General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Karen C. Timberlake Lecture Presentation.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Karen C. Timberlake Lecture Presentation Chapter 10 Reaction Rates and Chemical Equilibrium

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Topics covered in this Chapter 10.1 Rates of Reactions 10.2 Chemical Equilibrium 10.3 Equilibrium Constants 10.4 Using Equilibrium Constants 10.5 Changing Equilibrium Conditions: Le Châtelier’s Principle 10.6 Predicting the direction of Reaction 10.7 Solubility Product Constant

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. A neonatal nurse works with newborns that are premature or have birth defects, cardiac malformations, and surgical problems. Most neonatal nurses care for infants from the time of birth until the time they are discharged from the hospital. Chapter 4 Reaction Rates and Chemical Equilibrium

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Chapter 10 Readiness Key Math Skills Solving Equations (1.4D) Writing Numbers in Scientific Notation (1.4F) Core Chemistry Skills Using Significant Figures in Calculations (2.3) Balancing a Chemical Equation (7.1) Calculating Concentration (9.4)

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.1 Rates of Reactions Reaction rates vary greatly for everyday processes. A banana ripens in a few days, silver tarnishes in a few months, while the aging process of humans takes many years. Learning Goal Describe how temperature, concentration, and catalysts affect the rate of a reaction.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rates of Reaction Reacting molecules must collide, have a minimum amount of energy, and have the proper orientation to form products.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Activation Energy Even when a collision has the proper orientation, there still must be sufficient energy to break the bonds between the atoms of the reactants. Three Conditions Required for a Reaction to Occur 1. Collision The reactants must collide. 2. OrientationThe reactants must align properly to break and form bonds. 3. EnergyThe collision must provide the energy of activation.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Activation Energy The activation energy is the minimum amount of energy required to break the bonds between atoms of the reactants.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rate of Reaction The rate (or speed) of a reaction is determined by measuring the amount of reactant used up in a certain period of time. product formed in a certain period of time.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Factors That Affect the Rate of a Reaction Reactions with low activation energies go faster than reactions with high activation energies. For any reaction, the rate is affected by changes in temperature. changes in reaction concentration. adding a catalyst.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rate of Reaction: Temperature At higher temperatures the increase in kinetic energy of the reactant molecules makes them move faster. makes them collide more often. makes them collide with more energy. For every 10 °C increase in temperature, most reaction rates approximately double.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rate of Reaction: Reactant Concentration When there are more reacting molecules, more collisions that form products can occur, and the reaction goes faster.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rate of Reaction: Catalysts Adding a catalyst speeds up the rate of the reaction by providing an alternative pathway that has a lower activation energy. When activation energy is lowered, more collisions provide sufficient energy for reactants to form product. During a reaction, a catalyst is not changed or consumed.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Indicate the effect of each factor listed on the rate of the reaction as (I) increases, (D) decreases, or (N) no change. 2CO(g) + O 2 (g)  2CO 2 (g) A. raising the temperature B. removing O 2 C. adding a catalyst D. lowering the temperature (I)Increases (D) Decreases (I)Increases (D) Decreases

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Solution Indicate the effect of each factor listed on the rate of the reaction as (I) increases, (D) decreases, or (N) no change. 2CO(g) + O 2 (g)  2CO 2 (g) A. raising the temperature B. removing O 2 C. adding a catalyst D. lowering the temperature (I) Increases (D) Decreases (I) Increases (D) decreases

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check State the effect of each on the rate of reaction as (I) increases, (D) decreases, or (N) no change. A. decreasing the temperature B. removing one of the reactants C. adding a catalyst D. placing the reaction flask in ice E. increasing the concentration of a reactant (D) Decreases (N) no change (D) Decreases (I) increases

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Catalytic converters are used in automobile engines to reduce pollutants such as carbon monoxide (CO), hydrocarbons such as octane (C8H18), and nitrogen oxide (NO). Chemistry Link to the Environment: Catalytic Converters

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. When pollutants pass through the surface, they react with the catalysts and are converted to CO 2, N 2, O 2, and H 2 O. Chemistry Link to the Environment: Catalytic Converters

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.2 Chemical Equilibrium In most chemical reactions, the reactants are not completely converted to products because a reverse reaction takes place in which products collide to form the reactants. Learning Goal Use the concept of reversible reactions to explain chemical equilibrium.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reversible Reactions When a reaction proceeds in both a forward and a reverse direction, it is said to be a reversible reaction. As the reactants, H 2 and I 2, collide, the forward reaction begins. HI molecules begin to form and collide with each other to form reactants in the reverse reaction. This reversible reaction is written with a double arrow. H 2 (g) + I 2 (g) 2HI(g) forward reverse

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reversible Reactions A reversible reaction occurs in both the forward and reverse direction at the same time. forward H 2 (g) + I 2 (g) 2HI (g) reverse has two rates, a rate for the forward reaction and a rate for the reverse reaction.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Write the forward and reverse reactions for the following: CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) The forward reaction is CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) The reverse reaction is CS 2 (g) + 4H 2 (g) CH 4 (g) + 2H 2 S(g)

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Rate of Reversible Reactions As the reaction progresses, the rate of the forward reaction decreases and that of the reverse reaction increases. At equilibrium, the rates of the forward and reverse reactions are equal.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Equilibrium and Reversible Reactions Equilibrium is reached when there are no further changes in the concentrations of reactants and products.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Equilibrium In the reaction H 2 (g) + I 2 (g) 2HI(g) the forward reaction is H 2 (g) + I 2 (g) 2HI(g). the reverse reaction is 2HI(g) H 2 (g) + I 2 (g). As HI product builds up, the rate of the reverse reaction increases, while the rate of the forward reaction decreases.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Equilibrium A reaction reaches equilibrium when no further changes take place in the concentration of the reactants and products. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. the forward and reverse reactions continue at the same rate.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Complete each of the following with is/are equal or is/are not equal, change(s) or do(es) not change. A.Before equilibrium is reached, the concentrations of the reactants and products ______. B.At equilibrium, the rate of the forward reaction ______ to the rate of the reverse reaction. changes equals

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Forward and Reverse Reactions If we start with reactants SO 2 and O 2, the reaction to form SO 3 takes place until equilibrium is reached. If we start with only the product SO 3, the reaction to form SO 2 and O 2 takes place until equilibrium is reached.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Complete each sentence with 1) equal 2) not equal 3) forward 4) reverse 5) changes 6) does not change A. Reactants form products in the _______ reaction. B. At equilibrium, the reactant concentration _______. C. Products form reactants in the _______ reaction. forward does not change. reverse

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.3 Equilibrium Constants At equilibrium, the number of people riding up the lift and the number of people skiing down the slope are constant. Learning Goal Calculate the equilibrium constant for a reversible reaction given the concentrations of reactants and products at equilibrium.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Equilibrium Constants An equilibrium constant for a reversible chemical reaction multiplies the concentrations of the products together and divides by the concentrations of the reactants. raises the concentration (moles/liter) of each species to a power that is equal to its coefficient in the balanced chemical equation. Core Chemistry Skill Writing the Equilibrium Constant Expression aA + bB cC + dD

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Write an equilibrium expression for the following reaction: 2CO(g) + O 2 (g) 2CO 2 (g) STEP 1 Write the balanced chemical equation. 2CO(g) + O 2 (g) 2CO 2 (g) STEP 2 Write the concentrations of the products as the numerator and the reactants as the denominator. STEP 3Write any coefficient in the equation as an exponent.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating Equilibrium Constants The equilibrium constant, K c, is the numerical value obtained by substituting experimentally measured molar concentrations at equilibrium into the expression. For example, the K c for the reaction of H 2 and I 2 is written as H 2 (g) + I 2 (g)2HI(g) the K c expression is written as Core Chemistry Skill Calculating an Equilibrium Constant

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating Equilibrium Constants For example, the concentrations of each species in Experiment 1 are [H 2 ] = 0.10 M, [I 2 ] = 0.20 M, and [HI] = 1.04 M, and the K c is in additional Experiments 2 and 3, the mixtures have different equilibrium concentrations for the system at equilibrium at the same temperature, but they have the same value of K c. Thus, a reaction at a specific temperature can have only one value for the equilibrium constant.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating Equilibrium Constants The units of K c depend on the specific equation. In this example, the units of [M 2 ]/[M 2 ] cancel out to give a value of 54. In this text, the numerical value will be given without any units.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check The decomposition of dinitrogen tetroxide forms nitrogen dioxide. N 2 O 4 (g) 2NO 2 (g) What is the numerical value of K c at 100 °C if a reaction mixture at equilibrium contains 0.45 M N 2 O 4 and 0.31 M NO 2 ? ANALYZE Given Need THE PROBLEM 0.45 M N 2 O 4 K c 0.31 M NO 2 Equation N 2 O 4 (g) 2NO 2 (g) ANALYZE Given Need THE PROBLEM 0.45 M N 2 O 4 K c 0.31 M NO 2 Equation N 2 O 4 (g) 2NO 2 (g)

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Given the following chemical reaction, H 2 (g) + I 2 (g) 2HI(g) what is the numerical value of K c at 443 °C if the equilibrium concentrations are as follows? [H 2 ] = 1.2 M [I 2 ] = 1.2 M [HI] = 0.35 M

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. In the reaction of SO 2 (g) and O 2 (g), the equilibrium mixture contains mostly product SO 3 (g), which results in a large K c. Learning Goal Use an equilibrium constant to predict the extent of the reaction and to calculate equilibrium concentrations. 10.4 Using Equilibrium Constants

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. The values of K c can be large or small, depending on whether equilibrium is reached with more products than reactants. more reactants than products. However, the size of the equilibrium constant does not affect how fast equilibrium is reached. Using Equilibrium Constants

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reactions with a large K c have large amounts of products produced from the forward reaction at equilibrium. The equilibrium constant for the reaction of SO 2 and O 2 has a large K c. At equilibrium, the reaction mixture contains mostly product and few reactants. 2SO 2 (g) + O 2 (g) 2SO 3 (g) Equilibrium with a Large K c

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reactions with a small K c have an equilibrium mixture with a low concentration of products and high concentration of reactants. The equilibrium constant for the reaction of N 2 and O 2 has a small K c. At equilibrium, the reaction mixture contains mostly reactants and few products. N 2 (g) + O 2 (g) 2NO(g) Equilibrium with a Small K c

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. The equilibrium mixture contains a very small amount of the product NO and a large amount of the reactants N 2 and O 2, which results in a small K c. Equilibrium with a Small K c

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. A few reactions have equilibrium constants close to 1, which means they have about equal concentrations of reactants and products at equilibrium. At equilibrium, a reaction with a large K c contains mostly products, whereas a reaction with a small K c contains mostly reactants. Equilibrium with a K c close to 1

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Guide to Using the Equilibrium Constant Core Chemistry Skill Calculating Equilibrium Concentrations

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. For the reaction of carbon dioxide and hydrogen, the equilibrium concentrations are 0.25 M CO 2, 0.80 M H 2, and 0.50 M H 2 O. What is the equilibrium concentration of CO(g)? CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K c = 0.11 STEP 1 State the given and needed quantities. Calculating Concentrations at Equilibrium ANALYZE Given Need THE 0.25 M CO 2, 0.80 M H 2, [CO] PROBLEM0.50 M H 2 O Equation CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K c = 0.11 ANALYZE Given Need THE 0.25 M CO 2, 0.80 M H 2, [CO] PROBLEM0.50 M H 2 O Equation CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K c = 0.11

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. For the reaction of carbon dioxide and hydrogen, the equilibrium concentrations are 0.25 M CO 2, 0.80 M H 2, and 0.50 M H 2 O. What is the equilibrium concentration of CO(g)? CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K c = 0.11 STEP 2 Write the K c expression for the equilibrium and solve for the needed concentration. Calculating Concentrations at Equilibrium

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. For the reaction of carbon dioxide and hydrogen, the equilibrium concentrations are 0.25 M CO 2, 0.80 M H 2, and 0.50 M H 2 O. What is the equilibrium concentration of CO(g)? CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K c = 0.11 STEP 3 Substitute the equilibrium (molar) concentrations and calculate the needed concentration. Calculating Concentrations at Equilibrium

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. For each K c, indicate whether the reaction mixture at equilibrium contains mostly reactants or products. A.H 2 (g) + F 2 (g) 2HF(g) K c = 1 × 10 95 Mostly Products A.3O 2 (g) 2O 3 (g) K c = 1.8 × 10 −7 Mostly reactants Study Check

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2 has the concentrations [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the concentration of PCl 5 at equilibrium? Study Check

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2 has the concentrations [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the concentration of PCl 5 at equilibrium? STEP 1 State the given and needed quantities. Solution ANALYZE Given Need THE [PCl 3 ] = [Cl 2 ] = 0.10 M [PCl 5 ] PROBLEM Equation PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2 ANALYZE Given Need THE [PCl 3 ] = [Cl 2 ] = 0.10 M [PCl 5 ] PROBLEM Equation PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2 has the concentrations [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the concentration of PCl 5 at equilibrium? STEP 2Write the K c expression for the equilibrium and solve for the needed concentration. Solution

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) K c = 4.2 × 10 −2 has the concentrations [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the concentration of PCl 5 at equilibrium? STEP 3Substitute the equilibrium (molar) concentrations and calculate the needed concentration. Solution

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.5 Changing Equilibrium Conditions: Le Châtelier’s Principle The transport of oxygen involves an equilibrium between hemoglobin (Hb), oxygen, and oxyhemoglobin (HbO 2 ). Hb(aq) + O 2 (g) HbO 2 (aq) For a person living at higher altitudes, hemoglobin concentration increases, causing a shift in the equilibrium back in the direction of HbO 2 product. Learning Goal Use Le Châtelier’s Principle to describe the changes made in equilibrium concentrations when the reaction conditions change. Hypoxia may occur at high altitudes where the oxygen concentration is lower.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Le Châtelier’s Principle When the conditions of a reaction at equilibrium are changed, the forward and reverse reactions will no longer be equal. Le Châtelier’s principle states that when a stress is placed on a reaction at equilibrium, the system responds by changing the rate of the forward or reverse reaction in the direction that relieves that stress.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Le Châtelier’s Principle The stress of adding water to tank A increases the rate of the forward direction to reestablish equal water levels and equilibrium. Core Chemistry Skill Using Le Châtelier’s Principle

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Concentration Change on Equilibrium: Adding Reactant Consider the following reaction at equilibrium: H 2 (g) + I 2 (g) 2HI(g) If more reactant (H 2 or I 2 ) is added, the rate of the forward reaction increases to form more product until the system is again at equilibrium. the equilibrium shifts toward the products. Add H 2

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Concentration Change on Equilibrium: Removing Reactant Consider the following reaction at equilibrium: H 2 (g) + I 2 (g) 2HI(g) If some of a reactant (H 2 or I 2 ) is removed, the rate of the reverse reaction increases to form more reactant until the equilibrium is reached. the equilibrium shifts toward the reactants. Remove H 2

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Concentration Change on Equilibrium: Adding Product Consider the following reaction at equilibrium: H 2 (g) + I 2 (g) 2HI(g) If more product (HI) is added, the rate of the reverse reaction increases to form more H 2 and I 2 reactants. the equilibrium shifts toward the reactants. Add HI

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Concentration Change on Equilibrium: Removing Product Consider the following reaction at equilibrium: H 2 (g) + I 2 (g) 2HI(g) When some of the product (HI) is removed, there is an decrease in collisions of HI molecules. the rate of the forward reaction increases and forms more product (HI). Remove HI

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of a Catalyst on Equilibrium Adding a catalyst speeds up a reaction by lowering the activation energy, thus increasing the rate of the forward and reverse reactions. The time to reach equilibrium is shorter; however, the same ratios of reactants and products are present. The addition of a catalyst does not change the equilibrium mixture.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Volume Change on Equilibrium: Decreasing Volume A change in the volume of a gas mixture at equilibrium will change the concentration of the gases in the mixture. 2CO(g) + O 2 (g) 2CO 2 (g) Decreasing the volume increases the concentration of the gases, and the system shifts in the direction of the smaller number of moles to compensate. Decrease Volume

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Volume Change on Equilibrium: Increasing Volume A change in the volume of a gas mixture at equilibrium will change the concentration of the gases in the mixture. 2CO(g) + O 2 (g) 2CO 2 (g) Increasing the volume decreases the concentration of the gases, and the system shifts in the direction of the larger number of moles to compensate. Increase volume

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Volume Change on Equilibrium (a) A decrease in the volume of the container causes the system to shift in the direction of fewer moles of gas. (b) An increase in the volume of the container causes the system to shift in the direction of more moles of gas.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Chemistry Link to Health: Hb Oxygen transport involves an equilibrium between hemoglobin (Hb), oxygen, and oxy-hemoglobin (HbO 2 ). Hb(aq) + O 2 (g) HbO 2 (aq) When there is a high concentration of O 2 in the alveoli of the lungs, the reaction shifts in the direction of oxy- hemoglobin. When the concentration of O 2 is low in the tissues, the reverse reaction releases O 2 from oxy-hemoglobin.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Chemistry Link to Health: Hb Given the reaction of hemoglobin, Hb(aq) + O 2 (g) HbO 2 (aq) At normal atmospheric pressure, oxygen diffuses into the blood because the partial pressure of oxygen in the alveoli is higher than that in the blood. At altitudes above 8000 ft, a decrease in atmospheric pressure results in a lower pressure of O 2.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Chemistry Link to Health: Hb At an altitude of 18,000 ft, a person will obtain 29% less oxygen and may experience hypoxia. Hypoxia may occur at high altitudes where the oxygen concentration is lower.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Chemistry Link to Health: Hb According to Le Châtelier’s principle, a decrease in oxygen shifts the equilibrium in the direction of the reactants. depletes the concentration of HbO 2, causing hypoxia. Hb(aq) + O 2 (g) HbO 2 (aq) Remove O 2

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Temperature Change on Equilibrium: Decreasing Temperature Decreasing the temperature of an endothermic reaction causes the system to respond by shifting the reaction toward more heat. shifts the reaction toward the reactants, increasing heat in system. N 2 (g) + O 2 (g) + heat 2NO(g) Decrease Temperature

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Temperature Change on Equilibrium: Increasing Temperature Increasing the temperature of an endothermic reaction causes the system to respond by shifting the reaction to remove heat. shifts the reaction toward the products, using up the heat. H 2 (g) + O 2 (g) + heat 2NO(g) Increase Temperature

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Temperature Change on Equilibrium: Decreasing Temperature Decreasing the temperature of an exothermic reaction causes the system to respond by shifting the reaction toward more heat. shifts the reaction toward the products, increasing heat in system. 2SO 2 (g) + O 2 (g) 2SO 3 (g) + heat Decrease Temperature

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Effect of Temperature Change on Equilibrium: Increasing Temperature Increasing the temperature of an exothermic reaction causes the system to respond by shifting the reaction toward removing heat. shifts the reaction toward the reactants, decreasing heat in system. 2SO 2 (g) + O 2 (g) 2SO 3 (g) + heat Increase Temperature

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Study Check Indicate the shift in equilibrium of each change. 2NO 2 (g) + heat 2NO(g) + O 2 (g) 1) toward products2) toward reactants A.adding more NO2) toward reactants B.decreasing the temperature2) toward reactants C.removing some O 2 1) toward products D.increasing the volume1) toward products E.removing some NO1) toward products

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.6 Predicting the direction of Reaction If a reaction has not yet reached equilibrium (or one that exceeded it) and you wanted to check how close it is to reaching equilibrium we use an expression called the Reaction Quotient (Q c ). The Reaction Quotient (Q c ) has the same form as the Equilibrium Constant Expression but whose concentration values are not necessarily those at equilibrium. Consider the following reaction where methane is made from carbon monoxide by hydrogenation: CO(g) + 3H 2 (g) CH 4 (g) +H 2 O(g) @ 1200K 81

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reaction Quotient - Q CO(g) + 3H 2 (g) CH 4 (g) +H 2 O(g) @ 1200K K c for the above reaction is 3.92 at 1200K For this reaction at a given instant of time, Q is: Q c = [CH 4 ] i [H 2 O] I [CO] i [H 2 ] 3 i where “i” is the concentration of reactants and products at a particular instance of time. If at this instance of time say 4 hours after reaction started, the concentrations of reactants and products were determined to be as follows: [CH 4 ] = [H 2 O] = 0.00100M and [CO] =[H 2 ] = 0.0200M Then Q c = (0.00100) i (0.00100) i = 6.25 (0.0200) i (0.0200) 3 i 82

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Reaction Quotient For the above mentioned reaction mixture to go from 6.25 (Q c value) to 3.92 (K c ) value, the reaction must proceed to the left generating more of the reactants CO and H 2. For a general reaction: aA +bB cC + dD Q c =[C] c i [D] d i [ A] a i [B] b i 83 If Q c > K c The reaction will go LEFT If Q c < K c The reaction will go RIGHT If Q c = K c Reaction is at EQUILIBRIUM

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Using Reaction Quotient A reaction vessel contains 0.0200M N 2 and 0.0600M H 2 as well as 0.0100M NH 3. Given the following reaction: N 2 (g) +3H 2 (g) 2NH 3 (g) K c =0.500 at 400 0 C. Can you predict whether more NH 3 will form or will some of it dissociate back to reactants to reach Equilibrium? 84 Q c = [NH 3 ] 2 i = (0.0100) 2 = 23.1 [N 2 ] i [H 2 ] 3 i (0.0200)(0.0600) 3 Q c > K c so Reaction goes LEFT, Ammonia will dissociate.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Using Reaction Quotient A reaction vessel contains 0.00015M CO 2 and 0.010M CO. If a small amount of carbon is added to this reaction vessel and the temperature raised to 1000 0 C, will more CO form? CO 2 (g) + C(s) 2CO(g) ; Kc = 1.17 @1000 0 C 85 Q c = [CO] 2 i = (0.010) 2 = 0.067 [CO 2 ] I (0.0015) Q c < K c so Reaction goes RIGHT, more CO will form.

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. 10.7 Solubility Product Constant There are many ionic compounds that are partially soluble in water. For the most part they are mostly insoluble in water, but a small fraction of the compound does dissolve in water and form the corresponding aqueous ions. An equilibrium is established between the insoluble compound and the corresponding aqueous ions as shown: CaC 2 O 4 (s) +H 2 O(l) Ca 2+ (aq) +C 2 O 4 2- (aq) The equilibrium constant for the above reaction is called K sp or SOLUBILITY PRODUCT CONSTANT. K sp = [Ca 2+ ][C 2 O 4 2- ] K sp is the Equilibrium Constant for a slightly soluble or nearly insoluble ionic compound. 86

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Writing Solubility Product Expression – How to write K sp Write the K sp expression for the following slightly soluble ionic compounds: a)AgCl b)Fe(OH) 3 c)PbI 2 d)BaSO 4 e)Pb 3 (AsO 4 ) 2 87 K sp = [Ag + ][Cl - ] K sp = [Fe 3+ ][OH - ] 3 K sp = [Pb 2+ ][I - ] 2 K sp = [Ba 2+ ][SO 4 2- ] K sp = [Pb 2+ ] 3 [AsO 4 3- ] 2

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating K sp from solubility and solubility from K sp Calculate K sp for the following ionic compounds: 0.000048M CaC 2 O 4 and 0.0012 M PbI 2 CaC 2 O 4 (s) Ca 2+ (aq) +C 2 O 4 2- (aq) K sp = [Ca 2+ ][C 2 O 4 2- ] = (0.000048) 2 = 2.3 x 10 -9 PbI 2 (s) Pb 2+ (aq) + 2I - (aq) K sp = [Pb 2+ ][I - ] 2 = (0.0012)(2x 0.0012) 2 = 6.9x10 -9 88

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating K sp from solubility and solubility from K sp Calculate K sp for the following ionic compounds: 0.000048M CaC 2 O 4 0.0012 M PbI 2 Calculate solubility (x) given the K sp values for the following ionic compounds: K sp Al(OH) 3 4.6 x 10 -33 CaF 2 3.4 x 10 -11 BaSO 4 1.1 x 10 -10 Cu(OH) 2 2.6 x 10 -19 89 Ksp =(x)(3x) 3 = 27x 4 Ksp = (x)(2x) 2 = 4x 3 Ksp = (x) 2 Ksp = (x) (2x) 2 = 4x 3

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Calculating K sp from solubility and solubility from K sp Calculating solubility (x) from K sp : K sp X (solubility) Al(OH) 3 4.6 x 10 -33 4 √4.6 x 10 -33 ÷ 27 = 3.6 x 10 -9 CaF 2 3.4 x 10 -11 3 √3.4 x 10 -11 ÷ 4 = 2.04 x 10 -4 BaSO 4 1.1 x 10 -10 √1.1 x 10 -10 = 1.0 x 10 -5 Cu(OH) 2 2.6 x 10 -19 3 √2.6 x 10 -19 ÷ 4 = 4.02 x 10 -7 90

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Summary of Chapter 10 Rates of Reaction are affected by three main factors: Concentration of Reactants, Temperature of Reaction, and Catalysts Some reactions especially molecular reactions are Reversible. Reversible Reactions come to a stage of Equilibrium. Equilibrium Constant is a ratio of Product concentration divided by Reactant concentration. Kc = [Product]/[Reactant]; small Kc favors reactants and large Kc favors Products. Le Châtelier’s Principle explains how shifts in concentration, volume and temperature can affect an Equilibrium. Q c predicts direction of reaction. K sp 91

General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Karen C. Timberlake Lecture Presentation.

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General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Karen C. Timberlake Lecture Presentation.

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Presentation on theme: "General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc. Karen C. Timberlake Lecture Presentation."— Presentation transcript:

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