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Chemistry 140 Chapter 8 Reaction Rates and Equilibrium Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1.

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Presentation on theme: "Chemistry 140 Chapter 8 Reaction Rates and Equilibrium Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1."— Presentation transcript:

1 Chemistry 140 Chapter 8 Reaction Rates and Equilibrium Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1

2 CHAPTER OUTLINE 8.1 Spontaneous and Nonspontaneous Processes 8.2 Reaction Rates 8.3 Molecular Collisions 8.4 Energy Diagrams 8.5 Factors that Influence Reaction Rates 8.6 Chemical Equilibrium 8.7 The Position of Equilibrium 8.8 Factors That Influence Equilibrium Position 2

3 LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1. Use the concepts of energy and entropy to predict the spontaneity of processes and reactions. (Section 8.1; Exercise 8.6) 2. Calculate reaction rates from experimental data. (Section 8.2; Exercise 8.14) 3. Use the concept of molecular collisions to explain reaction characteristics. (Section 8.3; Exercise 8.20) 4. Represent and interpret the energy relationships for reactions by using energy diagrams. (Section 8.4; Exercise 8.26) 5. Explain how factors such as reactant concentrations, temperature, and catalysts influence reaction rates. (Section 8.5; Exercise 8.30) 6. Relate experimental observations to the establishment of equilibrium. (Section 8.6; Exercise 8.38) 7. Write equilibrium expressions based on reaction equations, and do calculations based on equilibrium expressions. (Section 8.7; Exercises 8.40 and 8.46) 8. Use Le Châtelier’s principle to predict the influence of changes in concentration and reaction temperature on the position of equilibrium for a reaction. (Section 8.8; Exercise 8.52) 3

4 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

5 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 The Law of Mass Action For jA + kB  lC + mD The law of mass action is represented by the equilibrium expression:

6 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Equilibrium Expression 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g)

7 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Notes on Equilibrium Expressions (EE) 4 The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. 4 When the equation for a reaction is multiplied by n, EE new = (EE original ) n 4 The units for K depend on the reaction being considered.

8 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 K v. K p For jA + kB  lC + mD K p = K(RT)  n  n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

9 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

10 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Reaction Quotient... helps to determine the direction of the move toward equilibrium. The law of mass action is applied with initial concentrations.

11 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Reaction Quotient (continued) H 2 (g) + F 2 (g)  2HF(g)

12 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Solving Equilibrium Problems 1.Balance the equation. 2.Write the equilibrium expression. 3.List the initial concentrations. 4.Calculate Q and determine the shift to equilibrium.

13 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Solving Equilibrium Problems (continued ) 5.Define equilibrium concentrations. 6.Substitute equilibrium concentrations into equilibrium expression and solve. 7.Check calculated concentrations by calculating K.

14 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Le Châtelier’s Principle... 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.

15 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Effects of Changes on the System 1.Concentration: The system will shift away from the added component. 2.Temperature: K will change depending upon the temperature (treat the energy change as a reactant).

16 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 Effects of Changes on the System (continued) 3.Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

17 Heat of Reaction heat of reaction - the quantity of heat released or absorbed during a chemical reaction (symbol -  H) H represents enthalpy - the heat content of a system at constant pressure  H - delta H - the change in enthalpy - the amount of heat absorbed or lost by a system in a process carried out at constant pressure -  H always has a “-” sign in exothermic reactions and a “+” sign in endothermic reactions.

18 Endothermic and Exothermic Reactions In the equation 2H 2 (g) + O 2 (g)  2H 2 O(g) + 483.6kJ  H = -483.6kJ exothermic (heat content of system decreases) 2H 2 O(g) + 483.6kJ  2H 2 (g) + O 2 (g)  H = 483.6kJ endothermic (heat content of system increases)

19 Heat of Formation molar heat of formation - the heat released or absorbed when one mole of a compound is formed by combination of its elements A “ 0 ” is added to  H to indicate standard heat of reaction. An “f” is added to indicate heat of formation. The symbol for standard heat of formation is  H f 0.

20 Thermochemical Equations Thermochemical equations for one mole of the substance given may look like: H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H f 0 = -285.8kJ Na(s) + 1/2Cl 2 (g)  NaCl(s)  H f 0 = -411.15kJ S(s) + O 2 (g)  SO 2 (g)  H f 0 = -296.83kJ Elements in their standard state are defined as having  H f 0 = 0. A negative  H f 0 indicates a substance that is more stable than the free elements. A compound with a high positive heat of formation is very unstable and may be explosive.

21 Heat of Combustion heat of combustion - the heat released by the complete combustion of one mole of a substance, symbol  H c. Heat of combustion is defined in terms of one mole of reactant. Heat of formation is defined in terms of one mole of product. Example: C(s) + O 2 (g)  CO 2 (g)  H c = -393.51kJ

22 Calculating Heats of Reaction Hess’s Law - The heat of a reaction is the same no matter how many intermediate steps make up the pathway from reactants to products

23 Driving Force of Reactions We will now look at what causes a reaction to occur or the driving force behind a reaction. Most chemical reactions are exothermic. This means the products have less energy than the original reactants. One would expect these reactions to occur without the assistance of an external agent. It should follow that endothermic reactions should not occur without the aid of an external agent. However, some endothermic reactions do take place without the aid of external agents. The production of water gas is such a reaction

24 Enthalpy and Reaction Tendency H 2 O(g) + C(s)  CO(g) + H 2 (g)  H = +131.3kJ The enthalpy change is positive, and therefore unfavorable, yet the reaction proceeds without external agents.

25 Entropy and Reaction Tendency If a process is not driven by decreasing energy, what does drive it? Natrual processes tend toward maximum disorder. entropy - the property that describes the disorder of a system A system that can go from one state to another without an enthalpy change does so by becoming more disordered. Entropy increases, for example, when a gas expands, a solid or liquid dissolves, or the number of particles in a system increases.

26 Free Energy The overall drive to spontaneous change depends on two factors: 1. toward lowest enthalpy 2. toward highest entropy free energy - the difference between the change in enthalpy,  H, and the product of the Kelvin temperature and the entropy change, which is defined as T  S

27 Reaction Rates reaction rate - the change in concentration of reactants per unit time as a reaction proceeds chemical kinetics - the branch of chemistry that is concerned with reaction rates and reaction mechanisms

28 Rate Influencing Factors 1. nature of reactants - the manner in which two substances react depends on the bonding capabilities between the substances 2. surface area - a powdered substance will react much quicker than a cubed substance 3. concentration - large increases in reaction rate are caused partly by the increase in collision frequency of reaction particles 4. temperature - the rates of many reactions roughly double or triple with a 10 o C rise in temperature 5. presence of catalysts - a catalyst is any substance which increases the reaction rate without being permanently consumed

29 Rate Law for Reactions R = k[A] n [B] m... R = the reaction rate k = the rate constant [A], [B], = molar concentrations of reactants n,m = experimentally determined powers


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