases their partial pressures. It does not change the concentrations of solutions! If their partial pressures increase, then the total pressure in the container will increase. According to Le Châtelier s Principle, the equilibrium should shift to remove that pressure. The way the system reduces the pressure is to reduce the number of gas molecules in the container. When the volume decreases, the equilibrium shifts to the side with fewer gas molecules. Disturbing Equilibrium: Changing the Volume After equilibrium is established, the container volume is decreased. How will it affect the concentration of solids, liquid, solutions, and gases? How will this affect the total pressure of solids, liquid, and gases? How will it affect the value of K? Disturbing Equilibrium: Reducing the Volume Decreasing the container volume will increase the total pressure. Boyle s law If the total pressure increases, the partial pressures of all the gases will increase Dalton s law of partial pressures. Because the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules. Shift toward the side with fewer gas molecules At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same. 7

8 The Effect of Volume Changes on Equilibrium The Effect of Volume Changes on Equilibrium Left side of figure Because there are more gas molecules on the reactants side of the reaction, when the pressure is increased, the position of equilibrium shifts toward the side with fewer molecules to decrease the pressure. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules the reactant side. Right side of figure The Effect of Temperature Changes on Equilibrium Position Exothermic reactions release energy and endothermic reactions absorb energy Writing heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, helps us use Le Châtelier s principle to predict the effect of temperature changes, even though heat is not matter and not written in a proper equation. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions For an exothermic reaction, heat is a product. Increasing the temperature is like adding heat. According to Le Châtelier s principle, the equilibrium will shift away from the added heat. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions Adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants. Adding heat to an exothermic reaction will decrease the value of K. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions For an endothermic reaction, heat is a reactant Increasing the temperature is like adding heat According to Le Châtelier s Principle, the equilibrium will shift away from the added heat How will decreasing the temperature affect the system? 8

9 The Effect of Temperature Changes on Equilibrium for Endothermic Reactions Adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products. Adding heat to an endothermic reaction will increase the value of K. How will decreasing the temperature affect the system? The Effect of Temperature Changes on Equilibrium Disturbing a Chemical Equilibrium Effect of Temperature Changes on Gas-Phase Equilibria Conclusion: Increasing the temperature of an endothermic reaction favors the products, equilibrium shifts to the right. Increasing the temperature of an exothermic reaction favors the reactants, equilibrium shifts to the left. Lowering temperature results in the reverse effects. Not Changing the Position of Equilibrium: The Effect of Catalysts Catalysts provide an alternative, more efficient mechanism. Catalysts work for both forward and reverse reactions. Catalysts affect the rate of the forward and reverse reactions by the same factor. Therefore, catalysts do not affect the position of equilibrium. Relationships between K and Chemical Equations When the reaction is written backward, the equilibrium constant is inverted. For the reaction aa + bb cc + dd the equilibrium constant expression is as follows: For the reaction cc + dd aa + bb the equilibrium constant expression is as follows: Relationships between K and Chemical Equations When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor. For the reaction aa + bb cc the equilibrium constant expression is as follows: For the reaction 2aA + 2bB 2cC the equilibrium constant expression is as follows: 9

10 Relationships between K and Chemical Equations When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations. For the reactions (1) aa bb and (2) bb cc the equilibrium constant expressions are as follows: For the reaction aa cc the equilibrium constant expression is as follows: Example 14.1 Expressing Equilibrium s for Chemical Equations Express the equilibrium constant for the chemical equation: Solution The equilibrium constant is the equilibrium concentrations of the products raised to their stoichiometric coefficients divided by the equilibrium concentrations of the reactants raised to their stoichiometric coefficients. For Practice 14.1 Express the equilibrium constant for the combustion of propane as shown by the balanced chemical equation: Example 14.2 Manipulating the Equilibrium to Reflect Changes in the Chemical Equation Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 C: Calculate the equilibrium constant for the following reaction at 25 C: Solution You want to manipulate the given reaction and value of K to obtain the desired reaction and value of K. You can see that the given reaction is the reverse of the desired reaction, and its coefficients are twice those of the desired reaction. Begin by reversing the given reaction and taking the inverse of the value of K. Next, multiply the reaction by and raise the equilibrium constant to the power. Example 14.2 Manipulating the Equilibrium to Reflect Changes in the Chemical Equation Calculate the value of K. For Practice 14.2 Consider the following chemical equation and equilibrium constant at 25 C: Calculate the equilibrium constant for the following reaction at 25 C: For More Practice 14.2 Predict the equilibrium constant for the first reaction shown here given the equilibrium constants for the second and third reactions: Equilibrium s for Reactions Involving Gases The concentration of a gas in a mixture is proportional to its partial pressure. Therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases. For aa(g) + bb(g) cc(g) + dd(g) the equilibrium constant expressions are as follows: K c and K p In calculating K p, the partial pressures are always in atm. The values of K p and K c are not necessarily the same because of the difference in units. K p = K c when Dn = 0 The relationship between them is as follows: or Dn is the difference between the number of moles of reactants and moles of products. 10

11 Deriving the Relationship between K p and K c Deriving the Relationship between K p and K c for aa(g) + bb(g) cc(g) + dd(g) substituting Heterogeneous Equilibria Heterogeneous Equilibria The concentrations of pure solids and pure liquids do not change during the course of a reaction. Because their concentration doesn t change, solids and liquids are not included in the equilibrium constant expression. For the reaction the equilibrium constant expression is as follows: The amount of C is different, but the amounts of CO and CO 2 remain the same. Therefore, the amount of C has no effect on the position of equilibrium. Calculating Equilibrium s from Measured Equilibrium Concentrations The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium. The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same, as long as the temperature is kept constant. The value of the equilibrium constant is independent of the initial amounts of reactants and products. Example 14.3 Relating K p and K c Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation: Find K c for this reaction. Sort You are given K p for the reaction and asked to find K c. Given: K p = Find: K c Strategize Use Equation 14.2 to relate K p and K c. Equation Solve Solve the equation for K c. Calculate n. K p = K c(rt) Δn Substitute the required quantities to calculate K c. The temperature must be in kelvins. The units are dropped when reporting K c as described previously. 11

12 Example 14.3 Relating K p and K c Solution Example 14.3 Relating K p and K c For Practice 14.3 Consider the following reaction and corresponding value of K c: What is the value of K p at this temperature? Check The easiest way to check this answer is to substitute it back into Equation 14.2 and confirm that you get the original value for K p. Example 14.4 Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid Write an expression for the equilibrium constant (K c) for this chemical equation: Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Consider the following reaction: Solution Since CaCO 3(s) and CaO(s) are both solids, omit them from the equilibrium expression. For Practice 14.4 Write an equilibrium expression (K c) for the equation: K c = [CO 2] A reaction mixture at 780 C initially contains [CO] = M and [H 2] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant? Procedure For Finding Equilibrium s from Experimental Concentration Measurements Step 1 Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reactants and products. Leave space in the middle of the table for determining the changes in concentration that occur during the reaction. Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Step 2 For the reactant or product whose concentration is known both initially and at equilibrium, calculate the change in concentration that occurs. Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations. Step 3 Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products. Since reactants are consumed during the reaction, the changes in their concentrations are negative. Since products are formed, the changes in their concentrations are positive. Step 5 Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to calculate K. For Practice 14.5 The reaction in Example 14.5 between CO and H 2 is carried out at a different temperature with initial concentrations of [CO] = 0.27 M and [H 2] = 0.49 M. At equilibrium, the concentration of CH 3OH is 0.11 M. Find the equilibrium constant at this temperature. 12

13 Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g) 2HI(g) at 445 C Calculating Equilibrium Concentrations Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration. Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction. The Reaction Quotient If a reaction mixture containing both reactants and products is not at equilibrium, how can we determine in which direction it will proceed? The answer is to compare the current concentration ratios to the equilibrium constant. The Reaction Quotient For the gas phase reaction aa + bb cc + dd the reaction quotient is as follows: The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q. The Reaction Quotient: Predicting the Direction of Change If Q > K, the reaction will proceed fastest in the reverse direction. The products will decrease and reactants will increase. The Reaction Quotient: Predicting the Direction of Change If Q < K, the reaction will proceed fastest in the forward direction. The products will increase and reactants will decrease. 13

14 The Reaction Quotient: Predicting the Direction of Change If Q = K, the reaction is at equilibrium The products and reactants will not change. If a reaction mixture contains just reactants, then Q = 0, and the reaction will proceed in the forward direction. If a reaction mixture contains just products, then Q =, and the reaction will proceed in the reverse direction. Finding Equilibrium Concentrations When Given the Equilibrium and Initial Concentrations or Pressures Step 1: Decide in which direction the reaction will proceed. Compare Q to K. Step 2: Define the changes of all materials in terms of x. Use the coefficient from the chemical equation as the coefficient of x. The x change is + for materials on the side the reaction is proceeding toward. The x change is for materials on the side the reaction is proceeding away from. Step 3: Solve for x. For second order equations, take square roots of both sides or use the quadratic formula. Simplify and approximate answer for very large or small equilibrium constants, if possible. Approximations to Simplify the Math When the equilibrium constant is very small, the position of equilibrium favors the reactants. For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium. assuming the reaction is proceeding forward The [X] equilibrium = ([X] initial ax) [X] initial We are approximating the equilibrium concentration of reactant to be the same as the initial concentration. Checking the Approximation and Refining as Necessary We can check our approximation by comparing the approximate value of x to the initial concentration. If the approximate value of x is less than 5% of the initial concentration, the approximation is valid. Example 14.4 Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid Write an expression for the equilibrium constant (K c) for this chemical equation: Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Consider the following reaction: Solution Since CaCO 3(s) and CaO(s) are both solids, omit them from the equilibrium expression. For Practice 14.4 Write an equilibrium expression (K c) for the equation: K c = [CO 2] A reaction mixture at 780 C initially contains [CO] = M and [H 2] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant? Procedure For Finding Equilibrium s from Experimental Concentration Measurements Step 1 Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reactants and products. Leave space in the middle of the table for determining the changes in concentration that occur during the reaction. 14

15 Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Step 2 For the reactant or product whose concentration is known both initially and at equilibrium, calculate the change in concentration that occurs. Example 14.5 Finding Equilibrium s from Experimental Concentration Measurements Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations. Step 3 Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products. Since reactants are consumed during the reaction, the changes in their concentrations are negative. Since products are formed, the changes in their concentrations are positive. Step 5 Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to calculate K. For Practice 14.5 The reaction in Example 14.5 between CO and H 2 is carried out at a different temperature with initial concentrations of [CO] = 0.27 M and [H 2] = 0.49 M. At equilibrium, the concentration of CH 3OH is 0.11 M. Find the equilibrium constant at this temperature. Example 14.6 Finding Equilibrium s from Experimental Concentration Measurements Consider the following reaction: A reaction mixture at 1700 C initially contains [CH 4] = M. At equilibrium, the mixture contains [C 2H 2] = M. What is the value of the equilibrium constant? Example 14.6 Finding Equilibrium s from Experimental Concentration Measurements Step 2 For the reactant or product whose concentration is known both initially and at equilibrium, calculate the change in concentration that occurs. Procedure For Finding Equilibrium s from Experimental Concentration Measurements Step 1 Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reactants and products. Leave space in the middle of the table for determining the changes in concentration that occur during the reaction. Step 3 Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products. Since reactants are consumed during the reaction, the changes in their concentrations are negative. Since products are formed, the changes in their concentrations are positive. Example 14.6 Finding Equilibrium s from Experimental Concentration Measurements Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations. Example 14.7 Predicting the Direction of a Reaction by Comparing Q and K Consider the reaction and its equilibrium constant: A reaction mixture contains = atm, = atm, and P ICl = atm. Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed? Solution To determine the progress of the reaction relative to the equilibrium state, first calculate Q. Step 5 Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to calculate K. Compare Q to K. For Practice 14.6 The reaction of CH 4 in Example 14.6 is carried out at a different temperature with an initial concentration of [CH 4] = M. At equilibrium, the concentration of H 2 is M. Find the equilibrium constant at this temperature. Q p = 10.8; K p = 81.9 Since Q p < K p, the reaction is not at equilibrium and will proceed to the right. 15

16 Example 14.7 Predicting the Direction of a Reaction by Comparing Q and K For Practice 14.7 Consider the reaction and its equilibrium constant: Example 14.8 Finding Equilibrium Concentrations When You Know the Equilibrium and All but One of the Equilibrium Concentrations of the Reactants and Products Consider the following reaction: A reaction mixture contains [NO 2] = M and [N 2O 4] = M. Calculate Q c and determine the direction in which the reaction will proceed. In an equilibrium mixture, the concentration of COF 2 is M and the concentration of CF 4 is M. What is the equilibrium concentration of CO 2? Sort You are given the equilibrium constant of a chemical reaction, together with the equilibrium concentrations of the reactant and one product. You are asked to find the equilibrium concentration of the other product. Given: Find: [CO 2] Strategize You can calculate the concentration of the product using the given quantities and the expression for K c. Example 14.8 Finding Equilibrium Concentrations When You Know the Equilibrium and All but One of the Equilibrium Concentrations of the Reactants and Products Conceptual Plan Example 14.8 Finding Equilibrium Concentrations When You Know the Equilibrium and All but One of the Equilibrium Concentrations of the Reactants and Products Check Check your answer by mentally substituting the given values of [COF 2] and [CF 4] as well as your calculated value for CO 2 back into the equilibrium expression. Solve Solve the equilibrium expression for [CO 2] and then substitute in the appropriate values to calculate it. Solution [CO 2] was found to be roughly equal to 1. [COF 2] and [CF 4] Therefore K c is approximately 2, as given in the problem. For Practice 14.8 Diatomic iodine [I 2] decomposes at high temperature to form I atoms according to the reaction: In an equilibrium mixture, the concentration of I 2 is 0.10 M. What is the equilibrium concentration of I? Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Consider the reaction: A reaction mixture at 2000 C initially contains [N 2] = M and [O 2] = M. Find the equilibrium concentrations of the reactants and product at this temperature. Procedure For Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 2 Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations. Compare Q to K to predict the direction in which the reaction will proceed. Q < K; therefore, the reaction will proceed to the right. Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. Step 3 Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x. It is usually most convenient to let x represent the change in concentration of the reactant or product with the smallest stoichiometric coefficient. 16

17 Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 4 Sum each column for each reactant and each product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x. In some cases, such as Example 14.9, you can take the square root of both sides of the expression to solve for x. In other cases, such as Example 14.10, you must solve a quadratic equation to find x. Remember the quadratic formula: Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two values for x, choose the value for x that gives a physically realistic answer. For example, reject the value of x that results in any negative concentrations. Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Since the calculated value of K c matches the given value (to within one digit in the least significant figure), the answer is valid. For Practice 14.9 The reaction in Example 14.9 is carried out at a different temperature at which K c = This time, however, the reaction mixture starts with only the product, [NO] = M, and no reactants. Find the equilibrium concentrations of N 2, O 2, and NO at equilibrium. Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that rounding errors could cause a difference in the least significant digit when comparing values of the equilibrium constant. Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Consider the reaction: A reaction mixture at 100 C initially contains [NO 2] = M. Find the equilibrium concentrations of NO 2 and N 2O 4 at this temperature. Procedure For Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 2 Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations. Compare Q to K to predict the direction in which the reaction will proceed. Q > K; therefore, the reaction will proceed to the left. Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. Step 3 Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x. It is usually most convenient to let x represent the change in concentration of the reactant or product with the smallest stoichiometric coefficient. 17

18 Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 4 Sum each column for each reactant and each product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x. In some cases, such as Example 14.9, you can take the square root of both sides of the expression to solve for x. In other cases, such as Example 14.10, you must solve a quadratic equation to find x. Remember the quadratic formula: Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two values for x, choose the value for x that gives a physically realistic answer. For example, reject the value of x that results in any negative concentrations. We reject the root x = because it gives a negative concentration for NO 2. Using x = 0.014, we get the following concentrations: Example Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Since the calculated value of K c matches the given value (to within one digit in the least significant figure), the answer is valid. For Practice The reaction in Example is carried out at the same temperature, but this time the reaction mixture initially contains only the reactant, [N 2O 4] = M, and no NO 2. Find the equilibrium concentrations of N 2O 4 and NO 2. Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that rounding errors could cause a difference in the least significant digit when comparing values of the equilibrium constant. Example Finding Equilibrium Partial Pressures When You Are Given the Equilibrium and Initial Partial Pressures Consider the reaction: A reaction mixture at 25 C initially contains = atm, = atm, and P ICl = atm. Find the equilibrium partial pressures of I 2, Cl 2, and ICl at this temperature. Solution Follow the procedure used in Example 14.5 and 14.6 (using partial pressures in place of concentrations) to solve the problem. 1. Using the balanced equation as a guide, prepare a table showing the known initial partial pressures of the reactants and products. Example Finding Equilibrium Partial Pressures When You Are Given the Equilibrium and Initial Partial Pressures 2. Use the initial partial pressures to calculate the reaction quotient (Q). Compare Q to K to predict the direction in which the reaction will proceed. Q < K; therefore, the reaction will proceed to the right. 3. Represent the change in the partial pressure of one of the reactants or products with the variable x. Define the changes in the partial pressures of the other reactants or products in terms of x. 18

19 Example Finding Equilibrium Partial Pressures When You Are Given the Equilibrium and Initial Partial Pressures 4. Sum each column for each reactant and product to determine the equilibrium partial pressures in terms of the initial partial pressures and the variable x. Example Finding Equilibrium Partial Pressures When You Are Given the Equilibrium and Initial Partial Pressures 6. Substitute x into the expressions for the equilibrium partial pressures of the reactants and products (from step 4) and calculate the partial pressures. 5. Substitute the expressions for the equilibrium partial pressures (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the expression for the variable x. 7. Check your answer by substituting the calculated equilibrium partial pressures into the equilibrium expression. The calculated value of K should match the given value of K. Since the calculated value of K p matches the given value (within the uncertainty indicated by the significant figures), the answer is valid. Example Finding Equilibrium Partial Pressures When You Are Given the Equilibrium and Initial Partial Pressures For Practice The reaction between I 2 and Cl 2 in Example is carried out at the same temperature, but with these initial partial pressures: = atm, = atm, P ICl = 0.00 atm. Find the equilibrium partial pressures of all three substances. Example Finding Equilibrium Concentrations from Initial Consider the reaction for the decomposition of hydrogen disulfide: A L reaction vessel initially contains mol of H 2S at 800 C. Find the equilibrium concentrations of H 2 and S 2. Procedure For Finding Equilibrium Concentrations from Initial Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. (In these examples, you must first calculate the concentration of H 2S from the given number of moles and volume.) Example Finding Equilibrium Concentrations from Initial Step 2 Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the direction in which the reaction will proceed. Example Finding Equilibrium Concentrations from Initial Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. By inspection, Q = 0; the reaction will proceed to the right. Step 3 Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products with respect to x. Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the resulting equation for the variable x. In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions are not usually simple. However, since the equilibrium constant is small, we know that the reaction does not proceed very far to the right. Therefore, x will be a small number and can be dropped from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small). 19

20 Example Finding Equilibrium Concentrations from Initial Example Finding Equilibrium Concentrations from Initial Check whether your approximation was valid by comparing the calculated value of x to the number it was added to or subtracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the approximation to be valid. If approximation is not valid, proceed to step 5a. Checking the x is small approximation: The x is small approximation is valid, proceed to step 6. Step 5a If the approximation is not valid, you can either solve the equation exactly (by hand or with your calculator), or use the method of successive approximations. In this case, we use the method of successive approximations. Substitute the value obtained for x in step 5 back into the original cubic equation, but only at the exact spot where x was assumed to be negligible, and then solve the equation for x again. Continue this procedure until the value of x obtained from solving the equation is the same as the one that is substituted into the equation. Example Finding Equilibrium Concentrations from Initial Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations. Example Finding Equilibrium Concentrations from Initial For Practice The reaction in Example is carried out at the same temperature with the following initial concentrations: [H 2S] = M, [H 2] = M, and [S 2] = 0.00 M. Find the equilibrium concentration of [S 2]. Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that the approximation method and rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium constant. The calculated value of K is close enough to the given value when we consider the uncertainty introduced by the approximation. Therefore the answer is valid. Example Finding Equilibrium Concentrations from Initial Consider the reaction for the decomposition of hydrogen disulfide: A L reaction vessel initially contains mol of H 2S at 800 C. Find the equilibrium concentrations of H 2 and S 2. Procedure For Finding Equilibrium Concentrations from Initial Example Finding Equilibrium Concentrations from Initial Step 2 Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the direction in which the reaction will proceed. By inspection, Q = 0; the reaction will proceed to the right. Step 3 Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products with respect to x. Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. (In these examples, you must first calculate the concentration of H 2S from the given number of moles and volume.) 20

21 Example Finding Equilibrium Concentrations from Initial Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Example Finding Equilibrium Concentrations from Initial Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the resulting equation for the variable x. In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions are not usually simple. However, since the equilibrium constant is small, we know that the reaction does not proceed very far to the right. Therefore, x will be a small number and can be dropped from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small). Example Finding Equilibrium Concentrations from Initial Check whether your approximation was valid by comparing the calculated value of x to the number it was added to or subtracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the approximation to be valid. If approximation is not valid, proceed to step 5a. Example Finding Equilibrium Concentrations from Initial Checking the x is small approximation: The approximation does not satisfy the <5% rule (although it is close). Step 5a If the approximation is not valid, you can either solve the equation exactly (by hand or with your calculator), or use the method of successive approximations. In this case, we use the method of successive approximations. Substitute the value obtained for x in step 5 back into the original cubic equation, but only at the exact spot where x was assumed to be negligible, and then solve the equation for x again. Continue this procedure until the value of x obtained from solving the equation is the same as the one that is substituted into the equation. Step 6 If we substitute this value of x back into the cubic equation and solve it, we get x = , which is nearly identical to Therefore, we have arrived at the best approximation for x. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations. Example Finding Equilibrium Concentrations from Initial Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that the approximation method and rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium constant. The calculated value of K is equal to the given value. Therefore the answer is valid. For Practice The reaction in Example is carried out at the same temperature with the following initial concentrations: [H 2S] = M, [H 2] = 0.00 M, and [S 2] = 0.00 M. Find the equilibrium concentration of [S 2]. Example The Effect of a Concentration Change on Equilibrium Consider the following reaction at equilibrium: What is the effect of adding additional CO 2 to the reaction mixture? What is the effect of adding additional CaCO 3? Solution Adding additional CO 2 increases the concentration of CO 2 and causes the reaction to shift to the left. Adding additional CaCO 3, however, does not increase the concentration of CaCO 3 because CaCO 3 is a solid and therefore has a constant concentration. Thus, adding additional CaCO 3 has no effect on the position of the equilibrium. (Note that, as we saw in Section 14.5, solids are not included in the equilibrium expression.) For Practice Consider the following reaction in chemical equilibrium: What is the effect of adding additional Br 2 to the reaction mixture? What is the effect of adding additional BrNO? 21

22 Example The Effect of a Volume Change on Equilibrium Consider the following reaction at chemical equilibrium: Example The Effect of a Temperature Change on Equilibrium The following reaction is endothermic: What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? Adding an inert gas at constant volume? Solution The chemical equation has 3 mol of gas on the right and zero moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles.) Adding an inert gas has no effect. For Practice Consider the following reaction at chemical equilibrium: What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature? Solution Since the reaction is endothermic, we can think of heat as a reactant: Raising the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. Lowering the temperature is equivalent to removing a reactant, causing the reaction to shift to the left. For Practice The following reaction is exothermic: What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature? 22