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CHAPTER SIXTEEN SPONTANEITY, ENTROPY, AND FREE ENERGY For Review 1. a. A spontaneous process is one that occurs without any outside intervention. b. Entropy is a measure of disorder or randomness. c. Positional probability is a type of probability that depends on the number of arrangements in space that yield a particular state. d. The system is the portion of the universe in which we are interested. e. The surroundings are everything else in the universe besides the system. f. The universe is everything; universe = system + surroundings. 2. Second law of thermodynamics: in any spontaneous process, there is always an increase in the entropy of the universe. Suniv = Ssys + Ssurr; When both Ssys and Ssurr are positive, Suniv must be positive (so process is spontaneous). Suniv is always negative (so process is nonspontaneous) when both Ssys and Ssurr are negative. When the signs of Ssys are opposite of each other [(Ssys (+), Ssurr () or vice versa], the process may or may not be spontaneous. 3. Ssurr is primarily determined by heat flow. This heat flow into or out of the surroundings comes from the heat flow out of or into the system. In an exothermic process (H < 0), heat flows into the surroundings from the system. The heat flow into the surroundings increases the random motions in the surroundings and increases the entropy of the surroundings (Ssurr > 0). This is a favorable driving force for spontaneity. In an endothermic reaction (H > 0), heat is transferred from the surroundings into the system. This heat flow out of the surroundings decreases the random motions in the surroundings and decreases the entropy of the surroundings (Ssurr < 0). This is unfavorable. The magnitude of Ssurr also depends on the temperature. The relationship is inverse; at low temperatures, a specific amount of heat exchange makes a larger percent change in the surroundings than the same amount of heat flow at a higher temperature. The negative sign in the Ssurr = H/T equation is necessary to get the signs correct. For an exothermic reaction where H is negative, this increases Ssurr so the negative sign converts the negative H value into a positive quantity. For an endothermic process where H is positive, the sign of Ssurr is negative and the negative sign converts the positive H value into a negative quantity. 632 CHAPTER 16 4. SPONTANEITY, ENTROPY, AND FREE ENERGY 633 Suniv = G/T (at constant T and P); When G is negative (Suniv > 0), the process is spontaneous. When G is positive (Suniv < 0), the process in nonspontaneous (the reverse process is spontaneous). When G = 0, the process is at equilibrium. G = H TS; See Table 16.5 for the four possible sign conventions and the temperature dependence for these sign combinations. When the signs for H and S are both the same, then temperature determines if the process is spontaneous. When H is positive (unfavorable) and S is positive (favorable), high temperatures are needed so the favorable S term dominates making the process spontaneous (G < 0). When H is negative (favorable) and S is negative (unfavorable), low temperatures are needed so the favorable H term dominates making the process spontaneous (G < 0). Note that if G is positive for a process, then the reverse process has a negative G value and is spontaneous. At the phase change temperature (melting point or boiling point), two phases are in equilibrium with each other so G = 0. The G = H TS equation reduces to H = TS at the phase change temperature. For the s → l phase change, G is negative above the freezing point because the process is spontaneous (as we know). For the l → g phase change, the sign of G is positive below the boiling point as the process is nonspontaneous (as we know). 5. Third law of thermodynamics: the entropy of a perfect crystal at 0 K is zero. Standard entropy values (S) represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. The equation to calculate S for a reaction using the standard entropy values is: ΔSorxn = ΣnpSoproducts ΣnrSoreactants This equation works because entropy is a state function of the system (it is not pathwaydependent). Because of this, one can manipulate chemical reactions with known Svalues to determine S for a different reaction. The entropy change for a different reaction equals the sum of the entropy changes for the reactions added together that yield the different reaction. This is utilizing Hess’s law. The superscript indicates conditions where T = 25C and P = 1 atm. To predict signs for gas phase reactions, you need to realize that the gaseous state represents a hugely more disordered state as compared to the solid and liquid states. Gases dominate sign predictions for reactions. Those reactions that show an increase in the number of moles of gas as reactants are converted to products have an increase in disorder which translates into a positive S value. S values are negative when there is a decrease in the moles of gas as reactants are converted into products. When the moles of gaseous reactants and products are equal, S is usually difficult to predict for chemical reactions. However, predicting signs for phase changes can be done by realizing the solid state is the most ordered phase (lowest S values), the liquid state is a slightly more disordered phase than the solid state, with the o Soliquid Sogas ) . gaseous state being the most disordered phase by a large margin (Ssolid Another process involving condensed phases whose sign is also easy to predict (usually) is the dissolution of a solute in a solvent. Here, the mixed up solution state is usually the more disordered state as compared to the solute and solvent separately. 6. Standard free energy change: the change in free energy that will occur for one unit of reaction if the reactants in their standard states are converted to products in their standard state. The standard free energy of formation (G of ) of a substance is the change in free energy that 634 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states. The equation that manipulates G of values to determining ΔG oreaction is: ΔG° = Σnp Δn of (products) Σnr ΔG of (reactants) . Because ΔG is a state function (path independent), chemical reactions with known ΔG values can be manipulated to determine ΔG for a different reaction. ΔG for the different reaction is the sum of ΔG for all the steps (reactions) added together to get the different reaction. This is Hess’s law. Another way to determine ΔG° for a reaction is to utilize the ΔG° = ΔH° TΔS° equation. Here, you need to know ΔH°, ΔS°, and the temperature, then you can use the above equation to calculate ΔG°. Of the functions ΔG, ΔH, and ΔS, ΔG has the greatest dependence on temperature. The temperature is usually assumed to be 25C. However, if other temperatures are used in a reaction, we can estimate ΔG° at that different temperature by assuming ΔH° and ΔS° are temperature independent (which is not always the best assumption). We calculate ΔH and ΔS° values for a reaction using Appendix 4 data, then use the different temperature in the ΔG° = ΔH TS° equation to determine (estimate) ΔG° at that different temperature. 7. No; When using G of values in Appendix 4, we have specified a temperature of 25°C. Further, if gases or solutions are involved, we have specified partial pressures of 1 atm and solute concentrations of 1 molar. At other temperatures and compositions, the reaction may not be spontaneous. A negative ΔG° value means the reaction is spontaneous under standard conditions. The free energy and pressure relationship is G = G° + RT ln (P). The RT ln P term corrects for nonstandard pressures (or concentrations if solutes are involved in the reaction). The standard pressure for a gas is 1 atm and the standard concentration for solutes is 1 M. The equation to calculate ΔG for a reaction at nonstandard conditions is: ΔG = ΔG° + RT ln Q where Q is the reaction quotient determined at the nonstandard pressures and/or concentrations of the gases and/or solutes in the reaction. The reaction quotient has the exact same form as the equilibrium constant K. The difference is that the partial pressures or concentrations used may or may not be the equilibrium concentrations. All reactions want to minimize their free energy. This is the driving force for any process. As long as ΔG is a negative, the process occurs. The equilibrium position represents the lowest total free energy available to any particular reaction system. Once equilibrium is reached, the system cannot minimize its free energy anymore. Converting from reactants to products or products to reactants will increase the total free energy of the system which reactions do not want to do. 8. At equilibrium, ΔG = 0 and Q = K (the reaction quotient equals the equilibrium constant value). From the ΔG° = RT ln K equation, when a reaction has K < 1, the ln K term is negative, giving a positive ΔG° value. When K > 1, the ln K term is positive so ΔG° is negative. When ΔG° = 0, this tell us that K for the process is equal to one (K = 1) because ln 1 = 0. CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 635 The sign of ΔG (positive or negative) tells us which reaction is spontaneous (the forward or reverse reaction). If ΔG < 0, then the forward reaction is spontaneous and if ΔG > 0, then the reverse reaction is spontaneous. If ΔG = 0, then the reaction is at equilibrium (neither the forward or reverse reactions are spontaneous). ΔG° gives the equilibrium position by determining K for a reaction utilizing the equation ΔG° = RT ln K. ΔG° can only be used to predict spontaneity when all reactants and products are present at standard pressures of 1 atm and/or standard concentrations of 1 M. 9. A negative ΔG value does not guarantee that a reaction will occur. It does say that it can occur (is spontaneous), but whether it will occur also depends on how fast the reaction is (depends on the kinetics). A process with a negative ΔG may not occur because it is too slow. The example used in the text is the conversion of diamonds into graphite. Thermodynamics says the reaction can occur (ΔG > 0), but the reaction is so slow that it doesn’t occur. The rate of a reaction is directly related to temperature. As temperature increases, the rate of a reaction increases. Spontaneity, however, does not necessarily have a direct relationship to temperature. The temperature dependence of spontaneity depends on the signs of ΔH and ΔS (see Table 16.5 of the text). For example, when ΔH and ΔS are both negative, the reaction becomes more favorable thermodynamically (ΔG becomes more negative) with decreasing temperature. This is just the opposite of the kinetics dependence on temperature. Other sign combinations of ΔH and ΔS have different spontaneity temperature dependence. 10. wmax = ΔG; When ΔG is negative, the magnitude of ΔG is equal to the maximum possible useful work obtainable from the process (at constant T and P). When ΔG is positive, the magnitude of ΔG is equal to the minimum amount of work that must be expended to make the process spontaneous. Due to waste energy (heat) in any real process, the amount of useful work obtainable from a spontaneous process is always less than wmax, and for a nonspontaneous reaction, an amount of work greater than wmax must be applied to make the process spontaneous. Reversible process: a cyclic process carried out by a hypothetical pathway, which leaves the universe the same as it was before. No real process is reversible. Questions 7. Living organisms need an external source of energy to carry out these processes. Green plants use the energy from sunlight to produce glucose from carbon dioxide and water by photosynthesis. In the human body, the energy released from the metabolism of glucose helps drive the synthesis of proteins. For all processes combined, ΔSuniv must be greater than zero (2nd law). 8. Dispersion increases the entropy of the universe because the more widely something is dispersed, the greater the disorder. We must do work to overcome this disorder. In terms of the 2nd law, it would be more advantageous to prevent contamination of the environment rather than to clean it up later. As a substance disperses, we have a much larger area that must be decontaminated. 636 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 9. It appears that the sum of the two processes has no net change. This is not so. By the second law of thermodynamics, ΔSuniv must have increased even though it looks as if we have gone through a cyclic process. 10. The introduction of mistakes is an effect of entropy. The purpose of redundant information is to provide a control to check the "correctness" of the transmitted information. 11. As a process occurs, Suniv will increase; Suniv cannot decrease. Time, like Suniv, only goes in one direction. 12. This reaction is kinetically slow but thermodynamically favorable (ΔG < 0). Thermodynamics only tells us if a reaction can occur. To answer the question will it occur, one also needs to consider the kinetics (speed of reaction). The ultraviolet light provides the activation energy for this slow reaction to occur. 13. Possible arrangements for one molecule: 1 way 1 way Both are equally probable. Possible arrangements for two molecules: 1 way 2 ways most probable 1 way Possible arrangement for three molecules: 1 way 3 ways 3 ways 1 way equally most probable 14. Ssurr = H/T; Heat flow (H) into or out of the system dictates Ssurr. If heat flows into the surroundings, the random motions of the surroundings increase and the entropy of the surroundings increase. The opposite is true when heat flows from the surroundings into the system (an endothermic reaction). Although the driving force described here really results from the change in entropy of the surroundings, it is often described in terms of energy. Nature tends to seek the lowest possible energy. CHAPTER 16 15. 16. 17. SPONTANEITY, ENTROPY, AND FREE ENERGY 637 Note that these substances are not in the solid state, but are in the aqueous state; water molecules are also present. There is an apparent increase in ordering when these ions are placed in water as compared to the separated state. The hydrating water molecules must be in a highly ordered arrangement when surrounding these anions. G = RTlnK = H TS; HX(aq) ⇌ H+(aq) + X(aq) Ka reaction; The value of Ka for HF is less than one while the other hydrogen halide acids have K a > 1. In terms of G, HF must have a positive G orxn value while the other H-X acids have Grxn < 0. The reason for the sign change in the Ka value between HF versus HCl, HBr, and HI is entropy. S for the dissociation of HF is very large and negative. There is a high degree of ordering that occurs as the water molecules associate (hydrogen bond) with the small F ions. The entropy of hydration strongly opposes HF dissociating in water, so much so that it overwhelms the favorable hydration energy making HF a weak acid. One can determine S and H for the reaction using the standard entropies and standard enthalpies of formation in Appendix 4, then use the equation G = H TS. One can also use the standard free energies of formation in Appendix 4. And finally, one can use Hess’s law to calculate G. Here, reactions having known G values are manipulated to determine G for a different reaction. For temperatures other than 25C, G is estimated using the G = H TS equation. The assumptions made are that the H and S values determined from Appendix 4 data are temperature independent. We use the same H and S values as determined when T = 25C, then plug in the new temperature in Kelvin into the equation to estimate G at the new temperature. 18. The sign of G tells us if a reaction is spontaneous or not at whatever concentrations are present (at constant T and P). The magnitude of G equals wmax. When G < 0, the magnitude tells us how much work, in theory, could be harnessed from the reaction. When G > 0, the magnitude tells us the minimum amount of work that must be supplied to make the reaction occur. G gives us the same information only when the concentrations for all reactants and products are at standard conditions (1 atm for gases, 1 M for solute). These conditions rarely occur. G = RTlnK; From this equation, one can calculate K for a reaction if G is known at that temperature. Therefore, G gives the equilibrium position for a reaction. To determine K at a temperature other than 25C, one needs to know G at that temperature. We assume H and S are temperature independent and use the equation G = H - TS to estimate G at the different temperature. For K = 1, we want G = 0, which occurs when H = TS. Again, assume H and S are temperature independent, then solve for T (=H/S). At this temperature, K = 1 because G = 0. This only works for reactions where the signs of H and S are the same (either both positive or both negative). When the signs are opposite, K will always be greater than one (when H is negative and S is positive) or K will always be less than one (when H is positive and S is negative). When the signs of H and S are opposite, K can never equal one. 638 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY Exercises Spontaneity, Entropy, and the Second Law of Thermodynamics: Free Energy 19. a, b and c; From our own experiences, salt water, colored water and rust form without any outside intervention. A bedroom, however, spontaneously gets cluttered. It takes an outside energy source to clean a bedroom. 20. c and d; It takes an outside energy source to build a house and to launch and keep a satellite in orbit. 21. We draw all of the possible arrangements of the two particles in the three levels. 2 kJ 1 kJ 0 kJ x Total E = xx x x x 0 kJ 1 kJ 2 kJ x x xx xx __ 2 kJ 3 kJ 4 kJ The most likely total energy is 2 kJ. 22. 2 kJ AB 1 kJ AB B B A A B A A B B A_ A B_ 0 kJ AB _ ET = 0 kJ 2 kJ 4 kJ 1 kJ 1 kJ 2 kJ 2 kJ 3 kJ 3 kJ The most likely total energy is 2 kJ. 23. a. H2 at 100°C and 0.5 atm; Higher temperature and lower pressure means greater volume and hence, greater positional probability. b. N2 at STP has the greater volume. c. H2O(l) is more positional probability than H2O(s). 24. Of the three phases (solid, liquid, and gas), solids are most ordered and gases are most disordered. Thus, a , b, and f (melting, sublimation, and boiling) involve an increase in the entropy of the system since going from a solid to a liquid or a solid to a gas or a liquid to a gas increases disorder. For freezing (process c), a substance goes from the more disordered liquid state to the more ordered solid state, hence, entropy decreases. Process d (mixing) involves an increase in disorder (entropy) while separation increases order (decreases the entropy of the system). So of all the processes, a, b, d, and f result in an increase in the entropy of the system. 25. a. To boil a liquid requires heat. Hence, this is an endothermic process. All endothermic processes decrease the entropy of the surroundings (ΔSsurr is negative). CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 639 b. This is an exothermic process. Heat is released when gas molecules slow down enough to form the solid. In exothermic processes, the entropy of the surroundings increases (ΔSsurr is positive). 26. 27. a. ΔSsurr = (2221kJ) ΔH = 7.45 kJ/K = 7.45 × 103 J/K T 298 K b. ΔSsurr = ΔH 112 kJ = 0.376 kJ/K = 376 J/K T 298 K ΔG = ΔH TΔS; When ΔG is negative, then the process will be spontaneous. a. ΔG = ΔH TΔS = 25 × 103 J (300. K)(5.0 J/K) = 24,000 J, Not spontaneous b. ΔG = 25,000 J (300. K)(100. J/K) = 5000 J, Spontaneous c. Without calculating ΔG, we know this reaction will be spontaneous at all temperatures. ΔH is negative and ΔS is positive (-TΔS < 0). ΔG will always be less than zero with these sign combinations for ΔH and ΔS. d. ΔG = 1.0 × 104 J (200. K)(40. J/K) = 2000 J, Spontaneous 28. ΔG = ΔH TΔS; A process is spontaneous when ΔG < 0. For the following, assume ΔH and ΔS are temperature independent. a. When ΔH and ΔS are both negative, ΔG will be negative below a certain temperature where the favorable ΔH term dominates. When ΔG = 0, then ΔH = TΔS. Solving for this temperature: T= ΔH 18,000 J = 3.0 × 102 K ΔS 60. J / K At T < 3.0 × 102 K, this process will be spontaneous (ΔG < 0). b. When ΔH and ΔS are both positive, ΔG will be negative above a certain temperature where the favorable ΔS term dominates. T= ΔH 18,000 J = 3.0 × 102 K ΔS 60. J / K At T > 3.0 × 102 K, this process will be spontaneous (ΔG < 0). c. When ΔH is positive and ΔS is negative, this process can never be spontaneous at any temperature because ΔG can never be negative. d. When ΔH is negative and ΔS is positive, this process is spontaneous at all temperatures because ΔG will always be negative. 640 29. CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY At the boiling point, ΔG = 0 so ΔH = TΔS. ΔS = ΔH 27.5 kJ / mol = 8.93 × 102 kJ/Kmol = 89.3 J/Kmol T (273 35) K ΔH 58.51 103 J / mol = 629.7 K ΔS 92.92 J / K mol 30. At the boiling point, ΔG = 0 so ΔH = TΔS. T = 31. a. NH3(s) → NH3(l); ΔG = ΔH TΔS = 5650 J/mol 200. K (28.9 J/Kmol) ΔG = 5650 J/mol 5780 J/mol = T = =130 J/mol Yes, NH3 will melt because ΔG < 0 at this temperature. b. At the melting point, ΔG = 0 so T = 32. ΔH 5650 J / mol = 196 K. ΔS 28.9 J / K mol C2H5OH(l) → C2H5OH(g); At the boiling point, G = 0 and Suniv = 0. For the vaporization process, S is a positive value while H is a negative value. To calculate Ssys, we will determine Ssurr from H and the temperature, then Ssys = Ssurr for a system at equilibrium. Ssurr = ΔH 38.7 103 J / mol = 110. J/Kmol T 351 K Ssys = Ssurr = (110.) = 110. J/K mol Chemical Reactions: Entropy Changes and Free Energy 33. a. Decrease in disorder; ΔS°() c. Decrease in disorder (Δn < 0); ΔS°() b. Increase in disorder; ΔS°(+) d. Increase in disorder (Δn > 0); ΔS°(+) For c and d, concentrate on the gaseous products and reactants. When there are more gaseous product molecules than gaseous reactant molecules (Δn > 0), then ΔS° will be positive (disorder increases). When Δn is negative, then ΔS° is negative (disorder decreases or order increases). 34. 35. a. Decrease in disorder (Δn < 0); ΔS°() b. Decrease in disorder (Δn < 0); ΔS°() c. Increase in disorder; ΔS°(+) d. Increase in disorder; ΔS°(+) a. Cgraphite(s); Diamond has a more ordered structure than graphite. b. C2H5OH(g); The gaseous state is more disordered than the liquid state. CHAPTER 16 36. SPONTANEITY, ENTROPY, AND FREE ENERGY 641 c. CO2(g); The gaseous state is more disordered than the solid state. a. He (10 K); S = 0 at 0 K b. N2O; More complicated molecule c. H2O(l): The liquid state is more disordered than the solid state. 37. a. 2 H2S(g) + SO2(g) → 3 Srhombic(s) + 2 H2O(g); Because there are more molecules of reactant gases as compared to product molecules of gas (Δn = 2 - 3 < 0), ΔS° will be negative. ΔS° = n pSoproducts n rSoreactants ΔS° = [3 mol Srhombic(s) (32 J/Kmol) + 2 mol H2O(g) (189 J/Kmol)] [2 mol H2S(g) (206 J/Kmol) + 1 mol SO2(g) (248 J/Kmol)] ΔS° = 474 J/K 660. J/K = 186 J/K b. 2 SO3(g) → 2 SO2(g) + O2(g); Because Δn of gases is positive (Δn = 3 2), ΔS° will be positive. ΔS = 2 mol(248 J/Kmol) + 1 mol(205 J/Kmol) [2 mol(257 J/Kmol)] = 187 J/K c. Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g); Because Δn of gases = 0 (Δn = 3 3), we can’t easily predict if ΔS° will be positive or negative. ΔS = 2 mol(27 J/Kmol) + 3 mol(189 J/Kmol) [1 mol(90. J/Kmol) + 3 mol (141 J/Kmol)] = 138 J/K 38. a. H2(g) + 1/2 O2(g) → H2O(l); Since Δn of gases is negative, then ΔS° will be negative. ΔS° = 1 mol H2O(l) (70. J/Kmol) [1 mol H2(g) (131 J/Kmol) + 1/2 mol O2(g) (205 J/Kmol)] ΔS° = 70. J/K 234 J/K = 164 J/K b. 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g); Because Δn of gases is positive, ΔS° will be positive. [2 mol (214 J/Kmol) + 4 mol (189 J/Kmol)] – [2 mol (240. J/Kmol) + 3 mol (205 J/Kmol)] = 89 J/K 642 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY c. HCl(g) → H+(aq) + Cl(aq); The gaseous state dominates predictions of ΔS°. Here the gaseous state is more disordered than the ions in solution so ΔS° will be negative. 39. ΔS° = 1 mol H+(0) + 1 mol Cl(57 J/Kmol) 1 mol HCl(187 J/Kmol) = 130. J/K C2H2(g) + 4 F2(g) → 2 CF4(g) + H2(g); ΔS° = 2 SoCF4 SoH 2 [SoC2H 2 4SoF2 ] 358 J/K = (2 mol) SoCF4 + 131 J/K [201 J/K + 4(203 J/K)], SoCF4 = 262 J/Kmol 40. 144 J/K = (2 mol) SoAlBr3 [2(28 J/K) + 3(152 J/K)], SoAlBr3 = 184 J/Kmol 41. a. Srhombic → Smonoclinic; This phase transition is spontaneous (ΔG < 0) at temperatures above 95°C. ΔG = ΔH TΔS; For ΔG to be negative only above a certain temperature, then ΔH is positive and ΔS is positive (see Table 16.5 of text). b. Because ΔS is positive, Srhombic is the more ordered crystalline structure. 42. Enthalpy is not favorable, so ΔS must provide the driving force for the change. Thus, ΔS is positive. There is an increase in disorder, so the original enzyme has the more ordered structure. 43. a. When a bond is formed, energy is released so ΔH is negative. There are more reactant molecules of gas than product molecules of gas (Δn < 0), so ΔS will be negative. b. ΔG = ΔH TΔS; For this reaction to be spontaneous (ΔG < 0), the favorable enthalpy term must dominate. The reaction will be spontaneous at low temperatures where the ΔH term dominates. 44. Because there are more product gas molecules than reactant gas molecules (Δn > 0), ΔS will be positive. From the signs of ΔH and ΔS, this reaction is spontaneous at all temperatures. It will cost money to heat the reaction mixture. Because there is no thermodynamic reason to do this, the purpose of the elevated temperature must be to increase the rate of the reaction, i.e., kinetic reasons. 45 a. CH4(g) + 2 O2(g) → CO2(g ) + 2 H2O(g) _________________________________________________________ ΔH of 75 kJ/mol 0 393.5 242 ΔG of 51 kJ/mol 0 394 229 Data from Appendix 4 S° 186 J/Kmol 205 214 189 __________________________________________________________ ΔH° = n pΔHof, products n r ΔHof, reactants ; ΔS° = n pSoproducts n rSoreactants ΔH° = 2 mol(242 kJ/mol) + 1 mol(393.5 kJ/mol) [1 mol(75 kJ/mol)] = 803 kJ CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔS° = 2 mol(189 J/K∙mol) + 1 mol(214 J/Kmol) [1 mol(186 J/Kmol) + 2 mol(205 J/Kmol)] = 4 J/K There are two ways to get ΔG°. We can use ΔG° = ΔH° TΔS° (be careful of units): ΔG° = ΔH° TΔS° = -803 × 103 J (298 K)( 4 J/K) = 8.018 × 105 J = 802 kJ or we can use ΔG of values where ΔG° = n pΔG of , products n r ΔG of, reactants : ΔG° = 2 mol(229 kJ/mol) + 1 mol(394 kJ/mol) [1 mol(51 kJ/mol)] ΔG° = 801 kJ (Answers are the same within round off error.) b. 6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g) ______________________________________________________ ΔH of 393.5 kJ/mol 286 1275 0 S° 214 J/Kmol 70. 212 205 ______________________________________________________ ΔH° = 1275 [6(286) + 6(393.5)] = 2802 kJ ΔS° = 6(205) + 212 [6(214) + 6(70.)] = 262 J/K ΔG° = 2802 kJ (298 K)( 0.262 kJ/K) = 2880. kJ c. P4O10(s) + 6 H2O(l) → 4 H3PO4(s) _____________________________________________ ΔH of 2984 286 1279 (kJ/mol) S° 229 70. 110. (J/Kmol) _____________________________________________ ΔH° = 4 mol(1279 kJ/mol) [1 mol(2984 kJ/mol) + 6 mol(286 kJ/mol)] = 416 kJ ΔS° = 4(110.) [229 + 6(70.)] = 209 J/K ΔG° = ΔH° TΔS° = 416 kJ (298 K)( 0.209 kJ/K) = 354 kJ d. HCl(g) + NH3(g) → NH4Cl(s) ___________________________________________ ΔH of 92 46 314 (kJ/mol) 643 644 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY S° (J/Kmol) 187 193 96 ____________________________________________ ΔH° = 314 [92 46] = 176 kJ; ΔS° = 96 [187 + 193] = 284 J/K 46. ΔG° = ΔH° TΔS° = 176 kJ (298 K)( 0.284 kJ/K) = 91 kJ a. ΔH° = 2(46 kJ) = 92 kJ; ΔS° = 2(193 J/K) [3(131 J/K) + 192 J/K] = 199 J/K ΔG° = ΔH° TΔS° = 92 kJ 298 K(0.199 kJ/K) = 33 kJ b. ΔG° is negative, so this reaction is spontaneous at standard conditions. c. ΔG° = 0 when T = 92 kJ ΔH o = 460 K o 0.199 kJ / K ΔS At T < 460 K and standard pressures (1 atm), the favorable ΔH° term dominates and the reaction is spontaneous (ΔG° < 0). 47. ΔG° = 58.03 kJ (298 K)( 0.1766 kJ/K) = 5.40 kJ ΔG° = 0 = ΔH° TΔS°, T = ΔH o 58.03 kJ = 328.6 K o 0.1766 kJ / K ΔS ΔG° is negative below 328.6 K where the favorable ΔH° term dominates. 48. H2O(l) → H2O(g); ΔG° = 0 at the boiling point of water at 1 atm and 100.C. ΔH° = TΔS°, ΔS° = ΔH o 40.6 103 J / mol = 109 J/Kmol T 373 K At 90.C: ΔG° = ΔH° TΔS° = 40.6 kJ/mol – (363 K)(0.109 kJ/Kmol) = 1.0 kJ/mol As expected, ΔG° > 0 at temperatures below the boiling point of water at 1 atm (process is nonspontaneous). At 110.C: ΔG° = ΔH° TΔS° = 40.6 kJ/mol – (383 K)(0.109 J/Kmol) = 1.1 kJ/mol When ΔG° < 0, the boiling of water is spontaneous at 1 atm and T > 100.C (as expected). 49. CH4(g) + CO2(g) → CH3CO2H(l) ΔH° = 484 [75 + (393.5)] = 16 kJ; ΔS° = 160 [186 + 214] = 240. J/K ΔG° = ΔH° TΔS° = 16 kJ (298 K)( 0.240 kJ/K) = 56 kJ This reaction is spontaneous only at temperatures below T = ΔH°/ΔS° = 67 K (where the favorable ΔH° term will dominate giving a negative ΔG° value). This is not practical. CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 645 Substances will be in condensed phases and rates will be very slow at this extremely low temperature. CH3OH(g) + CO(g) → CH3CO2H(l) ΔH° = -484 [110.5 + (201)] = 173 kJ; ΔS° = 160 [198 + 240.] = 278 J/K ΔG° = 173 kJ (298 K)( 0.278 kJ/K) = 90. kJ This reaction also has a favorable enthalpy and an unfavorable entropy term. This reaction is spontaneous at temperatures below T = ΔH°/ΔS° = 622 K. The reaction of CH3OH and CO will be preferred. It is spontaneous at high enough temperatures that the rates of reaction should be reasonable. 50. C2H4(g) + H2O(g) → CH3CH2OH(l) ΔH° = 278 (52 242) = 88 kJ; ΔS° = 161 (219 + 189) = 247 J/K When ΔG° = 0, ΔH° = TΔS°, T = ΔH o 88 103 J = 360 K 247 J / K ΔSo Because the signs of ΔH° and ΔS° are both negative, this reaction will be spontaneous at temperatures below 360 K (where the favorable ΔH° term will dominate). C2H6(g) + H2O(g) → CH3CH2OH(l) + H2(g) ΔH° = 278 (84.7 242) = 49 kJ; ΔS° = 131 + 161 (229.5 + 189) = 127 J/K This reaction can never be spontaneous because of the signs of ΔH° and ΔS°. Thus the reaction C2H4(g) + H2O(g) → C2H5OH(l) would be preferred. CH4(g) → 2 H2(g) + C(s) 51. 2 H2(g) + O2(g) → 2 H2O(l) ΔG° = (51 kJ) ΔG° = 2(237 kJ) C(s) + O2(g) → CO2(g) ΔG° = 394 kJ ______________________________________________________________ CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g) 52. ΔG° = 817 kJ 6 C(s) + 6 O2(g) → 6 CO2(g) ΔG° = 6(394 kJ) 3 H2(g) + 3/2 O2(g) → 3 H2O(l) ΔG° = 3(237 kJ) 6 CO2(g) + 3 H2O(l) → C6H6(l) + 15/2 O2(g) ΔG° = 1/2 (6399 kJ) ______________________________________________________________ 6 C(s) + 3 H2(g) → C6H6(l) 53. ΔG° = n pΔG of , products n r ΔG of, reactants , ΔG° = 125 kJ 374 kJ = 1105 kJ ΔG of , SF4 646 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔG of , SF4 = 731 kJ/mol 54. 5490. kJ = 8(394 kJ) + 10(237 kJ) 2 ΔG of , C4H10 , ΔG of , C4H10 = 16 kJ/mol 55. ΔG° = n pΔG of , products n r ΔG of, reactants ΔG° = [57.37 kJ + (68.85 kJ) + 3(95.30 kJ)] – [3(0) + 2(50.72 kJ)] = 310.68 kJ For a temperature change from 25°C to ~20°C (room temperature), the magnitude of ΔG° will not change much. Therefore, ΔG° will be a negative value at room temperature (~20°C), so the reaction will be spontaneous. 56. a. ΔG° = 2(270. kJ) 2(502 kJ) = 464 kJ b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions at 298 K. c. ΔG° = ΔH° TΔS°, ΔH° = ΔG° + TΔS° = 464 kJ + 298 K(0.179 kJ/K) = 517 kJ We need to solve for the temperature when ΔG° = 0: ΔH o 517 kJ ΔG° = 0 = ΔH° TΔS°, T = = 2890 K o 0.179 kJ / K ΔS This reaction will be spontaneous at standard conditions (ΔG° < 0) when T > 2890 K. Here the favorable entropy term will dominate. Free Energy: Pressure Dependence and Equilibrium 57. ΔG = ΔG° + RT ln Q; For this reaction: ΔG = ΔG° + RT ln PNO 2 PO 2 PNO PO3 ΔG° = 1 mol(52 kJ/mol) + 1 mol(0) [1 mol(87 kJ/mol) + 1 mol(163 kJ/mol)] = 198 kJ ΔG = 198 kJ + 8.3145 J / K mol (1.00 107 ) (1.00 103 ) (298 K) ln 1000 J / kJ (1.00 106 ) (1.00 106 ) ΔG = 198 kJ + 9.69 kJ = 188 kJ 58. ΔG° = 3(0) + 2(229) [2(-34) + 1(300.)] = 90. kJ ΔG = ΔG° + RT ln PH2 O 2 PH2 S PSO 2 2 = 90. kJ + (8.3145) (298) (0.030) 2 kJ ln 4 1000 (1.0 10 ) (0.010) CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 647 ΔG = 90. kJ + 39.7 kJ = 50. kJ 59. ΔG = ΔG° + RT ln Q = ΔG° + RT ln PN 2O 4 2 PNO 2 ΔG° = 1 mol(98 kJ/mol) 2 mol(52 kJ/mol) = 6 kJ a. These are standard conditions so ΔG = ΔG° because Q = 1 and ln Q = 0. Because ΔG° is negative, the forward reaction is spontaneous. The reaction shifts right to reach equilibrium. b. ΔG = 6 × 103 J + 8.3145 J/Kmol (298 K) ln 0.50 (0.21) 2 ΔG = 6 × 103 J + 6.0 × 103 J = 0 Because ΔG = 0, this reaction is at equilibrium (no shift). c. ΔG = 6 × 103 J + 8.3145 J/Kmol (298 K) ln 1.6 (0.29) 2 ΔG = 6 × 103 J + 7.3 × 103 J = 1.3 × 103 J = 1 × 103 J Because ΔG is positive, the reverse reaction is spontaneous, and the reaction shifts to the left to reach equilibrium. 60. a. ΔG = ΔG° + RT ln ΔG = 34 kJ + 2 PNH 3 PN 2 PH2 2 ; ΔG° = 2 ΔG of , NH3 = 2(17) = 34 kJ (8.3145 J / K mol) (298 K) (50.)2 ln 1000 J / kJ (200.) (200.)3 ΔG = 34 kJ 33 kJ = 67 kJ b. ΔG = 34 kJ (8.3145 J / K mol) (298 K) (200.)2 ln 1000 J / kJ (200.) (600.)3 ΔG = 34 kJ 34.4 kJ = 68 kJ 61. At 25.0°C: ΔG° = ΔH° − TΔS° = −58.03 × 103 J/mol − (298.2 K)(−176.6 J/Kmol) = −5.37 × 103 J/mol ΔG° = −RT ln K, ln K = K = e2.166 = 8.72 ΔG o (5.37 103 J / mol) = 2.166 exp RT (8.3145 J / K mol) (298.2 K ) 648 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY At 100.0°C: ΔG° = −58.03 × 103 J/mol − (373.2 K)(−176.6 J/Kmol) = 7.88 × 103 J/mol ln K = 62. (7.88 103 J / mol) = −2.540, K = e-2.540 = 0.0789 (8.3145 J / K mol) (373.2 K ) Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round off error. a. ΔH° = 2 mol(−92 kJ/mol) − [1 mol(0) + 1 mol(0)] = −184 kJ ΔS° = 2 mol(187 J/Kmol) − [1 mol(131 J/Kmol) + 1 mol(223 J/Kmol)] = 20. J/K ΔG° = ΔH° − TΔS° = −184 × 103 J − 298 K (20. J/K) = −1.90 × 105 J = −190. kJ ΔG° = −RT ln K, ln K = ΔG o (1.90 105 J) = 76.683 RT 8.3145 J / K mol(298 K) K = e76.683 = 2.01 × 1033 b. These are standard conditions so ΔG = ΔG° = −190. kJ. When ΔG is negative, the forward reaction is spontaneous so the reaction shifts right to reach equilibrium. 63. When reactions are added together, the equilibrium constants are multiplied together to determine the K value for the final reaction. H2(g) +O2(g) ⇌ H2O2(g) K = 2.3 × 106 H2O(g) ⇌ H2(g) + 1/2O2(g) K = (1.8 × 1037)1/2 _______________________________________________________________________ H2O(g) + 1/2O2(g) ⇌ H2O2(g) K = 2.3 × 106(1.8 × 1037) 1/2 = 5.4 × 1013 ΔG° = −RT ln K = −8.3145 J/Kmol (600. K) ln(5.4 × 1013 ) = 1.4 × 105 J/mol = 140 kJ/mol 64. ΔH of (kJ/mol) a. NH3(g) O2(g) NO(g) H2O(g) NO2(g) HNO3(l) H2O(l) −46 0 90. −242 34 −174 −286 S° (J/Kmol) 193 205 211 189 240. 156 70. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔH° = 6(−242) + 4(90.) − [4(−46)] = −908 kJ ΔS° = 4(211) + 6(189) − [4(193) + 5(205)] = 181 J/K ΔG° = −908 kJ − 298 K (0.181 kJ/K) = −962 kJ CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔG° = − RT ln K, ln K = 649 ΔG o (962 103 J) = 388 RT 8 . 3145 J / K mol 298 K ln K = 2.303 log K, log K = 168, K = 10168 (an extremely large number) 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = 2(34) − [2(90.)] = −112 kJ; ΔS° = 2(240.) − [2(211) + (205)] = −147 J/K ΔG° = −112 kJ − (298 K)( −0.147 kJ/K) = −68 kJ K = exp ΔG o (68,000 J) = e27.44 = 8.3 × 1011 exp RT 8 . 3145 J / K mol ( 298 K ) Note: When determining exponents, we will round off after the calculation is complete. 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g) ΔH° = 2(−174) + (90.) − [3(34) + (−286)] = −74 kJ ΔS° = 2(156) + (211) − [3(240.) + (70.)] = −267 J/K ΔG° = −74 kJ − (298 K)( −0.267 kJ/K) = 6 kJ K = exp ΔG o 6000 J = e 2.4 = 9 × 102 exp RT 8.3145 J / K mol (298 K) b. ΔG° = −RT ln K; T = 825°C = (825 + 273) K = 1098 K; We must determine ΔG° at 1098 K. o = ΔH° − TΔS° = −908 kJ − (1098 K)(0.181 kJ/K) = −1107 kJ ΔG1098 K = exp ΔG o (1.107 106 J) = e121.258 = 4.589 × 1052 exp RT 8 . 3145 J / K mol ( 1098 K ) c. There is no thermodynamic reason for the elevated temperature because ΔH° is negative and ΔS° is positive. Thus, the purpose for the high temperature must be to increase the rate of the reaction. 65. K= 2 PNF 3 PN 2 PF32 (0.48) 2 = 4.4 × 104 3 0.021(0.063) o ΔG 800 = −RT ln K = −8.3145 J/Kmol (800. K) ln (4.4 × 104) = −7.1 × 104 J/mol = −71 kJ/mol 66. 2 SO2(g) + O2(g) → 2 SO3(g); ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ 650 CHAPTER 16 ΔG° = −RT ln K, ln K = SPONTANEITY, ENTROPY, AND FREE ENERGY ΔG o (142.000 J) = 57.311 RT 8.3145 J / K mol (298 K) K = e57.311 = 7.76 × 1024 K = 7.76 × 1024 = 2 PSO 3 2 PSO PO 2 2 (2.0) 2 , PSO 2 = 1.0 × 1012 atm 2 PSO ( 0 . 50 ) 2 From the negative value of ΔG°, this reaction is spontaneous at standard conditions. There are more molecules of reactant gases than product gases, so ΔS° will be negative (unfavorable). Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS° are both negative, the reaction will be spontaneous at relatively low temperatures where the favorable ΔH° term dominates. 67. ΔH o 1 ΔSo is in the form of a straight line equation R T R (y = mx + b). A graph of ln K vs. 1/T will yield a straight line with slope = m = ΔH°/R and a y-intercept = b = ΔS°/R. The equation ln K = From the plot: slope = Δy 0 40. = −1.3 × 104 K 3 1 Δx 3.0 10 K 0 −1.3 × 104 K = −ΔH°/R, ΔH° = 1.3 × 104 K × 8.3145 J/Kmol = 1.1 × 105 J/mol y-intercept = 40. = ΔS°/R, ΔS° = 40. × 8.3145 J/Kmol = 330 J/Kmol As seen here, when ΔH° is positive, the slope of the ln K vs. 1/T plot is negative. When ΔH° is negative as in an exothermic process, then the slope of the ln K vs. 1/T plot will be positive (slope = −ΔH°/R). 68. The ln K vs. 1/T plot gives a straight line with slope = H/R and y-intercept = S/R. 1.352 × 104 K = H/R, H = (8.3145 J/ Kmol) (1.352 × 104 K) H = 1.124 × 105 J/mol = 112.4 kJ/mol 14.51 = S/R, S = (14.51)(8.3145 J/ Kmol) = 120.6 J / Kmol Note that the signs for ΔH° and ΔS° make sense. When a bond forms, ΔH° < 0 and ΔS° < 0. Additional Exercises 69. From Appendix 4, S = 198 J/Kmol for CO(g) and S = 27 J/Kmol for Fe(s). CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 651 Let S ol = Sfor Fe(CO)5(l) and S og = S for Fe(CO)5(g). S = 677 J/K = 1 mol(S ol ) – [1 mol (27 J/Kmol) + 5 mol(198 J/ Kmol] S ol = 340. J/Kmol S = 107 J/K = 1 mol (S og ) – 1 mol (340. J/Kmol) S og = S for Fe(CO)5(g) = 447 J/Kmol 70. When an ionic solid dissolves, one would expect the disorder of the system to increase, so ΔSsys is positive. Because temperature increased as the solid dissolved, this is an exothermic process and ΔSsurr is positive (ΔSsurr = ΔH/T). Because the solid did dissolve, the dissolving process is spontaneous, so ΔSuniv is positive. 71. ΔS will be negative because 2 mol of gaseous reactants form 1 mol of gaseous product. For ΔG to be negative, ΔH must be negative (exothermic). For exothermic reactions, K decreases as T increases. Therefore, the ratio of the partial pressure of PCl5 to the partial pressure of PCl3 will decrease when T is raised. 72. At boiling point, ΔG = 0 so ΔS = ΔH vap T ; For methane: ΔS = 8.20 103 J / mol 112 K = 73.2 J/molK For hexane: ΔS = Vmet = 28.9 103 J / mol = 84.5 J/molK 342 K nRT nRT 1.00 mol (0.08206) (112 K) = = 9.19 L; Vhex = = R(342 K) = 28.1 L P P 1.00 atm Hexane has the larger molar volume at the boiling point so hexane should have the larger entropy. As the volume of a gas increases, positional disorder increases. 73. solid I → solid II; Equilibrium occurs when ΔG = 0. ΔG = ΔH TΔS, ΔH = TΔS, T = ΔH/ΔS = 74. 743.1 J / mol = 43.7 K = 229.5°C 17.0 J / K mol a. ΔG° = RT ln K = (8.3145 J/K mol)(298 K) ln 0.090 = 6.0 × 103 J/mol = 6.0 kJ/mol b. H‒O‒H + Cl‒O‒Cl → 2 H‒O‒Cl On each side of the reaction there are 2 H‒O bonds and 2 O‒Cl bonds. Both sides have the same number and type of bonds. Thus, ΔH ≈ ΔH° ≈ 0. c. ΔG° = ΔH° TΔS°, ΔS° = ΔH o ΔG o 0 6.0 103 J = 20. J/K T 298 K 652 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY d. For H2O(g), ΔH of = 242 kJ/mol and S° = 189 J/Kmol ΔH° = 0 = 2 ΔH of , HOCl [1 mol(242 kJ/mol) + 1 mol(80.3 J/K/mol)], ΔH of , HOCl = 81 kJ/mol 20. J/K = 2 SoHOCl [1 mol(189 J/Kmol) + 1 mol(266.1 J/K∙mol)], SoHOCl = 218 J/Kmol o e. Assuming ΔH° and ΔS° are T independent: Δ G 500 = 0 (500. K)(20. J/K) = 1.0 × 104 J ΔG o 1.0 104 2.41 exp ΔG° = RT ln K, K = exp = 0.090 (8.3145) (500.) = e RT f. 2 PHOCl ; From part a, ΔG° = 6.0 kJ/mol. ΔG = ΔG° + RT ln PH 2O PCl2O We should express all P’s in atm. However, we perform the pressure conversion the same number of times in the numerator and denominator, so the factors of 760 torr/atm will all cancel. Thus, we can use the pressures in units of torr. ΔG = 75. 6.0 kJ / mol (8.3145 J / K mol) (298 K ) (0.10) 2 = 6.0 20. = 14 kJ/mol ln 1000 J / kJ ( 18 ) ( 2 . 0 ) HgbO2 → Hgb + O2 ΔG° = (70 kJ) Hgb + CO → HgbCO ΔG° = 80 kJ ____________________________________________ HgbO2 + CO → HgbCO + O2 ΔG° = 10 kJ ΔG o (10 103 J) exp = 60 ΔG° = RT ln K, K = exp RT 8.3145 J / K mol (298 K ) 76. Ba(NO3)2(s) ⇌ Ba2+(aq) + 2 NO3(aq) K = Ksp; ΔG° = 561 + 2(109) (797) = 18 kJ ΔG° = RT ln Ksp, ln Ksp = ΔG o 18,000 J = 7.26, Ksp = e 7.26 RT 8.3145 J / K mol (298 K) = 7.0 × 104 77. HF(aq) ⇌ H+(aq) + F(aq); ΔG = ΔG° + RT ln [H ][F ] [HF] ΔG° = RT ln K = (8.3145 J/Kmol) (298 K) ln (7.2 × 104 ) = 1.8 × 104 J/mol a. The concentrations are all at standard conditions so ΔG = ΔG = 1.8 × 104 J/mol CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 653 (Q = 1.0 and ln Q = 0). Because ΔG° is positive, the reaction shifts left to reach equilibrium. b. ΔG = 1.8 × 104 J/mol + (8.3145 J/Kmol) (298 K) ln ΔG = 1.8 × 104 J/mol 1.8 × 104 J/mol = 0 (2.7 102 ) 2 0.98 ΔG = 0, so the reaction is at equilibrium (no shift). c. ΔG = 1.8 × 104 J/mol + 8.3145 (298 K) ln (1.0 105 ) 2 = 1.1 × 104 J/mol; shifts right 1.0 105 7.2 104 (0.27) = 1.8 × 104 1.8 × 104 = 0; 0.27 at equilibrium 3 1.0 10 (0.67) e. ΔG = 1.8 × 104 + 8.3145 (298) ln = 2 × 103 J/mol; shifts left 0.52 [K ] K+ (blood) ⇌ K+ (muscle) ΔG° = 0; ΔG = RT ln m ; ΔG = wmax [ K ]b 8.3145 J 0.15 3 ΔG = (310. K) ln , ΔG = 8.8 × 10 J/mol = 8.8 kJ/mol 0 . 0050 K mol d. ΔG = 1.8 × 104 + 8.3145 (298) ln 78. At least 8.8 kJ of work must be applied. Other ions will have to be transported in order to maintain electroneutrality. Either anions must be transported into the cells, or cations (Na+) in the cell must be transported to the blood. The latter is what happens: [Na+] in blood is greater than [Na+] in cells as a result of this pumping. 79. (30,500 J) = 2.22 × 105 a. ΔG° = RT ln K, K = exp(ΔG°/RT) = exp 8 . 3145 J / K mol 298 K b. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔG° = 6 mol(394 kJ/mol) + 6 mol(-237 kJ/mol) 1 mol(911 kJ/mol) = 2875 kJ 2875 kJ 1 mol AT P 94.3 mol AT P ; 94.3 molecules ATP/molecule glucose mol glucose 30.5 kJ mol glucose This is an overstatement. The assumption that all of the free energy goes into this reaction is false. Actually only 38 moles of ATP are produced by metabolism of one mole of glucose. c. From Exercise 17.78, ΔG = 8.8 kJ in order to transport 1.0 mol K+ from the blood to the muscle cells. 654 CHAPTER 16 8.8 kJ × 80. SPONTANEITY, ENTROPY, AND FREE ENERGY 1 mol AT P = 0.29 mol ATP 30.5 kJ a. ΔG° = RT ln K ln K = b. ΔG o 14,000 J = 5.65, K = e 5.65 = 3.5 × 103 RT (8.3145 J / K mol) (298 K) Glutamic acid + NH3 → Glutamine + H2O ΔG° = 14 kJ ATP + H2O → ADP + H2PO4 ΔG° = 30.5 kJ _______________________________________________________________________ Glutamic acid + ATP + NH3 → Glutamine + ADP + H2PO4 ΔG° = 14 30.5 = 17 kJ o ΔG (17,000 J) ln K = = 6.86, K = e6.86 = 9.5 × 102 RT 8.3145 J / K mol (298 K) 81. ΔS is more favorable for reaction two than for reaction one, resulting in K2 > K1. In reaction one, seven particles in solution are forming one particle. In reaction two, four particles form one particle which results in a smaller decrease in disorder than for reaction one. 82. A graph of ln K vs. 1/T will yield a straight line with slope equal to -ΔH°/R and y-intercept equal to ΔS°/R. Temp (°C) 0 25 35 40. 50. T(K) 273 298 308 313 323 1000/T (K-1) 3.66 3.36 3.25 3.19 3.10 Kw ln Kw 1.14 × 1015 1.00 × 1014 2.09 × 1014 2.92 × 1014 5.47 × 1014 34.408 32.236 31.499 31.165 30.537 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 655 1 The straight line equation (from a calculator) is: ln K = 6.91 × 103 9.09 T o ΔH 3 Slope = 6.91 × 103 K = , ΔH° = (6.91 × 10 K × 8.3145 J/Kmol) R = 5.75 × 104 J/mol y-intercept = 9.09 = 83. ΔS o , ΔS° = 9.09 × 8.3145 J/Kmol = 75.6 J/Kmol R ΔG° = RT ln K; When K = 1.00, ΔG° = 0 since ln 1.00 = 0. ΔG° = 0 = ΔH° TΔS° ΔH° = 3(242 kJ) [-826 kJ] = 100. kJ; ΔS° = [2(27 J/K) + 3(189 J/K)] [90. J/K + 3(131 J/K)] = 138 J/K ΔH° = TΔS°, T = ΔH o 100. kJ = 725 K o 0.138 kJ / K ΔS Challenge Problems 84. The liquid water will evaporate at first and eventually an equilibrium will be reached (physical equilibrium). Because evaporation is an endothermic process, ΔH is positive. Because H2O(g) is more disordered (greater positional probability), ΔS is positive. The water will become cooler (the higher energy water molecules leave), thus ΔTwater will be negative. The vessel is insulated (q = 0), so ΔSsurr = 0. Because the process occurs, it is spontaneous, so ΔSuniv is positive. 85. 3 O2(g) ln K = ⇌ 2 O3(g); ΔH° = 2(143 kJ) = 286 kJ; ΔG° = 2(163 kJ) = 326 kJ ΔG o 326 103 J = 131.573, K = e 131.573 = 7.22 × 1058 RT (8.3145 J / K mol) (298 K ) We need the value of K at 230. K. From Section 16.8 of the text: ΔG o ΔSo ln K = RT R For two sets of K and T: 656 CHAPTER 16 ln K1 = SPONTANEITY, ENTROPY, AND FREE ENERGY ΔH o 1 ΔS ΔH o 1 ΔSo ; ln K2 = R T1 R R T2 R Subtracting the first expression from the second: ln K2 ln K1 = K ΔH o 1 1 ΔH o 1 1 or ln 2 R T1 T2 K1 R T1 T2 Let K2 = 7.22 × 1058 , T2 = 298 K; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J ln 7.22 1058 286 103 1 1 = 34.13 K 230 8.3145 230. 298 7.22 1058 = e34.13 = 6.6 × 1014, K230 = 1.1 × 1072 K 230 K230 = 1.1 × 1072 = PO23 PO3 3 PO23 3 (1.0 10 atm) 3 , PO3 = 3.3 × 1041 atm The volume occupied by one molecule of ozone is: V = nRT P = (1 / 6.022 1023 mol) (0.08206L atm/ mol K) (230. K) = 9.5 × 1017 L 41 (3.3 10 atm) Equilibrium is probably not maintained under these conditions. When only two ozone molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these conditions, Q > K and the reaction shifts left. But with only 2 ozone molecules in this huge volume, it is extremely unlikely that they will collide with each other. At these conditions, the concentration of ozone is not large enough to maintain equilibrium. 86. Arrangement I and V: S = k ln W; W = 1; S = k ln 1 = 0 Arrangement II and IV: W = 4; S = k ln 4 = 1.38 × 1023 J/K ln 4, S = 1.91 × 1023 J/K Arrangement III: 87. W = 6; S = k ln 6 = 2.47 × 1023 J/K a. From the plot, the activation energy of the reverse reaction is Ea + (ΔG°) = Ea ΔG° (ΔG° is a negative number as drawn in the diagram). (E a ΔG o ) k f Ea , kf = A exp and kr = A exp k RT RT r Ea A exp RT (E a ΔG o ) A exp RT CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY If the A factors are equal: 657 E ΔG o (E ΔG o ) kf = exp exp a a RT kr RT RT ΔG o k ; Because K and f are both equal to the From ΔG° = RT ln K, K = exp kr RT same expression, K = kf . kr b. A catalyst will lower the activation energy for both the forward and reverse reaction (but not change ΔG°). Therefore, a catalyst must increase the rate of both the forward and reverse reactions. 88. At equilibrium: PH 2 1.10 1013 molecules 0.08206L atm (298 K ) 6.022 1023 molecules/ mol mol K nRT V 1.00 L PH 2 = 4.47 × 1010 atm The pressure of H2 decreased from 1.00 atm to 4.47 × 1010 atm. Essentially all of the H2 and Br2 has reacted. Therefore, PHBr = 2.00 atm because there is a 2:1 mole ratio between HBr and H2 in the balanced equation. Because we began with equal moles of H2 and Br2, we will have equal moles of H2 and Br2 at equilibrium. Therefore, PH 2 PBr2 = 4.47 × 1010 atm. 2 PHBr (2.00) 2 = 2.00 × 1019; Assumptions good. 10 2 PH 2 PBr2 (4.47 10 ) K= ΔG° = RT ln K = (8.3145 J/Kmol)(298 K) ln (2.00 × 1019) = 1.10 × 105 J/mol ΔS° = 89. ΔH o ΔG o 103,800 J / mol (1.0 105 J / mol) = 20 J/Kmol T 298 K a. ΔG° = G oB G oA = 11,718 8996 = 2722 J ΔG o 2722 J = exp = 0.333 K = exp (8.3145 J / K mol) (298 K ) RT b. When Q = 1.00 > K, the reaction shifts left. Let x = atm of B(g) which reacts to reach equilibrium. A(g) Initial Equil. 1.00 atm 1.00 + x ⇌ B(g) 1.00 atm 1.00 x 658 CHAPTER 16 K= SPONTANEITY, ENTROPY, AND FREE ENERGY PB 1.00 x = 0.333, 1.00 x = 0.333 + 0.333 x, x = 0.50 atm PA 1.00 x PB = 1.00 0.50 = 0.50 atm; PA = 1.00 + 0.50 = 1.50 atm c. ΔG = ΔG° + RT ln Q = ΔG° + RT ln (PB/PA) ΔG = 2722 J + (8.3145)(298) ln (0.50/1.50) = 2722 J 2722 J = 0 (carrying extra sig. figs.) 90. ΔH o ΔSo . For K at two temperatures T1 and T2, the RT R K ΔH o 1 1 equation can be manipulated to give (see Exercise 16.77): ln 2 K1 R T1 T2 From Exercise 16.67, ln K = 3.25 102 1 ΔH o 1 ln 8.84 348 K 8.3145 J / K mol 298 K 5.61 = (5.8 ×10-5 mol/J) (ΔH°), ΔH° = 9.7 × 104 J/mol For K = 8.84 at T = 25°C: ln 8.84 = (9.7 104 J / mol) ΔSo ΔSo = 37 , (8.3145 J / K mol) (298 K) 8.3145 J / K mol 8.3145 ΔS° = 310 J/Kmol We get the same value for ΔS° using K = 3.25 × 102 at T = 348 K data. ΔG° = RT ln K; When K = 1.00, then ΔG° = 0 since ln 1.00 = 0. Assuming ΔH° and ΔS° do not depend on temperature: ΔG° = 0 = ΔH° TΔS°, ΔH° = TΔS°, T = 91. ΔH o 9.7 104 J / mol = 310 K 310 J / K mol ΔSo K = PCO2 ; To insure Ag2CO3 from decomposing, PCO2 should be greater than K. ΔH o ΔSo . For two conditions of K and T, the equation is: RT R K ΔH o 1 1 ln 2 K1 R T1 T2 From Exercise 16.67, ln K = Let T1 = 25°C = 298 K, K1 = 6.23 × 103 torr; T2 = 110.°C = 383 K, K2 = ? CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY ln K2 79.14 103 J / mol 1 1 3 8.3145J / K mol 298 K 383 K 6.23 10 torr ln K2 K2 = e7.1 = 1.2 × 103, K2 = 7.5 torr 7.1, 3 3 6.23 10 6.23 10 659 To prevent decomposition of Ag2CO3, the partial pressure of CO2 should be greater than 7.5 torr. 92. L From the problem, χ CL 6H6 χ CCl = 0.500. We need the pure vapor pressures (Po) in order to 4 calculate the vapor pressure of the solution. Using the thermodynamic data: C6H6(l) ⇌ C6H6(g) K = PC6H6 PCo6H6 at 25°C ΔG orxn ΔG of , C6H6 (g ) ΔG of , C6H6 (l) = 129.66 kJ/mol 124.50 kJ/mol = 5.16 kJ/mol ΔG° = RT ln K, ln K = ΔG o 5.16 103 J / mol = exp = 2.08 RT (8.3145 J / K mol) (298 K ) K = PCo6 H 6 = e 2.08 = 0.125 atm For CCl4: ΔG orxn ΔG of , CCl4 (g ) ΔG of , CCl4 (l) = 60.59 kJ/mol (65.21 kJ/mol) = 4.62 kJ/mol ΔG o 4620 J / mol o K = PCCl = exp 4 RT = exp 8.3145 J / K mol 298 K = 0.155 atm o PC6H6 χ CL 6H6 PCo6H6 = 0.500 (0.125 atm) = 0.0625 atm; PCCl = 0.500 (0.155 atm) 4 = 0.0775 atm χ CV6 H 6 = PC6 H 6 Ptot = 0.0625 0.0625atm = = 0.446 0.0625atm 0.0775atm 0.1400 V χ CCl = 1.000 0.446 = 0.554 4 93. Use the thermodynamic data to calculate the boiling point of the solvent. At boiling point: ΔG = 0 = ΔH TΔS, T = ΔH 33.90 103 J / mol = 353.3 K ΔS 95.95 J / K mol 660 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 2 .1 = 0.84 mol/kg 2 .5 ΔT = Kbm, (355.4 K 353.3 K) = 2.5 K kg/mol (m), m = 0.879 g 1 kg = 0.132 kg mL 1000 g mass solvent = 150. mL × 0.84 mol solute 142 g = 15.7 g = 16 g solute kg solvent mol mass solute = 0.132 kg solvent × 94. ΔSsurr = ΔH/T = qP/T q = heat loss by hot water = moles × molar heat capacity × ΔT q = 1.00 × 103 g H2O × ΔSsurr = 1 mol H 2 O 75.4 J × (298.2 – 363.2) = 2.72 × 105 J 18.02 g K mol (2.72 105 J) = 912 J/K 298.2 K 95. HX Initial Equil. ⇌ 0.10 M 0.10 – x H+ + X ~0 x Ka = [H ][X ] [HX] 0 x From problem, x = [H+] = 105.83 = 1.5 × 106 ; Ka = (1.5 106 ) 2 = 2.3 1011 0.10 1.5 106 ΔG = RT ln K = 8.3145 J/Kmol (298 K) ln(2.3 × 1011 ) = 6.1 × 104 J/mol = 61 kJ/mol 96. NaCl(s) ⇌ Na+(aq) + Cl(aq) K = Ksp = [Na+][Cl] ΔG = [(262 kJ) + (131 kJ)] – (384 kJ) = 9 kJ = 9000 J (9000 J) ΔG = = RT ln Ksp, Ksp = exp = 38 = 40 8.3145 J / K mol 298 K NaCl(s) ⇌ Initial s = solubility (mol/L) Equil. Na+(aq) + Cl(aq) 0 s 0 s Ksp = 40 = s(s), s = (40)1/2 = 6.3 = 6 M = [Cl] Ksp = 40 CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY 661 Integrative Problems 97. Because the partial pressure of C(g) decreased, the net change that occurs for this reaction to reach equilibrium is for products to convert to reactants. A(g) + 2 B(g) ⇌ C(g) Initial 0.100 atm 0.100 atm 0.100 atm Change +x +2x x Equil. 0.100 + x 0.100 + 2x 0.100 x From the problem, PC = 0.040 atm = 0.100 – x, x = 0.060 atm The equilibrium partial pressures are: PA = 0.100 + x = 0.100 + 0.060 = 0.160 atm, PB = 0.100 + 2((0.60) = 0.220 atm, and PC = 0.040 atm K= 0.040 = 5.2 0.160(0.220) 2 G = RT ln K = 8.3145 J/Kmol (298 K) ln 5.2 = 4.1 × 103 J/mol = 4.1 kJ/mol 98. G = H TS = 28.0 × 103 J – 298 K(175 J/K) = 24,200 J G = RT ln K, ln K = ΔG o 24,000 J = 9.767 RT 8.3145 J / K mol 298 K K = e 9.767 = 5.73 × 105 B Initial Change Equil. + 0.125 M x 0.125 x Kb = 5.73 × 105 = H2O ⇌ BH+ 0 +x x + OH K = Kb = 5.73 × 105 ~0 +x x [BH ][OH ] x2 x2 = , x = [OH] = 2.68 × 103 M [B] 0.125 x 0.125 pH = log(2.68 × 103 ) = 2.572; pOH = 14.000 – 2.572 = 11.428; Assumptions good Marathon Problem 99. a. ΔS° will be negative because there is a decrease in the number of moles of gas. b. Because ΔS° is negative, ΔH° must be negative for the reaction to be spontaneous at some temperatures. Therefore, ΔSsurr is positive. 662 CHAPTER 16 c. Ni(s) + 4 CO(g) ⇌ SPONTANEITY, ENTROPY, AND FREE ENERGY Ni(CO)4(g) ΔH° = 607 [4(110.5)] = 165 kJ; ΔS° = 417 [4(198) + (30.)] = 405 J/K d. ΔG° = 0 = ΔH° TΔS°, T = ΔH o 165 103 J = 407 K or 134°C 405 J / K ΔSo e. T = 50.°C + 273 = 323 K o ΔG 323 = 165 kJ (323 K)(0.405 kJ/K) = 34 kJ ln K = f. ΔG o (34,000 J) = 12.66, K = e12.66 = 3.1 × 105 RT 8.3145J / K mol (323 K ) T = 227°C + 273 = 500. K o ΔG 500 = 165 kJ (500. K)( 0.405 kJ/K) = 38 kJ ln K = 38,000 J = 9.14, K = e 9.14 = 1.1 × 104 (8.3145 J / K mol)(500. K ) g. The temperature change causes the value of the equilibrium constant to change from a large value favoring formation of Ni(CO)4 to a small value favoring the decomposition of Ni(CO)4 into pure Ni and CO. This is exactly what is wanted in order to purify a nickel sample. h. Ni(CO)4(l) ⇌ Ni(CO)4(g) K = PNi ( CO) 4 At 42°C (the boiling point): ΔG° = 0 = ΔH° TΔS° ΔS° = ΔH o 29.0 103 J = 92.1 J/K T 315 K o At 152°C: ΔG 152 = ΔH° TΔS° = 29.0 × 103 J 425 K (92.1 J/K) = 10,100 J ΔG° = RT ln K, ln K = (10,100 J) = 2.858, Kp = e2.858 = 17.4 8.3145 J / K mol(425 K) A maximum pressure of 17.4 atm can be attained before Ni(CO)4(g) will liquify.