*17Chemical Equilibrium

*Chapter GoalsBasic ConceptsThe Equilibrium Constant Variation of Kc with the Form of the Balanced Equation The Reaction Quotient Uses of the Equilibrium Constant, KcDisturbing a System at Equilibrium: PredictionsThe Haber Process: A Commercial Application of EquilibriumDisturbing a System at Equilibrium: CalculationsPartial Pressures and the Equilibrium ConstantRelationship between Kp and KcHeterogeneous EquilibriaRelationship between Gorxn and the Equilibrium ConstantEvaluation of Equilibrium Constants at Different Temperatures

*Basic ConceptsChemical reactions that can occur in either direction are called reversible reactionReversible reactions do not go to completion Reactants are not completely converted to products. ()They can occur in either directionSymbolically, this is represented as:

aA(g)+bB(g ) cC(g)+dD(g)When A and B react to form C and D at the same rate at which C and D react to form A and B, the system is equilibrium (ABCDCDAB)

*Basic Concepts Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate ()Chemical equilibria are dynamic equilibria ()Molecules are continually reacting, even though the overall composition of the reaction mixture does not change

*Basic ConceptsOne example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution. (131, ) 1. place solid PbI2* in a saturated PbI2 solution PbI2(s)* Pb2+(aq)+2I-(aq) 2. Stir for a few minutes, then filter the solution some of the radioactive iodine will go into solutionPbI2,,

*Basic ConceptsGraphically, this is a representation of the rates for the forward and reverse reactions for this general reactionaA(g)+bB(g ) cC(g)+dD(g)Equilibrium isestablished

*Basic ConceptsOne of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction2SO2(g)+ O2(g ) 2SO3(g)0.02M

*Basic Concepts2SO2(g)+ O2(g ) 2SO3(g)2SO2(g)+ O2(g ) 2SO3(g)2: 1: 22: 1: 2

*Basic Concepts2SO2(g)+ O2(g ) 2SO3(g)2SO2(g)+ O2(g ) 2SO3(g)2: 1: 22: 1: 2In 1.00 liter container,

*The Equilibrium ConstantFor a simple one-step mechanism reversible reaction such as:The rates of the forward and reverse reactions can be represented as:Forward rate (): Ratef = kf[A][B] Reverse rate (): Rater = kr[C][D]

A(g)+B(g ) C(g)+D(g)

*The Equilibrium ConstantWhen system is at equilibrium Ratef = Rater =which represents the forward rate kf[A][B] = kr[C][D] which rearranges to kf [C][D] kr [A][B] Because the ratio of two constants is a constant we can define a new constant as follows :

*The Equilibrium ConstantSimilarly, for the general reaction:we can define a constant: Kc

aA(g)+bB(g ) cC(g)+dD(g)Products Reactants This expression is valid for all reactions

*The Equilibrium ConstantKc is the equilibrium constant .Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.Kc

*The Equilibrium ConstantExample 17-1: Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC. PCl5 PCl3 + Cl2H2 + l2 2HI4NH3 + 5O2 4NO + 6H2 O

The Equilibrium Constant*Example 17-1: Calculation of Kc Some nitrogen and hydrogen are placed in an empty 5.00-liter container at 500oC. When equilibrium is established, 3.01mol of N2, 2.10 mol of H2, and 0.565 mol of NH3 are present. Evaluate Kc for the following reaction at 500oC.

N2(g) + 3H2(g) 2NH3(g) [N2]: 3.01mol/5L = 0.602 M[H2]: 2.10mol/5L = 0.420 M[NH3]: 0.565mol/5L = 0.113 M=(0.602)(0.420)3(0.113)2=0.286

*The Equilibrium ConstantExample 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.PCl5 PCl3 + Cl2One liter

*The Equilibrium ConstantExample 17-3: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.PCl5 PCl3 + Cl2-0.60MChange 0.60M0.60MAt another temperature=0.90

*The Equilibrium ConstantExample 17-4: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.N2 + 3H2 2NH3 -0.10MChange -0.30M0.20M=0.26N2: 0.8mole/1Liter = 0.8M H2: 0.9mol/1Liter = 0.9MNH3: 0.2mol/1Liter = 0.2M

The Equilibrium Constant*Example 17-2: Calculation of Kc We put 10.0 mol of N2O into a 2-L container at some temperature, where it decomposes according to

At equilibrium, 2.20 moles of N2O remain, Calculate the value of Kc for the reaction

2N2O (g) 2N2(g) + O2(g) Initial [N2O]: 10.0mol/2L = 5.0 M=(1.1)2(3.9)2(1.95)=24.5equili [N2O]: 2.20mol/2L = 1.1 M2N2O (g) 2N2(g) + O2(g) -3.9MChange +3.9M+1.95M1.1MEquilibrium concn3.9M1.95M

*Variation of Kc with the Form of the Balanced EquationThe value of Kc depends upon how the balanced equation is written.From example 17-2 we have this reaction:This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53PCl5 PCl3 + Cl2

*Variation of Kc with the Form of the Balanced EquationExample 17-5: Calculate the equilibrium constant for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.Equil. []s 0.172M 0.086M 0.028 MThe concentrations are from Example 17-2.PCl3 + Cl2 PCl5

*Variation of Kc with the Form of the Balanced EquationLarge equilibrium constants indicate that most of the reactants are converted to products. ()Small equilibrium constants indicate that only small amounts of products are formed. ()=1.9or=1.9

(A,B) (C,D)------Kc ([A][B][C][D])KcKc------

*The Reaction Quotient The mass action expression or reaction quotienthas the symbol Q. Q has the same form as Kc (QKc)The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values. (Q)For this general reactionaA(g)+bB(g ) cC(g)+dD(g)Not necessarily equilibrium concentrations

*The Reaction QuotientWhy do we need another equilibrium constant that does not use equilibrium concentrations?Q will help us predict how the equilibrium will respond to an applied stress. Q To make this prediction we compare Q with Kc.

*The Reaction QuotientWhen:QKc Reverse reaction predominates until equilibrium is established ()

*The Reaction QuotientExample 17-6: The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

The concentrations given in the problem are not necessarily equilibrium []s. We can calculate QH2 + l2 2HI=9.0Q=9.0 but Kc=49Q

*Uses of the Equilibrium Constant, KcExample 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?SO2(g) + NO2(g) SO3(g) +NO(g)=3.01.73=0.865-1.73x=x x=0.316M=[SO3]=[NO][SO2]=[NO2]=0.5-0.316=0.184M

*Uses of the Equilibrium Constant, KcExample 17-8: The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?H2(g) + I2(g) 2HI(g)Kc = 497.0x=1.0-2x x=0.11M=[H2]=[I2][HI]=1.0-(2x0.11)=0.78M

*Disturbing a System at Equilibrium: PredictionsLeChateliers Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibriumSome possible stresses to a system at equilibrium are:Changes in concentration of reactants or products.Changes in pressure or volume (for gaseous reactions)Changes in temperature

*Disturbing a System at Equlibrium: PredictionsFor convenience we may express the amount of a gas in terms of its partial pressure rather than its concentration. To derive this relationship, we must solve the ideal gas equation.PV=nRTP=(n/V)RT because (n/V) has the units mol/L ()P=MRTThus at constant T the partial pressure of a gas is directly proportional to its concentration,

*Disturbing a System at Equlibrium: PredictionsChanges in Concentration of Reactants and/or Products Also true for changes in pressure for reactions involving gases.Look at the following system at equilibrium at 450oC.H2(g) + I2(g) 2HI(g)If some H2 is added, QKc (,)This favors the reverse reaction ()Equilibrium will shift to the left or reactant side

*Disturbing a System at Equlibrium: PredictionsChanges in Volume () (and pressure for reactions involving gases)Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:2NO2(g) N2O4(g)If the volume is decreased, which increased the pressure (,), (V, P, [NO2] and [N2O4])Q= (2[N2O4])/(2[NO2])2 = (2/4)Kc= (1/2) KcQKcThis favors the reactants or the reverse reaction ()

*Disturbing a System at Equlibrium: PredictionsChanging the Reaction Temperature ()Consider the following reaction at equilibrium:2SO2(g) + O2(g) 2SO3(g) Horxn=-198kJ/mol Is heat a reactant or product in this reaction?Heat is a product of this reaction! ()

Increasing the reaction temperature () stresses the productsThis favors the reactant or reverse reaction ()Decreasing the reaction temperature stresses the reactantsThis favors the product or forward reaction ()2SO2(g) + O2(g) 2SO3(g) +198kJ/mol

Disturbing a System at Equlibrium: Predictions,,A+BC+D+ heatA+B+ heat C+D

*Disturbing a System at Equlibrium: PredictionsIntroduction of a Catalyst Catalysts decrease the activation energy of both the forward and reverse reaction equally ()Catalysts do not affect the position of equilibrium. ()The concentrations of the products and reactants will be the same whether a catalyst is introduced or notEquilibrium will be established faster with a catalyst ()

*Disturbing a System at Equlibrium: PredictionsExample 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?N2(g) + 3H2(g) 2NH3(g) Horxn=-92kJ/molFactorsEffect on reaction procedurea. Increasing the reaction temperatureb. Decreasing the reaction temperaturec. Increasing the pressure by decreasing the volumed. Increasing the concentration of H2e. Decreasing the concentration of NH3f. Introduction a platinum catalyst

Left Right Right

Right RightNo effect

Disturbing a System at Equlibrium: PredictionsExample 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?FactorsEffect on equilibriuma. H2(g) + I2(g) 2HI(g)b. 4NH3(g)+ 5O2(g) 4NO(g)+6H2O(g)c. PCl3(g) + Cl2(g) PCl5(g)d. 2H2(g) + O2(g) 2H2O(g)

No effect Left Right Right,Q (2)2[HI]2(2)[H2]x(2)[I2]=a. = KcQ (2)4[NH3]4x(2)5[O2]5(2)4[NO] 4x(2)6[H2O]6=b. = 2KcQ (2)[PCl5](2)[PCl3]x(2)[Cl2]=c. = 0.5KcQ (2)2[H2O]2(2)2[H2]2x(2)[O2]=d. = 0.5Kc

*Disturbing a System at Equlibrium: PredictionsExample 17-11: How will an increase in temperature affect each of the following reactions?

FactorsEffect on equilibriuma. 2NO2 (g) 2N2O4(g) Horxn

*The Haber Process: A Practical Application of EquilibriumThe Haber process is used for the commercial production of ammonia:This is an enormous industrial process in the US and many other countries.Ammonia is the starting material for fertilizer production.Look at Example 17-9. What conditions did we predict would be most favorable for the production of ammonia?

*The Haber Process: A Practical Application of EquilibriumN2 is obtained from liquid air; H2 obtain from coal gasThis reactions is run at T=450oC and P of N2 =200 to 1000atmG

*The Haber Process: A Practical Application of EquilibriumThis diagram illustrates the commercial system devised for the Haber process.

*Disturbing a System at Equilibrium: CalculationsTo help with the calculations, we must determine the direction that the equilibrium will shift by comparing Q with Kc.Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C. What is the value of Kc for this reaction?A(g) B(g) + C(g) = (0.2)(0.3)(0.3)=0.45

*Disturbing a System at Equilibrium: CalculationsIf the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?1. Calculate Q, after the volume has been doubledA(g) B(g) + C(g) = (0.10)(0.15)(0.15)=0.22,Q

*Disturbing a System at Equilibrium: CalculationsSince Q

*Disturbing a System at Equilibrium: Calculationsx2+0.75x-0.0225=0ax2+bx+c=0x= -b b2-4ac 2ax= -0.75 (0.075)2-4(1)(-0.0225) 2x1x= -0.75 0.081 2x= -0.78 and 0.03M Since 0

*Disturbing a System at Equilibrium: CalculationsExample 17-13: Refer to example 17-12. If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M.A(g) B(g) + C(g) = (0.40)(0.6)(0.6)=0.90,Q>Kc thus the equilibrium shifts to the left or reactant side

*Disturbing a System at Equilibrium: CalculationsSet up the algebraic expressions to determine the equilibrium concentrationsA(g) B(g) + C(g) =0.450.18+0.45x=0.36-1.2x+x2x2-1.65x+0.18=0x= 1.65 (-1.65)2-4(1)(0.18) 2x1x= 1.65 1.42 2x= 1.5 and 0.12M Since 0

*Disturbing a System at Equilibrium: CalculationsExample 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant. CO(g) + Cl2(g) COCl2(g)= (0.30)(0.10)(0.6)Kc=20

*Disturbing a System at Equilibrium: CalculationsAn additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established.

CO(g) + Cl2(g) COCl2(g)Q

*Partial Pressures and the Equilibrium ConstantFor gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. (,)For gases, the pressure is proportional to the concentration.()We can see this by looking at the ideal gas law.PV = nRTP = nRT/V n/V = MP= MRT and M = P/RT

*Partial Pressures and the Equilibrium ConstantConsider this system at equilibrium at 5000C.2Cl2(g) + 2H2O (g) 4HCl(g) + O2(g)

*Partial Pressures and the Equilibrium Constantso for this reactionKc=Kp(RT)-1 or Kp=Kc(RT)1

Must use R = 0.0821 L atm/mol K

P= MRT and M = P/RT

*Relationship Between Kp and KcFrom the previous slide we can see that the relationship between Kp and Kc is: Kp=Kc(RT)n or Kc=Kp(RT)-n

n= (# of moles of gaseous products) (# of moles of gaseous reactants)

n= () ()

Kp=Kc(RT)1or Kc=Kp(RT)-1

2Cl2(g) + 2H2O (g) 4HCl(g) + O2(g)n= (4+1)-(2+2)=1

*Relationship Between Kp and KcExample 17-15: Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?2NOBr(g) 2NO(g) + Br2(g)Ptot=PNOBr + PNO+ PBr20.25atm = (x-0.34x)atm + 0.34x atm + 0.17x atm0.25atm = 1.17x atm, thus x=0.21atm

*Relationship Between Kp and KcBecause NOBr is 34% dissociated,It is 66% undissociatedPNOBr=(x-0.34x) = 0.66xPNOBr=(0.66)(0.21atm) = 0.14atmPNO= 0.34x = (0.34) x (0.21atm) = 0.071 atmPBr2= 0.17x = (0.17) x (0.21atm) = 0.036 atm

=9.3x10-3The numerical value of Kc for this reaction can be determined from the relationship of Kp and Kc.Kp=Kc(RT)n or Kc=Kp(RT)-n n=1Kc= (9.3x10-3)[(0.0821)(298)]-1 = 3.8 x10-42NOBr(g) 2NO(g) + Br2(g)

*Relationship Between Kp and KcExample 17-16: Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel, (a)How many moles of I2 remain unreacted at equilibrium?H2(g) + I2 (g) 2HI(g) =499x=2.31 x=0.256M[H2]=[I2]=0.33-0.256=0.074[HI]=2x0.256=0.51M?mol I2 = 3.0L x0.074 =0.21 mol

*Relationship Between Kp and Kc(b) What are the equilibrium partial pressures of H2, I2 and HI?(c) What is the total pressure in the reaction vessel?PH2=PI2=MRT=(0.074mol/L)(0.0821Latm/molK)(723K) =4.4 atmPHI=MRT=(0.051mol/L)(0.0821Latm/molK)(723K) =30 atm

Ptot=PH2+PI2+PHI = 4.4 + 4.4 + 30 =38.8 atm

*The Equilibrium Constant(activity)activity1,(),Homogeneous equilibria (),Heterogeneous equilibria ()

*Heterogeneous EqulibriaHomogeneous equilibria : only single phaseHeterogeneous equilibria have more than one phase present.For example, a gas and a solid or a liquid and a gas.

How does the equilibrium constant differ for heterogeneous equilibria?Pure solids and liquids have activities of unity.Solvents in very dilute solutions have activities that are essentially unity.The Kc and Kp for the reaction shown above are:CaCO3(s) CaO(s) + CO2 (g) at 500oC Kc = [CO2] Kp = PCO2 ,

*Heterogeneous Equlibria What are the forms of Kc and Kp?

SO2(g) + H2O(l) H2SO3(aq) at 25oC H2O(l) is the solvent

*Heterogeneous EqulibriaWhat are Kc and Kp for this reaction?CaF2(s) Ca2+(aq) + 2F1-(aq) at 25oC Kc = [Ca2+][F-]2 Kp is not defined; no gas involved3Fe(s) + 4H2O(g) Fe3O4 (s) + 4H2(g) at 500oC

Heterogeneous Equlibria*Example 17-15: Kc and Kp for Heterogeneous EqulibriaWrite both Kc and Kp for the following reversible reactions.2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)2NH3(g) + H2SO4(l) (NH4)SO4(s) S(s) + H2SO3(aq) H2S2O3(aq)

a. b. c. Kp: undefined, no gases involved

*Spontaneity of Physical and Chemical Changes (Ch15, p575)Spontaneous changes happen without any continuing outside influences. A spontaneous change has a natural direction.For example the rusting of iron occurs spontaneously. ()Have you ever seen rust turn into iron metal without man made interference?The melting of ice at room temperature occurs spontaneously. ()Will water spontaneously freeze at room temperature?

*The Two Aspects of Spontaneity An exothermic reaction does not ensure spontaneity.()For example, the freezing of water is exothermic but spontaneous only below 0oC.An increase in disorder of the system also does not insure spontaneity.()It is a proper combination of exothermicity and disorder that determines spontaneity.()

*Entropy, SEntropy is a measure of the disorder or randomness of a system.As with H, entropies have been measured and tabulated in Appendix K as So298.KWhen:S > 0 disorder increases (which favors spontaneity).,S < 0 disorder decreases (does not favor spontaneity). , (The Second Law of Thermodynamics),()

*Entropy, SFrom the Second Law of Thermodynamics, for a spontaneous process to occur:In general for a substance in its three states of matter:Sgas > Sliquid > SsolidSuniverse = Ssystem + Ssurroundings >0

Entropy increase (Ssysytem>0), WhenTemperature increaseVolume increaseMixing of substanceIncreasing particle numberMolecular size and complexityIonic compounds with similar formulas but different charges

*Example Without doing a calculation, predict whether the entropy change will be positive or negativeC2H6(g) +7/2 O2(g) 3H2O(g) + 2 CO2(g)3C2H2(g) C6H6(l)C6H12O6(s) + 6 O2(g) 6 CO2(g) + 2 H2O(l)

a) S0>0b) S00d) Hg(l), Hg(s), Hg(g)e) C2H6(g), CH4(g) , C3H8(g) f) CaS(s), CaO(s)

d) Hg(l)< Hg(s)

*Entropy, SThe Third Law of Thermodynamics states, The entropy of a pure, perfect, crystalline solid at 0 K is zero.This law permits us to measure the absolute values of the entropy for substances. ()To get the actual value of S, cool a substance to 0 K, or as close as possible, then measure the entropy increase as the substance heats from 0 to higher temperatures.Notice that Appendix K has values of S not S.(K)

*Entropy, SEntropy changes for reactions can be determined similarly to H for reactions. H

*Entropy, SExample 15-15: Calculate the entropy change for the following reaction at 25oC. Use appendix K.2NO2(g) N2O4(g)= (304.2 J/molK) 2(240.0 J/molK)= -175.8J/molKThe negative sign of S indicates that the system is more ordered. (S)If the reaction is reversed the sign of S changes.For the reverse reaction So298= +0.1758 kJ/K The + sign indicates the system is more disordered ().

*Entropy, SExample 15-16: Calculate So298 for the reaction below. Use appendix K.3NO(g) N2O(g) + NO2(g)= (219.7 + 240.0) 3(210.4) J/molK= -172.4J/molKChanges in S are usually quite small compared to E and H. ()Notice that S has units of only a fraction of a kJ while E and H values are much larger numbers of kJ.

*The Second Law of ThermodynamicsThe second law of thermodynamics states, In spontaneous changes the universe tends towards a state of greater disorder.,Spontaneous processes have two requirements:The free energy change of the system must be negative. ()The entropy of universe must increase.()Fundamentally, the system must be capable of doing useful work on surroundings for a spontaneous process to occur.

*Free Energy Change, G,and SpontaneityIn the mid 1800s J. Willard Gibbs determined the relationship of enthalpy, H, and entropy, S, that best describes the maximum useful energy obtainable in the form of work from a process at constant temperature and pressure.The relationship also describes the spontaneity of a system.The relationship is a new state function, G, the Gibbs Free Energy .G=H-TS (at constant T and P)

*Free Energy Change, G, and SpontaneityThe change in the Gibbs Free Energy, G, is a reliable indicator of spontaneity of a physical process or chemical reaction.G does not tell us how quickly the process occurs. ()Chemical kinetics, the subject of Chapter 16, indicates the rate of a reaction.Sign conventions for G.G > 0 reaction is nonspontaneous ()G = 0 system is at equilibrium ()G < 0 reaction is spontaneous ()

*Free Energy Change, G, and SpontaneityChanges in free energy obey the same type of relationship we have described for enthalpy, H, and entropy, S, changes.

*Free Energy Change, G, and SpontaneityExample 15-17: Calculate Go298 for the reaction in Example 15-8. Use appendix K.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)= [3(-394.4)+4(-237.3)] [(-23.49)+5(0)] kJ/mol= -2108.9 kJ/molGo298 < 0, so the reaction is spontaneous at standard state conditions.If the reaction is reversed:Go298 > 0, and the reaction is nonspontaneous at standard state conditions.

*Relationship Between DGorxn and the Equilibrium Constant DGorxn is the standard free energy change DGorxn is defined for the complete conversion of all reactants to all products. ()DGrxn is the free energy change at nonstandard conditions (DGrxn ) For example, concentrations other than 1 M or pressures other than 1 atm.(1M1)DGrxn is related to DGorxn by the following relationship.Grxn=Gorxn + RT lnQ orG=Go + 2.303 RT logQR= universal gas constant (8.314J/molK)T= absolute temperature ()Q= reaction quotient ()

*Relationship Between DGorxn and the Equilibrium Constant, DG=0 and Q=Kc. Then we can derive this relationship:0=Go + RT lnK or0=Go + 2.303 RT logKWhich rearranges to:Go = -RT lnK orGo = -2.303 RT log K

, K Kp;, K Kc;,

*Relationship Between DGorxn and the Equilibrium ConstantFor the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:aA(g)+bB(g ) cC(g)+dD(g)Whereaa is the activity of A ab is the activity of B ac is the activity of C ad is the activity of D

*Relationship Between DGorxn and the Equilibrium ConstantThe relationships among DGorxn, K, and the spontaneity of a reaction are:

GorxnKSpontaneity at unit concentration< 0> 1Forward reaction spontaneous= 0= 1System at equilibrium> 0< 1Reverse reaction spontaneous

*Relationship Between DGorxn and the Equilibrium Constant

*Relationship Between DGorxn and the Equilibrium ConstantExample 17-17: Calculate the equilibrium constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.Note: this is a gas phase reaction.N2O4(g) 2NO2(g) 1. Calculate DGorxn DGorxn =2 DGofNO2(g) - DGofN2O4(g)DGorxn =2 (51.30kJ) (97.82kJ)DGorxn =4.78 kJ/mol rxnDGorxn =4.78x103 j/mol rxnThis reaction is nonspontaneous

*Relationship Between DGorxn and the Equilibrium Constant2. Calculate K from DGorxn =-RT lnKplnKp =DGorxn/-RT =(4.78x103J/mol)/-(8.314J/molK)(298K) =-1.93Kp = e-1.93 =0.145= (PNO2)2/(PN2O4)

*Relationship Between DGorxn and the Equilibrium ConstantKp for the reverse reaction at 25oC can be calculated easily, it is the reciprocal of the above reaction.2NO2(g) N2O4(g)DGorxn =-4.78 kJ/molKp = 1/Kp =1/0.145 = 6.9 = (PN2O4)/(PNO2)2

*Relationship Between Gorxn and the Equilibrium ConstantExample 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate Gorxn at 25oC.

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