# Chapter 19 Principles of Reactivity: Entropy & Free Energy

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Chapter 19 Principles of Reactivity: Entropy & Free Energy
Chem 106, Chapter 19 Chapter 19 Principles of Reactivity: Entropy & Free Energy Copyright 2007, David R. Anderson

Spontaneity Spontaneous processes Factors affecting spontaneity Entropy Entropy changes The third law of thermodynamics Calculating entropy changes in chemical reactions The second law of thermodynamics Gibbs Free Energy Definition Calculating free energy changes in chemical reactions Standard free energy of formation Free energy and temperature Free energy and equilibrium

Chemical Thermodynamics
Chapter 6: Thermochemistry heat of reaction, DH first law: conservation of energy Thermodynamics: energy changes entropy (DS), free energy (DG) second and third laws  spontaneity of reaction, position of equilibirum I. Spontaneity A. Spontaneous process = one that, once started, continues on its own without input of energy nonspontaneous = needs continual input of energy

I. Spontaneity B. Factors affecting spontaneity
1. decrease in potential energy (DH < 0) apple falling water flowing downhill 2H2 + O2  2H2O HCl + NaOH  NaCl + H2O thermite 2. increase in disorder (tendency toward higher probability) deck of cards (6 x 109 combinations) gas molecules in a container salt dissolving in water ice melting, water evaporating

II. Entropy, S = degree of disorder (probability)
Chem 106, Chapter 19 II. Entropy, S = degree of disorder (probability) A. Entropy changes Entropy change, DS = Sf - Si 1. examples expansion of a gas DS dissolution of a solid DS H2O(s)  H2O(l)  H2O(g) DS H2O(g)  H2O(l)  H2O(s) DS H2CO3  CO2+ H2O DS 2H2 + O2  2H2O DS CaO + SO2  CaSO3 DS remove partition NaCl(s) NaCl(aq) H2O Copyright 2007, David R. Anderson

II. Entropy A. Entropy changes
2. calculating DS for fusion and vaporization DS  q increase KE  increase in disorder DS = q T 10 K  20 K DS1 q DS  1 T DS1 > DS2 300 K  310 K DS2 q e.g., For H2O(s)  H2O(l), DHfus = kJ/mol at 0ºC and for H2O(l)  H2O(g), DHvap = kJ/mol at 100ºC. What are DSfus and DSvap?

II. Entropy B. The third law of thermodynamics
“For a pure, crystalline substance at 0 K, S = 0” Increase T  increase in: translational motion vibrational motion rotational motion increase in S Standard state entropy, Sº = entropy of a substance at 25ºC, 1 atm (Table 19.1, Appendix L) (Sº  0 for elements!)

II. Entropy C. Calculating entropy changes in chemical reactions
Standard entropy change, DSº DSºrxn = Sºproducts - Sºreactants e.g., 2H2 + O2  2H2O DSºrxn = [2Sº(H2O)] - [2Sº(H2) + Sº(O2)] = [2(188.8)] - [2(130.6) + (205.0)] = J/K (note units)

II. Entropy C. Calculating entropy changes in chemical reactions
e.g., Calculate the standard entropy change for the reaction, 2Al(s) + 3Cl2(g)  2AlCl3(s) Sº(Al) = 28.3 J/mol·K Sº(Cl2) = J/mol·K Sº(AlCl3) = J/mol·K

II. Entropy D. The second law of thermodynamics
“In any spontaneous process, the entropy of the universe must increase (DSuniverse > 0). system surroundings universe DSuniverse = (DSsystem + DSsurroundings) > 0 e.g., erecting a building 6CO2 + 6H2O  C6H12O6 + 6O2 CFCs

III. Gibbs Free Energy A. Definition
DSuniverse = DSsystem + DSsurroundings DSsurroundings = – qsystem T = – DHsystem DSuniverse = DSsystem – DHsystem T TDSuniverse = DHsystem – TDSsystem DGsystem = DHsystem – TDSsystem Gibbs free energy state function relates spontaneity to heat of reaction (DH) and entropy (DS) for a process to be spontaneous, DSuniverse > 0 or DGsystem < 0

III. Gibbs Free Energy A. Definition DG < 0: exergonic, spontaneous
DG > 0: endergonic, nonspontaneous DG = 0: system is at equilibrium (no free energy remains) G  Reactants Products Equilibrium DGrxn < 0 “spontaneous” R  E DG < 0 spontaneous P  E DG < 0 spontaneous E  R DG > 0 nonspontaneous E  P DG > 0 nonspontaneous

III. Gibbs Free Energy B. Calculating free energy changes in chemical reactions e.g., C(s) + ½O2(g) + 2H2(g)  CH3OH(l) at 25º DHºrxn = DHfº(products) – DHfº(reactants) = –238.7 kJ/mol DSºrxn = Sºproducts – Sºreactants = [Sº(CH3OH)] – [Sº(C) + ½Sº(O2) + 2Sº(H2)] = [126.8] – [(5.7) + ½(205.1) + 2(130.7)] = –242.9 J/mol·K DGºrxn = DHºrxn – TDSºrxn = (–238.7 kJ/mol) – (298 K)(–242.9 J/mol·K) = –166.3 kJ/mol = DGfº(CH3OH)

III. Gibbs Free Energy C. Using standard free energies of formation
DGºrxn = DGfº(products) - DGfº(reactants) standard free energies of formation (Table 19.3, Appendix L; note: DGfº(element) = 0) e.g., CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DGºrxn = [DGfº(CO2) + 2DGfº(H2O)] - [DGfº(CH4)] = [(-394.4) + 2(-237.1)] - [(-50.7)] = kJ

III. Gibbs Free Energy C. Using standard free energies of formation
e.g., If DGºrxn for the following reaction is kJ, what is DGºf for TiCl2? TiCl2(s) + Cl2(g)  TiCl4(s) (DGºf(TiCl4) = kJ/mol)

III. Gibbs Free Energy D. Free energy and temperature DG = DH – TDS
DH DS DG – – at all temperatures + – + at all temperatures – at higher temperatures, + at lower temperatures – – – at lower temperatures, + at higher temperatures e.g., C12H22O O2  12CO2 + 11H2O (produces heat) DH < 0, DS > 0  DG < 0 at any temperature (for reverse reaction, DG > 0 at any temperature; requires input of energy) NH4NO3(s)  NH4+(aq) + NO3–(aq) (gets cold) DH > 0, DS > 0  DG < 0 at higher temperatures H2O

III. Gibbs Free Energy D. Free energy and temperature
e.g., CH3–CH3  CH2=CH2 + H2 DHº = +137 kJ DSº = J/mol·K at 25ºC, DGº = kJ nonspontaneous Above what temperature does the reaction become spontaneous?

III. Gibbs Free Energy E. Free energy and equilibrium DG = DGº + RTlnQ
At equilibrium, DG = 0, Q = K 0 = DGº + RTlnK DGº = –RTlnK DGº < 0  K > 1 equilibrium lies to the right DGº > 0  K < 1 equilibrium lies to the left DGº = 0  K = 1 equilibrium lies in the middle G  R P DGº > 0 DGº < 0 DGº = 0

III. Gibbs Free Energy E. Free energy and equilibrium DGº = –RTlnK
or K = e–Gº/RT e.g., 2SO2+ O2 2SO3 What is K at 25ºC? DGºf(SO2) = kJ/mol DGºf(SO3) = kJ/mol