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12/20/16 Creative Commons License Ch. 16: Thermodynamics ! Images and tables in this file have been used from the following sources: ! OpenStax: Creative Commons Attribution License 4.0. ! ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at [email protected]. Geysers are a dramatic display of thermodynamic principles in nature. As water inside the earth heats up, it rises to the surface through small channels. Pressure builds up until the water turns to steam, and steam is expelled forcefully through a hole at the surface. (credit: modification of work by Yellowstone National Park) Chemistry: OpenStax ! Principles of General Chemistry (CC BY-NC-SA 3.0): http://2012books.lardbucket.org/pdfs/principles-of-generalchemistry-v1.0.pdf 2 1 Thermodynamics 16.1 Spontaneity • ! Recall thermochemistry (study of heat changes) from CHM 150/151: Chapter 5 sections: ! 1 (Conservation of energy), ! 2 (Energy change, system, surroundings) ! 3 (Enthalpies of Physical and Chemical Changes), ! 3 (Calculating ΔHorxn using ΔHof – Standard Enthalpy • • Thermodynamics allows us to determine IF a reaction will occur (is spontaneous) as well as telling us the direction in which (and extent to which) a reaction will occur (like Q vs K). A spontaneous process is one that proceeds on its own without any external influence. A spontaneous process always moves a reaction mixture toward equilibrium. of Formation). ! We’ll introduce the concept of Entropy this semester. Figure 16.4: An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks. 3 4 16.1 Spontaneity • • • • • • Spontaneity Other spontaneous processes include: Ice melting at room temperature Solid sodium reacting violently with water A ball rolling downhill An egg boiling in hot water Heat flowing from a hot object to a cold one ! Spontaneity is determined by enthalpy and entropy of a reaction. ! A system tends to move toward a lower state of energy. ! The enthalpy (ΔH in kJ/mol) of many spontaneous processes are exothermic (e.g., combustion reactions), but some spontaneous processes can be endothermic under the right conditions (e.g., melting of ice above 0oC). ! A system tends to become more disordered. ! The entropy (ΔS in J/mol·K) of a spontaneous system tends to increase. There is an increase in the disorder or randomness of a system (i.e., your room or office naturally becoming messier over time). Figure 16.5: When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object. 5 6 1 12/20/16 Predicting Sign of ΔHorxn 150/151 Thermochemistry Review ! Predict signs of enthalpy for the following processes: ! 1st Law of Thermodynamics: Energy is conserved, it cannot be created or destroyed ! If a system / reaction gives off heat, the surroundings / universe must absorb it (and vice versa). ! Heat flow (enthalpy, ΔH) is always defined with reference to the system: ! System absorbs heat, ΔH > 0 ! Associated with breaking bonds or Intermolecular forces. ! Heat energy goes into a system to break the bonds or pull Decomposition Combination Acid-Base Neut. Combustion • atoms/molecules apart. • ! System gives off heat, ΔH < 0 ! Associated with making bonds or Intermolecular forces Melting/fusion + Boiling/vaporization + Freezing Condensation + Sublimation Deposition In Chemistry reference manuals, only ΔHofusion and ΔHovaporization are listed. The opposite processes have the same values but are opposite in sign. ΔHosublimation = ΔHofusion + ΔHovaporization + - 7 8 Defining ΔHof; Calculating ΔHorxn Defining ΔHof, Calculating ΔHorxn ! Heat of formation, ΔHof , is the enthalpy change when 1 ! Write the equation for the heat of formation of water. mole of a substance is made from its elements in their standard state (defined in section 5.3) H2 (g) + ½ O2 (g) ! H2O (l) ! The symbol refers to the standard state, 1.00 atm pressure, o 25.0oC, 1.00 M for solutions (see section 16.3) 0 0 -285.83 kJ/mol ! This is how the values in Appendix G are obtained. ΔHorxn = Σ n·ΔHof (products) - Σ n·ΔHof (reactants) ! We then use these values to calculate the enthalpy of a reaction. ! n is moles (coefficient in balanced equation) ! ΔHof values, Appendix G, reported in kJ/mol ! ΔHof for elements is 0 (no energy needed!) ΔHorxn = Σ n·ΔHof (products) - Σ n·ΔHof (reactants) 9 Calculating ΔHorxn Group Work, Calculating ΔHorxn ! Calculate ΔHorxn for the following reaction: ! Calculate ΔHorxn for the following reaction: 4 NH3(g) + 5 O2(g) ! 4NO(g) + 6 H2O(g). What 1C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2 H2O(g) information do you need? ! Find ΔHof values in Appendix G. ! NH3(g) = - 45.9 kJ/mol ! O2(g) = 0 kJ/mol ! NO(g) = 90.25 kJ/mol ! H2O(g) = - 241.82 kJ/mol Predict sign of ΔHorxn first. ! Find ΔHof values in Appendix G. ! C2H4(g) = 52.4 kJ/mol ! O2(g) = 0 kJ/mol ! CO2 (g) = - 393.51 kJ/mol ! H2O(g) = - 241.82 kJ/mol ! Answer: -906.3 kJ (sig figs by decimal places!) ! Watch physical states when looking up values ! (l) versus (g) has a different value! 10 ! Answer: -1323.1 kJ 11 12 2 12/20/16 16.2 Entropy Entropy of Molecules To predict spontaneity, both the enthalpy and the entropy of a process must be known. Entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is (measured in J/mol·K). ΔS (entropy change for a system) = Sfinal - Sinitial If randomness increases, ΔS is positive. If randomness decreases, ΔS is negative. The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. 13 Entropy of a System Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry 14 -v1.0/s19-chemical-equilibrium.html, CC-BY-NC-SA 3.0 license Which phase has greatest entropy? ! Solid, liquid or gas? (ice, water, or steam) Figure 16.8: The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are collected into five distributions—(a), (b), (c), (d), and (e)—based on the numbers of particles in each box. Scenario c is most probable because it is most disperse. Which side has greater entropy? 16 Entropy and Temperature Figure 16.11: As temperature increases, entropy increases because of increased kinetic energy and molecular motion. Entropy increases as a substance melts (s ! l); even more increase is seen as it boils (l ! g). 18 3 12/20/16 Entropy of Dissolution Entropy Trends Dissolving a substance can lead to an increase or a decrease in entropy, depending on the nature of the solute. Molecular solutes (i.e. sugar): entropy increases Entropy increases as the number of gas particles increases: N2O4(g) ! 2 NO2(g) Comparing entropy of molecules – Entropy increases as: 1) temperature increases. H2O at 10oC vs H2O at 100oC 2) mass of an atom or molecule increases (more electrons moving around) F2 (158 J/K"mol) < Cl2 (165 J/K"mol) 3) molecular complexity increases (i.e., number of atoms, number of heavier atoms, etc.) CH4 (186 J/K"mol) < C2H6 (230 J/K"mol) Ionic compounds: entropy can decrease or increase NaCl (s): ΔSsoln = +43.4 J/K AlCl3 (s): ΔSsoln = -253.2 J/K Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry-19 v1.0/s22-chemical-thermodynamics.html#averill_1.0-ch18_s03_s01_f01, CC-BY-NC-SA 3.0 license 20 Group work: Which has greater entropy? Greater entropy? ! Determine which example in each scenario will have more ! Determine which example in each scenario will have more entropy. entropy. ! Temperature: C(s) at ! Volume: 1 mol O2(g) in 1 L ! Phase: O2(g) O2(l) ! Molar mass: He(g) Ne(g) 25oC ! Molecular Complexity: C(s) at 35oC 1 mol O2(g) in 2 L C4H10(g) ! Temperature: C(s) at 25oC ! Volume: 1 mol O2(g) in 1 L ! Phase: O2(g) O2(l) ! Molar mass: He(g) Ne(g) ! Molecular Complexity: C3H6(g) C4H10(g) C(s) at 35oC 1 mol O2(g) in 2 L C3H6(g) 21 Processes that Increase Entropy 22 Predict the sign of ΔS ! Does entropy increase (ΔS > 0) or decrease (ΔS < 0) for Entropy usually increases for the following: the following? Why? ! 2SO2 (g) + O2 (g) → 2SO3 (g) • Melting, vaporization, or sublimation ! 2O3 (g) → 3O2 (g) • Temperature increase ! CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) ! I2 (s) → I2 (g) • Reaction resulting in a greater number of gas molecules (or greater molar mass of solid) ! C2H4(g) + 3O2 (g) → 2CO2 (g) + 2H2O(g) ! 4 Al(s) + 3 O2(g) → 2Al2O3(s) • Decomposition reactions Example 16.3 23 24 4 12/20/16 16.3 2nd and 3rd Laws of Thermo Standard Entropy Values (So) Which phase has the lowest entropy? Highest? ! 3rd Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. ! This allows us to calculate entropy (So: (J/mol•K)) and changes in entropy (ΔSo); unlike enthalpy where we can only measure change! ! Use Standard Molar Entropies, So, (Table 16.2 and Appendix G) - entropy of 1 mole of a pure substance at 1 atm and 25oC ΔSorxn =Σ n·So (products) - Σ n·So (reactants) 25 Substance So (J mol-1 K-1) Substance So (J mol-1 K-1) C (s, diamond) 2.38 H (g) 114.6 C (s, graphite) 5.740 H2 (g) 130.57 Al (s) 28.3 CH4 (g) 186.3 Al2O3 (s) 50.92 HCl (g) 186.8 BaSO4 (s) 132.2 H2O (g) 188.8 CO (g) 197.7 H2O (l) 70.0 CO2 (g) 213.8 Hg (l) 75.9 C2H4 (g) 219.3 H2O2 (l) 109.6 C2H6 (g) 229.5 Br2 (l) 152.23 CH3OH (g) 239.9 CCl4 (l) 214.4 CCl4 (g) 309.7 Note: The standard entropy for a pure element in its most stable 26 form is NOT zero (like for ΔHof). Entropy and the 2nd Law Calculate Entropy of Reaction The second law of thermodynamics states that in any spontaneous process, the total entropy of a system and its surrounding always increases. ΔSuniverse (or ΔStotal) = ΔSsys + ΔSsurr > 0 ΔSsys (or ΔSrxn) (that we just calculated) ΔSsurr (or ΔSH2O) is entropy change for surroundings ! Calculate ΔSorxn for: ! 1 C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2 H2O(g) ! C2H4(g) = 219.3 J/mol•K, ! O2(g) = 205.2 J/mol•K, ! CO2(g) = 213.8 J/mol•K ! H2O(g) = 188.8 J/mol•K ΔStotal > 0; reaction is spontaneous ΔStotal < 0; reaction is nonspontaneous ΔStotal = 0; system is at equilibrium ! ΔSorxn = - 29.7 J/K (close to zero because same number of moles of gas on both sides) Examples 16.5, 16.6 27 28 Entropy Changes in Surroundings Entropy Changes in Surroundings ! ΔSsurr is related to the enthalpy of the reaction; if ! Temperature is inversely related to ΔSsurr: heat enters surroundings, molecules have more energy and entropy increases. ! If surroundings have a low temperature, adding heat results in a substantial change in randomness. (Like throwing a huge rock into a calm lake.) ! Conversely, if surroundings have a high temperature, adding Entropy Entropy Heat heat makes a marginal difference in change in randomness. (Like throwing a pebble into the ocean during a hurricane.) ! Putting it all together we get: Heat ΔSsurr = -ΔH sys T 29 5 12/20/16 Total Entropy Changes 16.4 Free Energy & Spontaneity • Gibbs free energy (G) or free energy can be used to express spontaneity from the perspective of the system. • ΔG (kJ/mol) is the maximum amount of energy available to do work on the surroundings and takes into account both enthalpy and entropy. • In a spontaneous process at constant temperature and pressure, the free energy of the system always decreases. ! ΔStotal = ΔSrxn + ΔSsurr ! ΔStotal = ΔSrxn – ΔHrxn/T ! Multiply both sides by –T ! -TΔStotal = ΔGrxn (note the opposite signs!) • The actual amount of work (wmax) obtained is always less than the maximum available because of energy lost in carrying out a process (given off as heat, light, sound, etc. energy). ! ΔGrxn = ΔHrxn – TΔSrxn ! ΔGrxn is Gibbs Free Energy; ΔG < 0 indicates spontaneous reaction ΔGsys = ΔHsys - TΔSsys 31 Predicting Spontaneity 32 Gibbs Free Energy and Temperature ΔGsys = ΔHsys - TΔSsys ΔG = ΔH – TΔS: Notice that ΔG is temperature dependent! Temperature plays a role in determining spontaneity. Also note the absence of standard conditions notation (o). ΔG < 0 The reaction is spontaneous (still moving forward). ΔG > 0 The reaction is nonspontaneous (moves in reverse direction). ΔG = 0 The reaction is at equilibrium Endothermic reactions will be spontaneous at higher temperatures (where TΔS > ΔH) Figure 16.12 Exothermic reactions will be spontaneous at lower temperatures (where ΔH > TΔS) 33 “Higher” and “lower” temps are reaction-specific. Every reaction has a cut-off temperature depending on values of ΔH and ΔS. Standard Free Energy Changes Standard Free Energy Changes The standard free energy of reaction (ΔGorxn) is freeenergy change for a reaction when it occurs under standard-state conditions. We can calculate ΔGorxn = ΔHorxn – TΔSorxn for the reaction: 1 C2H4(g) + 3O2 (g) ! 2 CO2 (g) + 2 H2O(g) ΔHorxn = -1323.1 kJ ΔSorxn = - 29.7 J/K Standard-state conditions are: • Solids, liquids, and gases in their pure form • 1 M solutions, 1 atm gases, 25oC (298 K) ΔGorxn = ΔHorxn – TΔSorxn ΔGorxn = -1314.2 kJ ΔG (non-standard) predicts spontaneity ΔGo (standard – at 25oC) predicts reactant- or productfavored at equilibrium; relates to K 35 Example 16.7 36 6 12/20/16 Gibbs Free Energy Temperature Dependence ΔG = Gibbs Free Energy (or free energy), kJ/mol: Uses both enthalpy and entropy to determine if a reaction will be spontaneous. ! ΔG = 0: 1) turning point between spontaneous and non- Calculate ΔG when ΔH = -227 kJ, ΔS = - 309 J/K, T = 1450 K. Are you calculating Standard or Non-Standard ΔG? Is this process spontaneous? ! 0 = ΔH - TΔS ΔG = ΔH – TΔS = -227 kJ – (1450 K * -0.309 kJ/K) = 221 kJ; Non-spontaneous at this temperature. spontaneous; 2) phase changes (occur at equilibrium) ! ΔG = 0 ! Solve for T ! T = ΔH / ΔS ! This temperature tells us the temperature cut-off for “high” and “low” temps. ! We can also use this to determine the temperature for a phase change (system at equilibrium). Note: this ΔG is the non-standard value! 38 Temperature of Spontaneity Temperature of Phase Change ! For the previous problem, we calculated ΔG = +221 kJ when ! 1) Calculate the melting point of ice? H2O(s) ! H2O(l) ΔH = -227 kJ, ΔS = - 309 J/K, and T = 1450 K. At what temperature (in oC) will the system become spontaneous? Will it be spontaneous above or below this temp? ! ΔHfusion = 6.01 kJ/mol; ΔSfusion = 22.0 J/K•mol ! For ice: ! 263K (-10oC): ΔG = +0.22 kJ/mol (non-spontaneous) ! 283K (10oC): ΔG = -0.22 kJ/mol (spontaneous) ! T = ΔH / ΔS = 734 K (461oC); spontaneous below this temperature ! 2) Calculate the boiling point (in oC) of ethanol if the entropy of vaporization is 111.9 J/K and the enthalpy of vaporization is 39.3 kJ ! T = ΔH / ΔS = 39.3 kJ / 0.1119 kJ/K = 351.206 K = 78.2oC Examples 16.9, 16.10 Predicting Spontaneity 40 Predicting Spontaneity ! ! Write the equation for the formation of rust. ! Calculate ΔHorxn and ΔSorxn using the table below. ! Is the reaction spontaneous or nonspontaneous at 500oC? ΔHo rxn = [2 mol * -824.2 kJ/mol] – [0] = -1648.4 kJ ! ΔSorxn = [2 mol * 87.40 J/K·mol] – [4 mol * 27.3 J/K·mol + 3 mol * 205.2 J/K·mol] = -550.0 J/K = -0.5500 kJ/K ! ΔG = ΔHo – TΔSo = -1648.4 kJ – (773.15 K * -0.5500 kJ/K) = -1223.2 kJ, spontaneous ΔHo Fe(s) O2(g) Fe2O3(s) Example 16.7 f (kJ/mol) 0 0 -824.2 So ! At what temperature does the reaction become (J/K·mol) 27.3 205.2 87.40 spontaneous? Above or below this temperature? ! T = ΔH / ΔS = 2997 K (FYI, surface of the sun is 5,778 K!) 41 7 12/20/16 Standard Free Energy of Formation ! Calculate ΔGorxn for the following reaction: Alternatively, we can use the equation 1C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2H2O(g) ΔGorxn = Σ n·ΔGof (products) - Σ n·ΔGof (reactants) to calculate ΔGo Practice Calculation of ΔGorxn rxn. ΔGof is the free energy change for the formation of 1 mole of a substance from elements in their standard states. ΔGof for any element in its most stable form at 1 atm is defined as zero (just like ΔHof values). Predict the sign of ΔGorxn first. ! Find ΔGof values in Appendix G. ! C2H4(g) = 68.4 kJ/mol ! O2(g) = 0 kJ/mol ! CO2 (g) = -394.36 kJ/mol ! H2O(g) = - 228.59 kJ/mol ! Answer: -1314.3 kJ (we calculated -1314.2 kJ using ΔG = ΔHo – TΔSo) Should they be the same or different? Example 16.8 Standard vs Non-Standard ΔG • • Free Energy and Equilibrium The sign of ΔG determines spontaneity (whether a reaction is still moving forward to reach equilibrium). ΔGo determines whether a reaction is reactant- or product-favored. The relationship between ΔG and ΔGo 44 ΔG is determined by the slope of the tangent at any point on the graph. Figure 16.14 is: ΔGo < 0 = Products – Reactants o ΔG = ΔG + RT ln Q What is K? R is the gas constant (8.314 J/K·mol). T is the Kelvin temperature. Q is the reaction quotient (from Ch. 13). 45 Free Energy and Equilibrium 46 Standard vs Non-Standard ΔG Left of equilibrium, ΔG < 0, spontaneous; Right of equilibrium, ΔG > 0, non-spontaneous; At equilibrium, ΔG = 0 ΔG is the actual free energy change under non-standard conditions. It changes as a reaction proceeds (as concentrations, pressures, etc. change). ΔG = ΔG o + RT ln Q ΔGo > 0 = Products – Reactants ΔG changes as a reaction proceeds (initially decreases). This value determines the spontaneity of a reaction at a given point in time. What is K? ΔGo is standard and does NOT change. It is a constant for a reaction (like ΔHo and ΔSo). ΔGo < 0 is spontaneous and product-favored. 47 48 8 12/20/16 Free Energy and Equilibrium Free Energy and Equilibrium What would the graph look like for a reaction where ΔGo = 0? At equilibrium, ΔG = 0 and Q = K: ΔG = ΔGo + RT ln K ΔGo = - RT ln K What is K? ΔGo K ΔGo < 0 K > 1 Equilibrium mixture contains mostly products. ΔGo > 0 K < 1 Equilibrium mixture contains mostly reactants. ΔGo = 0 K = 1 Equilibrium mixture contains appreciable amount of reactants and products at equilibrium Comment 49 Free Energy and Equilibrium 50 Free Energy and Equilibrium ΔGorxn for the formation of ethanol is -24.7 kJ/mol. Consider the following equilibrium: H2(g) + I2(g) ⇌ 2HI(g). ΔGo at 25oC = 2.6 kJ/mol. Calculate ΔG when PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 3.0 atm. C(s) + O2(g) + H2(g) ! CH3OH(g) Calculate the equilibrium constant, K, at 25oC. ln K = ΔGo/-RT = -24.7 kJ / -(0.008314 kJ/mol·K * 298K) Qp = ln K = 9.969; K = 2.14 x 104 ΔG = (PHI )2 (3.0)2 = = 2.25 (PH2 )(PI2 ) (2.0)(2.0) 2.6 kJ ⎛ 8.314 ×10−3 kJ ⎞ +⎜ ⎟ 298 K ln2.25 = 4.6 kJ/mol mol ⎝ K mol ⎠ ( )( ) Example 16.11 Example 16.12 52 Practice Calculations Practice Calculations 1) Calculate ΔGorxn for the following reaction at 25oC: P4 (g) + 6 Cl2 (g) ! 4 PCl3 (g) 2) ! 1) Calculate ΔGorxn for the following reaction at 25oC: for the melting of gold is 12.55 kJ/mol. fusion for the melting of gold is 9.38 J/mol·K. What is the melting point (in oC) of gold? What is ΔGo for at 25oC? Is ! P4 (g) + 6 Cl2 (g) ! 4 PCl3 (g) this reactant- or product-favored? ! ΔGof (Cl2 (g)) = 0 kJ/mol ΔHofusion ΔSo 3) Calculate ΔG for S(s) + O2(g) ! SO2(g) when [O2] = 0.140 M and [SO2] = 1.24 M at 25oC. In which direction will the reaction proceed? ! ΔGof (P4 (g)) = 24.4 kJ/mol ! ΔGof (PCl3 (g)) = -269.6 kJ/mol ! ΔGorxn = [4 mol(-269.6 kJ/mol] – [1 mol(24.4 kJ/mol) + 0] = - 1102.8 kJ 4) If K for the reaction CuO(s) ! Cu(s) + O2(g) at 25oC is 2.66 x 10-46, what is ΔGo? 53 54 9 12/20/16 Practice Calculations Practice Calculations ! 2) ΔHofusion for the melting of gold is 12.55 kJ/mol. ! 3) Calculate ΔG for S(s) + O2(g) ! SO2(g) when [O2] ! T = ΔH / ΔS = 12.55 kJ/mol / 0.00938 kJ/molK ! Need to calculate ΔGo and Q. ! T = 1337.95 K – 273 = 1060 oC ! ΔGorxn = -300.2 kJ ! What is ΔGo for at 25oC? Is this reactant- or product- ! Q = 1.24 M / 0.140 M = 8.8571 ΔSofusion for the melting of gold is 9.38 J/mol·K. What is the melting point (in oC) of gold? = 0.140 M and [SO2] = 1.24 M at 25oC. In which direction will the reaction proceed? ! ΔGof (SO2(g)) = -300.2 kJ/mol favored? ! ΔGo = ΔHo – TΔSo = 12.55 kJ/mol – (298K)(0.00938 kJ/ molK) = 9.75 kJ/mol; reactant-favored 55 ! ΔG = -294.8 kJ; spontaneous in the forward direction. 56 Practice Calculations ! 4) If K for the reaction CuO(s) ! Cu(s) + O2(g) at 25oC is 2.66 x 10-46, what is ΔGo? ΔGo = -RT ln K = -(8.314 J/K·mol)(298)(ln 2.66x10-46) ΔGo = 2.60 x 105 J/mol = 260. kJ/mol 57 10