Chapter 18 Entropy, Free Energy and Equilibrium Copyright McGraw-Hill 2009

18.1 Spontaneous Processes Spontaneous: process that does occur under a specific set of conditions Nonspontaneous: process that does not occur under a specific set of conditions Copyright McGraw-Hill 2009

Spontaneous Processes Often spontaneous processes are exothermic, but not always…. Methane gas burns spontaneously and is exothermic Ice melts spontaneously but this is an endothermic process… There is another quantity! Copyright McGraw-Hill 2009

Spontaneous vs Nonspontaneous Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 18.2 Entropy Entropy (S): Can be thought of as a measure of the disorder of a system In general, greater disorder means greater entropy Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Probability- what is the chance that the order of the cards can be restored by reshuffling? Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Entropy is a state function just as enthalpy S = Sfinal  Sinitial Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Standard Entropy Standard entropy: absolute entropy of a substance at 1 atm (typically at 25C) Complete list in Appendix 2 of text What do you notice about entropy values for elements and compounds? Units: J/K·mol Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Entropies Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Trends in Entropy Entropy for gas phase is greater than that of liquid or solid of same substance I2 (g) has greater entropy than I2 (s) More complex structures have greater entropy C2H6 (g) has greater entropy than CH4 (g) Allotropes - more ordered forms have lower entropy Diamond has lower entropy than graphite Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Ssolid < Sliquid Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Sliquid < Svapor Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Spure < Saqueous Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Slower temp < Shigher temp Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Sfewer moles < Smore moles Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative Determine the sign of S for the following (qualitatively) 1. Liquid nitrogen evaporates 2. Two clear liquids are mixed and a solid yellow precipitate forms 3. Liquid water is heated from 22.5 C to 55.8 C Copyright McGraw-Hill 2009

18.3 The Second and Third Laws of Thermodynamics System: the reaction Surroundings: everything else Both undergo changes in entropy during physical and chemical processes Copyright McGraw-Hill 2009

Second Law of Thermodynamics Entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Equilibrium process: caused to occur by adding or removing energy from a system that is at equilibrium Copyright McGraw-Hill 2009

Second Law of Thermodynamics Mathematically speaking: Spontaneous process: Suniverse = Ssystem + Ssurroundings > 0 Equilibrium process: Suniverse = Ssystem + Ssurroundings = 0 Copyright McGraw-Hill 2009

Entropy Changes in the System Entropy can be calculated from the table of standard values just as enthalpy change was calculated. Srxn = nS products  mS reactants Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Standard Entropy Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively) 2NH3(g)  N2(g) + 3H2(g) Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Standard Entropy 2NH3(g)  N2(g) + 3H2(g) Srxn = nS products  mS reactants = [(1)(191.5 J/K · mol) + (3)(131.0 J/K · mol)] - [(2)(193.0 J/K · mol)] = 584.5 J/K · mol - 386 J/K · mol Srxn = 198.5 J/K · mol (Entropy increases) (2 mol gas  4 mol gas) Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Your Turn! Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively) 2H2(g) + O2(g)  2H2O (g) Copyright McGraw-Hill 2009

Third Law of Thermodynamics Entropy of a perfect crystalline substance is zero at absolute zero. Importance of this law: it allows us to calculate absolute entropies for substances Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 18.4 Gibbs Free Energy G = H – T S The Gibbs free energy, expressed in terms of enthalpy and entropy, refers only to the system, yet can be used to predict spontaneity. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Gibbs Free Energy If G < 0,negative, the forward reaction is spontaneous. If G = 0, the reaction is at equilibrium. If G > 0, positive, the forward reaction is nonspontaneous Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Predicting Sign of G Copyright McGraw-Hill 2009

Predicting Temperature from Gibbs Equation Set G = 0 (equilibrium condition) 0 = H – T S Rearrange equation to solve for T- watch for units! This equation will also be useful to calculate temperature of a phase change. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Example For a reaction in which H = 125 kJ/mol and S = 325 J/Kmol, determine the temperature in Celsius above which the reaction is spontaneous. 385 K  273 = 112C Copyright McGraw-Hill 2009

Standard Free Energy Changes Free energy can be calculated from the table of standard values just as enthalpy and entropy changes. Grxn = nG products  mG reactants Copyright McGraw-Hill 2009

Standard Free Energy Changes Calculate the standard free-energy change for the following reaction. 2KClO3(s)  2KCl(s) + 3O2(g) Grxn = nG products  mG reactants = [2(408.3 kJ/mol) + 3(0)]  [2(289.9 kJ/mol)] = 816.6  (579.8) = 236.8 kJ/mol (spont) Copyright McGraw-Hill 2009

18.5 Free Energy and Chemical Equilibrium Reactions are almost always in something other than their standard states. Free energy is needed to determine if a reaction is spontaneous or not. How does free energy change with changes in concentration? Copyright McGraw-Hill 2009

Free Energy and Equilibrium G = G° + RT ln Q G = non-standard free energy G° = standard free energy (from tables) R = 8.314 J/K·mole T = temp in K Q = reaction quotient Copyright McGraw-Hill 2009

Free Energy and Equilibrium Consider the reaction, H2(g) + Cl2(g)  2 HCl(g) How does the value of G change when the pressures of the gases are altered as follows at 25 C? H2 = 0.25 atm; Cl2 = 0.45 atm; HCl = 0.30 atm Copyright McGraw-Hill 2009

Free Energy and Equilibrium First, calculate standard free energy: H2(g) + Cl2(g)  2 HCl(g) G° = [2(95.27 kJ/mol)]  [0 + 0] = 190.54 kJ/mol Second, find Q: Copyright McGraw-Hill 2009

Free Energy and Equilibrium Solve: G = G° + RT ln Q G = 190,540 J/mol + (8.314J/K·mol)(298 K) ln (0.80) G =  191.09 kJ/mol (the reaction becomes more spontaneous - free energy is more negative) Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Your Turn! Consider the reaction, O2(g) + 2CO(g)  2CO2(g) How does the value of G change when the pressures of the gases are altered as follows at 25 C? O2 = 0.50 atm; CO = 0.30 atm; CO2 = 0.45 atm Copyright McGraw-Hill 2009

Relationship Between G° and K At equilibrium, G = 0 and Q = K The equation becomes: 0 = G° + RT ln K or G° = – RT ln K Copyright McGraw-Hill 2009

Relationship Between G° and K Copyright McGraw-Hill 2009

Relationship Between G° and K Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C. 2HCl(g) H2(g) + Cl2(g) Copyright McGraw-Hill 2009

Relationship Between G° and K First, calculate the G°: = [0 + 0]  [2(95.27 kJ/mol)] = 190.54 kJ/mol (non-spontaneous) Substitute into equation: 190.54 kJ/mol =  (8.314 x 103 kJ/K·mol)(298 K) ln KP 76.90 = ln KP = 3.98 x 1034 K < 1 reactants are favored Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Your Turn! Calculate the value of the equilibrium constant, KP, for the following reaction at 25 C. 2NO2(g) N2O4(g) Copyright McGraw-Hill 2009

18.6 Thermodynamics in Living Systems Coupled reactions Thermodynamically favorable reactions drives an unfavorable one Enzymes facilitate many nonspontaneous reactions Copyright McGraw-Hill 2009

Thermodynamics in Living Systems Examples: C6H12O6 oxidation: G° = 2880 kJ/mol ADP  ATP G° = 31 kJ/mol Synthesis of proteins: (first step) alanine + glycine  alanylglycine G° = 29 kJ/mol Copyright McGraw-Hill 2009

Thermodynamics in Living Systems Consider the coupling of two reactions: ATP + H2O + alanine + glycine  ADP + H3PO4 + alanylglycine Overall free energy change: G° = 31 kJ/mol + 29 kJ/mol = 2 kJ/mol Protein synthesis is now favored. Copyright McGraw-Hill 2009

ATP-ADP Interconversions Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Key Points Spontaneous vs nonspontaneous Relate enthalpy, entropy and free energy Entropy Predict qualitatively Calculate from table of standard values Calculate entropy for universe Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Key Points Free energy Calculate from table of standard values Calculate from Gibbs equation Calculate non-standard conditions Calculate in relationship to equilibrium constant Copyright McGraw-Hill 2009