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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 18 Entropy, Free Energy, and Equilibrium THERMODYNAMICS IS AN EXTENSIVE AND FAR-REACHING SCIENTIFIC DIS- 18.1 THE THREE LAWS OF THERMODYNAMICS CIPLINE THAT DEALS WITH THE INTERCONVERSION OF HEAT AND OTHER FORMS OF ENERGY. THERMODYNAMICS 18.2 SPONTANEOUS PROCESSES AND ENTROPY ENABLES US TO USE INFORMA- 18.3 THE SECOND LAW OF THERMODYNAMICS TION GAINED FROM EXPERIMENTS ON A SYSTEM TO DRAW CONCLUSIONS ABOUT OTHER ASPECTS OF THE SAME SYSTEM WITHOUT FURTHER EX- 18.4 GIBBS FREE ENERGY PERIMENTATION. FOR EXAMPLE, WE SAW IN CHAPTER 6 THAT IT IS POS- SIBLE TO CALCULATE THE HEAT OF REACTION FROM THE STANDARD EN- 18.5 FREE ENERGY AND CHEMICAL EQUILIBRIUM THALPIES OF FORMATION OF THE REACTANT AND PRODUCT MOLECULES. 18.6 THERMODYNAMICS IN LIVING SYSTEMS THIS THE CHAPTER INTRODUCES THE SECOND LAW OF THERMODYNAMICS AND GIBBS FREE-ENERGY FUNCTION. SHIP BETWEEN GIBBS IT ALSO DISCUSSES THE RELATION- FREE ENERGY AND CHEMICAL EQUILIBRIUM. 725 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 726 ENTROPY, FREE ENERGY, AND EQUILIBRIUM 18.1 THE THREE LAWS OF THERMODYNAMICS In Chapter 6 we encountered the first of three laws of thermodynamics, which says that energy can be converted from one form to another, but it cannot be created or destroyed. One measure of these changes is the amount of heat given off or absorbed by a system during a constant-pressure process, which chemists define as a change in enthalpy ( H ). The second law of thermodynamics explains why chemical processes tend to favor one direction. The third law is an extension of the second law and will be examined briefly in Section 18.3. 18.2 A spontaneous reaction does not necessarily mean an instantaneous reaction. SPONTANEOUS PROCESSES AND ENTROPY One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration). This knowledge is important whether one is synthesizing compounds in a research laboratory, manufacturing chemicals on an industrial scale, or trying to understand the intricate biological processes in a cell. A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous. We observe spontaneous physical and chemical processes every day, including many of the following examples: • • • • • • • A waterfall runs downhill, but never up, spontaneously. A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form. Water freezes spontaneously below 0°C, and ice melts spontaneously above 0°C (at 1 atm). Heat flows from a hotter object to a colder one, but the reverse never happens spontaneously. The expansion of a gas in an evacuated bulb is a spontaneous process [Figure 18.1(a)]. The reverse process, that is, the gathering of all the molecules into one bulb, is not spontaneous [Figure 18.1(b)]. A piece of sodium metal reacts violently with water to form sodium hydroxide and hydrogen gas. However, hydrogen gas does not react with sodium hydroxide to form water and sodium. Iron exposed to water and oxygen forms rust, but rust does not spontaneously change back to iron. These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction. If we assume that spontaneous processes occur so as to decrease the energy of a system, we can explain why a ball rolls downhill and why springs in a clock unwind. Similarly, a large number of exothermic reactions are spontaneous. An example is the combustion of methane: CH4(g) Back Forward Main Menu TOC 2O2(g) 88n CO2(g) Study Guide TOC 2H2O(l ) H° Textbook Website 890.4 kJ MHHE Website 18.2 727 SPONTANEOUS PROCESSES AND ENTROPY FIGURE 18.1 (a) A spontaneous process. (b) A nonspontaneous process. (a) (b) Another example is the acid-base neutralization reaction: H (aq) OH (aq) 88n H2O(l ) H° 56.2 kJ But consider a solid-to-liquid phase transition such as H2O(s) 88n H2O(l ) H° 6.01 kJ In this case, the assumption that spontaneous processes always decrease a system’s energy fails. Experience tells us that ice melts spontaneously above 0°C even though the process is endothermic. Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: H2O NH4NO3(s) 88n NH4 (aq) NO3 (aq) H° 25 kJ This process is spontaneous, and yet it is also endothermic. The decomposition of mercury(II) oxide is an endothermic reaction that is nonspontaneous at room temperature, but it becomes spontaneous when the temperature is raised: 2HgO(s) 88n 2Hg(l ) When heated, HgO decomposes to give Hg and O2. O2(g) H° 90.7 kJ From a study of the examples mentioned and many more cases, we come to the following conclusion: Exothermicity favors the spontaneity of a reaction but does not guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is possible for an exothermic reaction to be nonspontaneous. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy. ENTROPY In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, which is nearly equivalent to E for most processes. The other is entropy (S), which is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system, the smaller its entropy. One way to illustrate order and dis- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 728 ENTROPY, FREE ENERGY, AND EQUILIBRIUM order is with a deck of playing cards. A new deck of cards is arranged in an ordered fashion (the cards are from ace to king and the suits are in the order of spades to hearts to diamonds to clubs). Once the deck has been shuffled, the cards are no longer in sequence by number or by suit. It is possible, but extremely unlikely, that by reshuffling the cards we can restore the original order. There are many ways for the cards to be out of sequence but only one way for them to be ordered according to our definition. For any substance, the particles in the solid state are more ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state. So for the same molar amount of a substance, we can write Ssolid Sliquid Sgas In other words, entropy describes the extent to which atoms, molecules, or ions are distributed in a disorderly fashion in a given region in space. One way to conceptualize order and disorder is in terms of probability. A probable event is one that can happen in many ways, and an improbable event is one that can happen in only one or a few ways. For example, referring to Figure 18.1(a), suppose that there is only one molecule in the left flask initially. Since the two flasks have equal volume, the probability of finding the molecule in either flask after the stopcock is opened is 1 . If two molecules are present to start with, the probability of finding 2 both molecules in the same flask after we open the stopcock becomes 1 1 , or 1 . Now 2 2 4 1 4 is a high ratio, and so it would not be surprising to find both molecules in the same flask after the stopcock has been opened. However, it is not difficult to see that as the number of molecules increases, the probability (P) of finding all the molecules in the same flask becomes progressively smaller: 1 2 P 1 2 1 2 1 2 .... N where N is the total number of molecules present. If N P TABLE 18.1 Standard Entropy Values (So) for Selected Substances SUBSTANCE S° (J/K mol) H2O(l ) H2O(g) Br2(l ) Br2(g) I2(s) I2(g) C(diamond) C(graphite) He(g) Ne(g) Back Forward 69.9 188.7 152.3 245.3 116.7 260.6 2.44 5.69 126.1 146.2 1 2 100 8 10 100, we have 31 If N is of the order of 6 1023, the number of molecules in 1 mole of a gas, the prob23 ability becomes ( 1 )6 10 , which is such a small number that, for all practical purposes, 2 it can be regarded as zero. On the basis of probability considerations, we would expect the gas to fill both flasks spontaneously and evenly. By the same token, we now understand that the situation depicted in Figure 18.1(b) is not spontaneous because it represents a highly improbable event. To summarize our discussion: An ordered state has a low probability of occurring and a small entropy, while a disordered state has a high probability of occurring and a large entropy. As we will see, it is possible to determine the absolute entropy of a substance, something we cannot do for energy or enthalpy. Standard entropy is the absolute entropy of a substance at 1 atm and 25°C. It is this value that is generally used in calculations. (Recall that the standard state refers only to 1 atm. The reason for specifying 25°C is that many processes are carried out at room temperature.) Table 18.1 lists standard entropies of a few elements and compounds; Appendix 3 provides a more extensive listing. The units of entropy are J/K or J/K mol for 1 mole of the substance. We use joules rather than kilojoules because entropy values are typically quite small. Entropies of elements and compounds are all positive (that is, S° 0). By contrast, the Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.2 State functions are properties that are determined by the state of the system (see Section 6.7). SPONTANEOUS PROCESSES AND ENTROPY 729 standard enthalpy of formation ( H°) for elements in their stable form is zero, and for f compounds it may be positive or negative. Referring to Table 18.1, we can make the following observations. The entropy of water vapor is greater than the entropy of liquid water, because a mole of water has a much smaller volume in the liquid state than it does in the gaseous state. In other words, water molecules are more ordered in the liquid state because there is less space for them to occupy. Similarly, bromine vapor has a higher entropy value than liquid bromine, and iodine vapor has a greater entropy value than solid iodine. For different substances in the same phase, the molecular complexity and molar mass determine which ones have higher entropy values. Both diamond and graphite are solids, for example, but diamond has the more ordered structure (see Figure 11.28). Therefore, diamond has a smaller entropy value than graphite. Both neon and helium are monatomic gases, but neon has a greater entropy value than helium because its molar mass is greater. Like energy and enthalpy, entropy is a state function. Consider a certain process in which a system changes from some initial state to some final state. The entropy change for the process, S, is S Sf Si where Sf and Si are the entropies of the system in the final and initial states, respectively. If the change results in an increase in randomness, or disorder, then Sf Si or S 0. Figure 18.2 shows several processes that lead to an increase in entropy. In each case you can see that the system changes from a more ordered state to a less ordered FIGURE 18.2 Processes that lead to an increase in entropy of the system: (a) melting: Sliquid Ssolid; (b) vaporization: Svapor Sliquid; (c) dissolving; (d) heating: ST2 ST1. Solid Liquid (a) Liquid Vapor (b) Solvent Solute Solution (c) System at T2 (T2 > T1) System at T1 (d) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 730 ENTROPY, FREE ENERGY, AND EQUILIBRIUM one. It is clear that both melting and vaporization processes have S 0. The solution process usually leads to an increase in entropy. When a sugar crystal dissolves in water, the highly ordered structure of the solid and part of the ordered structure of water break down. Consequently, the solution is more disordered than the pure solute and pure solvent. When an ionic solid dissolves in water, there are two contributions to entropy increase; the solution process and the dissociation of the compound into ions: H2O NaCl(s) 88n Na (aq) Cl (aq) However, we must also consider hydration, which causes the water molecules to become more ordered around the ions. This process decreases entropy. For small, highly charged ions such as Al3 and Fe3 , the decrease in entropy due to hydration can outweigh the increase in entropy due to mixing and dissociation so that the entropy change for the overall process can actually be negative. Heating also increases the entropy of a system. In addition to translational motion (that is, motion from one point to another), molecules can also execute rotational motion (Figure 18.3) and vibrational motion (see Figure 17.13). The higher the temperature, the greater the molecular motion. This means an increase in randomness at the molecular level, which corresponds to an increase in entropy. The following example deals with the entropy changes of a system resulting from physical changes. EXAMPLE 18.1 Predict whether the entropy change is greater than or less than zero for each of the following processes: (a) freezing ethanol, (b) evaporating a beaker of liquid bromine at room temperature, (c) dissolving sucrose in water, (d) cooling nitrogen gas from 80°C to 20°C. (a) This is a liquid-to-solid phase transition. The system becomes more ordered, so that S 0. (b) This is a liquid-to-vapor phase transition. The system becomes more disordered, and S 0. (c) A solution is invariably more disordered than its components (the solute and solvent). Therefore, S 0. (d) Cooling decreases molecular motion; therefore, S 0. Answer Bromine is a fuming liquid at room temperature. PRACTICE EXERCISE How does the entropy of a system change for each of the following processes? (a) condensing water vapor, (b) forming sucrose crystals from a supersaturated solution, (c) heating hydrogen gas from 60°C to 80°C, (d) subliming dry ice Similar problem: 18.5. FIGURE 18.3 Rotational motion of a molecule. A water molecule can rotate in three different ways about the x-, y-, and z-axes. Back Forward z z x y Main Menu y TOC z x Study Guide TOC x y Textbook Website MHHE Website 18.3 18.3 Just talking about entropy increases its value in the universe. THE SECOND LAW OF THERMODYNAMICS 731 THE SECOND LAW OF THERMODYNAMICS The connection between entropy and the spontaneity of a reaction is expressed by the second law of thermodynamics: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Since the universe is made up of the system and the surroundings, the entropy change in the universe ( Suniv) for any process is the sum of the entropy changes in the system ( Ssys) and in the surroundings ( Ssurr). Mathematically, we can express the second law of thermodynamics as follows: For a spontaneous process: Suniv Ssys Ssurr 0 (18.1) For an equilibrium process: Suniv Ssys Ssurr 0 (18.2) For a spontaneous process, the second law says that Suniv must be greater than zero, but it does not place a restriction on either Ssys or Ssurr. Thus it is possible for either Ssys or Ssurr to be negative, as long as the sum of these two quantities is greater than zero. For an equilibrium process, Suniv is zero. In this case Ssys and Ssurr must be equal in magnitude, but opposite in sign. What if for some process we find that Suniv is negative? What this means is that the process is not spontaneous in the direction described. Rather, it is spontaneous in the opposite direction. ENTROPY CHANGES IN THE SYSTEM To calculate Suniv, we need to know both Ssys and Ssurr. Let us focus first on Ssys. Suppose that the system is represented by the following reaction: aA bB 88n cC dD As is the case for the enthalpy of a reaction [see Equation (6.8)], the standard entropy of reaction S° is given by rxn Sr° xn [cS°(C) dS°(D)] [aS°(A) bS°(B)] (18.3) or, in general, using to represent summation and m and n for the stoichiometric coefficients in the reaction, Sr° xn nS°(products) mS°(reactants) (18.4) The standard entropy values of a large number of compounds have been measured in J/K mol. To calculate S rxn (which is Ssys), we look up their values in Appendix 3 ° and proceed according to the following example. EXAMPLE 18.2 From the absolute entropy values in Appendix 3, calculate the standard entropy changes for the following reactions at 25°C. (a) CaCO3(s) 88n CaO(s) CO2(g) (b) N2(g) 3H2(g) 88n 2NH3(g) (c) H2(g) Cl2(g) 88n 2HCl(g) Answer (a) Srxn ° Back Forward Main Menu TOC We can calculate S° by using Equation (18.4). [S °(CaO) S °(CO2)] [S °(CaCO3)] [(1 mol)(39.8 J/K mol) (1 mol)(213.6 J/K mol)] 160.5 J/K Study Guide TOC (1 mol)(92.9 J/K mol) Textbook Website MHHE Website 732 ENTROPY, FREE ENERGY, AND EQUILIBRIUM Thus, when 1 mole of CaCO3 decomposes to form 1 mole of CaO and 1 mole of gaseous CO2, there is an increase in entropy equal to 160.5 J/K. (b) S rxn [2S°(NH3)] [S°(N2) 3S°(H2)] ° (2 mol)(193 J/K mol) [(1 mol)(192 J/K mol) (3 mol)(131 J/K mol)] 199 J/K This result shows that when 1 mole of gaseous nitrogen reacts with 3 moles of gaseous hydrogen to form 2 moles of gaseous ammonia, there is a decrease in entropy equal to 199 J/K. (c) S rxn [2S°(HCl)] [S°(H2) S°(Cl2)] ° (2 mol)(187 J/K mol) [(1 mol)(131 J/K mol) (1 mol)(223 J/K mol)] 20 J/K Thus the formation of 2 moles of gaseous HCl from 1 mole of gaseous H2 and 1 mole of gaseous Cl2 results in a small increase in entropy equal to 20 J/K. Similar problems: 18.11 and 18.12. PRACTICE EXERCISE Calculate the standard entropy change for the following reactions at 25°C: (a) 2CO(g) O2(g) 88n 2CO2(g) (b) 3O2(g) 88n 2O3(g) (c) 2NaHCO3(s) 88n Na2CO3(s) H2O(l ) CO2(g) The results of Example 18.2 are consistent with results observed for many other reactions. Taken together, they support the following general rules: If a reaction produces more gas molecules than it consumes [Example 18.2(a)], S° is positive. • If the total number of gas molecules diminishes [Example 18.2(b)], S° is negative. • If there is no net change in the total number of gas molecules [Example 18.2(c)], then S ° may be positive or negative, but will be relatively small numerically. • These conclusions make sense, given that gases invariably have greater entropy than liquids and solids. For reactions involving only liquids and solids, predicting the sign of S ° is more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase in entropy. The following example shows how knowing the nature of reactants and products makes it possible to predict entropy changes. EXAMPLE 18.3 Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H2(g) O2(g) 88n 2H2O(l ) (b) NH4Cl(s) 88n NH3(g) HCl(g) (c) H2(g) Br2(g) 88n 2HBr(g) (a) Two gases combine to form a liquid. Therefore, S is negative. (b) Since the solid is converted to two gaseous products, S is positive. (c) We see that the same number of moles of gas is involved in the reactants as in the product. Therefore we cannot predict the sign of S, but we know the change must be quite small. Answer Similar problems: 18.13 and 18.14. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.3 THE SECOND LAW OF THERMODYNAMICS 733 PRACTICE EXERCISE Discuss qualitatively the sign of the entropy change expected for each of the following processes: (a) I2(g) 88n 2I(g) (b) 2Zn(s) O2(g) 88n 2ZnO(s) (c) N2(g) O2(g) 88n 2NO(g) ENTROPY CHANGES IN THE SURROUNDINGS Next we see how the Ssurr is calculated. When an exothermic process takes place in the system, the heat transferred to the surroundings enhances motion of the molecules in the surroundings. Consequently, there is an increase in disorder at the molecular level, and the entropy of the surroundings increases. Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the surroundings because molecular motion decreases (Figure 18.4). For constantpressure processes the heat change is equal to the enthalpy change of the system, Hsys. Therefore, the change in entropy of the surroundings, Ssurr, is proportional to Hsys: Ssurr Ssurr Surroundings System Heat Forward Main Menu TOC (18.5) Surroundings (a) Back Hsys T Study Guide TOC Heat System Entropy FIGURE 18.4 (a) An exothermic process transfers heat from the system to the surroundings and results in an increase in the entropy of the surroundings. (b) An endothermic process absorbs heat from the surroundings and thereby decreases the entropy of the surroundings. The minus sign is used because if the process is exothermic, Hsys is negative and Ssurr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, Hsys is positive and the negative sign ensures that the entropy of the surroundings decreases. The change in entropy for a given amount of heat also depends on the temperature. If the temperature of the surroundings is high, the molecules are already quite energetic. Therefore, the absorption of heat from an exothermic process in the system will have relatively little impact on molecular motion and the resulting increase in entropy will be small. However, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a more drastic increase in molecular motion and hence a larger increase in entropy. By analogy, someone coughing in a crowded restaurant will not disturb too many people, but someone coughing in a library definitely will. From the inverse relationship between Ssurr and temperature T (in kelvins)—that is, the higher the temperature, the smaller the Ssurr and vice versa— we can rewrite the above relationship as Entropy This equation, which can be derived from the laws of thermodynamics, assumes that both the system and the surroundings are at temperature T. Hsys (b) Textbook Website MHHE Website 734 ENTROPY, FREE ENERGY, AND EQUILIBRIUM Let us now apply the procedure for calculating Ssys and Ssurr to the synthesis of ammonia and ask whether the reaction is spontaneous at 25°C: N2(g) 3H2(g) 88n 2NH3(g) From Example 18.2(b) we have Ssys in Equation (18.5), we obtain 92.6 kJ 199 J/K, and substituting Hsys ( 92.6 kJ) ( 92.6 1000) J 298 K Ssurr H° 311 J/K The change in entropy of the universe is Suniv Ssys Ssurr 199 J/K 311 J/K 112 J/K Because Suniv is positive, we predict that the reaction is spontaneous at 25°C. It is important to keep in mind that just because a reaction is spontaneous does not mean that it will occur at an observable rate. The synthesis of ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur. Reaction rates are the subject of chemical kinetics (see Chapter 13). THE THIRD LAW OF THERMODYNAMICS AND ABSOLUTE ENTROPY Finally it is appropriate to consider the third law of thermodynamics briefly in connection with the determination of entropy values. So far we have related entropy to molecular disorder — the greater the disorder or freedom of motion of the atoms or molecules in a system, the greater the entropy of the system. The most ordered arrangement of any substance with the least freedom of atomic or molecular motion is a perfect crystalline substance at absolute zero (0 K). It follows, therefore, that the lowest entropy any substance can attain is that of a perfect crystal at absolute zero. According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at the absolute zero of temperature. As the temperature increases, the freedom of motion also increases. Thus the entropy of any substance at a temperature above 0 K is greater than zero. Note also that if the crystal is impure or if it has defects, then its entropy is greater than zero even at 0 K because it would not be perfectly ordered. The important point about the third law of thermodynamics is that it allows us to determine the absolute entropies of substances. Starting with the knowledge that the entropy of a pure crystalline substance is zero at 0 K, we can measure the increase in entropy of the substance when it is heated to, say, 298 K. The change in entropy, S, is given by S Sf Si Sf The entropy increase can be calculated from the temperature change and heat capacity of the substance. Back Forward since Si is zero. The entropy of the substance at 298 K, then, is given by S or Sf, which is called the absolute entropy because this is the true value and not a value derived using some arbitrary reference. Thus the entropy values quoted so far are all absolute entropies. In contrast, we cannot have the absolute energy or enthalpy of a substance because the zero of energy or enthalpy is undefined. Figure 18.5 shows the change (increase) in entropy of a substance with temperature. At absolute zero, it has a zero entropy value (assuming that it is a perfect crystalline substance). As it is heated, Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.4 FIGURE 18.5 Entropy increase of a substance as the temperature rises from absolute zero. Solid Liquid GIBBS FREE ENERGY 735 Gas S° (J/K• mol) Boiling Melting Temperature (K) its entropy increases gradually because of greater molecular motion. At the melting point, there is a sizable increase in entropy as the more random liquid state is formed. Further heating increases the entropy of the liquid again due to enhanced molecular motion. At the boiling point there is a large increase in entropy as a result of the liquid to gas transition. Beyond that temperature, the entropy of the gas continues to rise with increasing temperature. 18.4 GIBBS FREE ENERGY The second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, Suniv 0. In order to determine the sign of Suniv for a reaction, however, we would need to calculate both Ssys and Ssurr. However, we are usually concerned only with what happens in a particular system, and the calculation of Ssurr can be quite difficult. Therefore, we generally rely on another thermodynamic function to help us determine whether a reaction will occur spontaneously if we consider only the system itself. From Equation (18.1), we know that for a spontaneous process, we have Suniv Substituting Forward Main Menu TOC Ssurr 0 Hsys T 0 Hsys/T for Ssurr, we write Suniv Back Ssys Study Guide TOC Ssys Textbook Website MHHE Website 736 ENTROPY, FREE ENERGY, AND EQUILIBRIUM Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back The Efficiency of Heat Engines An engine is a machine that converts energy to work; a heat engine is a machine that converts thermal energy to work. Heat engines play an essential role in our technological society; they range from automobile engines to the giant steam turbines that run generators to produce electricity. Regardless of the type of heat engine, its efficiency level is of great importance; that is, for a given amount of heat input, how much useful work can we get out of the engine? The second law of thermodynamics helps us answer this question. The figure below shows a simple form of a heat engine. A cylinder fitted with a weightless piston is initially at temperature T1. Next, the cylinder is heated to a higher temperature T2. The gas in the cylinder expands and pushes up the piston. Finally the cylinder is cooled down to T1 and the apparatus returns to its original state. By repeating this cycle, the up-and-down movement of the piston can be made to do mechanical work. A unique feature of heat engines is that some heat must be given off to the surroundings when they do work. With the piston in the up position, no further work can be done if we do not cool the cylinder back to T1. The cooling process removes some of the thermal energy that could otherwise be converted to work and thereby places a limit on the efficiency of heat engines. The figure on p. 737 shows the heat transfer processes in a heat engine. Initially, a certain amount T1 T2 T1 (a) (b) of heat flows from the heat reservoir (at temperature Th) into the engine. As the engine does work, some of the heat is given off to the surroundings, or the heat sink (at temperature Tc). By definition, the efficiency of a heat engine is efficiency useful work output energy input 100% Analysis based on the second law shows that efficiency can also be expressed as efficiency Tc Th Tc 1 Th Th 100% 100% Thus the efficiency of a heat engine is given by the difference in temperatures between the heat reservoir and the heat sink (both in kelvins), divided by the temperature of the heat reservoir. In practice we can make (Th Tc) as large as possible, but since Tc cannot be zero and Th cannot be infinite, the efficiency of a heat engine can therefore never be 100 percent. At a power plant, superheated steam at about 560°C (833 K) is used to drive a turbine for electricity generation. The temperature of the heat sink is about 38°C (or 311 K). The efficiency is given by efficiency 833 K 311 K 833 K 63% 100% A simple heat engine. (a) The engine is initially at T1. (b) When heated to T2, the piston is pushed up due to gas expansion. (c) When cooled to T1, the piston returns to its original position. (c) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.4 Chemistry in Action Chemistry in Action Heat reservoir Th Work Heat engine GIBBS FREE ENERGY 737 In practice, because of friction, heat loss, and other complications, the maximum efficiency of a steam turbine is only about 40 percent. Therefore for every ton of coal used at a power plant, 0.40 ton generates electricity while the rest of it ends up warming the surroundings! Heat sink Tc Heat transfer during the operation of a heat engine. Multiplying both sides of the equation by T gives T Suniv The change in unequal sign when we multiply the equation by 1 follows from the fact that 1 0 and 1 0. Hsys T Ssys 0 Now we have a criterion for a spontaneous reaction that is expressed only in terms of the properties of the system ( Hsys and Ssys) and we can ignore the surroundings. For convenience, we can change the above equation, multiplying it throughout by 1 and replacing the sign with : T Suniv Hsys T Ssys 0 This equation says that for a process carried out at temperature T, if the changes in enthalpy and entropy of the system are such that Hsys T Ssys is less than zero, the process must be spontaneous. In order to express the spontaneity of a reaction more directly, we can use another thermodynamic function called Gibbs† free energy (G), or simply free energy: G H TS (18.6) All quantities in Equation (18.6) pertain to the system, and T is the temperature of the system. You can see that G has units of energy (both H and TS are in energy units). Like H and S, G is a state function. The change in free energy ( G) of a system for a constant-temperature process is G The word “free” in the term “free energy” does not mean without cost. H TS (18.7) In this context free energy is the energy available to do work. Thus, if a particular reaction is accompanied by a release of usable energy (that is, if G is negative), this fact alone guarantees that it is spontaneous, and there is no need to worry about what happens to the rest of the universe. Note that we have merely organized the expression for the entropy change of the universe, eliminating Suniv and equating the free-energy change of the system ( G) with T Suniv, so that we can focus on changes in the system. We can now summa† Josiah Willard Gibbs (1839–1903). American physicist. One of the founders of thermodynamics, Gibbs was a modest and private individual who spent almost all of his professional life at Yale University. Because he published most of his works in obscure journals, Gibbs never gained the eminence that his contemporary and admirer James Maxwell did. Even today, very few people outside of chemistry and physics have ever heard of Gibbs. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 738 ENTROPY, FREE ENERGY, AND EQUILIBRIUM rize the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of G as follows: G 0 The reaction is spontaneous in the forward direction. G 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction. G 0 The system is at equilibrium. There is no net change. STANDARD FREE-ENERGY CHANGES TABLE 18.2 Conventions for Standard States STATE OF STANDARD MATTER STATE Gas Liquid Solid Elements* Solution 1 atm pressure Pure liquid Pure solid G°(element) 0 f 1 molar concentration *The most stable allotropic form at 25°C and 1 atm. The standard free-energy of reaction ( Grxn ) is the free-energy change for a reaction ° when it occurs under standard-state conditions, when reactants in their standard states are converted to products in their standard states. Table 18.2 summarizes the conventions used by chemists to define the standard states of pure substances as well as solutions. To calculate Grxn we start with the equation ° aA bB 88n cC dD The standard free-energy change for this reaction is given by G° rxn [c Gf (C) ° d Gf (D)] ° [a Gf (A) ° b Gf (B)] ° or, in general, Grxn ° Σn Gf (products) ° Σm Gf (reactants) ° (18.8) where m and n are stoichiometric coefficients. The term G ° is the standard free enf ergy of formation of a compound, that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states. For the combustion of graphite: C(graphite) O2(g) 88n CO2(g) the standard free-energy change [from Equation (18.8)] is G° rxn G°(CO2) f [ G°(C, graphite) f G°(O2)] f As in the case of the standard enthalpy of formation (p. 216), we define the standard free energy of formation of any element in its stable form as zero. Thus G°(C, graphite) f 0 and G°(O2) f 0 Therefore the standard free-energy change for the reaction in this case is numerically equal to the standard free energy of formation of CO2: G° rxn Gf (CO2) ° Note that G ° is in kJ, but G ° is in kJ/mol. The equation holds because the coefrxn f ficient in front of G ° (1 in this case) has the unit “mol.” Appendix 3 lists the values f of G ° for a number of compounds. f Calculations of standard free-energy changes are handled as shown in the following example. EXAMPLE 18.4 Calculate the standard free-energy changes for the following reactions at 25°C. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.4 (a) CH4(g) 2O2(g) 88n CO2(g) (b) 2MgO(s) 88n 2Mg(s) O2(g) GIBBS FREE ENERGY 739 2H2O(l ) (a) According to Equation (18.8), we write Answer G° rxn [ G°(CO2) f 2 G°(H2O) f [ G°(CH4) f 2 G°(O2)] f We insert the appropriate values from Appendix 3: G° rxn [(1 mol)( 394.4 kJ/mol) (2 mol)( 237.2 kJ/mol)] [(1 mol)( 50.8 kJ/mol) (2 mol)(0 kJ/mol)] 818.0 kJ (b) The equation is G° rxn [2 Gf (Mg) ° Gf (O2)] ° [2 Gf (MgO)] ° From data in Appendix 3 we write G° rxn Similar problems: 18.17 and 18.18. [(2 mol)(0 kJ/mol) (1 mol)(0 kJ/mol)] [(2 mol)( 569.6 kJ/mol] 1139 kJ PRACTICE EXERCISE Calculate the standard free-energy changes for the following reactions at 25°C: (a) H2(g) Br2(g) 88n 2HBr(g) (b) 2C2H6(g) 7O2(g) 88n 4CO2(g) 6H2O(l ) In the preceding example the large negative value of Grxn for the combustion of ° methane in (a) means that the reaction is a spontaneous process under standard-state conditions, whereas the decomposition of MgO in (b) is nonspontaneous because Grxn ° is a large, positive quantity. Remember, however, that a large, negative Grxn does not ° tell us anything about the actual rate of the spontaneous process; a mixture of CH4 and O2 at 25°C could sit unchanged for quite some time in the absence of a spark or flame. APPLICATIONS OF EQUATION (18.7) In order to predict the sign of G, according to Equation (18.7) we need to know both H and S. A negative H (an exothermic reaction) and a positive S (a reaction that results in an increase in disorder of the system) tend to make G negative, although temperature may influence the direction of a spontaneous reaction. The four possible outcomes of this relationship are: If both H and S are positive, then G will be negative only when the T S term is greater in magnitude than H. This condition is met when T is large. • If H is positive and S is negative, G will always be positive, regardless of temperature. • If H is negative and S is positive, then G will always be negative regardless of temperature. • If H is negative and S is negative, then G will be negative only when T S is smaller in magnitude than H. This condition is met when T is small. • The temperatures that will cause G to be negative for the first and last cases depend on the actual values of H and S of the system. Table 18.3 summarizes the effects of the possibilities just described. We will now consider two specific applications of Equation (18.7). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 740 ENTROPY, FREE ENERGY, AND EQUILIBRIUM TABLE 18.3 H Factors Affecting the Sign of G in the Relationship S G H TS G EXAMPLE Reaction proceeds spontaneously at high temperatures. At low temperatures, reaction is spontaneous in the reverse direction. G is always positive. Reaction is spontaneous in the reverse direction at all temperatures. G is always negative. Reaction proceeds spontaneously at all temperatures. Reaction proceeds spontaneously at low temperatures. At high temperatures, the reverse reaction becomes spontaneous. H2(g) I2(g) 88n 2HI(g) 3O2(g) 88n 2O3(g) 2H2O2(l ) 88n 2H2O(l ) NH3(g) O2(g) HCl(g) 88n NH4Cl(s) Temperature and Chemical Reactions Calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic substance used in steelmaking, production of calcium metal, the paper industry, water treatment, and pollution control. It is prepared by decomposing limestone (CaCO3) in a kiln at a high temperature (Figure 18.6): CaCO3(s) 34 CaO(s) Le Chatelier ’s principle predicts that the forward, endothermic reaction is favored by heating. CO2(g) The reaction is reversible, and CaO readily combines with CO2 to form CaCO3. The pressure of CO2 in equilibrium with CaCO3 and CaO increases with temperature. In the industrial preparation of quicklime, the system is never maintained at equilibrium; rather, CO2 is constantly removed from the kiln to shift the equilibrium from left to right, promoting the formation of calcium oxide. The important information for the practical chemist is the temperature at which the decomposition of CaCO3 becomes appreciable (that is, the temperature at which the reaction becomes spontaneous). We can make a reliable estimate of that temperature as follows. First we calculate H° and S° for the reaction at 25°C, using the data in Appendix 3. To determine H° we apply Equation (6.8): H° [ Hf (CaO) ° Hf (CO2)] ° [(1 mol)( 635.6 kJ/mol) [ Hf (CaCO3)] ° (1 mol)( 393.5 kJ/mol)] [(1 mol)( 1206.9 kJ/mol)] 177.8 kJ Next we apply Equation (18.4) to find S° S° [S °(CaO) S °(CO2)] [(1 mol)(39.8 J/K mol) [S °(CaCO3)] (1 mol)(213.6 J/K mol)] [(1 mol)(92.9 J/K mol)] 160.5 J/K For reactions carried out under standard-state conditions, Equation (18.7) takes the form G° H° T S° so we obtain G° FIGURE 18.6 The production of CaO in a rotary kiln. Back Forward 177.8 kJ (298 K)(160.5 J/K) 1 kJ 1000 J 130.0 kJ Since G° is a large positive quantity, we conclude that the reaction is not favored at Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.4 GIBBS FREE ENERGY 741 25°C (or 298 K). In order to make G ° negative, we first have to find the temperature at which G° is zero; that is, 0 or T T H ° T S° H° S° (177.8 kJ)(1000 J/1 kJ) 160.5 J/K 1108 K or 835°C At a temperature higher than 835°C, G° becomes negative, indicating that the decomposition is spontaneous. For example, at 840°C, or 1113 K, Go Ho T So 177.8 kJ (1113 K)(160.5 J/K) 1 kJ 1000 J 0.8 kJ The equilibrium constant of this reaction is KP PCO2. Two points are worth making about such a calculation. First, we used the H° and S ° values at 25°C to calculate changes that occur at a much higher temperature. Since both H ° and S ° change with temperature, this approach will not give us an accurate value of G°, but it is good enough for “ball park” estimates. Second, we should not be misled into thinking that nothing happens below 835°C and that at 835°C CaCO3 suddenly begins to decompose. Far from it. The fact that G° is a positive value at some temperature below 835°C does not mean that no CO2 is produced, but rather that the pressure of the CO2 gas formed at that temperature will be below 1 atm (its standard-state value; see Table 18.2). As Figure 18.7 shows, the pressure of CO2 at first increases very slowly with temperature; it becomes easily measurable above 700°C. The significance of 835°C is that this is the temperature at which the equilibrium pressure of CO2 reaches 1 atm. Above 835°C, the equilibrium pressure of CO2 exceeds 1 atm. (In Section 18.5 we will see how G° is related to the equilibrium constant of a given reaction.) FIGURE 18.7 Equilibrium pressure of CO2 from the decomposition of CaCO3, as a function of temperature. This curve is calculated by assuming that H and S of the reaction do not change with temperature. 3 PCO2 (atm) 2 1 835°C 0 Back Forward Main Menu 100 TOC 200 300 400 500 600 t (°C) Study Guide TOC 700 800 900 Textbook Website MHHE Website 742 ENTROPY, FREE ENERGY, AND EQUILIBRIUM If a system is at equilibrium, then there is no tendency for spontaneous change in either direction. The condition G 0 applies to any phase transition. Phase Transitions At the temperature at which a phase transition occurs (the melting point or boiling point) the system is at equilibrium ( G 0), so Equation (18.7) becomes 0 S The melting of ice is an endothermic process ( H is positive), and the freezing of water is exothermic ( H is negative). H TS H T Let us first consider the ice-water equilibrium. For the ice n water transition, H is the molar heat of fusion (see Table 11.8), and T is the melting point. The entropy change is therefore 6010 J/mol 273 K Sice n water 22.0 J/K mol Thus when 1 mole of ice melts at 0°C, there is an increase in entropy of 22.0 J/K. The increase in entropy is consistent with the increase in disorder from solid to liquid. Conversely, for the water n ice transition, the decrease in entropy is given by 6010 J/mol 273 K Swater n ice 22.0 J/K mol In the laboratory we normally carry out unidirectional changes, that is, either ice to water or water to ice. We can calculate entropy change in each case using the equation S H/T as long as the temperature remains at 0°C. The same procedure can be applied to the water-steam transition. In this case H is the heat of vaporization and T is the boiling point of water. Example 18.5 examines the phase transitions in benzene. EXAMPLE 18.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid n liquid and liquid n vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. At the melting point, the system is at equilibrium. Therefore G the entropy of fusion is given by Answer 0 Liquid and solid benzene in equilibrium. Hfus Hfus Tf Sfus 0 and Tf Sfus (10.9 kJ/mol)(1000 J/1 kJ) (5.5 273) K 39.1 J/K mol Similarly, at the boiling point, G Back Forward Main Menu TOC 0 and we have Study Guide TOC Textbook Website MHHE Website 18.5 FREE ENERGY AND CHEMICAL EQUILIBRIUM 743 Hvap Tbp Svap (31.0 kJ/mol)(1000 J/1 kJ) (80.1 273) K 87.8 J/K mol Comment Since vaporization creates more disorder than the melting process, Svap Sfus. Similar problem: 18.59. PRACTICE EXERCISE The molar heats of fusion and vaporization of argon are 1.3 kJ/mol and 6.3 kJ/mol, and argon’s melting point and boiling point are 190°C and 186°C, respectively. Calculate the entropy changes for fusion and vaporization. 18.5 FREE ENERGY AND CHEMICAL EQUILIBRIUM Suppose we start a reaction in solution with all the reactants in their standard states (that is, all at 1 M concentration). As soon as the reaction starts, the standard-state condition no longer exists for the reactants or the products since their concentrations are different from 1 M. Under conditions that are not standard state, we must use G rather than G° to predict the direction of the reaction. The relationship between G and G° is This equation can be derived from thermodynamics. G G° RT ln Q (18.9) where R is the gas constant (8.314 J/K mol), T is the absolute temperature of the reaction, and Q is the reaction quotient, that is, a number equal to the ratio of product concentrations, each raised to the power of its stoichiometric coefficient at some point other than equilibrium (see p. 574). We see that G depends on two quantities: G° and RT ln Q. For a given reaction at temperature T the value of G° is fixed but that of RT ln Q is not, because Q varies according to the composition of the reacting mixture. Let us consider two special cases: Case 1: If G° is a large negative value, the RT ln Q term will not become positive enough to match the G° term until a significant amount of product has formed. Case 2: If G° is a large positive value, the RT ln Q term will be more negative than G° is positive only as long as very little product formation has occurred and the concentration of the reactant is high relative to that of the product. Sooner or later a reversible reaction will reach equilibrium. At equilibrium, by definition, G Thus 0 and Q 0 G° K, where K is the equilibrium constant. RT ln K or G° RT ln K (18.10) In this equation, KP is used for gases and Kc for reactions in solution. Note that the larger the K is, the more negative G° is. For chemists, Equation (18.10) is one of the most important equations in thermodynamics because it allows us to find the equilib- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ENTROPY, FREE ENERGY, AND EQUILIBRIUM G°(reactants) ∆ G° = G°(products) – G°(reactants) < 0 G°(products) Equilibrium position Equilibrium mixture of reactants and products Pure reactants Free energy (G ) of the reacting system Free energy (G ) of the reacting system 744 G°(products) ∆ G° = G°(products) – G°(reactants) > 0 G°(reactants) Equilibrium position Equilibrium mixture of reactants and products Pure reactants Pure products Extent of reaction (b) Extent of reaction (a) FIGURE 18.8 (a) Go products Go G° 0. At equilibreactants, so rium, there is a significant conversion of reactants to products. (b) Go Go Go products reactants, so 0. At equilibrium, reactants are favored over products. Pure products rium constant of a reaction if we know the change in standard free energy and vice versa. It is significant that Equation (18.10) relates the equilibrium constant to the standard free energy change G° rather than to the actual free energy change G. The actual free energy of the system changes as the reaction proceeds and becomes zero at equilibrium. On the other hand, both G° and K are constants for a particular reaction at a given temperature. Figure 18.8 shows plots of free energy of a reacting system versus the extent of the reaction for two types of reactions. As you can see, if G° 0, the products are favored over reactants at equilibrium. Conversely, if G° 0, there will be more reactants than products at equilibrium. Table 18.4 summarizes the three possible relations between G° and K, as predicted by Equation (18.10). For reactions having very large or very small equilibrium constants, it is generally very difficult, if not impossible, to measure the K values by monitoring the concentrations of all the reacting species. Consider, for example, the formation of nitric oxide from molecular nitrogen and molecular oxygen: N2(g) O2(g) 34 2NO(g) At 25°C, the equilibrium constant Kc is Kc [NO]2 [N2][O2] 4.0 10 31 The very small value of Kc means that the concentration of NO at equilibrium will be exceedingly low. In such a case the equilibrium constant is more conveniently obtained from G°. (As we have seen, G° can be calculated from H° and S°.) On the other hand, the equilibrium constant for the formation of hydrogen iodide from molecular TABLE 18.4 Relation between G° RT ln K K ln K 1 1 1 Back Forward Main Menu G° G° and K as Predicted by the Equation COMMENTS Positive 0 Negative Negative 0 Positive Products are favored over reactants at equilibrium. Products and reactants are equally favored at equilibrium. Reactants are favored over products at equilibrium. TOC Study Guide TOC Textbook Website MHHE Website 18.5 FREE ENERGY AND CHEMICAL EQUILIBRIUM 745 hydrogen and molecular iodine is near unity at room temperature: H2(g) I2(g) 34 2HI(g) For this reaction it is easier to measure K and then calculate Go using Equation (18.10) than to measure H° and S°. It is important to remember when solving Equation (18.10) that the unit of Go is J or kJ and those of RT ln K are J/mol. The reason for the difference is that in deriving Equation (18.10), the unit “mol” was absorbed into the logarithmic term (ln K ). Thus in using this equation, we will add (1 mol) to the RT ln K term to make the units consistent on both sides of the equation. The following examples illustrate the use of Equations (18.9) and (18.10). EXAMPLE 18.6 Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C: 2H2O(l ) 34 2H2(g) Answer O2(g) According to Equation (18.8) G°xn r [2 G°(H2) f G°(O2)] f [(2 mol)(0 kJ/mol) [2 G°(H2O)] f (1 mol)(0 kJ/mol)] [(2 mol)( 237.2 kJ/mol)] 474.4 kJ Using Equation (18.10) G °xn r 474.4 kJ As stated earlier, it is difficult to measure very large (and very small) values of K directly because the concentrations of some of the species are very small. Similar problems: 18.23 and 18.26. RT ln KP 1000 J 1 kJ (1 mol)(8.314 J/K mol)(298 K) ln KP ln KP 191.5 e 191.5 7 KP 10 84 This extremely small equilibrium constant is consistent with the fact that water does not decompose into hydrogen and oxygen gases at 25°C. Thus a large positive G° favors reactants over products at equilibrium. Note that we have added the (1 mol) term on the right side of Equation (18.10) here to make the units consistent. Comment PRACTICE EXERCISE Calculate the equilibrium constant (KP) for the reaction 2O3(g) 88n 3O2(g) at 25°C. EXAMPLE 18.7 In Chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility product of silver chloride at 25°C (1.6 10 10), calculate G° for the process AgCl(s) 34 Ag (aq) Back Forward Main Menu TOC Study Guide TOC Cl (aq) Textbook Website MHHE Website 746 ENTROPY, FREE ENERGY, AND EQUILIBRIUM Because this is a heterogeneous equilibrium, the solubility product is the equilibrium constant. Answer Ksp [Ag ][Cl ] 1.6 10 10 Using Equation (18.10) we obtain G° (1 mol)(8.314 J/K mol)(298 K) ln 1.6 10 10 4 5.6 10 J 56 kJ Similar problem: 18.25. Comment The large, positive G° indicates that AgCl is slightly soluble and that the equilibrium lies mostly to the left. PRACTICE EXERCISE Calculate G° for the following process at 25°C: BaF2(s) 34 Ba2 (aq) 2F (aq) 6 The Ksp of BaF2 is 1.7 10 . EXAMPLE 18.8 The standard free-energy change for the reaction N2(g) 3H2(g) 34 2NH3(g) is 33.2 kJ and the equilibrium constant KP is 6.59 105 at 25°C. In a certain experiment, the initial pressures are PH2 0.250 atm, PN2 0.870 atm, and PNH3 12.9 atm. Calculate G for the reaction at these pressures, and predict the direction of reaction. Answer Equation (18.9) can be written as G° RT ln QP G° G RT ln P2 3 NH P3 2PN2 H 33.2 1000 J 33.2 103 J 9.9 3 10 J (1 mol)(8.314 J/K mol)(298 K) 23.3 ln (12.9)2 (0.250)3(0.870) 103 J 9.9 kJ Since G is negative, the net reaction proceeds from left to right. Similar problems: 18.27 and 18.28. As an exercise, confirm this prediction by calculating the reaction quotient QP and comparing it with the equilibrium constant KP. Comment PRACTICE EXERCISE The G° for the reaction H2(g) I2(g) 34 2HI(g) is 2.60 kJ at 25°C. In one experiment, the initial pressures are PH2 4.26 atm, PI2 0.024 atm, and PHI 0.23 atm. Calculate G for the reaction and predict the direction of reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 18.6 18.6 747 THERMODYNAMICS IN LIVING SYSTEMS THERMODYNAMICS IN LIVING SYSTEMS Many biochemical reactions are nonspontaneous (that is, they have a positive G° value), yet they are essential to the maintenance of life. In living systems these reactions are coupled to an energetically favorable process, one that has a negative G° value. The principle of coupled reactions is based on a simple concept: we can use a thermodynamically favorable reaction to drive an unfavorable one. Consider an industrial process. Suppose we wish to extract zinc from the ore sphalerite (ZnS). The following reaction will not work since it is highly nonspontaneous: ZnS(s) 88n Zn(s) S(s) G° 198.3 kJ On the other hand, the combustion of sulfur to form sulfur dioxide is favored because of its large negative G° value: S(s) The price we pay for this procedure is acid rain. O2(g) 88n SO2(g) G° 300.1 kJ By coupling the two processes we can bring about the separation of zinc from zinc sulfide. In practice, this means heating ZnS in air so that the tendency of S to form SO2 will promote the decomposition of ZnS: ZnS(s) 88n Zn(s) S(s) Go S(s) G O2(s) 88n Zn(s) SO2(s) 300.1 kJ Go O2(g) 88n SO2(g) ZnS(s) 198.3 kJ o 101.8 kJ Coupled reactions play a crucial role in our survival. In biological systems, enzymes facilitate a wide variety of nonspontaneous reactions. For example, in the human body, food molecules, represented by glucose (C6H12O6), are converted to carbon dioxide and water during metabolism with a substantial release of free energy: C6H12O6(s) 6O2(g) 88n 6CO2(g) 6H2O(l ) G° 2880 kJ In a living cell, this reaction does not take place in a single step (as burning glucose in a flame would); rather, the glucose molecule is broken down with the aid of enzymes in a series of steps. Much of the free energy released along the way is used to synthesize adenosine triphosphate (ATP) from adenosine diphosphate (ADP) and phosphoric acid (Figure 18.9): ADP H3PO4 88n ATP H2O G° 31 kJ The function of ATP is to store energy until it is needed by cells. Under appropriate FIGURE 18.9 Structure of ATP and ADP in ionized forms. The adenine group is in blue, the ribose group in black, and the phosphate group in red. Note that ADP has one fewer phosphate group than ATP. NH2 NH2 C N C C N N HC C N N O O O B B B OO P OOO P OOO P OOCH2 O A A A O O O H H H H HO OH Adenosine triphosphate (ATP) Back Forward Main Menu C HC C N CH TOC Study Guide TOC CH N N O O B B OO P OOO P OOCH2 O A A O O H H H H HO OH Adenosine diphosphate (ADP) Textbook Website MHHE Website 748 ENTROPY, FREE ENERGY, AND EQUILIBRIUM conditions, ATP undergoes hydrolysis to give ADP and phosphoric acid, with a release of 31 kJ of free energy, which can be used to drive energetically unfavorable reactions, such as protein synthesis. Proteins are polymers made of amino acids. The stepwise synthesis of a protein molecule involves the joining of individual amino acids. Consider the formation of the dipeptide (a two-amino-acid unit) alanylglycine from alanine and glycine. This reaction represents the first step in the synthesis of a protein molecule: Alanine Glycine 88n Alanylglycine G° 29 kJ As you can see, this reaction is not spontaneous, and so only a little of the dipeptide would be formed at equilibrium. However, with the aid of an enzyme, the reaction is coupled to the hydrolysis of ATP as follows: ATP H2O Alanine Glycine 88n ADP H3PO4 Alanylglycine The overall free-energy change is given by Go 31 kJ 29 kJ 2 kJ, which means that the coupled reaction is spontaneous and an appreciable amount of alanylglycine will be formed under this condition. Figure 18.10 shows the ATP-ADP interconversions that act as energy storage (from metabolism) and free energy release (from ATP hydrolysis) to drive essential reactions. FIGURE 18.10 Schematic representation of ATP synthesis and coupled reactions in living systems. The conversion of glucose to carbon dioxide and water during metabolism releases free energy. This release of free energy is used to convert ADP into ATP. The ATP molecules are then used as an energy source to drive nonspontaneous reactions, such as the synthesis of proteins from amino acids. Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back The Thermodynamics of a Rubber Band We all know how useful a rubber band can be. But not everyone is aware that a rubber band has some very interesting thermodynamic properties based on its structure. You can easily perform the following experiments with a rubber band that is at least 0.5 cm wide. Quickly stretch the rubber band and then press it against your lips. You will feel a slight warming effect. Next, reverse the process. Stretch a rubber band and hold it in position for a few seconds. Then quickly re- Forward Main Menu TOC lease the tension and press the rubber band against your lips. This time you will feel a slight cooling effect. A thermodynamic analysis of these two experiments can tell us something about the molecular structure of rubber. Rearranging Equation (18.7) ( G H T S) gives TS H G The warming effect (an exothermic process) due to Study Guide TOC Textbook Website MHHE Website SUMMARY OF FACTS AND CONCEPTS Chemistry in Action Chemistry in Action Che stretching means that H 0, and since stretching is nonspontaneous (that is, G 0 and G 0), T S must be negative. Since T, the absolute temperature, is always positive, we conclude that S due to stretching must be negative, and therefore rubber in its natural state is more disordered than when it is under tension. When the tension is removed, the stretched rubber band spontaneously snaps back to its original shape; that is, G is negative and G is positive. The cooling effect means that it is an endothermic process ( H 0), so that T S is positive. Thus the entropy of the rubber band increases when it goes from the stretched state to the natural state. SUMMARY OF KEY EQUATIONS (a) (b) (a) Rubber molecules in their normal state. Note the high degree of disorder (high entropy). (b) Under tension, the molecules line up and the arrangement becomes much more ordered (low entropy). 0 (18.1) • Suniv Ssys Ssurr 0 (18.2) S° rxn ΣnS°(products) ΣmS°(reactants) (18.4) G H G° rxn • G • Main Menu Ssurr • Forward Ssys • Back Suniv • SUMMARY OF FACTS AND CONCEPTS • G° TS n Gf (products) ° G° 749 (18.7) m Gf (reactants) ° RT ln Q RT ln K (18.8) (18.9) (18.10) The second law of thermodynamics (spontaneous process). The second law of thermodynamics (equilibrium process). Standard entropy change of a reaction. Free-energy change at constant temperature. Standard free-energy change of a reaction. Relationship between freeenergy change and standard free-energy change and reaction quotient. Relationship between standard free-energy change and the equilibrium constant. 1. Entropy is a measure of the disorder of a system. Any spontaneous process must lead to a net increase in entropy in the universe (second law of thermodynamics). 2. The standard entropy of a chemical reaction can be calculated from the absolute entropies of reactants and products. 3. The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at 0 K. This law enables us to measure the absolute entropies of substances. 4. Under conditions of constant temperature and pressure, the free-energy change G is less than zero for a spontaneous process and greater than zero for a nonspontaneous process. For an equilibrium process, G 0. 5. For a chemical or physical process at constant temperature and pressure, G H T S. This equation can be used to predict the spontaneity of a process. 6. The standard free-energy change for a reaction, G°, can be found from the standard free energies of formation of reactants and products. 7. The equilibrium constant of a reaction and the standard free-energy change of the reaction are related by the equation G° RT ln K. TOC Study Guide TOC Textbook Website MHHE Website 750 ENTROPY, FREE ENERGY, AND EQUILIBRIUM 8. Most biological reactions are nonspontaneous. They are driven by the hydrolysis of ATP, for which G° is negative. KEY WORDS Second law of thermodynamics, p. 731 Standard entropy of reaction, p. 731 Entropy (S), p. 727 Free energy (G), p. 737 Gibbs free energy (G), p. 737 Standard free energy of reaction ( G°), p. 738 Standard free energy of formation ( Gf ), p. 738 ° Third law of thermodynamics, p. 734 QUESTIONS AND PROBLEMS SPONTANEOUS PROCESSES AND ENTROPY 18.8 State the third law of thermodynamics and explain its usefulness in calculating entropy values. Review Questions 18.1 Explain what is meant by a spontaneous process. Give two examples each of spontaneous and nonspontaneous processes. 18.2 Which of the following processes are spontaneous and which are nonspontaneous? (a) dissolving table salt (NaCl) in hot soup; (b) climbing Mt. Everest; (c) spreading fragrance in a room by removing the cap from a perfume bottle; (d) separating helium and neon from a mixture of the gases 18.3 Which of the following processes are spontaneous and which are nonspontaneous at a given temperature? HO 2 (a) NaNO3(s) 88n NaNO3(aq) saturated soln HO 2 (b) NaNO3(s) 88n NaNO3(aq) unsaturated soln HO 2 (c) NaNO3(s) 88n NaNO3(aq) supersaturated soln 18.4 Define entropy. What are the units of entropy? 18.5 How does the entropy of a system change for each of the following processes? (a) A solid melts. (b) A liquid freezes. (c) A liquid boils. (d) A vapor is converted to a solid. (e) A vapor condenses to a liquid. (f) A solid sublimes. (g) Urea dissolves in water. Problems 18.6 Referring to the setup in Figure 18.1(a), calculate the probability of all the molecules ending up in the same flask if the number is (a) 6, (b) 60, (c) 600. THE SECOND LAW OF THERMODYNAMICS Review Questions 18.7 State the second law of thermodynamics in words and express it mathematically. Back Forward Main Menu TOC Problems 18.9 For each pair of substances listed here, choose the one having the larger standard entropy value at 25°C. The same molar amount is used in the comparison. Explain the basis for your choice. (a) Li(s) or Li(l ); (b) C2H5OH(l ) or CH3OCH3(l ) (Hint: Which molecule can hydrogen-bond?); (c) Ar(g) or Xe(g); (d) CO(g) or CO2(g); (e) O2(g) or O3(g); (f) NO2(g) or N2O4(g) 18.10 Arrange the following substances (1 mole each) in order of increasing entropy at 25°C: (a) Ne(g), (b) SO2(g), (c) Na(s), (d) NaCl(s), (e) NH3(g). Give the reasons for your arrangement. 18.11 Using the data in Appendix 3, calculate the standard entropy changes for the following reactions at 25°C: (a) S(s) O2(g) 88n SO2(g) (b) MgCO3(s) 88n MgO(s) CO2(g) 18.12 Using the data in Appendix 3, calculate the standard entropy changes for the following reactions at 25°C: (a) H2(g) CuO(s) 88n Cu(s) H2O(g) (b) 2Al(s) 3ZnO(s) 88n Al2O3(s) 3Zn(s) (c) CH4(g) 2O2(g) 88n CO2(g) 2H2O(l ) 18.13 Without consulting Appendix 3, predict whether the entropy change is positive or negative for each of the following reactions. Give reasons for your predictions. (a) 2KClO4(s) 88n 2KClO3(s) O2(g) (b) H2O(g) 88n H2O(l ) (c) 2Na(s) 2H2O(l ) 88n 2NaOH(aq) H2(g) (d) N2(g) 88n 2N(g) 18.14 State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) PCl3(l ) Cl2(g) 88n PCl5(s) (b) 2HgO(s) 88n 2Hg(l ) O2(g) (c) H2(g) 88n 2H(g) (d) U(s) 3F2(g) 88n UF6(s) Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS GIBBS FREE ENERGY 2H2O(g) 34 2H2(g) Review Questions 18.15 Define free energy. What are its units? 18.16 Why is it more convenient to predict the direction of a reaction in terms of Gsys instead of Suniv? Under what conditions can Gsys be used to predict the spontaneity of a reaction? Problems 18.17 Calculate Go for the following reactions at 25°C: (a) N2(g) O2(g) 88n 2NO(g) (b) H2O(l ) 88n H2O(g) (c) 2C2H2(g) 5O2(g) 88n 4CO2(g) 2H2O(l ) (Hint: Look up the standard free energies of formation of the reactants and products in Appendix 3.) 18.18 Calculate Go for the following reactions at 25°C: (a) 2Mg(s) O2(g) 88n 2MgO(s) (b) 2SO2(g) O2(g) 88n 2SO3(g) (c) 2C2H6(g) 7O2(g) 88n 4CO2(g) 6H2O(l ) See Appendix 3 for thermodynamic data. 18.19 From the values of H and S, predict which of the following reactions would be spontaneous at 25°C: Reaction A: H 10.5 kJ, S 30 J/K; reaction B: H 1.8 kJ, S 113 J/K. If either of the reactions is nonspontaneous at 25°C, at what temperature might it become spontaneous? 18.20 Find the temperatures at which reactions with the following H and S values would become spontaneous: (a) H 126 kJ, S 84 J/K; (b) H 11.7 kJ, S 105 J/K. 751 O2(g) 18.27 (a) Calculate G° and KP for the following equilibrium reaction at 25°C. The G° values are 0 for f Cl2(g), 286 kJ/mol for PCl3(g), and 325 kJ/mol for PCl5(g). PCl5(g) 34 PCl3(g) Cl2(g) (b) Calculate G for the reaction if the partial pressures of the initial mixture are PPCl5 0.0029 atm, PPCl3 0.27 atm, and PCl2 0.40 atm. 18.28 The equilibrium constant (KP) for the reaction H2(g) CO2(g) 34 H2O(g) CO(g) o is 4.40 at 2000 K. (a) Calculate G for the reaction. (b) Calculate G for the reaction when the partial pressures are PH2 0.25 atm, PCO2 0.78 atm, PH2O 0.66 atm, and PCO 1.20 atm. 18.29 Consider the decomposition of calcium carbonate: CaCO3(s) 34 CaO(s) CO2(g) Calculate the pressure in atm of CO2 in an equilibrium process (a) at 25°C and (b) at 800°C. Assume that H° 177.8 kJ and S° 160.5 J/K for the temperature range. 18.30 The equilibrium constant KP for the reaction CO(g) Cl2(g) 34 COCl2(g) 35 is 5.62 10 at 25°C. Calculate G° for COCl2 at f 25°C. 18.31 At 25°C, G° for the process H2O(l ) 34 H2O(g) FREE ENERGY AND CHEMICAL EQUILIBRIUM Review Questions 18.21 Explain the difference between G and Go. 18.22 Explain why Equation (18.10) is of great importance in chemistry. is 8.6 kJ. Calculate the “equilibrium constant” for the process. 18.32 Calculate G° for the process C(diamond) 88n C(graphite) Is the reaction spontaneous at 25°C? If so, why is it that diamonds do not become graphite on standing? Problems 18.23 Calculate KP for the following reaction at 25°C: H2(g) G° I2(g) 34 2HI(g) 2.60 kJ Review Questions 18.24 For the autoionization of water at 25°C, H2O(l ) 34 H (aq) OH (aq) Kw is 1.0 10 14. What is G° for the process? 18.25 Consider the following reaction at 25°C: Fe(OH)2(s) 34 Fe2 (aq) Forward Main Menu TOC 18.33 What is a coupled reaction? What is its importance in biological reactions? 18.34 What is the role of ATP in biological reactions? Problems 2OH (aq) Calculate Go for the reaction. Ksp for Fe(OH)2 is 1.6 10 14. 18.26 Calculate G° and KP for the following equilibrium reaction at 25°C. Back THERMODYNAMICS IN LIVING SYSTEMS 18.35 Referring to the metabolic process involving glucose on p. 747, calculate the maximum number of moles of ATP that can be synthesized from ADP from the breakdown of one mole of glucose. 18.36 In the metabolism of glucose, the first step is the con- Study Guide TOC Textbook Website MHHE Website 752 ENTROPY, FREE ENERGY, AND EQUILIBRIUM version of glucose to glucose 6-phosphate: glucose H3PO4 88n glucose 6-phosphate G° H2O 13.4 kJ Since G° is positive, this reaction is not spontaneous. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction and estimate the equilibrium constant for the coupled process. ADDITIONAL PROBLEMS 18.37 Explain the following nursery rhyme in terms of the second law of thermodynamics. Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King’s horses and all the King’s men Couldn’t put Humpty together again. 18.38 Calculate G for the reaction H2O(l ) 34 H (aq) 18.39 18.40 18.41 18.42 18.43 18.44 18.45 Back OH (aq) at 25°C for the following conditions: (a) [H ] 1.0 10 7 M, [OH ] 1.0 10 7 M (b) [H ] 1.0 10 3 M, [OH ] 1.0 10 4 M (c) [H ] 1.0 10 12 M, [OH ] 2.0 10 8 M (d) [H ] 3.5 M, [OH ] 4.8 10 4 M Which of the following thermodynamic functions are associated only with the first law of thermodynamics: S, E, G, and H? A student placed 1 g of each of three compounds A, B, and C in a container and found that after one week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that A, B, and C are totally miscible liquids. Give a detailed example of each of the following, with an explanation: (a) a thermodynamically spontaneous process; (b) a process that would violate the first law of thermodynamics; (c) a process that would violate the second law of thermodynamics; (d) an irreversible process; (e) an equilibrium process. Predict the signs of H, S, and G of the system for the following processes at 1 atm: (a) ammonia melts at 60°C, (b) ammonia melts at 77.7°C, (c) ammonia melts at 100°C. (The normal melting point of ammonia is 77.7°C.) Consider the following facts: Water freezes spontaneously at 5°C and 1 atm, and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy. Ammonium nitrate (NH4NO3) dissolves spontaneously and endothermically in water. What can you deduce about the sign of S for the solution process? Calculate the equilibrium pressure of CO2 due to the decomposition of barium carbonate (BaCO3) at 25°C. Forward Main Menu TOC 18.46 (a) Trouton’s rule states that the ratio of the molar heat of vaporization of a liquid ( Hvap) to its boiling point in kelvins is approximately 90 J/K mol. Use the following data to show that this is the case and explain why Trouton’s rule holds true: tbp(°C) Benzene Hexane Mercury Toluene Hvap(kJ/mol) 80.1 68.7 357 110.6 31.0 30.8 59.0 35.2 (b) Use the values in Table 11.6 to calculate the same ratio for ethanol and water. Explain why Trouton’s rule does not apply to these two substances as well as it does to other liquids. 18.47 Referring to the above problem, explain why the ratio is considerably smaller than 90 J/K mol for liquid HF. 18.48 Carbon monoxide (CO) and nitric oxide (NO) are polluting gases contained in automobile exhaust. Under suitable conditions, these gases can be made to react to form nitrogen (N2) and the less harmful carbon dioxide (CO2). (a) Write an equation for this reaction. (b) Identify the oxidizing and reducing agents. (c) Calculate the KP for the reaction at 25°C. (d) Under normal atmospheric conditions, the partial pressures are PN2 0.80 atm, PCO2 3.0 10 4 atm, PCO 5.0 10 5 atm, and PNO 5.0 10 7 atm. Calculate QP and predict the direction toward which the reaction will proceed. (e) Will raising the temperature favor the formation of N2 and CO2? 18.49 For reactions carried out under standard-state conditions, Equation (18.7) takes the form G° H° T S°. (a) Assuming H° and S° are independent of temperature, derive the equation ln K2 K1 H° R T2 T1 T1T2 where K1 and K2 are the equilibrium constants at T1 and T2, respectively. (b) Given that at 25°C Kc is 4.63 10 3 for the reaction N2O4(g) 34 2NO2(g) H° 58.0 kJ calculate the equilibrium constant at 65°C. 18.50 The Ksp of AgCl is given in Table 16.2. What is its value at 60°C? [Hint: You need the result of Problem 18.49(a) and the data in Appendix 3 to calculate Ho.] 18.51 Under what conditions does a substance have a standard entropy of zero? Can a substance ever have a negative standard entropy? 18.52 Water gas, a mixture of H2 and CO, is a fuel made by reacting steam with red-hot coke (a by-product of Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS coal distillation): H2O(g) C(s) 34 CO(g) duce Ni(CO)4, which is a gas and can therefore be separated from solid impurities: H2(g) From the data in Appendix 3, estimate the temperature at which the reaction becomes favorable. 18.53 Consider the following Brønsted acid-base reaction at 25°C: HF(aq) Cl (aq) 34 HCl(aq) F (aq) (a) Predict whether K will be greater or smaller than unity. (b) Does S° or H° make a greater contribution to G°? (c) Is H° likely to be positive or negative? 18.54 Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see p. 468). What can you deduce about the signs of S and H? 18.55 Consider the thermal decomposition of CaCO3: CaCO3(s) 34 CaO(s) 18.56 18.57 18.58 18.59 Hf ° S° 60.78 kJ/mol 174.7 J/K mol 18.60 The molar heat of vaporization of ethanol is 39.3 kJ/mol and the boiling point of ethanol is 78.3°C. Calculate S for the vaporization of 0.50 mol ethanol. 18.61 A certain reaction is known to have a G° value of 122 kJ. Will the reaction necessarily occur if the reactants are mixed together? 18.62 In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to pro- Main Menu TOC 1 2 O2(g) 18.65 Comment on the statement: “Just talking about entropy increases its value in the universe.” 18.66 For a reaction with a negative G° value, which of the following statements is false? (a) The equilibrium constant K is greater than one, (b) the reaction is spontaneous when all the reactants and products are in their standard states, and (c) the reaction is always exothermic. 18.67 Consider the reaction N2(g) O2(g) 34 2NO(g) Given that G° for the reaction at 25°C is 173.4 kJ, (a) calculate the standard free energy of formation of NO, and (b) calculate KP of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 1100°C, estimate KP for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why? 18.68 Heating copper(II) oxide at 400°C does not produce any appreciable amount of Cu: CuO(s) 34 Cu(s) 77.4 J/K mol S° Forward Given that the standard free energies of formation of CO(g) and Ni(CO)4(g) are 137.3 kJ/mol and 587.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 80°C. Assume that G° is temperature independent. f 18.63 Calculate G° and KP for the following processes at 25°C: (a) H2(g) Br2(l ) 34 2HBr(g) (b) 1 H2(g) 1 Br2(l ) 34 HBr(g) 2 2 Account for the differences in G° and KP obtained for (a) and (b). 18.64 Calculate the pressure of O2 (in atm) over a sample of NiO at 25°C if G° 212 kJ for the reaction 0 (by definition) ° Hf 4CO(g) 34 Ni(CO)4(g) NiO(s) 34 Ni(s) The equilibrium vapor pressures of CO2 are 22.6 mmHg at 700°C and 1829 mmHg at 950°C. Calculate the standard enthalpy of the reaction. [Hint: See Problem 18.49(a).] A certain reaction is spontaneous at 72°C. If the enthalpy change for the reaction is 19 kJ, what is the minimum value of S (in joules per kelvin) for the reaction? Predict whether the entropy change is positive or negative for each of the following reactions: (a) Zn(s) 2HCl(aq) 34 ZnCl2(aq) H2(g) (b) O(g) O(g) 34 O2(g) (c) NH4NO3(s) 34 N2O(g) 2H2O(g) (d) 2H2O2(l ) 34 2H2O(l ) O2(g) The reaction NH3(g) HCl(g) 88n NH4Cl(s) proceeds spontaneously at 25°C even though there is a decrease in disorder in the system (gases are converted to a solid). Explain. Use the following data to determine the normal boiling point, in kelvins, of mercury. What assumptions must you make in order to do the calculation? Hg(g): Back Ni(s) CO2(g) Hg(l ): 753 1 2 O2(g) G° 127.2 kJ However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process and calculate the equilibrium constant for the coupled reaction. 18.69 The internal engine of a 1200 kg car is designed to run on octane (C8H18), whose enthalpy of combustion is 5510 kJ/mol. If the car is moving up a slope, calculate the maximum height (in meters) to which the car can be driven on 1.0 gallon of the fuel. Assume that the engine cylinder temperature is 2200°C and the exit temperature is 760°C, and neglect all forms Study Guide TOC Textbook Website MHHE Website 754 ENTROPY, FREE ENERGY, AND EQUILIBRIUM 18.70 18.71 18.72 18.73 of friction. The mass of 1 gallon of fuel is 3.1 kg. [Hint: See the Chemistry in Action essay on p. 736. The work done in moving the car over a vertical distance is mgh, where m is the mass of the car in kg, g the acceleration due to gravity (9.81 m/s2), and h the height in meters.] A carbon monoxide (CO) crystal is found to have entropy greater than zero at absolute zero of temperature. Give two possible explanations for this observation. (a) Over the years there have been numerous claims about “perpetual motion machines,” machines that will produce useful work with no input of energy. Explain why the first law of thermodynamics prohibits the possibility of such a machine’s existing. (b) Another kind of machine, sometimes called a “perpetual motion of the second kind,” operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then extracting heat from the water, converting the heat to electric power to run the ship, and dumping the water back into the ocean. This process does not violate the first law of thermodynamics, for no energy is created —energy from the ocean is just converted to electrical energy. Show that the second law of thermodynamics prohibits the existence of such a machine. The activity series in Chapter 4 shows that reaction (a) is spontaneous while reaction (b) is nonspontaneous at 25°C: (a) Fe(s) 2H 88n Fe2 (aq) H2(g) (b) Cu(s) 2H 88n Cu2 (aq) H2(g) Use the data in Appendix 3 to calculate the equilibrium constant for these reactions and hence confirm that the activity series is correct. The rate constant for the elementary reaction 2NO(g) O2(g) 88n 2NO2(g) is 7.1 109/M s at 25°C. What is the rate constant for the reverse reaction at the same temperature? 18.74 The following reaction was described as the cause of sulfur deposits formed at volcanic sites (see p. 705): 2H2S(g) SO2(g) 34 3S(s) 2H2O(g) Forward Main Menu TOC Calculate the equilibrium constant (KP) for the reaction. In view of the magnitude of the equilibrium constant, explain why this reaction is not considered a major cause of ozone depletion in the absence of manmade pollutants such as the nitrogen oxides and CFCs? Assume the temperature of the stratosphere to ° be 30°C and Gf to be temperature independent. 18.77 A 74.6-g ice cube floats in the Arctic Sea. The temperature and pressure of the system and surroundings are at 1 atm and 0°C. Calculate Ssys, Ssurr, and Suniv for the melting of the ice cube. What can you conclude about the nature of the process from the value of Suniv? (The molar heat of fusion of water is 6.01 kJ/mol.) 18.78 Comment on the feasibility of extracting copper from its ore chalcocite (Cu2S) by heating: Cu2S(s) 88n 2Cu(s) S(s) Calculate the G° for the overall reaction if the above process is coupled to the conversion of sulfur to sul86.1 kJ/mol. fur dioxide. Given that G°(Cu2S) f 18.79 Active transport is the process in which a substance is transferred from a region of lower concentration to one of higher concentration. This is a nonspontaneous process and must be coupled to a spontaneous process, such as the hydrolysis of ATP. The concentrations of K ions in the blood plasma and in nerve cells are 15 mM and 400 mM, respectively (1 m M 1 10 3 M). Use Equation (18.9) to calculate G for the process at the physiological temperature of 37°C: K (15 mM) 88n K (400 mM) In this calculation, the G° term can be set to zero. What is the justification for this step? 18.80 Large quantities of hydrogen are needed for the synthesis of ammonia. One preparation of hydrogen involves the reaction between carbon monoxide and steam at 300°C in the presence of a copper-zinc catalyst: CO(g) It may also be used to remove SO2 from power-plant stack gases. (a) Identify the type of redox reaction it is. (b) Calculate the equilibrium constant (KP) at 25°C and comment on whether this method is feasible for removing SO2. (c) Would this procedure become more or less effective at a higher temperature? 18.75 Describe two ways that you could measure G° of a reaction. 18.76 The following reaction represents the removal of ozone in the stratosphere: Back 2O3(g) 34 3O2(g) H2O(g) 34 CO2(g) H2(g) Calculate the equilibrium constant (KP) for the reaction and the temperature at which the reaction favors the formation of CO and H2O. Will a larger KP be attained at the same temperature if a more efficient catalyst is used? 18.81 Consider two carboxylic acids (acids that contain the OCOOH group): CH3COOH (acetic acid, Ka 1.8 10 5) and CH2ClCOOH (chloroacetic acid, Ka 1.4 10 3). (a) Calculate G° for the ionization of these acids at 25°C. (b) From the equation Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS G° H° T S°, we see that the contributions to the G° term are an enthalpy term ( H°) and a temperature times entropy term (T S°). These contributions are listed below for the two acids: H° (kJ) T S° (kJ) 0.57 4.7 27.6 21.1 CH3COOH CH2ClCOOH Which is the dominant term in determining the value of G° (and hence Ka of the acid)? (c) What processes contribute to H°? (Consider the ionization of the acids as a Brønsted acid-base reaction.) (d) Explain why T S° is more negative for CH3COOH. 18.82 Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula of C4H10 (see Problem 11.19 on p. 457). Calculate the mole percent of these molecules in an equilibrium mixture at 25°C, given that the standard free energy of formation of butane is 15.9 kJ/mol and that of isobutane is 18.0 kJ/mol. Does your result support the notion that straight-chain hydrocarbons (that is, hy- Back Forward Main Menu TOC 755 drocarbons in which the C atoms are joined along a line) are less stable than branch-chain hydrocarbons? 18.83 A rubber band is stretched vertically by attaching a weight to one end and holding the other end by hand. On heating the rubber band with a hot-air blower, it is observed to shrink slightly in length. Give a thermodynamic analysis for this behavior. (Hint: See the Chemistry in Action essay on p. 748.) 18.84 One of the steps in the extraction of iron from its ore (FeO) is the reduction of iron(II) oxide by carbon monoxide at 900°C: FeO(s) CO(g) 34 Fe(s) CO2(g) If CO is allowed to react with an excess of FeO, calculate the mole fractions of CO and CO2 at equilibrium. State any assumptions. Answers to Practice Exercises: 18.1 (a) Entropy decreases, (b) entropy decreases, (c) entropy increases, (d) entropy increases. 18.2 173.6 J/K, (b) 139.8 J/K, (c) 215.3 J/K. 18.3 (a) S 0, (b) S 0, (c) S 0. 18.4 (a) 106.4 kJ, (b) 2935.0 kJ. 18.5 Sfus 16 J/K mol; Svap 72 J/K mol. 18.6 1.93 1057. 18.7 33 kJ. 18.8 G 0.97 kJ; direction is from right to left. Study Guide TOC Textbook Website MHHE Website ...
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