© 2013 Pearson Education, Inc. Chapter 9, Section 1 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake Chapter 9 © 2013 Pearson Education, Inc. Reaction Rates and Chemical Equilibrium 9.1 Rates of Reactions Lectures

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Collision Theory of Reactions A chemical reaction occurs when  collisions between reactant molecules have sufficient energy to break their bonds.  molecules collide with the proper orientation.  bonds between atoms of the reactants (N 2 and O 2 ) are broken and new bonds (NO) form. 2

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Collision Theory of Reactions A chemical reaction does not take place if the  collisions between reactant molecules do not have sufficient energy to break their bonds, or if  molecules are not properly aligned. 3

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Activation Energy Activation energy is the minimum energy needed for a reaction to take place or break the bonds between atoms of reactants. 4

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Reaction Rate The rate of reaction is determined by measuring the amount of reactant used up or product formed in a certain period of time. The rate or reaction is affected by  temperature changes.  changes in concentrations.  the presence of a catalyst. 5

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Reaction Rate and Temperature At higher temperatures,  the increase in kinetic energy causes the reacting molecules to move faster,  more collisions occur, and  more colliding molecules have sufficient energy to react and form products. For example, we refrigerate perishable foods to slow down the reactions that cause spoilage. 6

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Reaction Rate and Concentration Increasing the concentration of reactants  increases the number of collisions and  increases the reaction rate. 7

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Reaction Rate and Catalysts A catalyst  speeds up the rate of a reaction.  lowers the energy of activation.  is not used up during the reaction. Insert reaction rate and catalysis top pg 5 8

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Learning Check State the effect of each change on the rate of reaction as increases, decreases or no change. 1. increasing the temperature 2. removing some of the reactants 3. adding a catalyst 4. placing the reaction flask in ice 5. increasing the concentration of a reactant 9

© 2013 Pearson Education, Inc. Chapter 9, Section 1 Learning Check Indicate the effect of each factor listed on the rate of the following reaction as increases, decreases, or does not change. A. raising the temperature B. removing some O 2 C. adding a catalyst D. lowering the temperature 10

© 2013 Pearson Education, Inc. Chapter 9, Section 2 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 9.2 Chemical Equilibrium Chapter 9 Reaction Rates and Chemical Equilibrium © 2013 Pearson Education, Inc. Lectures

© 2013 Pearson Education, Inc. Chapter 9, Section 2 Reversible Reactions A reversible reaction proceeds in both the forward and reverse directions. As a result there are two reaction rates: the rate of the forward reaction and the rate of the reverse reaction.  When molecules begin to react, the rate of the forward reaction is faster than the rate of the reverse reaction.  As reactants are consumed and products accumulate, the rate of the forward reaction decreases, whereas the rate of the reverse reaction increases. 12

© 2013 Pearson Education, Inc. Chapter 9, Section 213 Reversible Reactions

© 2013 Pearson Education, Inc. Chapter 9, Section 214 Reversible Reactions Suppose SO 2 and O 2 are present initially. As they collide, the forward reaction begins. 2SO 2 (g) + O 2 (g) 2SO 3 (g) As SO 3 molecules form, they also collide in the reverse reaction that forms reactants. This reversible reaction is written with a double arrow. forward 2SO 2 (g) + O 2 (g) 2SO 3 (g) reverse

© 2013 Pearson Education, Inc. Chapter 9, Section 2 Chemical Equilibrium At equilibrium,  the rate of the forward reaction becomes equal to the rate of the reverse reaction.  the forward and reverse reactions continue at equal rates.  no further changes occur in the concentration of reactants and products. Insert bottom page 7 right column, reaction diagram. 15

© 2013 Pearson Education, Inc. Chapter 9, Section 2 Equilibrium Given the reaction, H 2 (g) + I 2 (g) 2HI(g) at equilibrium, the forward and reverse reactions occur at the same time and are shown together using a double arrow.  Forward reaction: H 2 (g) + I 2 (g) 2HI(g)  Reverse reaction: 2HI(g) H 2 (g) + I 2 (g) H 2 (g) + I 2 (g) 2HI(g) 16

© 2013 Pearson Education, Inc. Chapter 9, Section 2 Equilibrium H 2 (g) + I 2 (g) 2HI(g)  Initially, the reaction flask contains only the reactants H 2 and I 2.  The forward reaction begins to produce HI.  As the reaction proceeds, there is less H 2 and I 2 and more HI, which increases the rate of the reverse reaction.  At equilibrium, the concentrations of reactants and product are constant.  The reaction continues, with the rate of the forward reaction equal to the rate of the reverse reaction. 17

© 2013 Pearson Education, Inc. Chapter 9, Section 218 Learning Check Write the forward and reverse reactions for the following. CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g)

© 2013 Pearson Education, Inc. Chapter 9, Section 219 Learning Check Complete with equal, not equal, forward, reverse, changes, or does not change. 1. Reactants form products in the ________ reaction. 2. At equilibrium, the reactant concentration _______. 3. When products form reactants, it is the _______ reaction. 4. At equilibrium, the rate of the forward reaction is ______ to the rate of the reverse reaction. 5. If the forward reaction is faster than the reverse, the amount of products ________.

© 2013 Pearson Education, Inc. Chapter 9, Section 3 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 9.3 Equilibrium Constants Chapter 9 Reaction Rates and Chemical Equilibrium © 2013 Pearson Education, Inc. Lectures

© 2013 Pearson Education, Inc. Chapter 9, Section 3 For the reaction,  the equilibrium constant expression, K c, gives the concentrations of the reactants and products at equilibrium,  the square brackets indicate the moles/liter of each substance, and  the coefficients b and a are written as superscripts that raise the moles/liter to a specific power. Equilibrium Constants 21

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Guide to Writing the K c Expression 22

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Writing a K c Expression Write the K c expression for the following reaction. Step 1 Write the balanced chemical equation. Step 2 Write the concentrations of the products as the numerator and the reactants as the denominator. 23

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Writing a K c Expression Write the K c expression for the following reaction. Step 3 Write any coefficient in the equation as an exponent. 24

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Learning Check Which of the following is the correctly written K c expression for the reaction shown below? A.B. C. D. 25

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Homogeneous, Heterogeneous Equilibrium  In a homogeneous equilibrium, all reactants and products in the reaction are in the same physical state.  In a heterogeneous equilibrium, the reactants and products are in two or more physical states. The concentration of solids and liquids are constant, and therefore omitted from the equilibrium constant expression. 26

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Learning Check Write the equilibrium constant expression for the following reaction at equilibrium. 27

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Guide to Calculating the K c 28

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Calculating Equilibrium Constants What is the K c for the following reaction at 427 ˚C if the equilibrium concentrations are [H 2 ] = 0.20 M, [I 2 ] = 0.20 M and [HI] = 1.47 M? Step 1 Write the K c expression for the equilibrium. 29

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Calculating Equilibrium Constants What is the K c for the following reaction at 427 ˚C if the equilibrium concentrations are [H 2 ] = 0.20 M, [I 2 ] = 0.20 M and [HI] = 1.47 M? Step 2 Substitute equilibrium (molar) concentrations and calculate Kc. 30

© 2013 Pearson Education, Inc. Chapter 9, Section 3 Learning Check Calculate the K c for the reaction, with the following equilibrium concentrations, 31

© 2013 Pearson Education, Inc. Chapter 9, Section 432 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 9.4 Using Equilibrium Constants Chapter 9 Reaction Rates and Chemical Equilibrium © 2013 Pearson Education, Inc. Lectures

© 2013 Pearson Education, Inc. Chapter 9, Section 433 Reaching Chemical Equilibrium The values of equilibrium constants can be large or small. The size of the K c depends on whether equilibrium is reached with  more products than reactants (K c is large), or  more reactants than products (K c is small).

© 2013 Pearson Education, Inc. Chapter 9, Section 434 K c is Large. When a reaction has a large equilibrium constant, it means  the forward reaction produced a large amount of products when equilibrium was reached.  the equilibrium mixture contains mostly products.

© 2013 Pearson Education, Inc. Chapter 9, Section 435 K c is Large.

© 2013 Pearson Education, Inc. Chapter 9, Section 436 K c is Small. When a reaction has a small equilibrium constant, it means  the reverse reaction converted a large amount of products back to reactants.  the equilibrium mixture contains mostly reactants.

© 2013 Pearson Education, Inc. Chapter 9, Section 437 K c is Small.

© 2013 Pearson Education, Inc. Chapter 9, Section 438 Summary of K c Values A reaction  that favors products has a large K c.  with about equal concentrations of products and reactants has a K c close to 1.  that favors reactants has a small K c.

© 2013 Pearson Education, Inc. Chapter 9, Section 439 Large and Small K c Values

© 2013 Pearson Education, Inc. Chapter 9, Section 440 Learning Check For each K c, indicate whether the reaction at equilibrium contains mostly reactants or products.

© 2013 Pearson Education, Inc. Chapter 9, Section 441 Guide to Using the K c Value

© 2013 Pearson Education, Inc. Chapter 9, Section 442 At equilibrium, the reaction, has a K c of and contains What is the equilibrium concentration of PCl 5 ? Problem Facts Calculating Equilibrium Concentration

© 2013 Pearson Education, Inc. Chapter 9, Section 443 At equilibrium, the reaction, has a K c of 4.2 x 10 –2 and contains What is the equilibrium concentration of PCl 5 ? Step 1 Write the K c expression for the equilibrium equation. Step 2 Solve the K c expression for the unknown concentration. Calculating Equilibrium Concentration

© 2013 Pearson Education, Inc. Chapter 9, Section 444 Calculating Equilibrium Concentration At equilibrium, the reaction, has a K c of and contains What is the equilibrium concentration of PCl 5 ? Step 3 Substitute the known values into the rearranged K c expression and calculate.

© 2013 Pearson Education, Inc. Chapter 9, Section 445 Learning Check The K c is 2.0 for the reaction, If the equilibrium concentrations are and [ what is the equilibrium concentration of Br 2 ? A M B M C. 1.3 M

© 2013 Pearson Education, Inc. Chapter 9, Section 546 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 9.5 Changing Equilibrium Conditions: Le Châtelier’s Principle Chapter 9 Reaction Rates and Chemical Equilibrium © 2013 Pearson Education, Inc. Lectures

© 2013 Pearson Education, Inc. Chapter 9, Section 547 Le Châtelier’s Principle Chemical equilibrium can be disturbed by a change in concentration, volume, or temperature. Altering any of these conditions puts the system under stress. Le Châtelier’s principle states that  when a system at equilibrium is disturbed, the system will shift in the direction that will reduce that stress.  there will be a change in the rate of the forward or reverse reaction to return the system to equilibrium.

© 2013 Pearson Education, Inc. Chapter 9, Section 548 Changing Concentrations For the following reaction at equilibrium,  the rate of the forward reaction is increased to relieve the stress when more H 2 or I 2 is added, or HI is removed.  the rate of the reverse reaction is increased to relieve the stress when H 2 or I 2 is removed or more HI is added.

© 2013 Pearson Education, Inc. Chapter 9, Section 549 Adding Reactant or Product The equilibrium shifts toward  products when H 2 (g) or I 2 (g) is added.  reactants when HI(g) is added. Add H 2 or I 2 Add HI

© 2013 Pearson Education, Inc. Chapter 9, Section 550 Removing Reactant or Product The system shifts toward  a reverse reaction when H 2 or I 2 is removed.  a forward reaction when HI(g) is removed.

© 2013 Pearson Education, Inc. Chapter 9, Section 551 Concentration Changes and Equilibrium (a) The addition of H 2 places stress on the equilibrium system of H 2 (b) To relieve the stress, the forward reaction converts some reactants, to product, HI(g). (c) A new equilibrium is established when the rates of the forward reaction and the reverse reaction become equal.

© 2013 Pearson Education, Inc. Chapter 9, Section 552 Effect of a Catalyst Adding a catalyst  lowers the activation energy of the forward reaction.  increases the rate of the forward reaction.  lowers the activation energy of the reverse reaction.  increases the rate of the reverse reaction.  decreases the time to reach equilibrium.  has no effect on the concentrations at equilibrium.

© 2013 Pearson Education, Inc. Chapter 9, Section 553 Learning Check Predict any shift in the forward or reverse reactions for each of the following changes on the reaction. 1. H 2 S(g) is added. 2. NH 3 (g) is removed. 3. A catalyst is added.

© 2013 Pearson Education, Inc. Chapter 9, Section 554 Effect of Volume Change Changing the volume of a gas mixture at equilibrium will change the concentrations of gases in the mixture, upsetting the equilibrium.  Decreasing the mixture volume will increase the concentration of gases.  Increasing the mixture volume will decrease the concentration of gases.

© 2013 Pearson Education, Inc. Chapter 9, Section 555 Effect of Volume Change Decreasing the volume of the gas mixture shifts the equilibrium towards the fewer number of moles. Increasing the volume of the gas mixture shifts the equilibrium toward the larger number of moles.

© 2013 Pearson Education, Inc. Chapter 9, Section 556 Volume Decrease and Equilibrium  A decrease in the volume of the container causes the system to shift in the direction of fewer moles of gas.  An increase in the volume of the container causes the system to shift in the direction of more moles of gas.

© 2013 Pearson Education, Inc. Chapter 9, Section 557 Temperature Change and Endothermic Reactions For an endothermic reaction at equilibrium, heat is a reactant.  A decrease in temperature (T) removes heat, shifting the equilibrium toward the reactants.  An increase in temperature adds heat, shifting the equilibrium toward the products.

© 2013 Pearson Education, Inc. Chapter 9, Section 558 Temperature Change and Exothermic Reactions For an exothermic reaction at equilibrium,  an increase in temperature adds heat and the equilibrium shifts toward the reactants.  a decrease in temperature (T) removes heat and the equilibrium shifts toward the products.

© 2013 Pearson Education, Inc. Chapter 9, Section 559 Effects of Changes on Equilibrium

© 2013 Pearson Education, Inc. Chapter 9, Section 560 Learning Check Indicate if each change on a reaction at equilibrium shifts towards the products or reactants or does not change the equilibrium. 1. adding NO(g) 2. raising the temperature 3. increasing the volume