Chemistry part 30, Julia Burdge,2e (2009) ppsx

734 CHAPTER 18 Entropy, Free Energy, and Equilibrium The reaction generally is the system. Therefore, t.S ~n is t.S ~ys . • Recall that here per mole means per mole of reaction as written [ H~ Section 5.6 ]. • • • • Think About It The results are consistent with the fact that production of gases causes an increase in entropy. In part (a), the increa se in moles of gas gives a large positive value for LlS :'xn ' In part (b), the decrease in mole s of gas gives a large negative value for LlS :'x n- In part (c), where there is no change in the number of moles of gas in the reaction, LlS ~xn is po sitive but is also fairly small. In general, in cases where there is no net change in the number of moles of gas in a reaction , we cannot predict whether LlS :'x n will be positive or negative-but we can predict that it will be a relatively small number. This make s s ense given that gases invariably have greater entropy than liquids and solids. For reactions involving only liquids a nd sol id s, predicting the sign of t. S" is more difficult, but in many such cas es an increase in the total number of molecules and/or ions is accompan ied by an increase in entropy. The standard entropy values of a large number of substances have been measured in 11K· mol. To · caiCi.iiate · tiie' s't'lliidarci"entrcipy ' di.ange · for ' li ' reaction (ilS~n)' we look up the standard entro- pies of the products and reactants and use Equation 18.7. Sample Problem 18.2 demonstrates this approach. • • • •• • • • • • • · • • · . . SampleProble l11 . J8.2 '.' From the s tandard entr o py values in Appendix 2, calculate the st andard entropy changes for the following reaction s at 25°C: (a) CaC0 3 (s) • CaO(s) + CO 2 (g) (b) N 2 (g) + 3H 2 (g) • 2NH 3 (g) (c) H 2 (g) + CI 2 (g) • 2HCl(g) : Strategy Look up standard entropy values a nd u se Equation 18.7 to calculate LlS ~xn ' Just as we did : when we calcula ted sta ndard enthalpies of r eac tion, we consider sto ichiometric coefficients to be · . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ' .' . . . . dimensionless-giving LlS :'x n unit s of J/K . mol. Setup From Appendix 2, SO [CaC0 3 (s)) = 92.9 J /K . mol, SO[CaO(s)] = 39.8 JIK . mol , SO[CO zeg) ] = 213.6 JIK . mol, SO[N 2 (g)] = 191.5 J/K . mol, SO[H 2(g) ] = 131.0 J/K . mol, SO [NH 3(g)] = 193.0 JIK . mol, SO[CI 2 (g)] = 223.0 JIK . mol , and SO [HCl (g)] = 187.0 JIK . mol. Solution (a) M:'xn = [SO( CaO ) + SO (C 0 2 )] - [SO( CaC0 3 ) ] = [(39.8 JIK . mol ) + (213.6 J/K . mol)] - (92.9 J/K . mol ) = 160 .5 JIK . mol (b) M :'x n = [2S0( NH 3 ) ] - [SO (N 2 ) + 3S 0(H2)] = (2)(193.0 JIK . mol ) - [(191.5 J/K . mol ) + (3)(131.0 J/K . mol)] = - 198.5 JIK· mol (c) M:'xn = [2S0( HCl )] - [SO (H 2 ) + SO(CI 2 )] = (2)(187.0 J/K . mol ) - [(131.0 J/K . mol ) + (223.0 JIK . mol)] = 20.0 JIK . mol Practice Problem Calculate the sta ndard entropy change for the following reaction s at 25°C. Predict first whether ea ch one will be po sitive, negative, or too close to call. (a) 2C0zCg) 2CO(g) + 0 2(g) (b) 30 2 (g) • 20 3 (g) • Entropy Changes in the Surroundings Next we discuss how ilS ~ urr is calculated. When an exothermic process takes place in the sys- tem, the heat transferred to the surroundings increases the motion of the molecules in the surroundings. Consequently, there is an increase in the number of microstates 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 by slowing molecular motion. Remember that for constant-pressure processes, the heat released or absorbed, q, is equal to the enthalpy change of the system, ilH sys [ ~~ Sectio n 5.3] . The change in entropy for the surroundings, ilS s um is directly proportional to ilH sy s: The minus sign indicates that a negative enthalpy change in the system (an exothermic process) corresponds to a positive entropy change in the surroundings. For an endothermic process, the enthalpy change in the system is a positive number and corresponds to a negative entropy change in the surroundings. SECTION 18.3 The Second and Third Laws of Thermodynamics 735 In addition to being directly proportional to b.H sys , b.S s urr is inversely propOltional to temperature: Combining the two expressions gives Equation 18.9 With Equations 18.8 and 18.9, we can calculate the entropy change in both the system and sur- roundings for a chemical reaction, and we can determine whether the reaction is spontaneous. Consider the synthesis of ammonia at 25 ° C: b.H~xn = -92 .6 kJ/mol From Sample Problem 18.2( b) , we have b.S ~ y s (- 92.6 kJ/mol) into Equation 18.9, we get -198.5 J/K . mol, and substituting b.H ~ y s A _ -(-92.6 X 1000) J/ mol _ . uS surr - 298 K - 311 J/K mol The entropy change for the universe is = -199 J/K· mol + 311 J/K· mol = 112 J/K . mol Because b.S ~niv is positive, the reaction will be spontaneous at 25° C. Keep in mind, though, 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 or not a reaction will occur spontaneously under specific conditions, but it does not tell us how fast it will occur. Third Law of Thermodynamics Finally, we consider the third law of thermodynamics briefly in connection with the determina- tion of standard entropy. So far we have related entropy to microsta te s the greater the number of microstates a system possesses, the larger is the entropy of the syste m. Consider a perfect crystal- line substance at absolute zero (0 K). Under these conditions, there is essentially no molecular motion and the number of microstates (W) is 1 (there is only one way to arrange the atoms or molecules to form a perfect crystal). From Equation 18.2, we write S = kin W =klnl=O According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero. As temperature increases, molecular motion increases, causing an increase in the number of microstates. Thus, the entropy of any substance at any temperature above 0 K is greater than zero. If the crystalline s ub stance is impure or imperfect in any way, then its entropy is greater than zero even at 0 K because without perfect order there is more than one microstate. The important point about the third law of thermodynamics is that it enables us to determine the absolute entropies of substances. Starting with the knowledge that the entropy of a pure crys- . . . . . . . . ., . . talline substance is zero at 0 K, we can measure the increase in entropy of the substance when it is heated. The change in entropy of a s ub stance, b.S, is the difference between the final and initial entropy values: b.S = Sf - Si where Si is zero if the substance starts at 0 K. Therefore, the measured change in entropy is equal to the absolute entropy at the new temperature. f Altho ugh the complete details of these measurements are beyond the scope of this book, entropy changes are determined in part by measur i ng the heat capacity of a substance [ ~~ Section 5.4] as a function of absolute temperature. 736 CHAPTER 18 Entropy, Free Energy, and Equilibrium Figure 18.4 Entropy increas es in a substance as temperature increases from absolute zero. • Solid Liquid Melting ( ~Sfu s) Boiling (~Sv a p) T em perature (K) Gas The entropy values arrived at in this way are ca lled absolute entropies because they are true values-unlike standard enthalpies of formation, which are derived using an arbitrary reference. Becau se the tabulated values are determined at I atm, we usually refer to absolute entropies as standard entropies, So. Figure 18.4 shows the increase in entropy of a substance as temperature increases from absolute zero. At 0 K, it ha s a zero entropy value (assuming that it is a perfect crystalline subs tance ). As it is heated, its entropy increases gradually at first because of greater molecular motion within the crystal. At the melting point, there is a large increase in entropy as the solid is transformed into the liquid. Further heating increases the entropy of the liquid again due to increa se d molecular motion. At the boiling point, there is a large increase in entropy as a result of the liquid-to-vapor transition. Beyond that temperature, the entropy of the gas continues to increase with increasing temperature. Checkpoint 18.3 The Second and Third Laws of Thermodynamics 18.3.1 U s ing data from Appendix 2, calculate t:.s o (in 1 1K · mol) for the following reaction: 18.3.2 Using data from Appendix 2, calculate t:.s o ( in 11 K· mol ) for the fo ll ow in g reactio n: 2NO(g) + 0 2(g) + . 2NOig) CH 4 (g) + 20ig) +. COig) + 2H 2 0(I) a) 145.3 11 K· mol b) -145.3IIK·mol c) 59.7 J/K . mol d) -59.7I I K· mol e) -421.2IIK· mol a) 107.7 11 K· mol b) -107.7I I K· mol c) 2.6 11 K· mol d) 242.8 11 K· mol e) -2 42.8 11 K· mol Gibbs Free Energy According to the se cond law of thermodynamics, D.S univ > 0 for a spontaneous process. We are usually concerned with and usually measure, however, the properties of the system rather than tho se of the surroundings or those of the universe overall. Ther efore, it is convenient to have a thermodynamic function that enables us to determine whether or not a process is spontaneous by considering the sys tem alone. We begin with Equation 18.5. For a spontaneous process, D.S u niv = D.S sys + D.S surr > 0 SECTION 18.4 Gibbs Free Energy 737 Substituting - fUi s y/T for ~SSUIT> we write ~H s ys ~Suniv = ~S s y s + - T > 0 Multiplying both sides of the equation by T gives T~Suni v = T~S sys - ~H s ys > 0 Now we have an equation that expresses the second law of thermodynamics (and predicts whether or not a proce ss is spontaneous) in terms of only the system. We no longer need to consider the surroundings. For convenience, we can rearrange the preceding equation, multiply through by -1, and replace the> sign with a < sign: - T~Suni v = ~H sys - T~S s y s < 0 According to this equation, a process carried out at constant pressure and temperature is spontaneous if the changes in enthalpy and entropy of the sys tem are such that fUi sys - T~Ssy s is less than zero. To express the spontaneity of a,process more directly, we introduce another thermodynamic function called the Gibbs! free energy (G), or simply free energy. G = H- TS Equation 18.10 Each of the term s in Equation 18.10 pertains to the system. G has units of energy ju st as Hand TS do. Furthermore, like enthalpy and entropy, free energy is a state function. The change in free energy, ~G, of a system for a process that occurs at constant temperature is ~G= ~H- T~S Equation 18.11 Equation 18.11 enables us to predict the spontaneity of a process using the change in enthalpy, the change in entropy, and the absolute temperature. At constant temperature and pressure, for pro- cesses that are spontaneous as written (in the forward direction), ~G is negative. For processes that are not spontaneous as written but that are spontaneous in the reverse direction, ~G is po sitive. For , , systems at equilibrium, ~G is zero. • ~G< 0 • ~G> 0 • ~G = 0 The reaction is spontaneous in the forward direction (and nonspontaneous in the reverse direction). The reaction is nonspontaneous in the forward direction (and spontaneous in the reverse direction). The system is at equilibrium. Often we can predict the sign of ~G for a process if we know the signs of ~H and ~S. Table 18.3 shows how we can use Equation 18.11 to make such predictions. Based on the information in Table 18.3, you may wonder what constitutes a "low" or a "high" temperature. For the example given in the table, O°C is the temperature that divides high from low. Water freezes spontaneously at temperatures below O°C, and ice melts spontaneously at temperatures above O°C. At O°C, a system of ice and water is at equilibrium. The temperature that In this context, free energy is the energy available to do work . Thus, if a particular pr oces s is accompanied by a release of usable energy ( i.e., if LiG is negative), this fact alone guarantees that it is spontaneous, and there is no need to consider what happens to the rest of the universe . When aH Is And as Is aG Will Be And the Process Is Example Negative Positive Negative Positive Positive Negative Negative Positive Negative Positive Negative when T~S < fUi Po sitive when T~S > ~H Negative when T~S > fUi Po sitive when T~S < fUi Always spo nt aneous Always nonspontaneous Spontaneous at low T Nonspontaneous at high T Spontaneous at high T Nonspontaneous at low T 2H 2 0 2 (aq) +. 2H 2 0(l) + °z(g) 30 z (s) • 20 3 (g) • Hz0(l) • HzO(s) (freezing of water) 2HgO(s) +. 2Hg(l) + 0 z(g) 1. Jo siah Willard Gibbs (1839-1903). American physicist. One of the founders of thermodynamics. Gibbs was a modest and private individual who spent almost a ll hi s professional life at Yale Universi ty. Because he published most of his work in obscure journals, Gibbs never ga ined the eminence that hi s contemporary and admirer Jam es Maxwell did. Even today, ve ry few people o ut side of chemistry and physics have ever heard of Gibbs. 738 CHAPTER 18 Entropy, Free Energy, and Equilibrium • Think About It Spontaneity is favored by a release of energy (tlH b ei ng negative) and by an increase in entropy (tlS being positive). When both quantities are positive, as in this cas e, only the entropy change favors spontaneit y. For an endothermic process such as thi s, which requires the input of he at, it should make sense that adding more heat by increas in g the temperature will shift the equilibrium to the right, thus making it "more spontaneous." divides "high" from "low" depends, though, on the individual reaction. To determine that tempera- ture, we must set t1G equal to 0 in Equation 18.11 (i.e., the equilibrium condition): 0= t1H - Tt1S Rearranging to solve for T yields T= t1H t1S The temperature that divides high from low for a particular reaction can now be calculated if the values of t1H and t1S are known. Sample Pr obl em 18 .3 demonstrates the use of this approach . Sample Problem 18.3 According to Table IS.3, a reaction will be spontaneous only at hi gh temperatures if both tlH and tlS are positive. For a reaction in w hi ch tlH = 199.5 kJ/mol and tlS = 476 J /K . mol, determine the temperature (in 0 c) above which the reaction is spontaneous. Strategy The temperature that divides high from low is the temperature at which tlH = TtlS ( tlG = 0). Therefore, we use Equation IS.11, substituting 0 for tlG and solving for T to determine temperature in kelvins; we th en convert to degrees Celsius. Setup Solution tlS = ( 476 J ) ( 1 kJ ) = 0.476 kJ/K . mol mol, K 1000 J T = D.H = 199.5 kl l mol = 419 K tlS 0.476 kllK . mol = (419 - 273) = 146°C Practice Problem A reaction will be spontaneous only at low temperatures if both tlH and tlS are negative. For a reaction in which tlH = - 3S0. 1 kJ/mol and tlS = -95.001IK· mol, determine the temperature (in 0 c) below which the reaction is spontaneous. Standard Free-Energy Changes The introduction of the term ~G O enabl es us to The " sia ' ndardfree~eneiijY ' of'riiaction ' (t1G~X ll ) is the free energy change for a reaction when it write Equation 18.11 as occurs under standard-state conditions-that is, when reactants in their standard states are con- ~G O = M ? - T/:,S verted to products in their standard states. The conventions used by chemists to define the standard states of pure substances and solutions are • Ga ses • Liquids • Solids • Elements • Solutions 1 atm pressure Pure liquid Pure solid The mo st stable allotropic form at 1 atm and 25 °C 1 molar concentration To calculate t1G ~xn ' we start with the general equation aA + bB -_. cC + dD The standard free-energy change for this reaction is given by Equation 18.12 t1G ~xn = [ct1G r (C) + dt1G r (D)) - [at1G r (A) + bt1G r (B)) Equation 18.12 can be generalized as follows: Equation 18.13 t1G ~xn = lnt1G r ( pr oducts) - lmt1G r (reactants) • • SECTION 18.4 Gibbs Free Energy 739 where m and n are stoichiometric coefficients. The term !1G J is the standard free energy of for- mation of a compound that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its constituent elements, each in its standard state. For the combustion of graphite, the standard free-energy change (from Equation 18.13) is !1G~xn = [!1G H C0 2 )] - [!1G ~ (C, graphite) + !1G ~ (02)] As with standard enthalpy of formation, the standard free energy of formation of any element (in its most stable allotropic form at 1 atm) is defined as zero. Thus, !1GJ?(C, graphite) = 0 and Therefore, the standard free-energy change for the reaction in this case is equal to the standard free energy of formation of CO 2 : !1G ~xn = !1Gf(C0 2 ) Appendix 2 lists the values of !1G J? at 25°C for a number of compounds. Sample Problem 18.4 demonstrates the calculation of standard free-energy changes. Sample Problem 18.4 Calculate the standard free-energy changes for the following re actions at 25°C: (a) CH 4 (g) + 20z(g) • COz(g) + 2H 2 0 (I) (b) 2MgO(s) • 2Mg(s) + Oz(g) Strategy Look up the !:J.G 'f values for the reactants and products in each equation, and use Equation 18.13 to solve for !:J.G ~x n' Setup From Appendix 2, we have the following values: !:J.Gf[CH 4 (g)] = -5 0.8 kllmol , !:J.Gf[COz(g)] = -3 94.4 kllmol, !:J.Gf[H 2 0(l)] = -237.2 kJ/mol, and !:J.Gf[MgO(s)] = - 569.6 kl lmo!. All the other s ubstance s are elements in their standard states and have, by definition, !:J.G'f = O. Solution (a) !:J.G ~n = (!:J.Gf[C0 2 (g)] + 2!:J.Gf[H 2 0( I)]) - (!:J.Gf[CHig)] + 2!:J.Gf[02(g)]) = [( -39 4.4 kllmol) + (2)( -237.2 kJ/mol)] - [( -50.8 kl lmol) + (2)(0 kl lmo l)] = - 818.0 kl l mol (b) !:J.G ~xn = (2!:J.G'f[Mg(s)] + !:J.Gf[02(g)]) - (2 !:J.G f[ MgO (s)]) = [(2)(0 kllmol ) + (0 kllmol)] - [(2)(-569.6 kJ/mol)] = 1139 kJ/mol Practice Problem Calculate the standard free-energy changes for the following reactions at 25°C: (a) H 2 (g) + Br2(l) +. 2HBr(g) (b) 2C 2 H 6 (g) + 70 2 (g) • 4COzeg) + 6H 2 0(I) Using L1G and L1G O to Solve Problems It is the sign of !1G, the free-energy change, not the sign of !1G o, the standard free-energy change, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . that indicates whether or not a process will occur spontaneously under a given set of conditions. What the sign of !1G o tells us is the same thing that the magnitude of the equilibrium constant (K) tells us [ ~~ Section 15.2]. A negative !1G o value corresponds to a large K value (products favored at equilibrium), whereas a positive !1G o value corresponds to a small K value (reactants favored at equilibrium). Like equilibrium constants, !1G o values change with temperature. One of the uses of Equa- tion 18.11 is to determine the temperature at which a particular equilibrium will begin to favor a desired product. For example, calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic substance with a variety of industrial uses, including water treatment and Think About It Note that, like st andard enthalpies of formation ( DJI f), standard free energies of formation (!:J.G7) depend on the state of matter. Using water as an example, !:J.G f[H 2 0(l)] = -237.2 kl l mol and !:J.G f[H 2 0(g)] = -228.6 kJ/mo!. Always double-check to make sure you have se lected the right value from the table. , The sign of !:J.G O does indicate whether or not a process is spontaneous when all reactants a na products are in their standard states, but this is very seldom the case. 740 CHAPTER 18 Entropy, Free Energy, and Equilibrium • pollution control. It is prepared by heating lime stone (CaC0 3 ), which decompose s at a high temperature: The reaction is reversible, and under the right conditions, CaO and CO 2 readily recombine to form CaC0 3 again. To pr event this from happening in the industrial preparation, the system is never maintained at equilibrium; rather, CO 2 is constantly removed as it forms, shifting the equilibrium from left to right, thus promoting the formation of calcium oxide. An important piece of information for the chemist responsible for maximizing CaO produc- tion is the temperature at which the decomposition equilibrium of CaC0 3 begins to favor products . We can mak e a reliable estimate of that temperature as follows. First we calculate b H o and b S o for the reaction at 2S o C, using the data in Appendix 2. To determine b H o, we apply Equation S .1 9: b W = [b H t' (CaO) + b Ht'(C0 2 )] - [b Ht'(CaC0 3 )] = [( -63S.6 kJ/mol) + (- 393.S kJ/mol)] - (-1206.9 kJ/mol) = 177.8 kJ/mol Next we apply Equation 18.8 to find b S o: b S o = [SO(CaO) + SOCO?)] - SO (CaC0 3 ) = [(39.8 JIK . mol) + (213.6 J/K . mol)] - (92.9 J/K . mol) = 160.S J/K . mol From Equation 18.11, we can write and we obtain . . . . . . . . . . . . . . . . . . . . . . . Be carefu l with units in problems of this type. b G o = (177.8 kJ/mol) - (298 K)(0.160S kJ/K . mol) 5" val ue s are ·tabulated using joules, whe r eas Ll.H \' value s are tabulated using kilojoules. = 130.0 kJ/mol Becau se b G o is a large positive numbe r, the reaction does not favor product formation at 2S oC (298 K). And, becau se b H o and b S o are both positive, we know that b G o will be negative (product formation will be favored) at high temperatures. We can determine what constitutes a high tem- perature for this reaction by calculating the temperature at which b G o is zero. or (177.8 kJ/mol)(1000 JlkJ) 0.160S kJ/K . mol = 1108 K (83S°C) • At temperatures higher than 83S o C, b G o becomes negative, indicating that the reaction would then favor the formation of CaO and CO? At 840°C (1113 K), for example, b G o = b H o - Tb S o = 177.8 kJ/mol - (1113 K)(0.160S kJ/K . mol) ( l~okg J = -0,8 kJ/mol At still higher temperatures, b G o becomes increasingly negative, thus favoring product formation even more. Note that in this example we used the b H o and b S o values at 2S oC to calculate changes to b G o at much higher temperatures. Because both b H o and b S o actually change with tempera- ture, this approach does not give us a truly accurate value for b G o, but it does give us a reasonably good estimate. Equation 18.11 can also be used to calculate the change in entropy that accompanies a phase change. At the temperature at which a phase change occurs (i.e., the melting point or boiling point of a substance), the system is at equilibrium (b G = 0). Therefore, Equation 18.11 becomes SECTION 18.4 G ibbs Free Energy 741 or 0= tlH - TtlS tlS = tlH T Consider the ice-water equilibrium. For the ice-to-water transition, tlH is the molar heat of fusion (see Table 12.8) and T is the melting point. The entropy change is therefore 6010 I lmo l tlS ic e +. wat er = 273 K = 22.0 I lK · mol Thus, when 1 mole of ice melts at OD C, there is an increase in entropy of 22.0 Il K· mol. The increase in entropy is consistent with the increase in microstates from solid to liquid. Conversel y, for the water-to-ice transition, the decrease in entropy is given by • tlS . = -6010 Il mol = -220 I lK · I water • Ice 273 K . mo The same approach can be applied to the water-to-steam transition. In this case, tlH is the heat of vaporization and T is the boiling point of water. Sample Problem 18.5 examines the phase transi- tions in benzene. Sample Problem 18.S The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ /mol, respectively. Calculate the entropy changes for the solid-to-liquid and liquid-to-vapor transitions for benzene. At 1 atm pr essure, benzene melts at 5.5°C and boils at 80.1 DC. Strategy The solid-liquid transition at the melting point and the liquid-vapor transition at the boiling point are equilibrium processes. Therefore, because D.G is zero at equilibrium, in each case we can use Equation 18.11, substituting 0 for D.G and solving for D.S, to determine the entropy change associated with the process. Setup The melting point of benzene is 5.5 + 273.15 = 278.7 K and the boiling point is 80 .1 + 273.15 = 353.3 K. Solution D.S fu s = D.H ru s T melting 10.9 kJ /mol 278.7 K = 0.0391 kJ /K . molar 39 .1 J/K . mol D.H v ap D. S y ap = - '- T boil in g 31.0 kJ/mol 353.3 K = 0.0877 kJ /K . mol or 87.7 J IK . mol Practice Problem The molar heats of fu sion and vap or ization of argon are 1.3 and 6.3 kJ /mol, respectivel y, and argon's melting point and boiling point are -190 °C a nd -186 °C, respectivel y. Calculate the entropy changes for the fusion and vaporization of argon. Checkpoint 18.4 Gibbs Free Energy 18.4.1 A reaction for which D.H and D.S are both negative is a) nonspontaneous at all temperatures b) spontaneous at a ll temperatures c) spontaneous at hi gh temperatures d) spontaneous at low temperatures e) at equilibrium 18.4.2 At what t empe rature (i n 0 c) does a re act ion go from being nonspontaneous to spontaneous if it has D.H = 171 kl lmol and D.S = 161 J/K . mol? a) 270°C b) 670°C c) 1100°C d) 790°C e) 28°C Think About It For the same substance, D.S v ap is always significantly larger than D.S fus ' The change in number of microstates is always bigger in a liquid-to-gas transition than in a solid-to-liquid transition. • 742 CHAPTER 18 Entropy, Free Energy, and Equilibrium • Eve n for a react ion that s tarts with all reactan ts and products in their standard states, as soon as the reaction begin s, the conc entra tions of all species change and standard- st ate conditions no longer exist. The Q used in Equation 18 .1 4 can be eith er Q, (for reactions that take plac e in so lution) or Qp (for reactions that take place in the gas ph as e ). • • • • • 18.4 .3 Using data from Appendix 2, calculate LlG o ( in kJ/mol) at 25°C for the reaction: CH 4 (g) + 20 2 (g) • CO 2 (g) + 2H 2 0(l ) a) -580 .8 kJ /mol b) 580.8 kJ/mol c) -572.0 kJ/mol d) 572.0 kJ/mol e) -818.0 kJ /mol 18.4.4 Calculate LlS vap (in J/K . mol) for the vaporization of bromine: Br 2 (l) -_. Br 2(g) LlH vap = 31 kJ /mol, and the boiling point of bromine is 59°C. a) 9311K'mol b) 0.53 11K· mol c) 11 11 K· mol d) l.0 X 10 4 J /K . mol e) 2. 1 X 10 2 11K . mol Free Energy and Chemical Equilibrium Reactants and products in a chemical reaction are almost always in something other than their standard state s- that is, solutions usually have concentrations other than 1 M and gases usually have pre ssures other than 1 atm. To determine whether or not a reaction is spontaneous, there- fore, we mu st take into account the actual concentrations and/or pressures of the species involved. And although we can determine t:: G o from tabulated values, we need to know t:: G to determine spontaneity. Relationship Between LlG and LlGO The relationship between t:: G and t:: G o, which is derived from thermodynamics, is Equation 18.14 t:: G = t:: G o + RTln Q where R is the gas constant (8.314 11K· mol or 8.314 X 10 - 3 kllK . mol), T is the absolute tem- , pe ' ratiiie ' at w hI ch' the ' reac"ii"on ' takes ' pi ace : and 'Q is the reaction quotient [ ~~ Secti on 15.2] . Thus, t:: G depends on two terms: t:: G o and RT In Q. For a given reaction at temperature T, the value of t:: G o is fixed but that of RT In Q can vary because Q varies according to the composition of the reaction mixture. Con sider the following equilibrium: Using Equation 18.13 and information from Appendix 2, we find that t:: G o for this reaction at 25°C is 2.60 kl lmol. The value of t:: G, however, depends on the pressures of all three gaseous spe- cies. If we start with a mixture of gases in which P H2 = 2.0 atm, PIz = 2.0 atm, and PHI = 3.0 atm, the reaction quotient, Qp, is (P HI )2 Qp = ( PHz)(PIz ) = 2.25 Using this value in Equation 18 .14 gives (3.0)2 9.0 (2.0)(2.0) 4.0 t:: G = 2.60 kl + mol 8.314 X 10 - 3 kl (298 K)(ln 2.25) K'mol = 4.3 kllmol Because t:: G is positive, we conclude that, starting with the se concentrations, the forward reaction will not occur spontaneously as written. Instead, the reverse reaction will occur spontaneously and the sys tem will reach equilibrium by consuming part of the HI initially present and producing more H 2 and I 2 . If, on the other hand, we start with a mixture of gases in which PH = 2.0 atm, PI = 2.0 atm, 2 2 and PHI = 1.0 atm, the reaction quotient, Qp, is (P HI )2 Qp = (PHz)(PIz) = 0.25 (1.0)2 (2.0)(2.0) 1 - - 4 SECTION 18.5 Free Energy and Chemical Equilibrium 743 Using this value in Equation 18.14 gives AG = 2.6 kJ + 8.314 X 10- 3 kJ (298 K)(ln 0.25) mol K . mol = - 0.8 kJ/mol With a negative value for AG, the reaction will be spontaneous as written in the forward direc- tion. In this case, the system will achieve equilibrium by consuming so me of the H2 and 12 to produce more HI. Sample Problem 18.6 uses AGo and the reaction quotient to determine in which direction a reaction is spontaneous . • Sample Problem 18.6 The equilibrium constant, K p , for th e reaction is 0.1l3 at 298 K, which corresponds to a standard free-energy change of 5.4 kllmo!. In a certain experiment, the initial pressures are P N 0 = 0.453 atm and P NO = 0.122 atm. Calculate 6.G for the 2 4 2 reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium. Strategy Use the partial pressures of N 2 0 4 and N0 2 to calculate the reaction quotient Qp, and then use Equation 18.14 to calculate 6.G. Setup The reaction quotient expression is Solution (P NO / (0. 122l Qp = = = 0.0329 P N 0 0.453 2 4 6.G = 6.Go + RT In Qp = 5.4 kl + (8.3 14 X 10- 3 kJ (298 K)(In 0.0329) mol K· mol = 5.4 kl lmol - 8.46 kl lmol = -3 .1 kJ/mol Because 6.G is negative, the reaction proceeds spontaneously fr om left to right to rea ch equilibrium. Practice Problem A 6.G o for the reaction HzCg) + 1 2 (g) :;:::, ~. 2HI(g) is 2.60 kllmol at 25°C. Calculate 6.G, and predict the direction in which the reaction is s pont aneous if the starting concentrations are P H2 = 3.5 atm, P I2 = 1.5 a tm , and PHI = 1.75 atm. Practice Problem B What is the minimum partial pressure of 12 required for the preceding reaction to be spontaneous in the forward direction at 25°C if the partial pressures of H2 and HI are 3.5 and 1.75 atm, respectively? Relationship Between ~Go and K By definition, AG = 0 and Q = K at equilibrium, where K is the equilibrium constant. Thus, AG = AGO + RTln Q (Equation 18.14) become s Think About It Remember, a reaction with a positive 6.G o value can be spontaneous if the starting concentrations of reactants and pr oducts are such that Q < K. • - . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 = AG O + RTlnK or AG O = - RTln K Equation 18.15 In this equation and the one that f ollows , K IS _ for reaction s that take place in solution an d Ko f or reactions that take place in the gas p hase . [...]... CI 2(g) (b) Calculate!::"G for the reaction if the partial press ures of the initial mixture are PPC1 5 = 0.0029 atm, PPC1, = 0.27 atm, and PC!, = 0.40 atm 18.40 The equilibrium constant (Kp) for the reaction H 2(g) + COzCg) :;:.==' H zO (g) + CO(g) is 4.40 at 2000 K (a) Calculate!::"GO for the reaction (b) Calculate!::"G for the reaction when the partial pressures are PH, = 0.25 atm, Peo2 = 0.78 atm,... nitrites to nitrates: Calculate!::"Go for the following reactions at 25 °C: Review Questions 18.32 Explain the difference between !::"G and !::"Go 18.33 Explain why Equation 18.15 is of great importance in chemistry 18.34 Fill in the missing entries in the following table: (a) 2Mg(s) + 02(g) - 2MgO(s) (b) 2S0zCg) + 02(g) • 2S0 3(g) (c) 2C 2H 6(g) + 70 z(g ) • 4C0 2(g) + 6H zO(I) K < 1 !::"Go InK 0 (See... to the actual free-energy change, !1G The actual free-energy change of the system varies as the reaction progresses and becomes zero at equilibrium On the other hand, !1Go, like K, is a constant for a particular reaction at a given temperature Figure 18.5 shows plots of the free energy of a reacting system versus the extent of the reaction for two reactions Table 18.4 summarizes the relationship between... carbon dioxide (C0 2 ) (a) Write an equation for this reaction (b) Identify the oxidizing and reducing agents (c) Calculate the Kp for the reaction at 25°e (d) Under normal atmospheric conditions, the partial pressures are P N2 = 0.80 atm, P eo, = 3.0 X 10- 4 atm, P eo = 5.0 X 10- 5 atm, and P NO = 5.0 X 10- 7 atm Calculate Qp, and predict the direction toward which the reaction will proceed (e) Will... Prom the data in Appendix 2, estimate the temperature at which the reaction begins to favor the formation of products 18.66 + Cnaq) 18.67 ' HBr(g) Account for the differences in !:1Go and Kp obtained for parts (a) and (b) ==' HCI (aq) + P-(aq) The pH of gastric juice is about 1.00 and that of blood plasma is 7.40 Calculate the Gibbs free energy required to secrete a mole of H + ions from blood plasma to... that the crystal does not have a perfect arrangement of the CO molecules (a) What would be the residual entropy if the arrangement were totally random? (b) Comment on the difference between the result in part (a) and 4.2 J/K mol (Hint: Assume that each CO molecule has two choices for Olientation, and use Equation 18.2 to calculate the residual entropy.) 18.11 0 Comment on the COlTectness of the analogy... signs of t::.H, t::.S, and t::.G when hydrogen gas is adsorbed onto the surface of Ni metal = 1173 K T J = 872 K 1 I I J liT PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES • In chemistry, the standard state for a solution is 1 M This means that each solute concentration expressed in molarity is divided by 1 M In biological systems, however, we define the standard state for the... , Electrolysis Electrolysis of Molten Sodium Chloride Electrolysis of Water Electrolysis of an Aqueous Sodium Chloride Solution Quantitative Applications of Electrolysis Corrosion • •• < The Electrochemistry of Dental Pain Anyone who accidentally bites on a piece of aluminum foil in such a way that the foil touches a dental filling will experience a momentary sharp pain The pain is actually the result... commonly used to fill cavities is known as dental amalgam (An amalgam is a substance made by combining mercury with one or more other metals.) Dental amalgam consists of liquid mercury mixed in roughly equal parts with an alloy powder containing silver, tin, copper, and sometimes smaller amounts of other metals such as zinc off the mark.com ~~(:,t tA'ic) y\! \ \~G What happens is this: the aluminum and the... flow This current stimulates the nerve of the tooth, causing a very unpleasant sensation o ALUt-1ft-lUM FOil o The pain caused by having aluminum foil contact an amalgam filling is the result of electrochemistry Used with permission of Atlantic Feature Syndicate for Mark Parisi In This Chapter, You Will learn how chemical reactions can produce electric energy and how reaction conditions affect the energy . gases causes an increase in entropy. In part (a), the increa se in moles of gas gives a large positive value for LlS :'xn ' In part (b), the decrease in mole s of. Practice Problem B What is the minimum partial pressure of 12 required for the preceding reaction to be spontaneous in the forward direction at 25°C if the partial pressures of H2 and HI are. these measurements are beyond the scope of this book, entropy changes are determined in part by measur i ng the heat capacity of a substance [ ~~ Section 5.4] as a function