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Chapter 17 Free Energy and Thermodynamics TMHsiung©2015 Chapter 17 Slide 1 of 66 Contents 1. 2. 3. 4. 5. 6. 7. Nature’s Heat Tax: You Can’t Win and You Can’t Break Even Spontaneous and Nonspontaneous Processes Entropy and the Second Law of Thermodynamics Heat Transfer and Changes in the Entropy of the Surroundings Gibbs Free Energy Entropy Changes in Chemical Reactions: Calculating ΔSorxn Free Energy Changes in Chemical Reactions: Calculating ΔGorxn TMHsiung©2015 Chapter 17 Slide 2 of 66 Continued 8. Free Energy Changes for Nonstandard States: The Relationship between ΔGorxn and ΔGrxn 9. Free Energy and Equilibrium: Relating ΔGorxn to the Equilibrium Constant (K) TMHsiung©2015 Chapter 17 Slide 3 of 66 ***** 1. Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 1) First Law of Thermodynamics The total energy of the universe is constant that energy cannot be created or destroyed, but you can transfer it from one place to another: ΔEuniverse = 0 = ΔEsystem + Δ Esurroundings There are two ways energy is “lost” from a system: ̶ Converted to heat, q ̶ Used to do work, w The energy change in the system: ΔE = q + w or ΔE = ΔH + P ΔV Internal energy change (ΔE) is a state function, depends only on the state of the system, not on how the system got to that state. TMHsiung©2015 Chapter 17 Slide 4 of 66 ***** 2) Heat Tax Conversion of energy to heat, which is “lost” by heating up the surroundings. TMHsiung©2015 Chapter 17 Slide 5 of 66 2. Spontaneous and Nonspontaneous Processes 1) Spontaneous and Nonspontaneous Spontaneous process: A process that occurs without outside intervention. Nonspontaneous: A process that occurs must be assisted by outside intervention. * If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction. TMHsiung©2015 Chapter 17 Slide 6 of 66 2) Chemical Thermodynamics Information Stability of particular substances The Spontaneity of a chemical reaction Equilibrium constant (Keq) of a chemical reaction Predict the proportions of products and reactants at equilibrium. The optimum temperature, pressure, and choice of solvent etc. for a particular reaction. TMHsiung©2015 Chapter 17 Slide 7 of 66 3) Kinetics versus Thermodynamics Thermodynamics tells us what processes are possible. Kinetics tells us whether the process is practical. (high activation energy can effectively block a reaction although that is thermodynamically favored) TMHsiung©2015 Chapter 17 Slide 8 of 66 4) Exothermic reaction versus Endothermic reaction Most spontaneous processes are exothermic, however, some spontaneous processes are endothermic. For example: - the melting of ice above 0 oC - the evaporation of liquid water to gaseous water - the dissolution of sodium chloride in water TMHsiung©2015 Chapter 17 Slide 9 of 66 5) Natural tendencies for spontaneous processes To achieve a lower energy state, i.e., decreasing enthalpy (H) - The enthalpy change, ΔH, is the difference in the sum of the internal energy and PV work energy of the reactants to the products. Toward a more disordered state, i.e., increasing entropy (S) - The entropy change, ΔS, is the difference in randomness of the reactants compared to the products. TMHsiung©2015 Chapter 17 Slide 10 of 66 ***** 3. Entropy and the Second Law of Thermodynamics 1) Entropy (S) A thermodynamic function that is proportional to the number of energetically equivalent ways to arrange the components of a system to achieve a particular state; a measure of the energy randomization or energy dispersal in a system. S = k ln W k = Boltzmann constant = 1.38 × 10−23 J/K W: the number of energetically equivalent ways a system can exist, unitless S, unit generally in J/mol The greater spreading of the energy of the microscopic particles, the greater the S of the system. TMHsiung©2015 Chapter 17 Slide 11 of 66 Continued Each system has the same total energy (4 J), but System A has only one possible microstate while System B has two possible microstate. TMHsiung©2015 Chapter 17 Slide 12 of 66 Continued The Spontaneous Mixing of Ideal Gases Intermolecular forces are negligible, no significant enthalpy change, total energy of the system remains unchanged However, the number of possibilities for the distribution of that energy increases TMHsiung©2015 Chapter 17 Slide 13 of 66 Continued Spontaneous Formation of An Ideal Solution Entropy of the mixture is higher than the entropies of the two substances separated. TMHsiung©2015 Chapter 17 Slide 14 of 66 Continued Vaporization of Water Evaporation is spontaneous because of the increase in entropy TMHsiung©2015 Chapter 17 Slide 15 of 66 2) Second Law of Thermodynamics ***** All spontaneous process are accompanied by an increase in entropy (S) of the universe. ***** Universe Spontaneous process: Surrounding ΔSuniv = ΔSsys + ΔSsurr > 0 System TMHsiung©2015 *ΔS = Sfinal − Sinitial Chapter 17 Slide 16 of 66 3) Absolute Entropy of a Substance The particular S of a substance is calculated from the amount of heat required to raise the temperature from 0 K. Standard molar entropy (So): The entropy of one mole (or 1 M) of a substance in its standard state. (Appendix IIB) In a particular phase, S slow increase with increasing T, the temperature dependence of S: S(T2) = S(T1) + Cp ln(T2 –T1) Cp: Heat capacity of the substance S sharply change at phase change/transition: S(phase 2) = S(phase 1) + ΔH/T example: ΔSfusion = ΔHfusion/Tm) TMHsiung©2015 Chapter 17 Slide 17 of 66 4) Entropy Change Associated with a Change in State TMHsiung©2015 Chapter 17 Slide 18 of 66 5) Predicting the Trends of Entropy ***** SSolid < SLiquid < SGas S(small atom or molecule) < S(large atom or molecule) S(simple molecule) < S(complex molecule) S(low temperature) < S(high temperature) S(solute) + S(solvent) < S(solution) S(less amount of gases) < S(more amount of gases) TMHsiung©2015 Chapter 17 Slide 19 of 66 ***** Example 17.1 Predicting the Sign of Entropy Change Predict the sign of ΔS for each process: a. H2O(g) → H2O(l) b. Solid carbon dioxide sublimes. c. 2 N2O(g) → 2 N2(g) + O2(g) Solution a. Since a gas has a greater entropy than a liquid, the entropy decreases and ΔS is negative. b. Since a solid has a lower entropy than a gas, the entropy increases and ΔS is positive. c. Since the number of moles of gas increases, the entropy increases and ΔS is positive. TMHsiung©2015 Chapter 17 Slide 20 of 66 4. Heat Transfer and Changes in the Entropy of the Surroundings 1) Entropy of the Surroundings (ΔSsurr) a) Observation: • Water freezes at temperatures below 0 oC, the entropy of the water decreases, the process is spontaneous. • Water vapor in air condenses into fog on a cold night, the entropy of the water decreases, the process is spontaneous. TMHsiung©2015 Chapter 17 Slide 21 of 66 b) Explanation: When the entropy change in a system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large in order to allow the process to be spontaneous. * An exothermic process increases the entropy of the surroundings. * An endothermic process decreases the entropy of the surroundings. TMHsiung©2015 Chapter 17 Slide 22 of 66 2) The Temperature Dependence of ΔSsurr a) Observation: Water freezes at temperatures below 0 oC spontaneously, however, freezing of water becomes nonspontaneous above 0 oC. b) Explanation: The greater the temperature, the smaller the increase in entropy for a given amount of energy dispersed into the surroundings. TMHsiung©2015 Chapter 17 Slide 23 of 66 Continued TMHsiung©2015 Chapter 17 Slide 24 of 66 ***** 3) Quantifying Entropy Changes in the Surroundings At Constant P and T Ssurr qsys Ssurr = Ssurr 1 T qsys T qsys H sys Ssurr = TMHsiung©2015 H sys Chapter 17 T Slide 25 of 66 ***** Example 17.2 Calculating Entropy Changes in the Surroundings Consider the combustion of propane gas: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔHrxn = –2044 kJ a. Calculate the entropy change in the surroundings associated with this reaction occurring at 25 oC. b. Determine the sign of the entropy change for the system. c. Determine the sign of the entropy change for the universe. Will the reaction be spontaneous? Solution a. TMHsiung©2015 Chapter 17 Slide 26 of 66 Continued b. An increase in the number of moles of gas implies a positive ΔSsys. c. ΔSuniv is positive and the reaction is spontaneous. TMHsiung©2015 Chapter 17 Slide 27 of 66 5. Gibbs Free Energy 1) The Equation ***** At Constant T and P: Since ΔSuniv = ΔSsurr + ΔSsys , therefore, ΔH sys ΔSuniv = ΔSsys T multiple by T, TΔSuniv = TΔSsys + ΔHsys = –(ΔHsys – TΔSsys) Multiple by –1 –TΔSuniv = ΔHsys – TΔSsys Let introduce another thermodynamic state function, Gibbs free energy, G, which is defined as: ΔG = –TΔSuniv TMHsiung©2015 Chapter 17 Slide 28 of 66 ***** Thus, ΔG = ΔHsys – TΔSsys Now, let’s memorize this Gibbs equation as: ΔG = ΔH – TΔS 2) A Summary ΔG < 0 : spontaneous process ΔG = 0 : at equilibrium ΔG > 0 : nonspontaneous process TMHsiung©2015 Chapter 17 ***** Slide 29 of 66 Continued TMHsiung©2015 Chapter 17 Slide 30 of 66 3) The Effect of ΔH, ΔS, and T on Spontaneity ***** ΔG = ΔH − TΔS TMHsiung©2015 Chapter 17 Slide 31 of 66 ***** Example 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from ΔH and ΔS Consider the reaction for the decomposition of carbon tetrachloride gas: CCl4(g) → C(s, graphite) + 2 Cl2(g) ΔH = + 95.7 kJ; ΔS = +142.2 J/K a. Calculate ΔG at 25 oC and determine whether the reaction is spontaneous. b. If the reaction is not spontaneous at 25 oC, determine at what temperature (if any) the reaction becomes spontaneous. Solution a. The reaction is not spontaneous. TMHsiung©2015 Chapter 17 Slide 32 of 66 ***** Continued b. Since ΔS is positive, ΔG becomes more negative with increasing temperature. The reaction is spontaneous above this temperature. TMHsiung©2015 Chapter 17 Slide 33 of 66 ***** 6. Entropy Changes in Chemical Reactions: Calculating ΔSrxn 1) Standard State • For Gas: pure gas at exactly 1 atm pressure. • For Solid or Liquid: 1 mole pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest. Usually 25 oC • Solution: Substance in a solution with concentration 1 M. TMHsiung©2015 Chapter 17 Slide 34 of 66 2) Third Law of Thermodynamics ***** For a perfect crystal at absolute zero (0 K), the absolute entropy = 0 J/mol ∙ K. S = k ln W = k ln 1 = 0 TMHsiung©2015 Chapter 17 Slide 35 of 66 3) Factors that affect the standard entropy • State of the substance • Molar mass of the substance • Particular allotrope • Molecular complexity • Dissolution TMHsiung©2015 Chapter 17 Slide 36 of 66 State of the substance Molar mass of the substance TMHsiung©2015 Chapter 17 Slide 37 of 66 Particular allotrope * Allotrope: Those elements can exist in two or more forms. • • Diamond: three-dimensional crystal structure. Graphite: the atoms bond together in sheets, but the sheets have freedom to slide past each other. TMHsiung©2015 Chapter 17 Slide 38 of 66 Molecular complexity ***** • Ar: only translational motion. • NO, translational motion, rotational motion, and vibrational motions of the molecules TMHsiung©2015 Chapter 17 Slide 39 of 66 Dissolution TMHsiung©2015 Chapter 17 Slide 40 of 66 4) The Standard Entropy Change, ΔS ***** The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions. ΔSºreaction = (∑npSºproducts) − (∑nrSºreactants) The standard enthalpy of formation, ΔHf°, of an element is 0 kJ/mol. The absolute entropy (So) at 25 oC is always positive. TMHsiung©2015 Chapter 17 Slide 41 of 66 Standard Absolute Entropies ***** * More complete list can be found in Appendix IIB ** Intensive property: A property such as density that is independent of the amount of a given substance. Extensive property: A property that depends on the amount of a given substance, such as mass. TMHsiung©2015 Chapter 17 Slide 42 of 66 ° ) Example 17.4 Calculating Standard Entropy Changes (ΔSrxn ***** Calculate°ΔSrxn for the balanced chemical equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Solution Appendix IIB. ° * ΔSrxn is positive, the number of moles of gas increases. OK. TMHsiung©2015 Chapter 17 Slide 43 of 66 7. Free Energy Changes in Chemical Reactions: Calculating ΔGorxn 1) Standard free energies of formation (Gof) The change in free energy when 1 mol (or 1 M) of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation listed in Appendix IIB . Same as standard formation enthalpy (Hfo), Gof of the pure elements in their standard states is zero. TMHsiung©2015 Chapter 17 Slide 44 of 66 ***** 2) Calculating Gorxn with Tabulated Values of Free Energies of Formation (Gof) ΔGorxn = ∑npΔGfo(products) − ∑nrΔGfo(reactants) TMHsiung©2015 Chapter 17 Slide 45 of 66 ***** ° Example 17.7 Ozone in the lower atmosphere is a pollutant that can form by the following reaction involving the oxidation of unburned hydrocarbons: CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g) Use the standard free energies of formation°to determine ΔGrxn for this reaction at 25 oC. Solution Appendix IIB * Gof of a pure element is zero. ° TMHsiung©2015 Chapter 17 Slide 46 of 66 ***** 3) Calculating Standard Free Energy Changes with ΔGorxn = ΔHorxn − TΔSorxn Example 17.5 At 25 oC One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant SO2 to SO3 by the reaction: Calculate ΔGrxn at 25 oC and determine whether the reaction is spontaneous. Solution Appendix IIB TMHsiung©2015 Chapter 17 Slide 47 of 66 Continued The reaction is spontaneous at this temperature because ΔGrxn is negative. TMHsiung©2015 Chapter 17 Slide 48 of 66 Example 17.6 Other than 25 oC ***** For the reaction in Example 17.5, estimate the value of ΔGorxn at 125 oC. Is the reaction more or less spontaneous at °this elevated temperature; that is, is the value of ΔGorxn more negative (more spontaneous) or more positive (less spontaneous)? Solution ΔGorxn at 125 oC is less negative (or more positive) than the value of ΔGorxn at 25 oC (which is –70.9 kJ), the reaction is less spontaneous. * The different temperature resulted in relative small variations of ΔSorxn and ΔHorxn (25oC). TMHsiung©2015 Chapter 17 Slide 49 of 66 4) Calculating ΔGorxn for a Stepwise Reaction ° for a Stepwise Reaction Example 17.8 Calculating ΔGrxn ° Find ΔG rxn for the reaction. 3 C(s) + 4 H2(g) → C3H8(g) Use the following reactions with known ΔG’s: Solution TMHsiung©2015 Chapter 17 Slide 50 of 66 5) Coupled Reactions Coupled reactions: Use a thermodynamically favorable reaction to drive an unfavorable one. Example1: TMHsiung©2015 Chapter 17 Slide 51 of 66 Example 2 (a biochemical cycle) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ADP + H3PO4 → ATP + H2O(l) C6H12O6(s) + 6O2(g) + ADP + H3PO4 → 6CO2(g) + ATP + 7H2O(l) Alnaine + Glycine → Alnanylglycine ATP + H2O → ADP + H3PO4 ATP + H2O + Alnaine + Glycine → Ananylglycine + ADP + H3PO4 TMHsiung©2015 Chapter 17 ΔGo = –2280 kJ ΔGo = +31 kJ ΔGo = –2249 kJ ΔGo = +29 kJ ΔGo = –31 kJ ΔGo = –2 kJ Slide 52 of 66 8. Free Energy Changes for Nonstandard States: The Relationship between ΔGorxn and ΔGrxn Q. Spilled water spontaneously evaporates even though ΔGorxn for the vaporization of water is positive. Why? A. Because ordinary conditions are not standard conditions and ΔGorxn applies only to standard conditions. Gorxn describes the free energy change at standard state. Grxn describes the free energy change at any specified state. TMHsiung©2015 Chapter 17 Slide 53 of 66 The Free Energy Change of a Reaction under Nonstandard Conditions ***** Grxn = Gorxn + RTlnQ Q: R: T: TMHsiung©2015 reaction quotient 8.314 Jmol–1K–1 K Chapter 17 Slide 54 of 66 Example Grxn = Gorxn + RTlnQ (a) Standard Conditions, PH2O = 1 atm, i.e., Q = 1 Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln(1) = +8.59 kJ/mol (spontaneous in the reverse direction) (b) At pressure of 0.0313 atm, i.e., Q = Kp = 0.0313 Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln (0.0313) = +8.59 kJ/mol - 8.59 kJ/mol = 0 (c) At pressure of 5.00x10-3 atm, i.e., Q = Kp = 5.00x10-3 atm Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln (5.00x10-3) = +8.59 kJ/mol – 13.1 kJ/mol = – 4.5 kJ/mol TMHsiung©2015 Chapter 17 Slide 55 of 66 Continued TMHsiung©2015 Chapter 17 Slide 56 of 66 Example 17.9 Calculating ΔGrxn under Nonstandard Conditions Consider the reaction at 298 K: ***** Calculate ΔGrxn under these conditions: PNO = 0.100 atm; PO2 = 0.100 atm; PNO2 = 2.00 atm Is the reaction more or less spontaneous under these conditions than under standard conditions? Solution ° The reaction is spontaneous under these conditions, but less spontaneous than it would be under standard conditions. ° TMHsiung©2015 ° Chapter 17 Slide 57 of 66 9. Free Energy and Equilibrium: Relating ΔGorxn ***** to the Equilibrium Constant ( K )Free En ergy 1) Equation and Explanation Grxn = Gorxn + RTlnQ At equilibrium, Grxn = 0 and Q = K 0 = Gorxn + RTlnQ = Gorxn + RTlnK Gorxn = -RTlnK = -2.303RTlogKeq lnK = -Go/RT TMHsiung©2015 Chapter 17 Slide 58 of 66 Gorxn = -RTlnK a) When K < 1 Gorxn is positive. Under standard conditions (when Q = 1 ), the reaction is spontaneous in the reverse direction. TMHsiung©2015 Chapter 17 Slide 59 of 66 Gorxn = -RTlnK b) When K > 1 Gorxn is negative. Under standard conditions (when Q = 1 ), the reaction is spontaneous in the forward direction. TMHsiung©2015 Chapter 17 Slide 60 of 66 Gorxn = -RTlnK c) When K = 1 Gorxn is zero. Under standard conditions (when Q = 1 ), the reaction is equilibrium. TMHsiung©2015 Chapter 17 Slide 61 of 66 ***** Example 17.10 The Equilibrium Constant and ΔGrxn Use tabulated free energies of formation to calculate the equilibrium constant for the reaction at 298 K: Solution Appendix IIB ° TMHsiung©2015 Chapter 17 Slide 62 of 66 2) The Temperature Dependence of the Equilibrium Constant For a particular reaction, H and S do not change much for different temperature. At standard state: Gorxn = – RTlnK (1) Gorxn = Horxn – TSorxn (2) Combing (1) and (2) – RTlnK = Horxn – TSorxn = Gorxn o o H rxn S 1 rxn ln K R T R y = mx + b TMHsiung©2015 Chapter 17 Slide 63 of 66 Example: For the reaction: CO(g) + H2O(g) → CO2(g) + H2(g) lnKeq vs. 1/T plot Slope 4810 = –Ho/R (R=8.3145 J mol–1K–1) Ho – 40 kJ (tabulated data is – 41.2 kJ). TMHsiung©2015 Chapter 17 Slide 64 of 66 Calculating K at different temperature H 1 S H 1 ln K Constant R T R R T o rxn o rxn o rxn van't Hoff equation: K 2 H ln K1 R o rxn TMHsiung©2015 Chapter 17 1 1 ( ) T1 T2 Slide 65 of 66 End of Chapter 17 TMHsiung©2015 Chapter 17 Slide 66 of 66