15.S THERMODYNAMICS AND EQUILIBRIUM ■ y is very small compared with 5.0 and 10.0 atm, so our approximations are valid, i ium pressures are: (Piro)«, = y = 2-2 X 10"6 atm (Po )« = 5-° aim + °-5^ = 5-° atm (Pno )« = 10-° atm - v - 10.0 atm 739 I "he sample problems illustrate the general approach to equilibrium problems. In ter 16 we apply these ideas to the important types of equilibria that occur in jus solutions. Box 15-2 discusses the carbon cycle and chemical equilibria. '2(g) '2(g) (ECTION EX£RC9S£S t Barium sulfate is a salt that is opaque to x rays. It is used as an imaging agent for medical radiographs of the gastrointestinal tract. (See Figure 4-11.) The concentration of Ba2+ ions in a saturated aqueous solution is 1.05 X 10~5 M. Determine Ksp for barium sulfate. I 1 A solution of acetic acid in water is described by the following equilibrium: CH3C02H(aq) + H20(1) Acetic acid CH,CO, l3*~v-'2 Acetate (aq) + H,(V (aq) The equilibrium constant for this acid-base reaction is Ka = 1.8 X 10"5 M. Calculate the equilibrium concentrations of acetic acid, acetate ion, and hydronium ion in a 2.5 M solution of acetic acid. í The key step in the synthesis of sulfuric acid is the combustion of S02 to give sulfur trioxide: Catalyst 2 S02 (gl + 02 (() < > 2 SO. 3 Ig) £eq = 5.60x 104atm" at 350 °C Sulfur dioxide and oxygen are mixed initially at 0.350 atm and 0.762 atm, respectively, at 350 °C. What are the partial pressures of the three gases when the mixture reaches equilibrium? 15.5 THERMODYNAMICS AND EQUILIBRIUM ■. librium constants have values covering a very large range. Why do some reac- i reach equilibrium when hardly any of the starting materials have been i limed, whereas others go virtually to completion? At the molecular level, bond .»ies and molecular organization are the determining factors. Macroscopically, ; translate into the thermodynamic properties of enthalpy and entropy. As ■ issed in Chapter 13, free energy (G) is a state function that combines these prop-"■ 5. Free energy provides information about the spontaneous direction of a r ion. In this section we establish the connection between free energy and the ■■ ibrium constant. THE CONCENTRATION QUOTIENT ■ n the reaction is at equilibrium, the ratio of product concentrations to reactant ■ entrations is equal to the equilibrium constant, Keq: aA + bB cC + dD K _ [CľM W\ 742 CHAPTER 15 PRINCIPLES OF CHEMICAL EQUILIBRIUM When a chemical system is not at equilibrium, the ratio of concentrations is- talk the concentration quotient, (0. The concentration quotient has the same lot in the equilibrium concentration ratio, but the concentrations are not equilihnu values: Consequently, Q can have any value. X_ T The subscript "eq" indicates that these are equilibrium concentrations. Q [Af[B]h f These arc not equilibrium concentrations. The N02/N204 system can be used to show how Q is related to Ktq. We have ^« that N02 gas reacts to produce N204 until an equilibrium is established between tl two gases. If a sample contained only N02, the partial pressure of N204 would \r zero and Q = 0. This system would react to form N204 and consume N02, und th value of Q would increase until Q = K^. At these concentrations, the system woul be at equilibrium. If, on the other hand, a sample contained only N204 but no NO the partial pressure of N02 would be zero and Q = °o. Under these conditions th system would react to form N02 and consume N204, and the value of Q vvonh decrease, again until Q = KQq. Thus we see that the direction in which a rcaciio proceeds depends on the relationship between Q and Aľeq: Q < Keg reaction goes to the right to make products. Q > Keq reaction goes to the left to make reactants. Notice that if the concentration of any product is zero, 0 = 0, and the rcattioi must go to the right, toward products. If any reactant concentration is zero, 0=« and the reaction must go to the left, toward reactants. The relationship between Q and Kcq signals the direction of a chemical reaction The free energy change, AG (introduced in Chapter 13), also signals the direction o a chemical reaction. These two criteria can be compared: Reaction goes right when AG„n < 0 Equilibrium when AG,., = 0 Reaction goes left when Q>Keq AG .„ > 0 The similarities suggest a link among Q, Keq, and AG. This link can be found fron the equations of thermodynamics. The concentration quotient appears in Et 13-12 as a link between the standard free energy change for a reaction and i ' * energy change under nonstandard conditions: AG = AG° + RTlnQ ilC-H- At equilibrium, AG = 0 and Q = Ktq. Substituting these equalities into Ei ■ i 13-12, we obtain: 0- AGa + RTlnK This rearranges to: AG°= -RTlnK •IS-5) 15.5 THERMODYNAMICS AND EQUILIBRIUM 743 I Ration 15-5 is an extremely important relationship, because it links thermody-|t ■ data with equilibrium constants. Because many values of AG° appear in tables, If n be calculated from Equation 15-5 for many reactions. This equation is h- d to the Haber process in Sample Problem 15-9. [IIMPLE PROBLEM 15-9 THERMODYNAMICS AND Keq í' standard thermodynamic data, find the value of Kcq at 298 K for the Haber reaction: N2(g) + 3H3(g)^2NHJ(g) Jl OD Equation 15-5 provides the link between thermodynamic data and Kiq. We must iculate AG° n from tabulated standard free energies of formation. AG° n - 2 (coeff) AG? (products) - 2 (coeff) AG? (reactants) áix E contains the appropriate values: AG? (kJ/mol): N3 (g), 0; H2 lg), 0; NH3 (g), -16.4 kJ/mol AG° = (2moINH3)(-16.4kJ/molNH3)-3(0)- 1(0)= -32.8 kJ :rmine the equilibrium constant, Equation 15-5 must be rearranged to isolate In K^. lnK„= - AG0 RT ^ = e' -(-32.8kJ)(103J/kJ) (8.314 J/K)(298K) ' = 5.6 x 105atm~2 = 13.24 ponential gives a dimensionless number, since ex is always a pure number. However, ign units to K as required by the concentration quotient. Remember that the super-'o" in AG0 refers to standard conditions that include concentrations of 1 M for solutes tm partialpressure for gases. I EQUILIBRIUM CONSTANTS AND TEMPERATURE s of the effect of temperature on equilibria reveal a consistent pattern. The i ■ mum constant of an exothermic reaction decreases with increasing tempera- vhereas the equilibrium constant of an endothermic reaction increases with ■sing temperature. Equation 15-5 provides a thermodynamic explanation for i ;havior. Recall that free energy is related to enthalpy and entropy through II ion 13-9: Remember that AG," for any element in its standard state is zero. AG0 - AW - TAS° (13-9) quality can be substituted into Equation 15-5 to show how In Keq depends on i: \S°, and T: AH°-TAS°= -RTlnK ■ rearranges to give: AH" AS0 (15-6) ' ii exothermic reaction has a negative Ar/°, making the first term on the right of [ I i on 15-6 positive. As T increases, this term decreases, causing Kcq to decrease. iothermic reaction, in contrast, has a positive A//0, making the first term on the i of Equation 15-6 negative. As T increases, this term becomes less negative, 744 CHAPTER 15 PRINCIPLES OF CHEMICAL EQUILIBRIUM causing Keq to increase. These variations in Keq with temperature, which car ■■ substantial, can be estimated using Equation 15-6 and standard thermodynamic fu i tions. Sample Problem 15-10 applies Equation 15-6 to the Haber synthesis. FIGURE 15-12 Rhizobium bacteria, which form colonies in nodules on the roots of leguminous plants, have an enzyme that allows them to manufacture ammonia from molecular nitrogen at ambient temperature. SAMPLE PROBLEM 15-10 Keq AND TEMPERATURE Use tabulated thermodynamic data to estimate Keq for the Haber reaction at 500 °C. METHOD: Values for AH0 and AS0 can be calculated using tabulated thermodynamic vak Then Equation 15-6 can be applied to determine the value of the equilibrium constant at 500 °C. First, we need AH°m and AS^n: AH°nn = 2 (coeff) AH? (products) - 2 (coeff) AH°f (reactants) AS°rxn = 2 (coeff) 5° (products) - 2 (coeff) S° (reactants) Appendix E contains the appropriate values: A/i°f(kJ/mol) N2(g):0 H2(g):0 NH3(g):-46 S° (J/mol K) N2(g): 191.6 H2(g): 130.7 NH3 (g): 192.45 AB°nn = (2)(-46 kJ/mol) - 3(0) - 0 = -92 kJ/mol AS°„n - (2X192.45 J/mol K) - (3)(130.7 J/mol K) - 191.6 J/mol K A5°rsn= -198.8J/molK Now calculate K^ at 500 °C using Equation 15-6: - (-92kJ/mol)(103J/kJ) " "» ~ (8.314 J/mol K)(500 + 273 K) (-198.8 J/molK) (8.314 J/mol K) lnKeq= 14.3 -23.9- -9.6 Taking the antiln, or e\ of -9.6 gives the estimated equilibrium constant at 500 °C: ÄL = 6.7 X 10"5 atm"2 Sample Problems 15-9 and 15-10 underscore the dilemma faced by indus i I chemists and engineers. At 298 K, the equilibrium position of the Haber reac ■ strongly favors the formation of ammonia. Why, then, is the Haber synthesis i i carried out at 298 K, where equilibrium strongly favors the desired product' Ii reason is that even with a catalyst, the reaction is much too slow to be useful at temperature. At low temperature, thermodynamics favors ammonia, but lm • prohibits the reaction from occurring at a discernible rate. To make the reac ' proceed at a practical rate, the temperature must be increased. Unfortunately ■"' equilibrium constant falls dramatically as temperature increases. At 773 K, a real temperature for the Haber reaction, the equilibrium position does not favor NH, N, + 3H,<=^2 NH, ^MK = 5.6X 10s atm"2 ^q.773K = 6.7xl0^atm"2 Nature has solved this problem with the enzyme nitrogenase. This enzyit found in bacteria that live in nodules among the roots of leguminous plants sue soybeans and clover (Figure 15-12). Nitrogenase catalyzes the formatioi ammonia from atmospheric nitrogen. Chemists are studying the mechanism of nn IB.g SHIFTS IN EQIMLIBBIOM enase, but its chemistry is so complicated that we still have incomplete knowledge f how the enzyme converts nitrogen to ammonia. Perhaps further research will neover its secret and make it possible to design new commercial catalysts for itrogen fixation that operate at 298 K. The Haber reaction is a practical example of the effect of temperature on an mthermic reaction. A practical example of an endothermic reaction is the use of íethaiie and steam to produce the molecular hydrogen needed for the Haber reac-on This reaction is highly endothermic, so it is carried out at temperatures much reater than 1000 K to force the equilibrium toward the products: CH, (g) + H20 (g) <=^ CO (() + 3 H2 fe, A//0 = +206 kJ ^2«K = 1-2 X lO^atm"2 K^ISOOK - 1.1 X ltfatnT2 SECTION EXERCISES '5 5.1 Use thermodynamic data to determine K^ for the oxidation of NO at 298 K: 2NO(g) + 02(g)^2N02(g) 552 Does AT for the reaction described in 15.5.1 increase or decrease when the temperature is raised above room temperature? Give your reasoning. i55J Use thermodynamic data to estimate AT at 1000 °C for the reaction in 15.5.1. 15.6 SHIFTS IN EQUILIBRIUM Ve call K the equilibrium constant because it has the same value regardless of the mounts of reactants and products that we put into the system. In other words, Keq .; independent of initial concentrations. No matter what value Q has before the eattion occurs, equilibrium will be reached when concentrations have changed so LECHÄTELIER'S PRINCIPLE Vhal happens if we change the conditions on a system that is already at equilibrium? luppose we vary the amounts or concentrations of one or more substances, or change he temperature of the system. How does a system that is at equilibrium respond to ■hese changes? Although the equations that describe equilibrium provide quantise answers to these questions, a simple principle can be applied to obtain a quick ualitative indication. This principle, which was first formulated by Henri-Louis Lc Chittelier, a French industrial chemist of the early twentieth century, states: When a change is imposed on a system at equilibrium, the system will react, if possible, in the direction that reduces the amount of change. According to Le Chätelier's principle, if we introduce more of one reactant, the ction will proceed in the direction that consumes this reactant. If we reduce the emperature, thereby removing heat from the system, the reaction will proceed in the xothermic direction, producing some heat.