Lecture Power Point Chemistry The Molecular Nature of

Lecture Power. Point Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 20 -1 Copyright The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 20 Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 20 -2

Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 20. 1 The Second Law of Thermodynamics: Predicting Spontaneous Change 20. 2 Calculating Entropy Change of a Reaction 20. 3 Entropy, Free Energy, and Work 20. 4 Free Energy, Equilibrium, and Reaction Direction 20 -3

Spontaneous Change A spontaneous change is one that occurs without a continuous input of energy from outside the system. All chemical processes require energy (activation energy) to take place, but once a spontaneous process has begun, no further input of energy is needed. A nonspontaneous change occurs only if the surroundings continuously supply energy to the system. If a change is spontaneous in one direction, it will be nonspontaneous in the reverse direction. 20 -4

The First Law of Thermodynamics Does Not Predict Spontaneous Change Energy is conserved. It is neither created nor destroyed, but is transferred in the form of heat and/or work. DE = q + w The total energy of the universe is constant: DEsys = -DEsurr or DEsys + DEsurr = DEuniv = 0 The law of conservation of energy applies to all changes, and does not allow us to predict the direction of a spontaneous change. 20 -5

DH Does Not Predict Spontaneous Change A spontaneous change may be exothermic or endothermic. Spontaneous exothermic processes include: • freezing and condensation at low temperatures, • combustion reactions, • oxidation of iron and other metals. Spontaneous endothermic processes include: • melting and vaporization at higher temperatures, • dissolving of most soluble salts. The sign of DH does not by itself predict the direction of a spontaneous change. 20 -6

Figure 20. 1 A spontaneous endothermic chemical reaction. Ba(OH)2 · 8 H 2 O(s) + 2 NH 4 NO 3(s) → Ba 2+(aq) + 2 NO 3 -(aq) + 2 NH 3(aq) + 10 H 2 O(l) DH°rxn = +62. 3 k. J water This reaction occurs spontaneously when the solids are mixed. The reaction mixture absorbs heat from the surroundings so quickly that the beaker freezes to a wet block. 20 -7

Freedom of Particle Motion All spontaneous endothermic processes result in an increase in the freedom of motion of the particles in the system. solid → liquid → gas crystalline solid + liquid → ions in solution less freedom of particle motion localized energy of motion more freedom of particle motion dispersed energy of motion A change in the freedom of motion of particles in a system is a key factor affecting the direction of a spontaneous process. 20 -8

Microstates and Dispersal of Energy • Just as the electronic energy levels within an atom are quantized, a system of particles also has different allowed energy states. • Each quantized energy state for a system of particles is called a microstate. – At any instant, the total energy of the system is dispersed throughout one microstate. • At a given set of conditions, each microstate has the same total energy as any other. – Each microstate is therefore equally likely. • The larger the number of possible microstates, the larger the number of ways in which a system can disperse its energy. 20 -9

Entropy The number of microstates (W) in a system is related to the entropy (S) of the system. S = k ln. W A system with fewer microstates has lower entropy. A system with more microstates has higher entropy. All spontaneous endothermic processes exhibit an increase in entropy. Entropy, like enthalpy, is a state function and is therefore independent of the path taken between the final and initial states. 20 -10

Figure 20. 2 Spontaneous expansion of a gas. When the stopcock is opened, the gas spontaneously expands to fill both flasks. Increasing the volume increases the number of translational energy levels the particles can occupy. The number of microstates – and the entropy – increases. 20 -11

Figure 20. 3 The entropy increase due to the expansion of a gas. Opening the stopcock increases the number of possible energy levels, which are closer together on average. More distributions of particles are possible. 20 -12

Figure 20. 4 Expansion of a gas and the increase in number of microstates. When the stopcock opens, the number of microstates is 2 n, where n is the number of particles. 20 -13

DS for a Reversible Process DS = qrev T A reversible process is one that occurs in such tiny increments that the system remains at equilibrium, and the direction of the change can be reversed by an infinitesimal reversal of conditions. 20 -14

Figure 20. 5 Simulating a reversible process. A sample of Ne gas is confined to a volume of 1 L by the “pressure” of a beaker of sand on the piston. We remove one grain of sand (an “infinitesimal” decrease in pressure), causing the gas to expand a tiny amount. The gas does work on its surroundings, absorbing a tiny increment of heat, q, from the heat reservoir. This simulates a reversible process, since it can be reversed by replacing the grain of sand. 20 -15

The Second Law of Thermodynamics The sign of ΔS for a reaction does not, by itself, predict the direction of a spontaneous reaction. If we consider both the system and the surroundings, we find that all real processes occur spontaneously in the direction that increases the entropy of the universe. For a process to be spontaneous, a decrease in the entropy of the system must be offset by a larger increase in the entropy of the surroundings. DSuniv = DSsys + DSsurr > 0 20 -16

Comparing Energy and Entropy The total energy of the universe remains constant. DEsys + DEsurr =DEuniv = 0 DH is often used to approximate. DE. For enthalpy there is no zero point; we can only measure changes in enthalpy. The total entropy of the universe increases. DSuniv =DSsys +DSsurr > 0 For entropy there is a zero point, and we can determine absolute entropy values. 20 -17

The Third Law of Thermodynamics A perfect crystal has zero entropy at absolute zero. Ssys = 0 at 0 K A “perfect” crystal has flawless alignment of all its particles. At absolute zero, the particles have minimum energy, so there is only one microstate. S = k ln. W = k ln 1 = 0 To find the entropy of a substance at a given temperature, we cool it as close to 0 K as possible. We then heat in small increments, measure q and T, and calculate DS for each increment. The sum of these DS values gives the absolute entropy at the temperature of interest. 20 -18

Standard Molar Entropies S° is the standard molar entropy of a substance, measured for a substance in its standard state in units of J/mol·K. The conventions for defining a standard state include: • 1 atm for gases • 1 M for solutions, and • the pure substance in its most stable form for solids and liquids. 20 -19

Factors Affecting Entropy • Entropy depends on temperature. – For any substance, S° increases as temperature increases. • Entropy depends on the physical state of a substance. – S° increases as the phase changes from solid to liquid to gas. • The formation of a solution affects entropy. • Entropy is related to atomic size and molecular complexity. – Remember to compare substances in the same physical state. 20 -20

Figure 20. 6 A Visualizing the effect of temperature on entropy. Computer simulations show each particle in a crystal moving about its lattice position. Adding heat increases T and the total energy, so the particles have greater freedom of motion, and their energy is more dispersed. S therefore increases. 20 -21

Figure 20. 6 B Visualizing the effect of temperature on entropy. At any T, there is a range of occupied energy levels and therefore a certain number of microstates. Adding heat increases the total energy (area under the curve), so the range of occupied energy levels becomes greater, as does the number of microstates. 20 -22

Figure 20. 6 C Visualizing the effect of temperature on entropy. A system of 21 particles occupy energy levels (lines) in a box whose height represents the total energy. When heat is added, the total energy increases and becomes more dispersed, so S increases. 20 -23

Figure 20. 7 20 -24 The increase in entropy during phase changes from solid to liquid to gas.

Figure 20. 8 The entropy change accompanying the dissolution of a salt. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. pure solid MIX pure liquid solution The entropy of a salt solution is usually greater than that of the solid and of water, but it is affected by the organization of the water molecules around each ion. 20 -25

Figure 20. 9 The small increase in entropy when ethanol dissolves in water. Ethanol (A) and water (B) each have many H bonds between their own molecules. In solution (C) they form H bonds to each other, so their freedom of motion does not change significantly. 20 -26

Figure 20. 10 20 -27 The entropy of a gas dissolved in a liquid.

Entropy and Atomic Size S° is higher for larger atoms or molecules of the same type. Li Atomic radius (pm) 152 Na K Rb Cs 186 227 248 265 Molar mass (g/mol) 6. 941 22. 99 39. 10 85. 47 132. 9 S°(s) 29. 1 51. 4 64. 7 69. 5 85. 2 HF HCl Molar mass (g/mol) S°(s) 20 -28 20. 01 173. 7 36. 46 186. 8 HBr 80. 91 198. 6 HI 127. 9 206. 3

Entropy and Structure For allotropes, S° is higher in the form that allows the atoms more freedom of motion. S° of graphite is 5. 69 J/mol·K, whereas S° of diamond is 2. 44 J/mol·K. For compounds, S° increases with chemical complexity. S°(s) S°(g) Na. Cl Al. Cl 3 72. 1 167 P 4 O 10 NO NO 2 N 2 O 4 211 240 304 229 These trends only hold for substances in the same physical state. 20 -29

Figure 20. 11 20 -30 Entropy, vibrational motion, and molecular complexity.

Sample Problem 20. 1 Predicting Relative Entropy Values PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: (a) 1 mol of SO 2(g) or 1 mol of SO 3(g) (b) 1 mol of CO 2(s) or 1 mol of CO 2(g) (c) 3 mol of O 2(g) or 2 mol of O 3(g) (d) 1 mol of KBr(s) or 1 mol of KBr(aq) (e) seawater at 2°C or at 23°C (f) 1 mol of CF 4(g) or 1 mol of CCl 4(g) PLAN: In general, particles with more freedom of motion have more microstates in which to disperse their kinetic energy, so they have higher entropy. Raising the temperature or having more particles increases entropy. 20 -31

Sample Problem 20. 1 SOLUTION: (a) 1 mol of SO 3(g). For equal numbers of moles of substances with the same types of atoms in the same physical state, the more atoms in the molecule the higher the entropy. (b) 1 mol of CO 2(g). For a given substance, entropy increases as the phase changes from solid to liquid to gas. (c) 3 mol of O 2(g). The two samples contain the same number of oxygen atoms but different numbers of molecules. Although each O 3 molecule is more complex than each O 2 molecule, the greater number of molecules dominates because there are many more microstates possible for 3 mol of particles than for 2. 20 -32

Sample Problem 20. 1 (d) 1 mol of KBr(aq). The two samples have the same number of ions, but their motion is more limited and their energy less dispersed in the solid than in the solution. An ionic substance in solution usually has a higher entropy than the solid. (e) Seawater at 23°C. Entropy increases with rising temperature. (f) 20 -33 1 mol of CCl 4(g). For similar compounds, entropy increases with molar mass.

Entropy Changes in the System The standard entropy of reaction, DS°rxn, is the entropy change that occurs when all reactants and products are in their standard states. DS°rxn = Sm. S°products - Sn. S°reactants where m and n are the amounts (mol) of products and reactants, given by the coefficients in the balanced equation. 20 -34

Predicting the Sign of DS°rxn We can often predict the sign of DS°rxn for processes that involve a change in the number of moles of gas. • DS°rxn is positive if the amount of gas increases; – e. g. , H 2(g) + I 2(s) → 2 HI(g). • DS°rxn is negative if the amount of gas decreases; – e. g. , N 2(g) + 3 H 2(g) • 20 -35 2 NH 3(g) DS° is likely to be positive if a new structure forms that has more freedom of motion.

Sample Problem 20. 2 Calculating the Standard Entropy of Reaction, DS°rxn PROBLEM: Predict the sign of DS°rxn and calculate its value for the combustion of 1 mol of propane at 25°C. C 3 H 8(g) + 5 O 2(g) → 3 CO 2(g) + 4 H 2 O(l) PLAN: From the change in the number of moles of gas (6 mol yields 3 mol), the entropy should decrease (DS°rxn < 0). To find DS°rxn, we apply Equation 20. 4, using the S° values from Appendix B. SOLUTION: ΔS°rxn = [(3 mol CO 2)(S° of CO 2) + (4 mol H 2 O)(S° of H 2 O)] - [(1 mol C 3 H 8)(S° of C 3 H 8) + (5 mol O 2)(S° of O 2) = [(3 mol)(213. 7 J/K·mol) + (4 mol)(69. 9 J/K·mol)] – [(1 mol)(269. 9 J/K·mol) + (5 mol)(205. 0 J/K·mol) = -374 J/K 20 -36

Entropy Changes in the Surroundings A decrease in the entropy of the system is outweighed by an increase in the entropy of the surroundings. The surroundings function as a heat source or heat sink. In an exothermic process, the surroundings absorbs the heat released by the system, and Ssurr increases. qsys < 0; qsurr > 0 and DSsurr > 0 In an endothermic process, the surroundings provides the heat absorbed by the system, and Ssurr decreases. qsys > 0; qsurr < 0 and DSsurr < 0 20 -37

Temperature at which Heat is Transferred Since entropy depends on temperature, DS°surr is also affected by the temperature at which heat is transferred. For any reaction, qsys = -qsurr. The heat transferred is specific for the reaction and is the same regardless of the temperature of the surroundings. The impact on the surroundings is larger when the surroundings are at lower temperature, because there is a greater relative change in Ssurr. DSsurr = - qsys T 20 -38 DSsurr = - DHsys for a process at constant P. T

Sample Problem 20. 3 Determining Reaction Spontaneity PROBLEM: At 298 K, the formation of ammonia has a negative DS°sys; N 2(g) + 3 H 2(g) → 2 NH 3(g); DS°sys = -197 J/K Calculate DS°univ, and state whether the reaction occurs spontaneously at this temperature. PLAN: For the reaction to occur spontaneously, DS°univ > 0, so DS°surr must be greater than +197 J/K. To find DS°surr we need DH°sys, which is the same as DH°rxn. We use the standard enthalpy values from Appendix B to calculate this value. SOLUTION: DH°rxn = [(2 mol)(DH°f of NH 3)] - [(1 mol)(DH°f of N 2) + (3 mol)(DH°f of H 2)] = [(2 mol)(-45. 9 k. J/mol)] - [(1 mol)(0 k. J/mol) + (3 mol)(0 k. J/mol)] = 91. 8 k. J 20 -39

Sample Problem 20. 3 DSsurr = - DHsys T -91. 8 k. J x 1000 J 1 k. J =298 K = 308 J/K DS°univ = DS°sys + DS°surr = -797 J/K + 308 J/K = 111 J/K Since DS°univ > 0, the reaction occurs spontaneously at 298 K. Although the entropy of the system decreases, this is outweighed by the increase in the entropy of the surroundings. 20 -40

Figure 20. 12 A whole-body calorimeter. In this room-sized apparatus, a person exercises while respiratory gases, energy input and output, and other physiological variables are monitored. Biological systems also obey the laws of thermodynamics. For all processes, however complex, the entropy of the university increases. 20 -41

DS and the Equilibrium State For any process approaching equilibrium, DS°univ > 0. At equilibrium, there is no further net change, and DS°sys is balanced by DS°surr. DS°univ = DS°sys + DS°surr = 0, so DS°sys = - DS°surr When a system reaches equilibrium, neither the forward nor the reverse reaction is spontaneous, so there is no net reaction in either direction. 20 -42

Figure 20. 13 AB Components of DS°univ for spontaneous reactions. exothermic For an exothermic reaction in which DSsys > 0, the size of DSsurr is not important since DSsurr > 0 for all exothermic reactions. The reaction will always be spontaneous. 20 -43 exothermic For an exothermic reaction in which DSsys < 0, DSsurr must be larger than DSsys for the reaction to be spontaneous.

Figure 20. 13 C Components of DS°univ for spontaneous reactions. endothermic For an endothermic reaction in which DSsys > 0, DSsurr must be smaller than DSsys for the reaction to be spontaneous. DSsurr is always < 0 for endothermic reactions. 20 -44

Gibbs Free Energy The Gibbs free energy (G) combines the enthalpy and entropy of a system; G = H – TS. The free energy change (DG) is a measure of the spontaneity of a process and of the useful energy available from it. DGsys = DHsys - T DSsys DG < 0 for a spontaneous process DG > 0 for a nonspontaneous process DG = 0 for a process at equilibrium 20 -45

Calculating DG°rxn can be calculated using the Gibbs equation: DG°sys = DH°sys - T DS°sys DG°rxn can also be calculated using values for the standard free energy of formation of the components. DG°rxn = Sm. DG°products - Sn. DG°reactants 20 -46

Calculating DG°rxn from Enthalpy and Entropy Values PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. +5 +7 -1 Δ 4 KCl. O 3(s) 3 KCl. O 4(s) + KCl(s) Sample Problem 20. 4 PLAN: To solve for DG°, we need values from Appendix B. We use DHf° to calculate DH°rxn, use S° values to calculate DS°rxn and then apply the Gibbs equation. SOLUTION: DH°rxn = Sm. DHf°(products) - Sn. DHf°(reactants) = [(3 mol KCl. O 4)(DHf° of KCl. O 4) + (1 mol KCl)(DHf° of KCl)] ‒ [(4 mol KCl. O 3)(DHf° of KCl. O 3)] = [(3 mol)(-432. 8 k. J/mol) + (1 mol)(-436. 7 k. J/mol) – [(4 mol)(-397. 7 k. J/mol)] = 144 k. J 20 -47

Sample Problem 20. 4 DS°rxn = Sm. S°(products) - SS°(reactants) = [(3 mol of KCl. O 4)(S° of KCl. O 4) + (1 mol KCl)(S° of KCl)] ‒ [(4 mol KCl. O 3)(S° of KCl. O 3)] = [(3 mol)(151. 0 J/mol·K) + (1 mol)(82. 6 J/mol·K) – [(4 mol)(-397. 7 k. J/mol)] = -36. 8 J/K DG°sys = DH°sys - T DS°sys = 144 k. J – (298 K)(-36. 8 J/K) 1000 J 1 k. J = -133 k. J 20 -48

Sample Problem 20. 5 Calculating DG°rxn from DG°f Values PROBLEM: Use DG°f values to calculate DG°rxn for the reaction in Sample Problem 20. 4: 4 KCl. O 3(s) Δ 3 KCl. O 4(s) + KCl(s) PLAN: We apply Equation 20. 8 and use the values from Appendix B to calculate DG°rxn. SOLUTION: DG°rxn = Sm. DG°products - Sn. DG°reactants = [(3 mol KCl. O 4)(DGf° of KCl. O 4) + (1 mol KCl)(DGf° of KCl)] ‒ [(4 mol KCl. O 3)(DGf° of KCl. O 3)] = [(3 mol)(-303. 2 k. J/mol) + (1 mol)(-409. 2 k. J/mol)] ‒ [(4 mol)(-296. 3 k. J/mol)] = -134 k. J 20 -49

DG° and Useful Work DG is the maximum useful work done by a system during a spontaneous process at constant T and P. In practice, the maximum work is never done. Free energy not used for work is lost to the surroundings as heat. DG is the minimum work that must be done to a system to make a nonspontaneous process occur at constant T and P. A reaction at equilibrium (DGsys = 0) can no longer do any work. 20 -50

Figure 20. 14 20 -51 An expanding gas lifting a weight.

Effect of Temperature on Reaction Spontaneity DGsys = DHsys - T DSsys Reaction is spontaneous at all temperatures If DH < 0 and DS > 0 DG < 0 for all T Reaction is nonspontaneous at all temperatures If DH > 0 and DS < 0 DG > 0 for all T 20 -52

Effect of Temperature on Reaction Spontaneity Reaction becomes spontaneous as T increases If DH > 0 and DS > 0 DG becomes more negative as T increases. Reaction becomes spontaneous as T decreases If DH < 0 and DS < 0 DG becomes more negative as T decreases. 20 -53

Table 20. 1 Reaction Spontaneity and the Signs of DH, DS, and DG DH DS -TDS DG – + – – Spontaneous at all T + – + + Nonspontaneous at all T + + – + or – Spontaneous at higher T; Description nonspontaneous at lower T – – + + or – Spontaneous at lower T; nonspontaneous at higher T 20 -54

Sample Problem 20. 6 Using Molecular Scenes to Determine the Signs of DH, DS, and DG PROBLEM: The following scenes represent a familiar phase change for water (blue spheres). (a) What are the signs of DH and DS for this process? Explain. (b) Is the process spontaneous at all T, no T, low T, or high T? Explain. PLAN: (a) From the scenes, we determine any change in the amount of gas, which indicates the sign of DS, and any change in the freedom of motion of the particles, which indicates whether heat is absorbed or released. (b) To determine reaction spontaneity we need to consider the sign of DG at different temperatures. 20 -55

Sample Problem 20. 6 SOLUTION: (a) The scene represents the condensation of water vapor, so the amount of gas decreases dramatically, and the separated molecules give up energy as they come closer together. DS < 0 and DH > 0 (b) Since DS is negative and DH is positive, the –TDS term is positive. In order for DG to be < 0, the temperature must be low. The process is spontaneous at low temperatures. 20 -56

Sample Problem 20. 7 Determining the Effect of Temperature on ΔG PROBLEM: A key step in the production of sulfuric acid is the oxidation of SO 2(g) to SO 3(g): 2 SO 2(g) + O 2(g) → 2 SO 3(g) At 298 K, DG = -141. 6 k. J; DH = -198. 4 k. J; and DS = -187. 9 J/K (a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how DG will change with increasing T. (b) Assuming DH and DS are constant with increasing T, is the reaction spontaneous at 900. ° C? PLAN: We note the sign of DG to see if the reaction is spontaneous and the signs of DH and DS to see the effect of T. We can then calculate whether or not the reaction is spontaneous at the higher temperature. 20 -57

Sample Problem 20. 7 SOLUTION: (a) DG < 0 at 209 K (= 25°C), so the reaction is spontaneous. With DS < 0, the term -TDS > 0 and this term will become more positive at higher T. DG will become less negative, and the reaction less spontaneous, with increasing T. (b) DG = DH - TDS Convert T to K: 900. + 273. 15 = 1173 K Convert S to k. J/K: -187. 9 J/K = 0. 1879 k. J/K DG = -198. 4 k. J – [1173 K)(-0. 187. 9 k. J/K) = 22. 0 k. J DG > 0, so the reaction is nonspontaneous at 900. °C. 20 -58

Figure 20. 15 The effect of temperature on reaction spontaneity. The sign of DG switches at T = DH ΔS 20 -59

Sample Problem 20. 8 Finding the Temperature at Which a Reaction Becomes Spontaneous PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is nonspontaneous DG = 8. 9 k. J). Calculate the temperature at which the reaction becomes spontaneous. PLAN: We need to calculate the temperature at which DG crosses over from a positive to a negative value. We set DG equal to zero, and solve for T, using the values for DH and DS from the text. SOLUTION: DH 58. 1 k. J = 352 K T= = DS 0. 165 k. J/K At any temperature above 352 K (= 79°C), the reaction becomes spontaneous. 20 -60

Chemical Connections Figure B 20. 1 The coupling of a nonspontaneous reaction to the hydrolysis of ATP. A spontaneous reaction can be coupled to a nonspontaneous reaction so that the spontaneous process provides the free energy required to drive the nonspontaneous process. The coupled processes must be physically connected. 20 -61

Chemical Connections Figure B 20. 2 The cycling of metabolic free energy. 20 -62

Chemical Connections Figure B 20. 3 ATP is a high-energy molecule. When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A) and the increase in resonance stabilization (B) causes a large amount of energy to be released. 20 -63

DG, Equilibrium, and Reaction Direction A reaction proceeds spontaneously to the right if Q < K; Q < 1 so ln Q < 0 and DG < 0 K K A reaction proceeds spontaneously to the left if Q > K; Q > 1 so ln Q > 0 and DG > 0 K K A reaction is at equilibrium if Q = K; Q = 1 so ln Q = 0 and DG = 0 K K 20 -64

DG, Q, and K DG = RT ln Q = RT ln. Q – RT ln. K K If Q and K are very different, DG has a very large value (positive or negative). The reaction releases or absorbs a large amount of free energy. If Q and K are nearly the same, DG has a very small value (positive or negative). The reaction releases or absorbs very little free energy. 20 -65

ΔG and the Equlibrium Constant For standard state conditions, Q = 1 and DG° = -RT ln. K A small change in DG° causes a large change in K, due to their logarithmic relationship. As DG° becomes more positive, K becomes smaller. As DG° becomes more negative, K becomes larger. To calculate DG for any conditions: DG = DG° + RT ln. Q 20 -66

Table 20. 2 The Relationship Between DG° and K at 298 K DG° (k. J) 9 x 10 -36 100 3 x 10 -18 50 2 x 10 -9 10 2 x 10 -2 1 7 x 10 -1 0 1 -1 1. 5 -10 5 x 101 -50 6 x 108 -100 3 x 1017 -200 1 x 1035 { Essentially no forward reaction; reverse reaction goes to completion { Forward and reverse reactions proceed to same extent { Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION 200 Significance FORWARD REACTION 20 -67 K

Using Molecular Scenes to Find DG for a Reaction at Nonstandard Conditions PROBLEM: These molecular scenes represent three mixtures in which A 2 (black) and B 2 (green) are forming AB. Each molecule represents 0. 10 atm. The equation is A 2(g) + B 2(g) 2 AB(g) DGo = -3. 4 k. J/mol Sample Problem 20. 9 (a) If mixture 1 is at equilibrium, calculate K. (b) Which mixture has the most negative DG, and which has the most positive? (c) Is the reaction spontaneous at the standard state, that is PA 2 = PB 2 = PAB = 1. 0 atm? 20 -68

Sample Problem 20. 9 PLAN: (a) Mixture 1 is at equilibrium, so we first write the expression for Q and then find the partial pressure of each substance from the number of molecules and calculate K. (b) To find DG, we apply equation 20. 13. We can calculate the value of T using K from part (a), and substitute the partial pressure of each substance to get Q. (c) Since 1. 0 atm is the standard state for gases, DG = DG°. SOLUTION: (a) A 2(g) + B 2(g) (0. 40)2 K= (0. 20) 20 -69 = 4. 0 2 AB(g) PAB 2 Q= PA 2 x PB 2

Sample Problem 20. 9 (b) DG° = -RT ln. K -3. 4 k. J 8. 314 J T ln 4. 0 =mol mol·K T= -3. 4 k. J 1000 J mol 1 k. J - 8. 314 J mol·K Mixture 1 is at equilibrium, so ΔG = 0; DG = DG° + RT ln. Q = -3400 J/mol + (8. 314 J/mol·K)(295 K)(ln 4. 0) = -3400 J + 3400 J = 0. 0 J Mixture 2: (0. 20)2 Q= = 0. 44 (0. 30) DG = DG° + RT ln. Q = -3400 J/mol + (8. 314 J/mol·K)(295 K)(ln 0. 44) = -3400 J – 2010 J = -5. 4 x 10 -3 J 20 -70 ln 4. 0 = 295 K

Sample Problem 20. 9 Mixture 3: (0. 60)2 Q= = 36 (0. 10) DG = DG° + RT ln. Q = -3400 J/mol + (8. 314 J/mol·K)(295 K)(ln 36) = -3400 J + 8800 J = +5. 4 x 10 -3 J Mixture 2 has the most negative DG, and mixture 3 has the most positive DG. (c) Under standard conditions, DG = DG° = -3. 4 k. J/mol. Yes, the reaction is spontaneous when the components are in their standard states. 20 -71

Sample Problem 20. 10 Calculating DG at Nonstandard Conditions PROBLEM: The oxidation of SO 2(g) is too slow at 298 K to be useful in the manufacture of sulfuric acid, so the reaction is run at high T. 2 SO 2(g) + O 2(g) → 2 SO 3(g) (a) Calculate K at 298 K and at 973 K, (DG° 298 = -141. 6 k. J/mol of reaction as written; using DH° and DS°values at 973 K, DG° 973 = -12. 12 k. J/mol for reaction as written. ) (b) Two containers are filled with 0. 500 atm of SO 2, 0. 0100 atm of O 2, and 0. 100 atm of SO 3; one is kept at 25°C and the other at 700. °C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate DG for the system in part (b) at each temperature. 20 -72

Sample Problem 20. 10 PLAN: (a) We know DG°, T, and R, so we can calculate the values of K at each temperature. (b) To determine if a net reaction will occur, we find Q from the given partial pressures and compare Q to K from part (a). of each substance to get Q. (c) These are not standard-state pressures, so we find DG at each temperature using the equation DG = DG° + RT ln. Q. SOLUTION: (a) DG° = -RT ln. K so K = e -(DG°/RT) -141. 6 x 103 J/mol = 57. 2 At 298 K, -(DG°)/RT = 8. 314 J/mol·K x 298 K K at 298 K = e 57. 2 = 7 x 1024 20 -73

Sample Problem 20. 10 -12. 12 x 103 J/mol = 1. 50 At 973 K, -(DG°)/RT = 8. 314 J/mol·K x 973 K K at 973 K = e 1. 50 = 4. 5 (b) Calculating the value of Q Q= 2 PSO 3 2 PSO 2 (0. 100)2 = = 4. 00 2 x PO (0. 500) (0. 0100) 2 Since Q < K at both temperatures, the reaction will proceed toward the products (to the right) in both cases until Q equals K. At 298 K, the system is far from equilibrium and will proceed far to the right. At 973 K the system is closer to equilibrium and will proceed only slightly toward the right. 20 -74

Sample Problem 20. 10 (c) DG 298 = DG° + RT ln Q = -141. 6 k. J/mol + (8. 314 x 10 -3 k. J/mol·K x 298 K x ln 4. 00) = -138. 2 k. J/mol DG 973 = DG° + RT ln Q = -12. 12 k. J/mol + (8. 314 x 10 -3 k. J/mol·K x 973 K x ln 4. 00) = -0. 9 k. J/mol 20 -75

Figure 20. 16 Free energy and the extent of reaction. Each reaction proceeds spontaneously (green curved arrows) from reactants or products to the equilibrium mixture, at which point DG = 0. After that, the reaction is nonspontaneous in either direction (red curved arrows). Free energy reaches a minimum at equilibrium. 20 -76