Chapter 17a Reaction Rates and Equilibrium

Chapter 17 Table of Contents Copyright © Cengage Learning. All rights reserved How Chemical Reactions Occur 17.2 Conditions That Affect Reaction Rates 17.3 The Equilibrium Condition 17.4 Chemical Equilibrium: A Dynamic Condition

Chapter 17 Table of Contents Copyright © Cengage Learning. All rights reserved 3 Rates of Chemical Reactions 4 C 3 H 5 N 3 O 9  6 N H 2 O + 12 CO 2 + O kJ Nitroglycerine Exothermic

Chapter 17 Table of Contents Copyright © Cengage Learning. All rights reserved 4 An Example of Reaction Rates Fast Reaction vs. Slow Reaction 35/97 people died in 1937

Section 17.1 How Chemical Reactions Occur Return to TOC Copyright © Cengage Learning. All rights reserved 5 Collision Model Molecules must collide in order for a reaction to occur. Rate depends on concentrations of reactants and temperature.

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 6 Concentration – increases rate because more molecules lead to more collisions. Temperature – increases rate.  Why?

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 7 How to Tame Allergic Reactions How can you slow down a histamine attack? Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 8 Activation Energy Minimum energy required for a reaction to occur.

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 9 What makes Switzerland unique?

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 10

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 11 Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 12 Catalyst A substance that speeds up a reaction without being consumed. Enzyme – catalyst in a biological system

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 13 Catalyst A substance that speeds up a reaction without being consumed. Enzyme – catalyst in a biological system

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 14 Catalyst A substance that speeds up a reaction without being consumed. Enzyme – catalyst in a biological system Chlorofluoro Carbons (CFC’s) are acting as catalysts to decompose the ozone (O 3 ) layer. The ozone layer is formed from cosmic radiation and protects us from UV light.

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 15 Depletion is measured by T.O.M.S. “Total Ozone Mapping Spectrometer” The below dark shaded are shows the amount of depletion around the Antarctica

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 16 An Amana refrigerator, one of many appliances that now use HFC-134a. This compound is replacing CFC’s, which lead to destruction of atmospheric ozone.

Conditions That Affect Reaction Rates Section 17.2 Return to TOC Copyright © Cengage Learning. All rights reserved 17 Use a catalytic converter to convert the polluting exhaust gases of burned lead-free gasoline into harmless gases. Platinum (Pt) is the catalysts used. Only a small amount is needed.

The Equilibrium Condition Section 17.3 Return to TOC Copyright © Cengage Learning. All rights reserved 18 Equilibrium The exact balancing of two processes, one of which is the opposite of the other.

The Equilibrium Condition Section 17.3 Return to TOC Copyright © Cengage Learning. All rights reserved 19 Chemical Equilibrium A dynamic state where the concentrations of all reactants and products remain constant.

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 20 Chemical Equilibrium On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Macroscopically static Microscopically dynamic

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 21 The Reaction of H 2 O and CO to Form CO 2 and H 2 as Time Passes Equal numbers of moles of H 2 O and CO are mixed in a closed container. The reaction begins to occur, and some products (H 2 and CO 2 ) are formed. The reaction continues as time passes and more reactants are changed to products. Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 22

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 23 Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 25 Concept Check Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved 26 Reactions Rates and Equilibrium Chapter 17b W

Section 17.4 Chemical Equilibrium: A Dynamic Condition Return to TOC Copyright © Cengage Learning. All rights reserved The Equilibrium Constant: An Introduction 17.6Heterogeneous Equilibria 17.7Le Châtelier’s Principle 17.8Applications Involving the Equilibrium Constant 17.9Solubility Equilibria

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 28 Consider the following reaction at equilibrium: jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). j l k m [B][A] [D] [C] K =

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 29 Example N 2 (g) + 3H 2 (g) 2NH 3 (g)

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 30 K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one value for K.  Equilibrium position is a set of equilibrium concentrations.

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 31 Concept Check Consider the following equilibrium reaction: HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 – (aq) Determine the equilibrium constant expression for the dissociation of acetic acid. a)b) c)d)

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 32 Exercise For the reaction below, calculate the value of the equilibrium constant, given the equilibrium concentrations. N 2 O 4 (g) 2NO 2 (g) [N 2 O 4 ] = M [NO 2 ] = M a)K = b)K = 0.92 c)K = 1.1 d)K = K = (0.060) 2 /0.055 = 0.065

Section 17.5 The Equilibrium Constant: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 33 What is the equilibrium expression for the following? CH 4 (g) + 2 H 2 S(g) CS 2 (g) + 4 H 2 (g) K eq = [CS 2 ][H 2 ] 4 [CH 4 ][H 2 S] 2 H 2 (g) + I 2 (g) 2 HI(g) K eq = [HI] 2 [H 2 ][I 2 ] Fe 3+ (aq) + SCN - (aq) Fe(SCN) +2 (aq) K eq = [Fe(SCN )+2 ] [Fe 3+ ][SCN - ]

Section 17.6 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 34 Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

Section 17.6 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 35 Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g)

Section 17.6 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 36 The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.  The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g)

Section 17.6 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 37 Concept Check Determine the equilibrium expression for the reaction: CaF 2 (s) Ca 2+ (aq) + 2F – (aq) a) b) c) d)

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 38 If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. O 3 (g) + Cl (g) O 2 (g) + OCl(g) Equilibrium shifts to counter a disturbance. Hills and Valleys!

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 39 Effect of a Change in Concentration When a reactant or product is added the system shifts away from that added component. If a reactant or product is removed, the system shifts toward the removed component.

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 40 Effect of a Change in Volume The system is initially at equilibrium. The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO 2 molecules, lowering the pressure again.

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 41 Effect of a Change in Volume Decreasing the volume  The system shifts in the direction that gives the fewest number of gas molecules.

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 42 Effect of a Change in Volume Increasing the volume  The system shifts in the direction that increases its pressure.

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 43 Effect of a Change in Temperature The value of K changes with temperature. We can use this to predict the direction of this change. Exothermic reaction – produces heat (heat is a product)  Adding energy shifts the equilibrium to the left (away from the heat term). Endothermic reaction – absorbs energy (heat is a reactant)  Adding energy shifts the equilibrium to the right (away from the heat term).

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 44 Effect of Temperature on Equilibrium NO 2 (g) N 2 O 4 (g)

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 45 Industrial Application-The Manufacture of Ammonia N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH = kJ mol -1 To increase production how would you manipulate the equilibrium? 2. Lower Temperature 1. Lower Volume 3. Remove Product

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 46 Concept Check Consider the reaction: 2CO 2 (g) 2CO(g) + O 2 (g) How many of the following changes would lead to a shift in the equilibrium position towards the reactant? I. The removal of CO gas. II. The addition of O 2 gas. III. The removal of CO 2 gas. IV. Increasing the pressure in the reaction by decreasing the volume of the container. a) 1 b)2 c)3 d)4

Section 17.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 47 Concept Check One method for the production of hydrogen gas can be described by the following endothermic reaction: CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) How many of the following changes would decrease the amount of hydrogen gas (H 2 ) produced? I. H 2 O(g) is added to the reaction vessel. II. The volume of the container is doubled. III.CH 4 (g) is removed from the reaction vessel. IV.The temperature is increased in the reaction vessel. a) 1 b)2 c)3 d)4

Section 17.8 Applications Involving the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 48 A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.  Reaction goes essentially to completion. The Extent of a Reaction

Section 17.8 Applications Involving the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 49 A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.  Reaction does not occur to any significant extent. The Extent of a Reaction

Section 17.8 Applications Involving the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 50 The value of K for a system can be calculated from a known set of equilibrium concentrations. Unknown equilibrium concentrations can be calculated if the value of K and the remaining equilibrium concentrations are known.

Section 17.8 Applications Involving the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 51 Concept Check If the equilibrium lies to the right, the value for K is __________. large (or >1) If the equilibrium lies to the left, the value for K is ___________. small (or <1)

Section 17.8 Applications Involving the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 52 Concept Check At a given temperature, K = 50 for the reaction: H 2 (g) + I 2 (g) 2HI(g) Calculate the equilibrium concentration of H 2 given: [I 2 ] = 1.5 × 10 –2 M and [HI] = 5.0 × 10 –1 M a) 1.5 × 10 –2 M b) 3.0 × 10 –2 M c) 5.0 × 10 –1 M d) 3.3 × 10 –1 M K = (HI) 2 /(H 2 )(I 2 ) 50 = (5.0 × 10 –1 ) 2 /(H 2 )(1.5 × 10 –2 ) (H 2 ) = 3.3 × 10 –1 M

Section 17.9 Solubility Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 53 The equilibrium conditions also applies to a saturated solution containing excess solid, MX(s).  K sp = [M + ][X  ] = solubility product constant  The value of the K sp can be calculated from the measured solubility of MX(s).

Section 17.9 Solubility Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 54 Solubility Equilibria Solubility product (K sp ) – equilibrium constant; has only one value for a given solid at a given temperature. Solubility – an equilibrium position. Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2– (aq)

Section 17.9 Solubility Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 55 Concept Check If a saturated solution of PbCl 2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb 2+ is determined to be 1.6 × 10 –2 M, what is the value of K sp ? a) 2.6 × 10 –4 b) 2.0 × 10 –4 c) 3.2 × 10 –2 d) 1.6 × 10 –5 K sp = [Pb 2+ ] [Cl – ] 2 = (1.6 × 10 – 2 ) (3.2 × 10 – 2 ) 2 = 1.6 × 10 – 5

Section 17.9 Solubility Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 56 Concept Check Calculate the solubility of silver chloride in water. K sp = 1.6 × 10 –10 a) 1.3 × 10 –5 M b)1.6 × 10 –10 M c)3.2 × 10 –10 M d)8.0 × 10 –11 M K sp = [Ag + ][Cl – ] 1.6 × 10 – 10 = (x)(x) = x 2 x = 1.3 × 10 – 5 M