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Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 19 Chemical Thermodynamics John D. Bookstaver St. Charles Community College Cottleville, MO Chemical Thermodynamics © 2009, Prentice-Hall, Inc. First Law of Thermodynamics • You will recall from Chapter 5 that energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. • Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. • Above 0 C it is spontaneous for ice to melt. • Below 0 C the reverse process is spontaneous. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.1 Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150 °C is added to water at 40 °C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g), (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 Chemical °C. Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.1 Solution Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gases spontaneously bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is spontaneous. (c) By definition, the normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. If the temperature were below 80.1 °C, condensation would be spontaneous. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.1 • Under 1 atm pressure CO2(s) sublimes at –78 °C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100 °C and 1 atm pressure? • Answer: No, the reverse process is spontaneous at this temperature. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Irreversible Processes • Irreversible processes cannot be undone by exactly reversing the change to the system. • Spontaneous processes are irreversible. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy • Entropy (S) is a term coined by Rudolph Clausius in the 19th century. • Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q . T Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy • Entropy can be thought of as a measure of the randomness of a system. • It is related to the various modes of motion in molecules. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy • Like total energy, E, and enthalpy, H, entropy is a state function. • Therefore, S = Sfinal Sinitial Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: qrev S = T Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.2 • The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9 °C, and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.2 • The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point? • Answer: –163 J/K Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics In other words: For reversible processes: Suniv = Ssystem + Ssurroundings = 0 For irreversible processes: Suniv = Ssystem + Ssurroundings > 0 Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • Ludwig Boltzmann described the concept of entropy on the molecular level. • Temperature is a measure of the average kinetic energy of the molecules in a sample. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • Molecules exhibit several types of motion: – Translational: Movement of the entire molecule from one place to another. – Vibrational: Periodic motion of atoms within a molecule. – Rotational: Rotation of the molecule on about an axis or rotation about bonds. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. – This would be akin to taking a snapshot of all the molecules. • He referred to this sampling as a microstate of the thermodynamic system. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • Each thermodynamic state has a specific number of microstates, W, associated with it. • Entropy is S = k lnW where k is the Boltzmann constant, 1.38 1023 J/K. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • The change in entropy for a process, then, is S = k lnWfinal k lnWinitial Wfinal S = k ln Winitial • Entropy increases with the number of Chemical microstates in the system. Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale • The number of microstates and, therefore, the entropy tends to increase with increases in – Temperature. – Volume. – The number of independently moving molecules. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy and Physical States • Entropy increases with the freedom of motion of molecules. • Therefore, S(g) > S(l) > S(s) Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Solutions Generally, when a solid is dissolved in a solvent, entropy increases. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes • In general, entropy increases when – Gases are formed from liquids and solids; – Liquids or solutions are formed from solids; – The number of gas molecules increases; – The number of moles increases. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.3 • Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature: Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.3 • Indicate whether each of the following processes produces an increase or decrease in the entropy of the system: • Answer: (a) increase, (b) decrease, (c) Chemical decrease, (d) decrease Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.4 • Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.4 Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. Answers: (a) 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of Chemical SO2(g) at STP, (d) 2 mol of NO2(g) at STP Thermodynamics © 2009, Prentice-Hall, Inc. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Entropies • These are molar entropy values of substances in their standard states. • Standard entropies tend to increase with increasing molar mass. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Entropies Larger and more complex molecules have greater entropies. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated: S = nS(products) — mS(reactants) where n and m are the coefficients in the balanced chemical equation. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.5 • Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K: • N2(g) + 3 H2(g) → 2 NH3(g) Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.5 • Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K: • Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g) • Answers: 180.39 J/K Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process: Ssurr = qsys T • At constant pressure, qsys is simply H for the system. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, Suniverse = Ssystem + Ssurroundings • For spontaneous processes Suniverse > 0 Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Change in the Universe • Since Ssurroundings = qsystem and qsystem = Hsystem T This becomes: Hsystem Suniverse = Ssystem + T Multiplying both sides by T, we get TSuniverse = Hsystem TSsystem Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Gibbs Free Energy • TSuniverse is defined as the Gibbs free energy, G. • When Suniverse is positive, G is negative. • Therefore, when G is negative, a process is spontaneous. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Gibbs Free Energy 1. If G is negative, the forward reaction is spontaneous. 2. If G is 0, the system is at equilibrium. 3. If G is positive, the reaction is spontaneous in the reverse direction. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.6 • Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K: • N2(g) + O2(g) → 2 NO(g) • given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances? Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.6 • A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG°. • Is the reaction spontaneous under these conditions? • Answers: ΔG° = –14.7 kJ • the reaction is spontaneous. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, G. f G = nGf (products) mG f(reactants) where n and m are the stoichiometric coefficients. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.7 • (a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: • P4(g) + 6 Cl2(g) → 4 PCl3(g) • (b) What is ΔG°for the reverse of the above reaction? Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.7 • By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane: • CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g). • Answer: –800.7 kJ Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.8 • In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K: • C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = –2220 kJ • (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. • (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.8 • Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K: • C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g). Would you expect ΔG° to be more negative or less negative than ΔH°? • Answer: more negative Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy Changes At temperatures other than 25°C, G° = H TS How does G change with temperature? Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Temperature • There are two parts to the free energy equation: H— the enthalpy term – TS — the entropy term • The temperature dependence of free energy, then comes from the entropy term. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Temperature Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.9 • The Haber process for the production of ammonia involves the equilibrium • Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.9 • (a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: 2 SO2(g) + O2(g) → 2 SO3(g). (b) Using the values obtained in part (a), estimate ΔG° at 400 K. • Answers: (a) ΔH° = –196.6 kJ, ΔS° Chemical = –189.6 J/K; (b) ΔG° = –120.8 kJ Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way: G = G + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.) Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium • At equilibrium, Q = K, and G = 0. • The equation becomes 0 = G + RT lnK • Rearranging, this becomes G = RT lnK or, -G K = e RT Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.10 • As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.12 to estimate the normal Chemical boiling point of CCl4. Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.10 • Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.) • Answers: 330 K Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.11 • We will continue to explore the Haber process for the synthesis of ammonia: • Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.11 • Calculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. • Answers: –26.0 kJ/mol Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.12 • Use standard free energies of formation to calculate the equilibrium constant, K, at 25 °C for the reaction involved in the Haber process: • The standard free-energy change for this reaction was calculated in Sample Exercise 19.9: • ΔG° = –33.3 kJ/mol = –33,300 J/mol. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.12 • Use data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction • Answers: ΔG° = –106.4 kJ/mol, K = 4 × 1018 Chemical Thermodynamics © 2009, Prentice-Hall, Inc.