CHAPTER 18 ENTROPY, FREE ENERGY, AND EQUILIBRIUM

Transcription

1 CHAPER 18 ENROPY, FREE ENERGY, AND EQUILIBRIUM 18.6 he probability (P) of finding all the molecules in the same flask becomes progressively smaller as the number of molecules increases. An equation that relates the probability to the number of molecules is given in the text. N 1 P where, N is the total number of molecules present. Using the above equation, we find: (a) P 0.0 P (c) P (a) his is easy. he liquid form of any substance always has greater entropy (more microstates). (c) his is hard. At first glance there may seem to be no apparent difference between the two substances that might affect the entropy (molecular formulas identical). However, the first has the OH structural feature which allows it to participate in hydrogen bonding with other molecules. his allows a more ordered arrangement of molecules in the liquid state. he standard entropy of CHOCH is larger. his is also difficult. Both are monatomic species. However, the Xe atom has a greater molar mass than Ar. Xenon has the higher standard entropy. (d) Same argument as part (c). Carbon dioxide gas has the higher standard entropy (see Appendix ). (e) (f) O has a greater molar mass than O and thus has the higher standard entropy. Using the same argument as part (c), one mole of NO4 has a larger standard entropy than one mole of NO. Compare values in Appendix. Use the data in Appendix to compare the standard entropy of one mole of NO4 with that of two moles of NO. In this situation the number of atoms is the same for both. Which is higher and why? In order of increasing entropy per mole at 5C: (c) < (d) < (e) < (a) < (c) Na(s): ordered, crystalline material. (d) NaCl(s): ordered crystalline material, but with more particles per mole than Na(s). (e) H: a diatomic gas, hence of higher entropy than a solid. (a) Ne(g): a monatomic gas of higher molar mass than H. SO(g): a polyatomic gas of higher molar mass than Ne Using Equation (18.7) of the text to calculate S rxn (a) Srxn S(SO ) [ S(O ) S (S)]

2 468 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM S rxn (1)(48.5 J/K mol) (1)(05.0 J/K mol) (1)(1.88 J/K mol) 11.6 J/K mol Srxn S(MgO) S(CO ) S (MgCO ) S rxn (1)(6.78 J/K mol) (1)(1.6 J/K mol) (1)(65.69 J/K mol) J/K mol 18.1 Strategy: o calculate the standard entropy change of a reaction, we look up the standard entropies of reactants and products in Appendix of the text and apply Equation (18.7). As in the calculation of enthalpy of reaction, the stoichiometric coefficients have no units, so S rxn is expressed in units of J/Kmol. Solution: he standard entropy change for a reaction can be calculated using the following equation. Srxn ns(products) ms (reactants) (a) S rxn S(Cu) S(HO) [ S(H ) S(CuO)] (1)(. J/Kmol) (1)(188.7 J/Kmol) [(1)(11.0 J/Kmol) (1)(4.5 J/Kmol)] 47.5 J/Kmol S rxn S(AlO ) S(Zn) [ S(Al) S(ZnO)] (1)(50.99 J/Kmol) ()(41.6 J/Kmol) [()(8. J/Kmol) ()(4.9 J/Kmol)] 1.5 J/Kmol (c) S rxn S(CO ) S(HO) [ S(CH 4) S(O )] (1)(1.6 J/Kmol) ()(69.9 J/Kmol) [(1)(186. J/Kmol) ()(05.0 J/Kmol)] 4.8 J/Kmol Why was the entropy value for water different in parts (a) and (c)? 18.1 All parts of this problem rest on two principles. First, the entropy of a solid is always less than the entropy of a liquid, and the entropy of a liquid is always much smaller than the entropy of a gas. Second, in comparing systems in the same phase, the one with the most complex particles has the higher entropy. (a) (c) (d) Positive entropy change (increase). One of the products is in the gas phase (more microstates). Negative entropy change (decrease). Liquids have lower entropies than gases. Positive. Same as (a). Positive. here are two gas-phase species on the product side and only one on the reactant side (a) S < 0; gas reacting with a liquid to form a solid (decrease in number of moles of gas, hence a decrease in microstates). (c) (d) S > 0; solid decomposing to give a liquid and a gas (an increase in microstates). S > 0; increase in number of moles of gas (an increase in microstates). S < 0; gas reacting with a solid to form a solid (decrease in number of moles of gas, hence a decrease in microstates).

3 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM Using Equation (18.1) of the text to solve for the change in standard free energy, (a) ΔG Gf (NO) Gf (N ) Gf (O ) ()(86.7 kj/mol) kj/mol ΔG Gf [HO( g)] Gf [HO( l)] (1)( 8.6 kj/mol) (1)( 7. kj/mol) 8.6 kj/mol (c) ΔG 4 Gf (CO ) Gf (HO) Gf (CH ) 5 Gf (O ) (4)(94.4 kj/mol) ()(7. kj/mol) ()(09. kj/mol) (5)(0) 470 kj/mol Strategy: o calculate the standard free-energy change of a reaction, we look up the standard free energies of formation of reactants and products in Appendix of the text and apply Equation (18.1). Note that all the stoichiometric coefficients have no units so G rxn is expressed in units of kj/mol. he standard free energy of formation of any element in its stable allotropic form at 1 atm and 5C is zero. Solution: he standard free energy change for a reaction can be calculated using the following equation. Grxn ngf (products) mg f (reactants) (a) Grxn Gf (MgO) [ Gf (Mg) G f (O )] G rxn ()( kj/mol) [()(0) (1)(0)] Grxn Gf (SO ) [ Gf (SO ) G f (O )] G rxn 119 kj/mol ()( 70.4 kj/mol) [()( 00.4 kj/mol) (1)(0)] kj/mol (c) Grxn 4 Gf [CO ( g)] 6 Gf [HO( l)] { Gf [CH 6( g)] 7 Gf [O ( g )]} G rxn (4)( 94.4 kj/mol) (6)( 7. kj/mol) [()(.89 kj/mol) (7)(0)] 95.0 kj/mol Reaction A: First apply Equation (18.10) of the text to compute the free energy change at 5C (98 K) G H S 10,500 J/mol (98 K)(0 J/Kmol) he 1560 J/mol shows the reaction is not spontaneous at 98 K. he G will change sign (i.e., the reaction will become spontaneous) above the temperature at which G 0. H J/mol S 0 J/K mol 50 K Reaction B: Calculate G. G H S 1800 J/mol (98 K)(11 J/Kmol) 5,500 J/mol he free energy change is positive, which shows that the reaction is not spontaneous at 98 K. Since both terms are positive, there is no temperature at which their sum is negative. he reaction is not spontaneous at any temperature.

4 470 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 18.0 Reaction A: Calculate G from H and S. G H S 16,000 J/mol (98 K)(84 J/Kmol) 151,000 J/mol he free energy change is negative so the reaction is spontaneous at 98 K. Since H is negative and S is positive, the reaction is spontaneous at all temperatures. Reaction B: Calculate G. G H S 11,700 J/mol (98 K)(105 J/Kmol) 19,600 J he free energy change is positive at 98 K which means the reaction is not spontaneous at that temperature. he positive sign of G results from the large negative value of S. At lower temperatures, the S term will be smaller thus allowing the free energy change to be negative. G will equal zero when H S. Rearranging, H J/mol S 105 J/K mol 111 K At temperatures below 111 K, G will be negative and the reaction will be spontaneous. 18. Find the value of K by solving Equation (18.14) of the text. G R J/mol (8.14 J/Kmol)(98 K) 1.05 K p e e e Strategy: According to Equation (18.14) of the text, the equilibrium constant for the reaction is related to the standard free energy change; that is, G R ln K. Since we are given the equilibrium constant in the problem, we can solve for G. What temperature unit should be used? Solution: he equilibrium constant is related to the standard free energy change by the following equation. G Rln K Substitute Kw, R, and into the above equation to calculate the standard free energy change, G. he temperature at which Kw is 5C 98 K. G Rln Kw G (8.14 J/molK)(98 K) ln ( ) J/mol kj/mol 18.5 Ksp [Fe ][OH ] G Rln Ksp (8.14 J/Kmol)(98 K)ln ( ) J/mol 79 kj/mol 18.6 Use standard free energies of formation from Appendix to find the standard free energy difference. Grxn Gf [H ( g)] Gf [O ( g)] Gf [HO( g )] G rxn ()(0) (1)(0) ()( 8.6 kj/mol) 5 G rxn 457. kj/mol J/mol

5 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 471 We can calculate Kp using the following equation. G Rln Kp J/mol (8.14 J/molK)(98 K) ln Kp 185 ln Kp aking the antiln of both sides, e 185 Kp Kp (a) We first find the standard free energy change of the reaction. G rxn Gf [PCl ( g)] Gf [Cl ( g)] Gf [PCl 5( g)] (1)(86 kj/mol) (1)(0) (1)(5 kj/mol) 9 kj/mol We can calculate Kp using Equation (18.14) of the text. G R 9 10 J/mol (8.14 J/Kmol)(98 K) 16 Kp e e e We are finding the free energy difference between the reactants and the products at their nonequilibrium values. he result tells us the direction of and the potential for further chemical change. We use the given nonequilibrium pressures to compute Qp. PPCl P Cl (0.7)(0.40) Qp 7 P PCl5 he value of G (notice that this is not the standard free energy difference) can be found using Equation (18.1) of the text and the result from part (a). G G Rln Q (9 10 J/mol) (8.14 J/Kmol)(98 K)ln (7) 48 kj/mol Which way is the direction of spontaneous change for this system? What would be the value of G if the given data were equilibrium pressures? What would be the value of Qp in that case? 18.8 (a) he equilibrium constant is related to the standard free energy change by the following equation. G Rln K Substitute Kp, R, and into the above equation to the standard free energy change, G. G Rln Kp G (8.14 J/molK)(000 K) ln (4.40) J/mol 4.6 kj/mol Strategy: From the information given we see that neither the reactants nor products are at their standard state of 1 atm. We use Equation (18.1) of the text to calculate the free-energy change under non-standardstate conditions. Note that the partial pressures are expressed as dimensionless quantities in the reaction quotient Qp.

6 47 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM Solution: Under non-standard-state conditions, G is related to the reaction quotient Q by the following equation. G G Rln Qp We are using Qp in the equation because this is a gas-phase reaction. Step 1: G was calculated in part (a). We must calculate Qp. Q p PH O PCO (0.66)(1.0) P P (0.5)(0.78) H CO 4.1 Step : Substitute G J/mol and Qp into the following equation to calculate G. G G Rln Qp G J/mol (8.14 J/molK)(000 K) ln (4.1) G ( J/mol) ( J/mol) G J/mol 1.10 kj/mol 18.9 he expression of Kp is: Kp P CO hus you can predict the equilibrium pressure directly from the value of the equilibrium constant. he only task at hand is computing the values of Kp using Equations (18.10) and (18.14) of the text. (a) At 5C, G H S ( J/mol) (98 K)(160.5 J/Kmol) J/mol G J/mol (8.14 J/Kmol)(98 K) 5.47 p R CO K e e e atm P At 800C, G H S ( J/mol) (107 K)(160.5 J/Kmol) J/mol P CO p G R J/mol (8.14 J/Kmol)(107 K) 0.65 K e e e 0.55 atm What assumptions are made in the second calculation? 18.0 We use the given Kp to find the standard free energy change. G Rln K G (8.14 J/Kmol)(98 K) ln ( ) J/mol 04 kj/mol he standard free energy of formation of one mole of COCl can now be found using the standard free energy of reaction calculated above and the standard free energies of formation of CO(g) and Cl(g). Grxn ngf (products) mg f (reactants) Grxn Gf [COCl ( g)] { Gf [CO( g)] Gf [Cl ( g )]} 04 kj/mol (1) Gf[COCl ( g )] [(1)( 17. kj/mol) (1)(0)] Gf[COCl ( g )] 41 kj/mol

7 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM he equilibrium constant expression is: K P p We are actually finding the equilibrium vapor pressure of water (compare to Problem 18.9). We use Equation (18.14) of the text. G J/mol (8.14 J/Kmol)(98 K).47 p R PHO K e e e.1 10 atm H O he positive value of G implies that reactants are favored at equilibrium at 5C. Is that what you would expect? 18. he standard free energy change is given by: Grxn Gf (graphite) G f (diamond) You can look up the standard free energy of formation values in Appendix of the text. G rxn (1)(0) (1)(.87 kj/mol).87 kj/mol hus, the formation of graphite from diamond is favored under standard-state conditions at 5C. However, the rate of the diamond to graphite conversion is very slow (due to a high activation energy) so that it will take millions of years before the process is complete C6H1O6 6O 6CO 6HO G 880 kj/mol ADP HPO4 AP HO G 1 kj/mol Maximum number of AP molecules synthesized: 1 AP molecule 880 kj/mol 1 kj/mol 9 AP molecules 18.6 he equation for the coupled reaction is: glucose AP glucose 6phosphate ADP G 1.4 kj/mol 1 kj/mol 18 kj/mol As an estimate: G ln K R ( J/mol) ln K 7. (8.14 J/K mol)(98 K) K When Humpty broke into pieces, he became more disordered (spontaneously). he king was unable to reconstruct Humpty.

8 474 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 18.8 In each part of this problem we can use the following equation to calculate G. or, (a) G G Rln Q G G Rln [H ][OH ] In this case, the given concentrations are equilibrium concentrations at 5C. Since the reaction is at equilibrium, G 0. his is advantageous, because it allows us to calculate G. Also recall that at equilibrium, Q K. We can write: G Rln Kw G (8.14 J/Kmol)(98 K) ln ( ) J/mol G G Rln Q G Rln [H ][OH ] G ( J/mol) (8.14 J/Kmol)(98 K) ln [( )( )] J/mol (c) G G Rln Q G Rln [H ][OH ] G ( J/mol) (8.14 J/Kmol)(98 K) ln [( )( )] J/mol (d) G G Rln Q G Rln [H ][OH ] G ( J/mol) (8.14 J/Kmol)(98 K) ln [(.5)( )] J/mol 18.9 Only E and H are associated with the first law alone One possible explanation is simply that no reaction is possible, namely that there is an unfavorable free energy difference between products and reactants (G > 0). A second possibility is that the potential for spontaneous change is there (G < 0), but that the reaction is extremely slow (very large activation energy). A remote third choice is that the student accidentally prepared a mixture in which the components were already at their equilibrium concentrations. Which of the above situations would be altered by the addition of a catalyst? (a) An ice cube melting in a glass of water at 0C. he value of G for this process is negative so it must be spontaneous. (c) (d) A "perpetual motion" machine. In one version, a model has a flywheel which, once started up, drives a generator which drives a motor which keeps the flywheel running at a constant speed and also lifts a weight. A perfect air conditioner; it extracts heat energy from the room and warms the outside air without using any energy to do so. (Note: this process does not violate the first law of thermodynamics.) Same example as (a). (e) A closed flask at 5C containing NO(g) and NO4(g) at equilibrium For a solid to liquid phase transition (melting) the entropy always increases (S > 0) and the reaction is always endothermic (H > 0). (a) Melting is always spontaneous above the melting point, so G < 0.

9 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 475 At the melting point (77.7C), solid and liquid are in equilibrium, so G 0. (c) Melting is not spontaneous below the melting point, so G > For a reaction to be spontaneous, G must be negative. If S is negative, as it is in this case, then the reaction must be exothermic (why?). When water freezes, it gives off heat (exothermic). Consequently, the entropy of the surroundings increases and Suniverse > If the process is spontaneous as well as endothermic, the signs of G and H must be negative and positive, respectively. Since G H S, the sign of S must be positive (S > 0) for G to be negative he equation is: BaCO(s) BaO(s) CO(g) G Gf (BaO) Gf (CO ) G f (BaCO ) G (1)(58.4 kj/mol) (1)(94.4 kj/mol) (1)(118.9 kj/mol) 16.1 kj/mol G Rln Kp J/mol ln K p 87. (8.14 J/K mol)(98 K) K p PCO e 1 10 atm (a) Using the relationship: H vap b.p. S 90 J/K mol vap benzene hexane mercury toluene Svap 87.8 J/Kmol Svap 90.1 J/Kmol Svap 9.7 J/Kmol Svap 91.8 J/Kmol Most liquids have Svap approximately equal to a constant value because the order of the molecules in the liquid state is similar. he order of most gases is totally random; thus, S for liquid vapor should be similar for most liquids. Using the data in able 11.6 of the text, we find: ethanol water Svap J/Kmol Svap J/Kmol Both water and ethanol have a larger Svap because the liquid molecules are more ordered due to hydrogen bonding (there are fewer microstates in these liquids) Evidence shows that HF, which is strongly hydrogen-bonded in the liquid phase, is still considerably hydrogen-bonded in the vapor state such that its Svap is smaller than most other substances.

10 476 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM (a) CO NO CO N he oxidizing agent is NO; the reducing agent is CO. (c) G Gf (CO ) Gf (N ) Gf (CO) G f (NO) G ()(94.4 kj/mol) (0) ()(17. kj/mol) ()(86.7 kj/mol) kj/mol G Rln Kp J/mol ln K p 78 (8.14 J/K mol)(98 K) Kp (d) N P CO (0.80)(0.00) COPNO 5 7 ( ) ( ) P 18 Q p P Since Qp << Kp, the reaction will proceed from left to right. (e) H Hf (CO ) Hf (N ) Hf (CO) H f (NO) H ()(9.5 kj/mol) (0) ()(110.5 kj/mol) ()(90.4 kj/mol) kj/mol Since H is negative, raising the temperature will decrease Kp, thereby increasing the amount of reactants and decreasing the amount of products. No, the formation of N and CO is not favored by raising the temperature (a) At two different temperatures 1 and, G1 H 1 S R ln K 1 (1) G H S R ln K () Rearranging Equations (1) and (), ln K 1 H R 1 S R () ln K H R S R (4) Subtracting equation () from equation (4) gives, H S H S ln K ln K1 R R R1 R K H 1 1 ln K1 R 1

11 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 477 K H 1 ln K1 R 1 Using the equation that we just derived, we can calculate the equilibrium constant at 65C. K K K? 8 K ln K J/mol 8 K 98 K J/K mol (8 K)(98 K) K ln aking the antiln of both sides of the equation, K K e.77 K > K1, as we would predict for a positive H. Recall that an increase in temperature will shift the equilibrium towards the endothermic reaction; that is, the decomposition of NO he equilibrium reaction is: AgCl(s) Ag (aq) Cl (aq) Ksp [Ag ][Cl ] We can calculate the standard enthalpy of reaction from the standard enthalpies of formation in Appendix of the text. H Hf (Ag ) Hf (Cl ) H f (AgCl) H (1)(105.9 kj/mol) (1)(167. kj/mol) (1)(17.0 kj/mol) 65.7 kj/mol From Problem 18.49(a): K H ln 1 K1 R 1 K K? 1 98 K K ln 4 K J K 98 K J/K mol ( K)(98 K) K ln

12 478 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM K e K he increase in K indicates that the solubility increases with temperature At absolute zero. A substance can never have a negative entropy Assuming that both H and S are temperature independent, we can calculate both H and S. H Hf (CO) Hf (H ) [ Hf (HO) H f (C)] H (1)(110.5 kj/mol) (1)(0)] [(1)(41.8 kj/mol) (1)(0)] H 11. kj/mol S S(CO) S(H) [S(HO) S(C)] S [(1)(197.9 J/Kmol) (1)(11.0 J/Kmol)] [(1)(188.7 J/Kmol) (1)(5.69 J/Kmol)] S 14.5 J/Kmol It is obvious from the given conditions that the reaction must take place at a fairly high temperature (in order to have redhot coke). Setting G 0 0 H S 1000 J 11. kj/mol H 1 kj 976 K 70C S 14.5 J/K mol he temperature must be greater than 70C for the reaction to be spontaneous (a) We know that HCl is a strong acid and HF is a weak acid. hus, the equilibrium constant will be less than 1 (K < 1). he number of particles on each side of the equation is the same, so S 0. herefore H will dominate. (c) HCl is a weaker bond than HF (see able 9.4 of the text), therefore H > For a reaction to be spontaneous at constant temperature and pressure, G < 0. he process of crystallization proceeds with more order (less disorder), so S < 0. We also know that G H S Since G must be negative, and since the entropy term will be positive (S, where S is negative), then H must be negative (H < 0). he reaction will be exothermic.

13 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM For the reaction: CaCO(s) CaO(s) CO(g) Kp P CO Using the equation from Problem 18.49: K H 1 1 H 1 ln K1 R 1 R 1 Substituting, Solving, 189 H 1 K 97 K ln J/K mol (97 K)(1 K) H J/mol 174 kj/mol For the reaction to be spontaneous, G must be negative. G H S Given that H 19 kj/mol 19,000 J/mol, then Solving the equation with the value of G 0 G 19,000 J/mol (7 K 7 K)(S) 0 19,000 J/mol (7 K 7 K)(S) S 55 J/Kmol his value of S which we solved for is the value needed to produce a G value of zero. he minimum value of S that will produce a spontaneous reaction will be any value of entropy greater than 55 J/Kmol (a) S > 0 S < 0 (c) S > 0 (d) S > he second law states that the entropy of the universe must increase in a spontaneous process. But the entropy of the universe is the sum of two terms: the entropy of the system plus the entropy of the surroundings. One of the entropies can decrease, but not both. In this case, the decrease in system entropy is offset by an increase in the entropy of the surroundings. he reaction in question is exothermic, and the heat released raises the temperature (and the entropy) of the surroundings. Could this process be spontaneous if the reaction were endothermic? At the temperature of the normal boiling point the free energy difference between the liquid and gaseous forms of mercury (or any other substances) is zero, i.e. the two phases are in equilibrium. We can therefore use Equation (18.10) of the text to find this temperature. For the equilibrium, Hg(l) Hg(g) G H S 0 H Hf[Hg( g)] Hf[Hg( l )] 60, 780 J/mol J/mol S S[Hg(g)] S[Hg(l)] J/Kmol 77.4 J/Kmol 97. J/Kmol

14 480 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM bp H J/mol 65 K S 97. J/K mol 5C What assumptions are made? Notice that the given enthalpies and entropies are at standard conditions, namely 5C and 1.00 atm pressure. In performing this calculation we have tacitly assumed that these quantities don't depend upon temperature. he actual normal boiling point of mercury is 56.58C. Is the assumption of the temperature independence of these quantities reasonable? Strategy: At the boiling point, liquid and gas phase ethanol are at equilibrium, so G 0. From equation (18.10) of the text, we have G 0 H S or S H/. o calculate the entropy change for the liquid ethanol gas ethanol transition, we write Svap Hvap/. What temperature unit should we use? Solution: he entropy change due to the phase transition (the vaporization of ethanol), can be calculated using the following equation. Recall that the temperature must be in units of Kelvin (78.C 51 K). S vap H vap b.p. 9. kj/mol S vap 0.11 kj/mol K 11 J/mol K 51 K he problem asks for the change in entropy for the vaporization of 0.50 moles of ethanol. he S calculated above is for 1 mole of ethanol. S for 0.50 mol (11 J/molK)(0.50 mol) 56 J/K here is no connection between the spontaneity of a reaction predicted by G and the rate at which the reaction occurs. A negative free energy change tells us that a reaction has the potential to happen, but gives no indication of the rate. Does the fact that a reaction occurs at a measurable rate mean that the free energy difference G is negative? 18.6 For the given reaction we can calculate the standard free energy change from the standard free energies of formation (see Appendix of the text). hen, we can calculate the equilibrium constant, Kp, from the standard free energy change. G Gf [Ni(CO) 4] [4 Gf (CO) G f (Ni)] G (1)(587.4 kj/mol) [(4)(17. kj/mol) (1)(0)] 8. kj/mol J/mol Substitute G, R, and (in K) into the following equation to solve for Kp. G Rln Kp ln K p G (.8 10 J/mol) R (8.14 J/K mol)(5 K) 4 Kp (a) G Gf (HBr) Gf (H ) G f (Br ) ()( 5. kj/mol) (1)(0) (1)(0) G kj/mol

15 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 481 G J/mol ln Kp 4.9 R (8.14 J/K mol)(98 K) Kp G G f G f G f G 5. kj/mol (HBr) (H ) (Br ) (1)( 5. kj/mol) ( )(0) ( )(0) G J/mol ln Kp 1.5 R (8.14 J/K mol)(98 K) Kp 10 9 he Kp in (a) is the square of the Kp in. Both G and Kp depend on the number of moles of reactants and products specified in the balanced equation he equilibrium constant is related to the standard free energy change by the following equation: G Rln Kp J/mol (8.14 J/molK)(98 K) ln Kp 85.6 ln Kp Kp We can write the equilibrium constant expression for the reaction. Kp P O P ( K ) his pressure is far too small to measure. O p 8 P O ( ) atm alking involves various biological processes (to provide the necessary energy) that lead to a increase in the entropy of the universe. Since the overall process (talking) is spontaneous, the entropy of the universe must increase Both (a) and apply to a reaction with a negative G value. Statement (c) is not always true. An endothermic reaction that has a positive S (increase in entropy) will have a negative G value at high temperatures (a) If G for the reaction is 17.4 kj/mol, then, 17.4 kj/mol G f 86.7 kj/mol G Rln Kp J/mol (8.14 J/Kmol)(98 K)ln Kp Kp

16 48 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM (c) H for the reaction is H f (NO) ()(86.7 kj/mol) 17.4 kj/mol Using the equation in Problem 18.49: ln K J/mol 17 K 98 K J/mol K (17 K)(98 K) K 10 7 (d) Lightning promotes the formation of NO (from N and O in the air) which eventually leads to the formation of nitrate ion (NO ), an essential nutrient for plants We write the two equations as follows. he standard free energy change for the overall reaction will be the sum of the two steps. CuO(s) Cu(s) 1 O(g) G 17. kj/mol C(graphite) 1 O(g) CO(g) G 17. kj/mol CuO C(graphite) Cu(s) CO(g) G 10.1 kj/mol We can now calculate the equilibrium constant from the standard free energy change, G. ln K G ( J/mol) R (8.14 J/K mol)(67 K) ln K 1.81 K Using the equation in the Chemistry in Action entitled he Efficiency of Heat Engines in Chapter 18: 1 47 K 10 K Efficiency K he work done by moving the car: mgh (100 kg)(9.81 m/s ) h heat generated by the engine. he heat generated by the gas:.1 kg 1000 g 1 mol J gal J 1 gal 1 kg 114. g 1 mol he maximum use of the energy generated by the gas is: (energy)(efficiency) ( J)(0.58) J Setting the (useable) energy generated by the gas equal to the work done moving the car: J (100 kg)(9.81 m/s ) h h m

17 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM Crystal structure has disorder.. here is impurity present in the crystal (a) he first law states that energy can neither be created nor destroyed. We cannot obtain energy out of nowhere. If we calculate the efficiency of such an engine, we find that h c, so the efficiency is zero! See Chemistry in Action on p. 777 of the text (a) G Gf (H ) Gf (Fe ) Gf (Fe) G f (H )] G (1)(0) (1)(84.9 kj/mol) (1)(0) ()(0) G 84.9 kj/mol G Rln K J/mol (8.14 J/molK)(98 K) ln K K G Gf (H ) Gf (Cu ) Gf (Cu) G f (H )] G kj/mol G Rln K J/mol (8.14 J/molK)(98 K) ln K K he activity series is correct. he very large value of K for reaction (a) indicates that products are highly favored; whereas, the very small value of K for reaction indicates that reactants are highly favored. kf 18.7 NO O NO kr G ()(51.8 kj/mol) ()(86.7 kj/mol) kj/mol G Rln K J/mol (8.14 J/molK)(98 K)ln K K M 1 K f k k r M M s k r kr M 1 s 1

18 484 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM (a) It is a reverse disproportionation redox reaction. G ()(8.6 kj/mol) ()(.0 kj/mol) (1)(00.4 kj/mol) G 90.8 kj/mol J/mol (8.14 J/molK)(98 K) ln K K Because of the large value of K, this method is efficient for removing SO. (c) H ()(41.8 kj/mol) ()(0) ()(0.15 kj/mol) (1)(96.1 kj/mol) H 147. kj/mol S ()(188.7 J/Kmol) ()(1.88 J/Kmol) ()(05.64 J/Kmol) (1)(48.5 J/Kmol) S J/Kmol G H S Due to the negative entropy change, S, the free energy change, G, will become positive at higher temperatures. herefore, the reaction will be less effective at high temperatures (1) Measure K and use G R ln K () Measure H and S and use G H S O O G Gf (O ) G f (O ) 0 ()(16.4 kj/mol) G 6.8 kj/mol J/mol (8.14 J/molK)(4 K) ln Kp Kp Due to the large magnitude of K, you would expect this reaction to be spontaneous in the forward direction. However, this reaction has a large activation energy, so the rate of reaction is extremely slow First convert to moles of ice. For a phase transition: 1 mol H O( s) 74.6 g H O( ) 4.14 mol H O( s) S S sys sys s 18.0 g HO( s) Hsys (4.14)(6060 J/mol) 7 K 91.1 J/K mol

19 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 485 S S surr surr Hsys (4.14)(6060 J/mol) 7 K 91.1 J/K mol Suniv Ssys Ssurr 0 his is an equilibrium process. here is no net change Heating the ore alone is not a feasible process. Looking at the coupled process: CuS Cu S G 86.1 kj/mol S O SO G 00.4 kj/mol CuS O Cu SO G 14. kj/mol Since G is a large negative quantity, the coupled reaction is feasible for extracting sulfur Since we are dealing with the same ion (K ), Equation (18.1) of the text can be written as: G G Rln Q 400 mm G 0 (8.14 J/mol K)(10 K)ln 15 m M G J/mol 8.5 kj/mol First, we need to calculate H and S for the reaction in order to calculate G. H 41. kj/mol S 4.0 J/Kmol Next, we calculate G at 00C or 57 K, assuming that H and S are temperature independent. G H S G J/mol (57 K)(4.0 J/Kmol) G J/mol Having solved for G, we can calculate K p. G Rln K p J/mol (8.14 J/Kmol)(57 K) ln K p ln K p.59 K p 6 Due to the negative entropy change calculated above, we expect that G will become positive at some temperature higher than 00C. We need to find the temperature at which G becomes zero. his is the temperature at which reactants and products are equally favored (K p 1). G H S 0 H S

20 486 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM H J/mol S 4.0 J/K mol 981 K 708C his calculation shows that at 708C, G 0 and the equilibrium constant K p 1. Above 708C, G is positive and K p will be smaller than 1, meaning that reactants will be favored over products. Note that the temperature 708C is only an estimate, as we have assumed that both H and S are independent of temperature. Using a more efficient catalyst will not increase K p at a given temperature, because the catalyst will speed up both the forward and reverse reactions. he value of K p will stay the same (a) G for CHCOOH: G (8.14 J/molK)(98 K) ln ( ) G J/mol 7 kj/mol G for CHClCOOH: G (8.14 J/molK)(98 K) ln ( ) G J/mol 16 kj/mol he S is the dominant term. (c) he breaking of the OH bond in ionization of the acid and the forming of the OH bond in HO. (d) he CHCOO ion is smaller than CHClCOO and can participate in hydration to a greater extent, leading to a more ordered solution butane isobutane G Gf(isobutane) G f(butane) G (1)(18.0 kj/mol) (1)(15.9 kj/mol) G.1 kj/mol For a mixture at equilibrium at 5C: G Rln K p.1 10 J/mol (8.14 J/molK)(98 K) ln K p K p. K p P isobutane P butane mol isobutane mol butane. mol isobutane mol butane

21 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM 487 his shows that there are. times as many moles of isobutane as moles of butane. Or, we can say for every one mole of butane, there are. moles of isobutane.. mol mol % isobutane 100% 70%. mol 1.0 mol By difference, the mole % of butane is 0%. Yes, this result supports the notion that straight-chain hydrocarbons like butane are less stable than branchedchain hydrocarbons like isobutane Heat is absorbed by the rubber band, so H is positive. Since the contraction occurs spontaneously, G is negative. For the reaction to be spontaneous, S must be positive meaning that the rubber becomes more disordered upon heating. his is consistent with what we know about the structure of rubber; he rubber molecules become more disordered upon contraction (See the Figure in the Chemistry in Action Essay on p. 787 of the text) We can calculate K p from G. G (1)(94.4 kj/mol) (0) (1)(17. kj/mol) (1)(55. kj/mol) G 1.9 kj/mol J/mol (8.14 J/molK)(117 K) ln K p K p 1. Now, from K p, we can calculate the mole fractions of CO and CO. P K P P CO p 1. CO 1. CO PCO P CO PCO 1 CO P P P 1.P CO CO CO CO CO We assumed that G calculated from G f values was temperature independent. he Gf values in Appendix of the text are measured at 5C, but the temperature of the reaction is 900C G Rln K and, G G Rln Q Substituting, G Rln K Rln Q G R(ln Q ln K) ln Q G R K

22 488 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM If Q > K, G > 0, and the net reaction will proceed from right to left (see Figure 14.4 of the text). If Q < K, G < 0, and the net reaction will proceed from left to right. If Q K, G 0. he system is at equilibrium For a phase transition, G 0. We write: G H S 0 H S sub S sub H Substituting H and the temperature, (78 7)K 195 K, gives Hsub J S sub 19 J/K mol 195 K his value of Ssub is for the sublimation of 1 mole of CO. We convert to the S value for the sublimation of 84.8 g of CO. 1 mol CO 19 J 84.8 g CO 49 J/K g CO K mol he second law of thermodynamics states that the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. herefore, the entropy of the universe is increasing with time, and thus entropy could be used to determine the forward direction of time First, let's convert the age of the universe from units of years to units of seconds days 4 h 600 s 17 (1 10 yr) s 1 yr 1 day 1 h he probability of finding all 100 molecules in the same flask is Multiplying by the number of seconds gives: ( )( s) 10 1 s Equation (18.10) represents the standard free-energy change for a reaction, and not for a particular compound like CO. he correct form is: G H S For a given reaction, G and H would need to be calculated from standard formation values (graphite, oxygen, and carbon dioxide) first, before plugging into the equation. Also, S would need to be calculated from standard entropy values. C(graphite) O(g) CO(g)

23 CHAPER 18: ENROPY, FREE ENERGY, AND EQUILIBRIUM We can calculate Ssys from standard entropy values in Appendix of the text. We can calculate Ssurr from the Hsys value given in the problem. Finally, we can calculate Suniv from the Ssys and Ssurr values. Ssys ()(69.9 J/Kmol) [()(11.0 J/Kmol) (1)(05.0 J/Kmol)] 7 J/Kmol S surr Hsys ( J/mol) 98 K 1918 J/K mol Suniv Ssys Ssurr (7 1918) J/Kmol 1591 J/Kmol H is endothermic. Heat must be added to denature the protein. Denaturation leads to more disorder (an increase in microstates). he magnitude of S is fairly large (1600 J/Kmol). Proteins are large molecules and therefore denaturation would lead to a large increase in microstates. he temperature at which the process favors the denatured state can be calculated by setting G equal to zero. G H S 0 H S H 51 kj/mol 0 K 47C S 1.60 kj/k mol 18.9 q, and w are not state functions. Recall that state functions represent properties that are determined by the state of the system, regardless of how that condition is achieved. Heat and work are not state functions because they are not properties of the system. hey manifest themselves only during a process (during a change). hus their values depend on the path of the process and vary accordingly (d) will not lead to an increase in entropy of the system. he gas is returned to its original state. he entropy of the system does not change Since the adsorption is spontaneous, G must be negative (G < 0). When hydrogen bonds to the surface of the catalyst, the system becomes more ordered (S < 0). Since there is a decrease in entropy, the adsorption must be exothermic for the process to be spontaneous (H < 0).