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

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

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Chemical Equilibrium

Chemistry 140Chapter 8Reaction Rates and EquilibriumCopyright2000 by Houghton Mifflin Company. All rights reserved. 1CHAPTER OUTLINE8.1 Spontaneous andNonspontaneous Processes8.2 Reaction Rates8.3 Molecular Collisions8.4 Energy Diagrams8.5 Factors that Influence ReactionRates8.6 Chemical Equilibrium8.7 The Position of Equilibrium8.8 Factors That InfluenceEquilibrium Position2LEARNING OBJECTIVES/ASSESSMENTWhen 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. (Section8.4; Exercise 8.26)5. Explain how factors such as reactant concentrations, temperature, and catalysts influence reactionrates. (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 equilibriumexpressions. (Section 8.7; Exercises 8.40 and 8.46)8. Use Le Chteliers principle to predict the influence of changes in concentration and reactiontemperature on the position of equilibrium for a reaction. (Section 8.8; Exercise 8.52)3Copyright2000 by Houghton Mifflin Company. All rights reserved. 4Chemical EquilibriumThe 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.Copyright2000 by Houghton Mifflin Company. All rights reserved. 5The Law of Mass ActionForjA + kB lC + mDThe law of mass action is represented by the equilibrium expression:

Copyright2000 by Houghton Mifflin Company. All rights reserved. 6Equilibrium Expression4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

Copyright2000 by Houghton Mifflin Company. All rights reserved. 7Notes on Equilibrium Expressions (EE)The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse.When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)nThe units for K depend on the reaction being considered.Copyright2000 by Houghton Mifflin Company. All rights reserved. 8K v. KpFor jA + kB lC + mDKp = K(RT)nn = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.Copyright2000 by Houghton Mifflin Company. All rights reserved. 9Heterogeneous Equilibria. . . are equilibria that involve more than one phase.CaCO3(s) CaO(s) + CO2(g)K = [CO2]The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.Copyright2000 by Houghton Mifflin Company. All rights reserved. 10Reaction Quotient. . . helps to determine the direction of the move toward equilibrium.The law of mass action is applied with initial concentrations.Copyright2000 by Houghton Mifflin Company. All rights reserved. 11Reaction Quotient (continued)H2(g) + F2(g) 2HF(g)

Copyright2000 by Houghton Mifflin Company. All rights reserved. 12Solving Equilibrium Problems1.Balance the equation.2.Write the equilibrium expression.3.List the initial concentrations.4.Calculate Q and determine the shift to equilibrium.

Copyright2000 by Houghton Mifflin Company. All rights reserved. 13Solving Equilibrium Problems(continued)5.Define equilibrium concentrations.6.Substitute equilibrium concentrations into equilibrium expression and solve.7.Check calculated concentrations by calculating K.Copyright2000 by Houghton Mifflin Company. All rights reserved. 14Le Chteliers 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.Copyright2000 by Houghton Mifflin Company. All rights reserved. 15Effects of Changes on the System1.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).Copyright2000 by Houghton Mifflin Company. All rights reserved. 16Effects 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.

Heat of Reactionheat of reaction - the quantity of heat released or absorbed during a chemical reaction(symbol - DH) H represents enthalpy - the heat content of a system at constant pressureDH - delta H - the change in enthalpy - the amount of heat absorbed or lost by a system in a process carried out at constant pressure - DH always has a - sign in exothermic reactions and a + sign in endothermic reactions.Endothermic and Exothermic ReactionsIn the equation2H2(g) + O2(g) 2H2O(g) + 483.6kJDH = -483.6kJ exothermic (heat content of system decreases)2H2O(g) + 483.6kJ 2H2(g) + O2(g)DH = 483.6kJ endothermic (heat content of system increases)Heat of Formationmolar heat of formation - the heat released or absorbed when one mole of a compound is formed by combination of its elementsA 0 is added to DH to indicate standard heat of reaction. An f is added to indicate heat of formation. The symbol for standard heat of formation is DHf0.Thermochemical EquationsThermochemical equations for one mole of the substance given may look like:H2(g) + 1/2O2(g) H2O(l) DHf0 = -285.8kJNa(s) + 1/2Cl2(g) NaCl(s) DHf0 = -411.15kJS(s) + O2(g) SO2(g) DHf0 = -296.83kJElements in their standard state are defined as having DHf0 = 0. A negative DHf0 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.Heat of Combustionheat of combustion - the heat released by the complete combustion of one mole of a substance, symbol DHc. 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) + O2(g) CO2(g)DHc = -393.51kJCalculating Heats of ReactionHesss Law - The heat of a reaction is the same no matter how many intermediate steps make up the pathway from reactants to productsDriving Force of ReactionsWe 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 reactionEnthalpy and Reaction TendencyH2O(g) + C(s) CO(g) + H2(g) DH = +131.3kJThe enthalpy change is positive, and therefore unfavorable, yet the reaction proceeds without external agents.Entropy and Reaction TendencyIf 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 systemA 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.Free EnergyThe overall drive to spontaneous change depends on two factors:1. toward lowest enthalpy2. toward highest entropyfree energy - the difference between the change in enthalpy, DH, and the product of the Kelvin temperature and the entropy change, which is defined as TDSReaction Ratesreaction rate - the change in concentration of reactants per unit time as a reaction proceedschemical kinetics - the branch of chemistry that is concerned with reaction rates and reaction mechanismsRate Influencing Factors1. nature of reactants - the manner in which two substances react depends on the bonding capabilities between the substances2. surface area - a powdered substance will react much quicker than a cubed substance3. concentration - large increases in reaction rate are caused partly by the increase in collision frequency of reaction particles4. temperature - the rates of many reactions roughly double or triple with a 10oC rise in temperature5. presence of catalysts - a catalyst is any substance which increases the reaction rate without being permanently consumedRate Law for ReactionsR = k[A]n[B]m...R = the reaction ratek = the rate constant[A], [B], = molar concentrations of reactants n,m = experimentally determined powers

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