Introductory Chemistry, 3rd Edition Nivaldo Tro Chapter 15 Chemical Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2009, Prentice Hall

Equilibrium vs. Disequilibrium When systems are at equilibrium with their surroundings, their conditions are the same as the surroundings and they stay that way. When systems are in disequilibrium with their surroundings, their conditions are not the same as the surroundings. Systems that are in disequilibrium tend to change until they reach equilibrium with their surroundings. Living things are in controlled disequilibrium with their environment—they are not at the same conditions as the environment and do not tend to change toward those conditions. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Reaction Rates Some chemical reactions proceed rapidly. Like the precipitation reactions studied in Chapter 7 where the products form practically the instant the two solutions are mixed. Other reactions proceed slowly. Like the decomposition of dye molecules of a sofa placed in front of a window. The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time. Generally moles of reactant used per second. Like miles per hour. Chemists study ways of controlling reaction rates. Tro's Introductory Chemistry, Chapter 15

Reaction Rates, Continued Initially, only reactants are present After 15 seconds, the left reaction is 60% complete, but the right reaction is only 20% complete After 30 seconds, the left reaction is complete, whereas the right reaction is only 40% done. After 45 seconds, the right reaction is still not complete Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 2 N2O5 (g) → 4 NO2(g) + O2(g) Over time, the concentrations of reactants decrease as products increase. Tro's Introductory Chemistry, Chapter 15

2 N2O5 (g) → 4 NO2(g) + O2(g): Rate vs. Time Because reactant concentrations decrease, the rates of reactions slow down over time. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Collision Theory In order for a reaction to take place, the reacting molecules must collide with each other. Once molecules collide they may react together or they may not, depending on two factors: Whether the collision has enough energy to “start to break the bonds holding reactant molecules together." Whether the reacting molecules collide in the proper orientation for new bonds to form. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Effective Collisions Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. The higher the frequency of effective collisions, the faster the reaction rate. There is a minimum energy needed for a collision to be effective. We call this the activation energy. The lower the activation energy, the faster the reaction will be. Tro's Introductory Chemistry, Chapter 15

Effective Collisions: Kinetic Energy Factor For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide, it can form the activated complex. Tro's Introductory Chemistry, Chapter 15

Effective Collisions: Orientation Effect Tro's Introductory Chemistry, Chapter 15

Reaction Energy Diagram Tro's Introductory Chemistry, Chapter 15

Factors Effecting Reaction Rate: Reactant Concentration The higher the concentration of reactant molecules, this increases the frequency of reactant molecule collisions the faster the reaction will generally go. Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time. Tro's Introductory Chemistry, Chapter 15

Effect of Concentration on Rate Low concentrations of reactant molecules leads to fewer effective collisions, therefore a slower reaction rate. High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate. Tro's Introductory Chemistry, Chapter 15

Factors Effecting Reaction Rate: Temperature Increasing the temperature increases the energy so that their collisions can overcome the activation energy. And, increasing the temperature also increases the frequency of collisions. Both these mean that increasing temperature increases the reaction rate. Tro's Introductory Chemistry, Chapter 15

Effect of Temperature on Rate Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster reaction rate.

Tro's Introductory Chemistry, Chapter 15 Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants. This is called the reverse reaction. Reactions that can proceed in both the forward and reverse directions are called reversible reactions. Tro's Introductory Chemistry, Chapter 15

Reaction Dynamics, Continued The forward reaction slows down as the amounts of reactants decreases. At the same time, the reverse reaction speeds up as the concentration of the products increases. Eventually, the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. Dynamic equilibrium is reached when the rates of two opposite processes are the same. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Chemical Equilibrium When a reaction reaches equilibrium, the production of both reactants and products is constant. The forward and reverse reactions still continue. Because they go at the same rate, the amounts of materials does not appear to change. Tro's Introductory Chemistry, Chapter 15

Equilibrium As the reaction proceeds, the forward reaction slows down as the reactants get used up. At the same time, the reverse reaction speeds up as product concentration increases. Initially, we only have reactant molecules in the mixture. The reaction can only proceed in the forward direction, making products. Once equilibrium is established, the concentrations of the reactants and products in the final mixture do not change, (unless conditions are changed). Eventually, the forward and reverse rates are equal. At this time equilibrium is established.

Equilibrium, Continued As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant. Initially, only the forward reaction takes place. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Rate Rate forward Rate reverse Time Tro's Introductory Chemistry, Chapter 15

Hypothetical Reaction 2 Red  Blue Time [Red] [Blue] 0.400 0.000 10 0.208 0.096 20 0.190 0.105 30 0.180 0.110 40 0.174 0.113 50 0.170 0.115 60 0.168 0.116 70 0.167 0.117 80 0.166 90 0.165 0.118 100 110 0.164 120 130 140 150 The reaction slows over time, but the red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the red and the blue molecules no longer change— equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of red molecules turning into blue is the same as the rate of blue molecules turning into red.

Hypothetical Reaction 2 Red  Blue, Continued Tro, Chemistry: A Molecular Approach

Tro's Introductory Chemistry, Chapter 15 Equilibrium  Equal 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. Tro's Introductory Chemistry, Chapter 15

Equilibrium, Continued As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant. Initially, only the forward reaction takes place. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Rate Rate forward Rate reverse Time Tro's Introductory Chemistry, Chapter 15

What occurs in a reaction at equilibrium. When the number of people moving up is the same as the number of people moving down, the number of people on each floor remains constant, and the two populations are in equilibrium. Equilibrium occurs when the forward and reverse reactions are the same. Chapter Seven

An Analogy: Population Changes When Narnians feel overcrowded, some will emigrate to Middle Earth. However, as time passes, emigration will occur in both directions at the same rate, leading to populations in Narnia and Middle Earth that are constant, though not necessarily equal. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Equilibrium Constant Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them. For the reaction H2(g) + I2(g)  2HI(g) at equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant. Tro's Introductory Chemistry, Chapter 15

Equilibrium Constant, Continued For the general equation aA + bB  cC + dD, the relationship is given below: The lowercase letters represent the coefficients of the balanced chemical equation. Always products over reactants. The constant is called the equilibrium constant, Keq. Tro's Introductory Chemistry, Chapter 15

Writing Equilibrium Constant Expressions For aA + bB  cC + dD, the equilibrium constant expression is: So for the reaction 2 N2O5  4 NO2 + O2, the equilibrium constant expression is: Tro's Introductory Chemistry, Chapter 15

Equilibrium Constants for Heterogeneous Equilibria Pure substances in the solid and liquid state have constant concentrations. Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution. Therefore, these substances are not included in the equilibrium constant expression. For the reaction CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l): Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Write the Equilibrium Constant Expressions, Keq, for Each of the Following: 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba+2(aq) + SO4-2(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Tro's Introductory Chemistry, Chapter 15

Write the Equilibrium Constant Expressions, Keq, for Each of the Following, Continued: 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba+2(aq) + SO4-2(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Keq = [CO]2•[O2] [CO2]2 Keq = [Ba+2]•[SO4-2] Keq = [CO2] [CH4]•[O2]2

What Does the Value of Keq Imply? When the value of Keq > > 1, we know that when the reaction reaches equilibrium, there will be many more product molecules present than reactant molecules. The position of equilibrium favors products. When the value of Keq < < 1, we know that when the reaction reaches equilibrium, there will be many more reactant molecules present than product molecules. The position of equilibrium favors reactants. Tro's Introductory Chemistry, Chapter 15

A Large Equilibrium Constant Tro's Introductory Chemistry, Chapter 15

A Small Equilibrium Constant Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following: 2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95 2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17 Tro's Introductory Chemistry, Chapter 15

Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following, Continued: 2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95 2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17 Favors reactants. Favors products. Favors reactants.

Tro's Introductory Chemistry, Chapter 15 Calculating Keq The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for Keq. Although you may have different amounts of reactants and products in the equilibrium mixture, the value of Keq will always be the same. The value of Keq depends only on the temperature. The value of Keq does not depend on the amounts of reactants or products with which you start. Tro's Introductory Chemistry, Chapter 15

Initial and Equilibrium Concentrations for H2(g) + I2(g)  2HI(g) Constant [H2] [I2] [HI] 0.50 0.0 0.11 0.78 0.055 0.39 0.165 1.17 1.0 0.5 0.53 0.033 0.934 Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example 15.3—Find the Value of Keq for the Reaction from the Given Concentrations: 2 CH4(g)  C2H2(g) + 3 H2(g). Given: Find: [CH4] = 0.0203 M, [C2H2] = 0.0451 M, [H2] = 0.112 M Keq Solution Map: Relationships: [CH4], [C2H2], [H2] Keq Solve: Check: Keq is unitless. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example 15.3: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2 CH4(g)  C2H2(g) + 3 H2(g) Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g) Write down the given quantity and its units. Given: [CH4] = 0.0203 M [C2H2] = 0.0451 M [H2] = 0.112 M Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g) Information: Given: [CH4] = 0.0203M [C2H2] = 0.0451M [H2] = 0.112M Write down the quantity to find and/or its units. Find: Keq Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g) Information: Given: [CH4] = 0.0203M [C2H2] = 0.0451M [H2] = 0.112M Find: Keq Collect needed equation: Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g) Information: Given: [CH4] = 0.0203M [C2H2] = 0.0451M [H2] = 0.112M Find: Keq Write a solution map: [CH4], [C2H2], [H2] Keq Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g) Information: Given: [CH4] = 0.0203M [C2H2] = 0.0451M [H2] = 0.112M Find: Keq Solution Map: [ ]s → Keq Apply the solution map: Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Practice—Calculate Keq for the Reaction 2 NO2(g)  N2O4(g) at 100 C if the Equilibrium Concentrations Are [NO2] = 0.0172 M and [N2O4] = 0.0014 M. Tro's Introductory Chemistry, Chapter 15

Practice—Find the Value of Keq for the Reaction from the Given Concentrations: 2 NO2(g)  N2O4(g). [NO2] = 0.0172 M, [N2O4] = 0.0014 M Keq Solution Map: Relationships: [NO2], [N2O4] Keq Solve: Check: Keq is unitless.

Tro's Introductory Chemistry, Chapter 15 Example 15.4—Find the Value of [HI] for the Reaction at Equilibrium from the Given Concentrations and Keq: H2(g) + I2(g)  2HI(g) Given: Find: [I2] = 0.020 M, [H2] = 0.020 M, Keq = 69 [HI] Solution Map: Relationships: [I2], [H2], Keq [HI] Solve: Tro's Introductory Chemistry, Chapter 15

H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Example 15.4: In an equilibrium mixture, the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Write down the given quantity and its units. Given: [H2] = 0.020 M [I2] = 0.020 M Keq = 69 Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Information: Given: [H2] = 0.020M [I2] = 0.020M; Keq = 69 Write down the quantity to find and/or its units. Find: concentration of HI, M Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Information: Given: [H2] = 0.020M [I2] = 0.020M; Keq = 69 Find: [HI] Collect needed equation: Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Information: Given: [H2] = 0.020M [I2] = 0.020M; Keq = 69 Find: [HI] Write a solution map: [H2], [I2] Keq [HI] Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: In an equilibrium mixture the concentrations of H2 and I2 are both 0.020 M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C Information: Given: [H2] = 0.020M [I2] = 0.020M; Keq = 69 Find: [HI] SM: [ ] & Keq → [ ] Apply the solution map: Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 A Sample of PCl5(g) Is Placed in a 0.500 L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, 0.203 Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq = 0.0635. Tro's Introductory Chemistry, Chapter 15

A Sample of PCl5(g) Is Placed in a 0 A Sample of PCl5(g) Is Placed in a 0.500 L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, 0.203 Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq = 0.0635, Continued. PCl5 Û PCl3 + Cl2 Equilibrium Concentration, M ? if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1

Disturbing and Re-Establishing Equilibrium 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 re-established. The new concentrations will be different, but the equilibrium constant will be the same. Unless you change the temperature. Tro's Introductory Chemistry, Chapter 15

Le Châtelier’s Principle Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change. “When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.” Tro's Introductory Chemistry, Chapter 15

An Analogy: Population Changes, due to a Gold Rush! When the populations of Narnia and Middle Earth are in equilibrium, the emigration rates between the two states are equal so the populations stay constant. When an influx of population enters Middle Earth from somewhere outside Narnia, it disturbs the equilibrium established between Narnia and Middle Earth. The result will be people moving from Middle Earth into Narnia faster than people moving from Narnia into Middle Earth. This will continue until a new equilibrium between the populations is established, However, the new populations will have different numbers of people than the old ones.

The Effect of Concentration Changes on Equilibrium Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found. That has the same Keq. Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. You can use to this to drive a reaction to completion! Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium. Tro's Introductory Chemistry, Chapter 15

The Effect of Concentration Changes on Equilibrium, Continued When NO2 is added, some of it combines to make more N2O4. Tro's Introductory Chemistry, Chapter 15

The Effect of Concentration Changes on Equilibrium, Continued When N2O4 is added, some of it decomposes to make more NO2. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems: 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba2+(aq) + SO42-(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Tro's Introductory Chemistry, Chapter 15

Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued: 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba2+(aq) + SO42-(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Shift right, removing some of the added CO2 and increasing the concentrations of CO and O2. Shift left, removing some of the added Ba2+ and reducing the concentration of SO42-. Shift right, removing some of the added CO2 and decreasing the O2, while increasing the concentration of CO2.

Effect of Volume Change on Equilibrium For solids, liquids, or solutions, changing the size of the container has no effect on the concentration. Changing the volume of a container changes the concentration of a gas. Same number of moles, but different number of liters, resulting in a different molarity. Tro's Introductory Chemistry, Chapter 15

Effect of Volume Change on Equilibrium, Continued Decreasing the size of the container increases the concentration of all the gases in the container. This increases their partial pressures. 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 to reduce the pressure is to reduce the number of molecules in the container. When the volume decreases, the equilibrium shifts to the side with fewer molecules. Tro's Introductory Chemistry, Chapter 15

The Effect of Volume Change on Equilibrium, Continued Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. 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. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced. 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba2+(aq) + SO42-(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Tro's Introductory Chemistry, Chapter 15

Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced, Continued. 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba2+(aq) + SO42-(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) Shift left because there are fewer gas molecules on the reactant side than on the product side. No effect because none of the substances are gases. Shift right because there are fewer gas molecules on the product side than on the reactant side.

The Effect of Temperature Changes on Equilibrium Exothermic reactions release energy and endothermic reactions absorb energy. If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes. However, heat is not matter and not written in a proper equation. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 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 concentrations of C and D will decrease and the concentrations of A and B will increase. The value of Keq will decrease. How will decreasing the temperature effect the system? aA + bB  cC + dD + heat Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 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. The concentrations of C and D will increase and the concentrations of A and B will decrease. The value of Keq will increase. How will decreasing the temperature effect the system? Heat + aA + bB  cC + dD Tro's Introductory Chemistry, Chapter 15

The Effect of Temperature Changes on Equilibrium Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced. Heat + 2 CO2(g) Û 2 CO(g) + O2(g) BaSO4(s) Û Ba2+(aq) + SO42-(aq) (endothermic) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) (exothermic) Tro's Introductory Chemistry, Chapter 15

Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced, Continued. Heat + 2 CO2(g) Û 2 CO(g) + O2(g) Heat + BaSO4(s) Û Ba2+(aq) + SO42-(aq) CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) + Heat Shift left, reducing the value of Keq. Shift left, reducing the value of Keq. Shift right, increasing the value of Keq.

Solubility and Solubility Product Even “insoluble” salts dissolve somewhat in water. Insoluble = less than 0.1 g per 100 g H2O. The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced. AnYm(s) Û n A+(aq) + m Y-(aq) Equilibrium constant for this process is called the solubility product constant, Ksp. Ksp = [A+]n[Y-]m If there is undissolved solid in equilibrium with the solution, the solution is saturated. Larger Ksp = more soluble. For salts that produce the same number of ions. Tro's Introductory Chemistry, Chapter 15

Example—Determine the Ksp of PbBr2 if its Solubility Is 1.44 x 10-2 M. PbBr2(s) Û Pb2+(aq) + 2 Br–(aq) init -- 0 0 equil -- 0.0144 0.0288 Ksp = [Pb2+][Br–]2 = (0.0144)(0.0288)2 = 1.19 x 10-5 Tro's Introductory Chemistry, Chapter 15

Example 15.9— Calculating Molar Solubility from Ksp Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example 15.9: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Write down the given quantity and its units. Given: Ksp = 1.07 x 10-10 Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Write down the quantity to find and/or its units. Find: molar solubility, M Tro's Introductory Chemistry, Chapter 15

Collect needed equations: BaSO4(s)  Ba2+(aq) + SO42-(aq) Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Find: [BaSO4], M Collect needed equations: BaSO4(s)  Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][SO42-] because for each BaSO4 dissolved you produce one Ba2+: [BaSO4] = [Ba2+] = [SO42-] Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Find: [BaSO4], M = [Ba2+] = [SO42-] Equation: Ksp = [Ba2+][SO42-] Write a solution map: Ksp [Ba2+] Ksp = [Ba2+][SO42-] Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Find: [BaSO4], M = [Ba2+] = [SO42-] Equation: Ksp = [Ba2+][SO42-] Solution Map: Ksp → [Ba2+] Apply the solution map: Tro's Introductory Chemistry, Chapter 15

Example: Calculate the molar solubility of BaSO4. Ksp = 1 Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Find: [BaSO4], M = [Ba2+] = [SO42-] Equation: Ksp = [Ba2+][SO42-] Solution Map: Ksp → [Ba2+] Check the answer: The molar solubility of BaSO4 is 1.03 x 10-5 M. The units, M, are correct. The magnitude makes sense as BaSO4 is classified insoluble, so its solubility should be very small. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Activation Energy The energy barrier that prevents any collision between molecules from being an effective collision is called the activation energy. The larger the activation energy of a reaction, the slower it will be. At a given temperature. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Exothermic Reaction Activation energy, large Activation energy, small Reactants Relative potential energy DHreaction Products Progress of reaction Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Endothermic Reaction Activation energy Products Relative potential energy DHreaction Reactants Progress of reaction Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Catalysts A catalyst is a substance that increases the rate of a reaction, but is not consumed in the reaction. Catalysts lower the activation energy of a reaction. Catalysts work by providing an easier pathway for the reaction. Tro's Introductory Chemistry, Chapter 15

Catalyst Effect on Activation Energy Tro's Introductory Chemistry, Chapter 15

Catalyst Effect on Activation Energy, Continued Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Enzymes Enzymes are protein molecules produced by living organisms that catalyze chemical reactions. The enzyme molecules have an active site to which organic molecules bind. When the organic molecule is bound to the active site, certain bonds are weakened. This allows a particular chemical change to occur with greater ease and speed. i.e., the activation energy is lowered. Tro's Introductory Chemistry, Chapter 15

Tro's Introductory Chemistry, Chapter 15 Sucrase Tro's Introductory Chemistry, Chapter 15