Chapter 4 Chemical Equilibrium

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Chapter 4 Chemical Equilibrium. Dynamical Equilibrium. concept introduced by A. W. Williamson in 1851 during a study of esterifications. at a fixed temperature (1862, Berthelot & Gilles)
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Chapter 4 Chemical EquilibriumDynamical Equilibriumconcept introduced by A. W. Williamson in 1851 during a study of esterifications.at a fixed temperature (1862, Berthelot & Gilles) the same after equilibrium has been reached.~all reaction has not ceased at equilibrium~forward reaction rate = reverse reaction rate~double arrow (↔) presents that chemical equilibrium, and reactions in the two opposite directions, are of interest.Equilibrium Constantratiois constant at a given temperature.In 1864 Guldberg & Waage gave a equilibrium expression as at equilibriumNote: The rate equations in fact depend on the mechanism. (chemical kinetic) The equilibrium expression is independent of the mechanism. In 1887, Van’t Hoff presented a derivation of the equilibrium constant based on thermodynamics.Gibbs energy & equilibrium constantGchemical equilibrium is a state of “minimum free energy (G , A)”extent of reaction4.1 Chemical Equilibrium Involving Ideal GasWhen the volume of n mol of an ideal gas changes from V1 to V2 at constant temperature T, the Gibbs energy change is For gas, standard state is 1 barP1=1 bardimensionlessIf the Gibbs energy at 1 bar is denoted as Goat PFor 1 mol of gaschemical potential ()If nA mol of A are present in a mixture of A, B, Y, etc., the chemical potential A of A is defined asConsider a gas-phase reactionall reactants and products are ideal gases.standard chemical potential of A(i.e. Gibbs energy of 1 mol of A at a pressure of 1 bar)standard Gibbs energy change, standard state is 1 bar pressure.At equilibriumindependent of pressurethermodynamic equilibrium constantNote: is a dimensionless quantity. has the same value as the pressure equilibrium constant KPProblem 4.17At 3000 K the equilibrium partial pressure of CO2, CO and O2 are 0.6, 0.4 and 0.2 atm, respectively. Calculate Go at 3000 K for the reactionSolutionAt 3000 KProblem 4.13One mole of HCl mixed with oxygen is brought into contact with a catalyst until the following equilibrium has been established:If y mol of HCl is formed, derive an expression for KP in terms of y and the partial pressure of oxygen.SolutionEquilibrium Constant in Concentration unitsFor an ideal gas:stoichiometric sumKcThe relationship could be obtainedis the numerical value of Kc and is thus the standard equilibrium constant in terms of concentration.Note: If the concentrations are in mol dm-3, the standard state for ∆Go is 1 mol dm-3. In general, the change of standard state will change the value of ∆Go, and it is important always to indicate the standard state.by analogyThe pressure PA of substance A is related to the total pressure P byKxThe standard state is unit mole fraction.Note:KP, Kc are functions of temperature only.Kx is pressure dependent unless ∑= 0 at constant temperature.Units of Equilibrium ConstantConsider a reaction of the typeThere is a decrease, by one, in the number of molecules.There is no change in the number of molecules.dimensionlessIn general, if the stoichiometric sum is ∑, the units areNote:dimensionlessProblem 4.7Nitrogen dioxide, NO2, exists in equilibrium with dinitrogen tetroxide, N2O4:at 25.0 C and a pressure of 0.597 bar the density of the gas is 1.477 g dm-3. Calculate the degree of dissociation under these conditions, and equilibrium constant Kc, KP and Kx.SolutionAssume that in 1 dm3 there are x mol of N2O4 and y mol of NO2x=0.00802; y=0.0161Problem 4.8At 25.0 C the equilibriumis rapidly established. When 1.1g of NOBr is present in a 1.0-dm3 vessel at 25.0 C the pressure is 0.355 bar. Calculate the equilibrium constants KP, Kc and Kx.SolutionConsiderconversion of 0.5 mol of A into Zconversion of 1 mol of A into 2ZExample 4.1The Gibbs energies of formation of NO2(g) and N2O4(g) are 51.30 and 102.00 kJ mol-1, respectively (standard state: 1 bar and 25 C).a. Assume ideal behavior and calculate, for the reaction N2O42NO2, KP (standard state; 1 bar) and Kc (standard state; 1 mol dm-3).b. Calculate Kx at 1 bar pressure.c. At what pressure is N2O4 50% dissociated?d. What is G if the standard state is 1 mol dm-3?SolutionFor the reaction∆G(standard state : 1 bar) = 251.30-102.0 = 0.6 kJ mol -1a.b.c.=0.5d.Example 4.2From the 1890s and for many years after the German physical chemist Max Bodenstein (1871-1942), who was a remarkably skillful experimentalist, carried out detailed studies on several reactions, including that between hydrogen and iodine,He studied both the equilibrium established and the rates in forward and reverse directions. In one investigation of the equilibrium constant at 731 K, he introduced 22.13 cm3 of hydrogen and 16.18 cm3 of iodine into a vessel. After waiting a sufficient time for equilibrium to be established, he chilled the reaction vessel rapidly so as to "freeze" the equilibrium, (i.e., prevent its shifting appreciably), and he then analyzed the products, finding that 28.98 cm3 of hydrogen iodide had been formed. (These volumes relate to STP which at the time meant a temperature of 0 C and a pressure of 1 atm; see also Appendix A, p.999). Calculate the equilibrium constant and the Gibbs energy change for the reaction.Solution4.2 Equilibrium in Nonideal Gaseous Systemsideal gas  P (pressure)real gas  fugacity (f) or activity (a)where is the dimensionless equilibrium constant in terms of activities:If pressure is low,4.3 Chemical Equilibrium in Solutionideal solution  a = mreal solution  a = m : activity coefficientm : molality (mol kg-1)dimensionlessNote:  is used for an activity coefficient used with concentration.f is used for an activity coefficient used with mole fraction. (Section 5.6)For uncharged species in solution → idealK in terms of molalities, concentrations, mole fraction, without the use of activity coefficient.For ions → realK in terms of activities (section 7.10, 8.5)For solvent species,a = xfx : mole fraction of the solventf : activity coefficient of the solventExample 4.3The equilibrium constant Kc for the reaction fructose-1.6-diphosphate  glyceraldehyde-3-phosphate + dihydroxyacetone phosphateis 8.910-5 M at 25 C, and we can assume the behavior to be ideal.a. calculate Go for the process (standard state : 1 M)b. suppose that we have a mixture that is initially 0.01 M in fructose-1.6-diphosphate and 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is G? In which direction will reaction occur?Solutiona.b.G < 0 → reaction proceeds from left to right.is less than the equilibrium constant 8.910-5 mol dm-34.4 Heterogeneous Equilibriumfor a given solid or pure liquid, the amount contained in a given volume is a fixed quantity.[CO2] = Kc→ aCO2 = KcNote that this procedure of incorporating the concentrations of pure solids and liquids into the equilibrium constants is equivalent to defining the activities of pure solids and liquid as unityasolid = 1 aliquid = 1Also note that if solids are present in solid solution or liquids are mixed with other liquids, their concentration are obviously not fixed, and they must not be incorporated into the equilibrium constant.For a salt and its saturated solutionExample 4.4The solubility of silver chloride in pure water at 25 C is 1.26510-5 mol dm-3.Calculate the solubility product and Go for the processSolutionProblem 4.24The solubility product of Cr(OH)3 is 3.010-29 mol4 dm-12 at 25C. What is the solubility of Cr(OH)3 in water at this temperature?SolutionC is solubility (mol dm-3)4.5 Tests for Chemical EquilibriumReaction may be procedure so slowly that no detectable change will occur over a long period of time.for this reaction, the equilibrium lies almost completely to the right except at extremely high temperature. We can add certain substances known as catalysts (such as powered platinum) that speeds up reaction. (section 10.9)Note that catalysts do not affect the position of equilibrium but merely decrease the time required for equilibrium to be attained.*addition of A (or B) will cause the equilibrium to shift to the right.*addition of Y (or Z) will cause a equilibrium shift to the left.4.6 Shifts of Equilibrium at Constant TemperatureLe Chatelier principle (勒沙特列原理)If there is a change in the condition of a system will adjust itself in such a way as to counteract as far as possible, the effect of that change.~when total pressure is increased, the equilibrium shifts in the direction to reduce the number of molecules.~An increase in temperature at constant pressure in a closed system shifts the equilibrium in the direction in which the system absorbs heat from the surroundings.Consider the equilibrium*A addition → equilibrium shifts to left.*AB addition → equilibrium shifts to right.*solvent addition → equilibrium shifts to right.Example 4.5The equilibrium constant KPfor dissociation of chlorine into atomshas been determined to be 0.57 bar at 2000 K. Suppose that 0.1 mole of chlorine is present in a volume of 5 dm3 at that temperature. What will be the degree of dissociation (the fraction present as chlorine atoms)? If the volume is expanded to 20 dm3, what will them be the degree of dissociation?SolutionEffect of pressure on equilibrium shifts(1) if  = 0 → Kx is independent of pressure → equilibrium doesn’t shift(2) if > 0 → Kx is lower when pressure is higher → equilibrium toward reactants(3) if  < 0 → Kx is higher when pressure is lower → equilibrium toward productsEffect of volume on equilibrium shifts(1) if  = 0 → Kn is independent of volume → equilibrium doesn’t shift(2) if  > 0 → Kn is lower when volume is lower→ equilibrium toward reactants(3) if  < 0 → Kn is higher when volume is lower→ equilibrium toward productsKnEffect of inert gas on equilibrium shifts(1) if  = 0 → Kn is equal to KP→ equilibrium doesn’t shift(2) if  > 0 → adding inert gas is to increase nt→ Kn increasing, equilibrium toward products(3) if  < 0 → adding inert gas is to increase nt→ Kn decreasing, equilibrium toward reactantsConsider that a dissociation reaction of water becomes O2 and H2 :(a) Will the dissociation increase or decrease if the pressure is reduced?(b) will the dissociation increase or decrease if argon gas is added, holding the total pressure equal to 1 bar?(c) will the dissociation increase or decrease if the pressure is raised by addition of argon at constant volume to the closed system?(d) will the dissociation increase or decrease if oxygen gas is added?(e) will the dissociation increase or decrease if temperature is raised?Solution(a)(b)(c)(d)(e)reduce the effect of temperature  increase4.7 Coupling of Reactionfor reaction 1for reaction 2for reaction 3If Go1 > 0  reaction 1 will not occur.If Go2 < 0, Go3 < 0  A + B → Y + ZNoteExample 4.6The aminotransferases, also known as transaminases, are enzymes which perform the function of transferring amino groups from one amino acid to another, as for example in the reaction-ketoglutarate + alanine  glutamine + pyruvateAt 250 C the Go for this reaction is 251 J mol-1. Calculate the equilibrium constant for the process. Suppose that pyruvate formed is oxidized in a process for which Go is -258.6 kJ mole-1; what effect does this have on the transamination process?SolutionThe equilibrium for the simple process is When the reaction is combined with the oxidation process, the overall Go is 251-258600 = -258360 J mol-1. The equilibrium constant isThe oxidation of the pyruvate has greatly enhanced the formation of glutamine in the process, effectively taking it to completion.Problem 4.22Suppose that there is a biological reactionfor which the Go value at 37 C is 23.8 kJ mol-1 (standard state = 1 mol dm-3). Suppose that an enzyme couples this reaction withfor which Go = -31.0 kJ mol-1. Calculate the equilibrium constant at 37.0 C for these two reactions and for the coupled reactionSolutionFor reaction (1)For reaction (2)For the coupled reaction (3)Coupling by a catalyst+catalystThe catalyst would couple the reaction 1 and 2. The type of coupling of reaction is quite common in biological systems.Example 4.7A good example is to be found in the synthesis of glutamine in living systems, by the processglutamate + NH4+ glutaminefor which Go is 15.7 kJ mol-1 at 37 C (standard state, 1 mol dm-1).a. Calculate the equilibrium constant for this reaction.b. Another reaction is the hydrolysis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) and phosphate (P): ATP  ADP + Pfor which the Go value may be taken as –31.0 kJ mol-1 (the value varies considerably with the ions present in the solution). Since these two reactants have no reactants or products in common they can only be coupled together by a catalyst.The enzyme glutamine synthetate in fact catalyzes the reaction glutamate + NH4+ +ATP  glutamine +ADP +Pand therefore couples the reactions together. Calculate the equilibrium constant for this combined reaction.Solutiona. The equilibrium constant for the first reaction isb. For the coupled reactionThe equilibrium constant for the coupled reaction is thereforeNote that the occurrence of the second reaction, coupled to the first, has therefore greatly enhanced the glutamine formation.4.8 Temperature Dependence of Equilibrium ConstantsOwing to that equilibrium constant KP is affected very little by the pressure.→ a plot of lnKPo against 1/T will have a slope equal to Ho/RNote that in general Kx is pressure dependent, it is necessary to express the condition that the pressure must be held constant.Note that Kco changes only slightly with pressure, it is usually not necessary to specify constant pressure conditions.Also note that for reactions in solution, the above equation is the one usually employed, but Uo and Ho are then very close to each other.Example 4.8The equilibrium constant for an association reaction A + B  A + Bis 1.8103 dm3 mol-1 at 25 C and 3.45103 dm3 mol-1 at 40 C. Assuming Ho to be independent of temperature, calculate Ho and So .SolutionProblem 4.25A gas reactionA  B+Cis endothermic and its equilibrium constant KP is 1 bar at 25 C.a. What is Go at 25 C (standard state: 1 bar)?b. Is So, with the same standard state, positive or negative? c. For the standard state of 1 M, what are Kc and Go?d. Will KP at 40 Cbe greater than or less than 1 bar?e. Will Go at 40 C (standard state: 1 bar) be positive or negative?Solutiona.b.c.d.e.Problem 4.32a. An equilibrium constant KP is increased by a factor of 3 when the temperature is raised from 25.0 C to 40.0 C. Calculate the standard enthalpy change.b. What is the standard enthalpy change if instead KP is decreased by a factor 3 under the same conditions?Solutiona.b.Ho is a function of temperatureFor a gas, the heat capacity at constant pressure can often be expressed asThe enthalpy change at temperature T2 is related to that at T1 by an equation of the formIf T1 is 25 C and Ho(25 C) refers to standard states, Ho at any temperature T is4.9 Pressure Dependence of Equilibrium Constantsfor the change Go in a chemical reactionIf Vo = 10 cm3 mol-1a pressure of 1/(4.0310-4) = 2480 bar will change lnKo by unit.Example 4.9The equilibrium constant for the reaction2NO2 N2O4in carbon tetrachloride solution at 22C is increased by a factor of 3.77 when the pressure is increased from 1 bar to 1500 bar. Calculate Vo, on the assumption that the volume change is independent of pressure.SolutionProblem 4.41The following diagram shows the variation with temperature of the equilibrium constant Kc for a reaction. Calculate Go, Ho , and So at 300 K.SolutionProblem 4.46At 1 bar pressure liquid bromine boils at 58.2 C, and 9.3 C its vapor pressure is 0.1334 bar. Assuming Ho and So to be temperature independent, calculate their values, and calculate the vapor pressure and Go at 25 C.SolutionFor the equilibrium At 25 C
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