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Free Energy (Symbol: G)

Free Energy (Symbol: G). G = H - TS.  G =  H - T  S (from the standpoint of the system) Divide by T:  G/T =  H/T -  S = -  S surr -  S (constant T and P) =   S Universe +  S univ = -  G T.

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Free Energy (Symbol: G)

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  1. Free Energy (Symbol: G) G = H - TS G = H-TS (from the standpoint of the system) Divide by T: G/T = H/T-S = -Ssurr -S (constant T and P) = SUniverse +Suniv = -G T

  2. A process (at constant T, P) is spontaneous in the direction in which free energy decreases: • G < 0 • Since Suniverse = - G / T at constant T and P • Therefore in a spontaneous process, the free • Energy (of the system) is decreasing

  3. H2O (s)  H2O(l) Calculation of Go = Ho -TSo for 3 different temperatures Process is Spontaneous if : Suniverse > 0 ; or G < 0 (Refer to Chapter 6, p. 246, for definition of standard state)

  4. Exothermicity is important at low temperatures, but not very important at high temperatures

  5. Entropy Changes in Chemical Reactions N2 (g) + 3 H2 (g)  2NH3 (g) S < 0 [4 molecules of gas  2 molecules of gas] Fewer molecules mean fewer possible configurations H2 (g)  2H (g) S > 0 [1 molecule of gas  2 atoms of gas] higher possible configuration, higher disorder 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O(g) S > 0 [9 molecules of gas  10 molecules of gas] higher possible configuration, higher disorder

  6. The change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O(g) S > 0 [9 molecules of gas  10 molecules of gas] higher possible configuration, higher disorder CaCO3 (s)  CaO (s) + CO2 (g) S > 0

  7. The Third Law of Thermodynamics • . . . the entropy of a perfect crystal at 0 K is zero. • Because S is explicitly known (= 0) at 0 K, S values at other temps can be calculated.

  8. The 3rd Law of Thermodynamics: • A perfect crystal of hydrogen chloride at 0 K. • (b) As the temperature rises above 0 K, lattice vibrations allow some dipoles to change their orientations, producing some disorder and an increase in entropy.

  9. Calculating Entropy Changes for Chemical Reactions: Soreaction = npSoproducts -  nrSoreactants (extensive property) So denotes Standard Entropy: the entropy when the material is in its standard state (P = 1 atm, a = 1 M), reported at 298 K in Appendix 4. Calculate Soreaction for the reduction of aluminum oxide by H2(g): Al2O3 (s) + 3H2 (g) 2 Al (s) + 3H2O(g) Given: So (Al2O3) = 51 J/K.mol; So (H2) = 131 J/K.mol; So (Al) = 28 J/K.mol; So (H2O) = 189 J/K.mol Soreaction = npSoproducts -  nrSoreactants= 2 SoAl + 3  SoH2O – So Al2O3 - 3  SoH2 = 179 J/K Note: 3 molecules of gas  3 molecules of gas yet Soreaction > 0

  10. The more complex is the molecule The higher is its standard entropy The H2O molecule can vibrate and rotate in several ways, this freedom of motion leads to higher entropy than the diatomic H2 molecule

  11. Free Energy and Chemical Reactions • Go = standard free energy change that occurs if reactants in their standard state are converted to products in their standard state. Example: N2 (g) + 3H2 (g)  2NH3 (g) Go=-33.3 Kj; when all gases are in their standard state, means at gas pressures of 1 atm

  12. Methods for calculating free energy changes: free energy cannot be measured directly --systems go to equilibrium state • From the values of Ho and So, since Go = Ho -T  So • Ho = npHf(products) nrHf(reactants) • Soreaction = npSo (products)-  nrSo(reactants) 2. Since Gis a state function: meaning it depends only on state and not on pathway to reach the state, we can use values of G for other reactions (like Hess’s law for enthalpy changes)

  13. Method 1. Consider the reaction: 2SO2 (g) + O2 (g) 2SO3 (g) Ho Kj/molSo J/K.mol SO2 (g) -297 248 SO3 (g) -396 257 O2 (g) 0 205 Calculate the standard free energy change at 25oC. H = npHf(products) nrHf(reactants)= =2mol(-396 kJ/mol)-2 mol(-297 kJ/mol)-0=-198 kJ Soreaction = npSo (products)-  nrSo(reactants)= =2mol(257 J/K.mol)-2 mol(248 J/K.mol)-1mol (205 J/K.mol)- =-187 J/K G = H - TS =-198 kJ+55.7 kJ=-142 kJ.

  14. Method 2. Consider the reaction: 2CO (g) + O2 (g) 2CO2 (g) Obtain G for this reaction from the following data: 2CH4(g) + 3O2 (g) 2CO (g)+4H2O(g) G =-1088 kJ (1) CH4 (g) + 2O2 (g) CO2 (g)+2H2O(g) G =-801 kJ (2) Reverse Reaction (1): 2CO (g)+4H2O(g) 2CH4(g) + 3O2 (g) G =1088 kJ Multiply Reaction (2) by 2: 2CH4 (g) + 4O2 (g) 2CO2 (g)+4H2O(g) G =2(-801 kJ) 2CO (g) + O2 (g) 2CO2 (g) G =1088 kJ+ 2(-801 kJ) = -514 kJ Add:

  15. Method 2. Consider the reaction: 2CO (g) + O2 (g) 2CO2 (g) Obtain G for this reaction from the following data: 2CH4(g) + 3O2 (g) 2CO (g)+4H2O(g) G =-1088 kJ (1) CH4 (g) + 2O2 (g) CO2 (g)+2H2O(g) G =-801 kJ (2) Reverse Reaction (1): 2CO (g)+4H2O(g) 2CH4(g) + 3O2 (g) G =1088 kJ Multiply Reaction (2) by 2: 2CH4 (g) + 4O2 (g) 2CO2 (g)+4H2O(g) G =2(-801 kJ) 2CO (g) + O2 (g) 2CO2 (g) G =1088 kJ+ 2(-801 kJ) = -514 kJ Add:

  16. Method 3 3. G can be calculated by using values of Gof the standard free energy of formation of reactants and products. Gof = defined as the change in free energy that accompanies the formation of 1 mol of that substance from its constituent elements with all reactants and products in their standard state: e.g. Formation of Glucose C6H12O6 (s), the reaction is: 6C(s) + 6H2 (g) + 3O2 (g)  C6H12O6 (s) Gof (glucose) G(reaction) = npGf(products)nrGf(reactants) note: the standard free energy of formation of an element is zero. e.g., O2 (g)  O2 (g)

  17. Method 3. Methanol is a high-octane fuel used in high-performance racing Engines. Calculate Go of the reaction given the Gof: 2CH3OH (g) + 3O2 (g) 2CO2 (g)+4H2O (g) Substance Gof (kJ/mol) CH3OH -163 O2 0 CO2 -394 H2O -229 G(reaction) = npGf(products)nrGf(reactants)= 2mol(-394 kJ/mol)+4 mol(-229 kJ/mol)- 2mol(-163 kJ/mol) -3mol  0 =-1378 kJ

  18. The dependence of free energy on pressure •A system at constant temperature and pressure will proceed spontaneously in the direction that lowers its free energy • Reactions therefore proceed until they reach equilibrium The Equilibrium Position is the lowest free energy value available for the reaction system G = f (pressure) Therefore G varies as the reaction proceeds with varying pressure (or concentrations)

  19. G = H-TS = f(P) • Assume an ideal gas: • H is not pressure dependent • But Entropy S is: • Slarge volume > Ssmall volume • Or, since pressure and volume are inversely related: • Slow pressure > Shigh pressure • S is a function of Pressure • G is a function of Pressure

  20. G = Go + RT ln (P) Pressure Gas Constant Temperature (K) Consider: N2 (g) + 3H2 (g)  2NH3 (g) G(reaction) = npG(products)nrG(reactants) G(reaction) = 2G(NH3)- G (N2)-3G (H2) (1) Where G(NH3) = Go (NH3)+ RTln(P NH3) G(N2) = Go (N2) + RTln(P N2) G(H2) = Go (H2) + RTln(P H2) Replacing into the equation (1) gives:

  21. Consider: N2 (g) + 3H2 (g)  2NH3 (g) • G= npG(products)nrG(reactants) • Greaction = 2G(NH3)- G (N2)-3G (H2) = • 2Go(NH3)- Go (N2)-3Go (H2) + RT[2 ln(P NH3) – ln (P N2) – 3 ln (P H2)] • Goreaction + RT ln P2NH3 P N2 P3H2 • G = G + RT ln(Q) • Q = reaction quotient from the law of mass action. • R 8.4135 J/K mol

  22. Consider: CO (g) + 2 H2 (g)  CH3OH (l) Calculate Gat 25oC for this reaction when CO (g) at 5.0 atm and H2 (g) at 3.0 atm are converted to liquid methanol G = G + RT ln(Q) Step 1: compute G from Gf G(reaction) = npGf(products)nrGf(reactants) = -166 kJ –(-137 kJ) –0 = -29 kJ Step 2: compute G = G + RT ln(Q) Q = 1 = 1/(5.0)(3.0)2 = 0.022 (P CO ) (P H22 ) G = G + RT ln(Q)= (-2.9  104 J/mol rxn)+(8.3145 J/mol.K)(298 K)ln(0.022) = -3.8  104 J/mol rxn = -38 kJ/mol rxn (note: more negative than Go therefore more spontaneous at these conditions)

  23. The meaning of G for a chemical reaction • The calculation showed that G < 0 for the formation of CH3OH (l) from CO (g) at 5 atm mixed with H2 (g) at 3.0 atm. • Does this mean that we will form 1 mol CH3OH in the flask • The answer is NO! • Because when the CO and H2 are mixed together at these pressures there is even a lower free energy state of the system than 1 mol CH3OH. • The system can achieve its lowest free energy by going to equilibrium and not by going to completion

  24. Schematic representations of balls rolling down two types of hills. Equilibrium Analogous to Phase Change Analogous to Chemical Reaction

  25. Consider A (g)  B (g) GA = GAo + RT ln (PA) GB = GBo + RT ln (PB) Total Free Energy = G = GA + GB • The initial free energies of A and B. • As A(g) changes to B(g), the free energy of A decreases and that of B increases. • Eventually, pressures of A and B are achieved such that GA = GB, the equilibrium position. G=0 PAe PBe

  26. The change in free energy to reach equilibrium, beginning with 1.0 mol A(g) at PA = 2.0 atm. • The change in free energy to reach equilibrium, beginning with 1.0 mol B(g) at PB = 2.0 atm. • (c) The free energy profile for A(g)  B(g) in a system containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm. Each point on the curve corresponds to the total free energy of the system for a given combination of A and B. 75% of A changed into B PA = (0.25)(2.0 atm) = 0.50 atm PB = (0.75)(2.0 atm) = 1.5 atm

  27. In summary: when substances undergo a chemical reaction, the reaction proceeds to the minimum free energy (Equilibrium) which corresponds to the point where: Gproducts = Greactants or G = 0

  28. The equilibrium constant K G = G + RT ln(Q) At equilibrium Q = K and G = 0 • G = RT ln(K) • K = equilibrium constant • This is so because G = 0 and Q = K at equilibrium.

  29. Discussion of the meaning of the sign of Go A (g)  B (g)

  30. Sample 16.14: Consider the ammonia synthesis reaction: • N2 (g) + 3H2 (g)  2 NH3 (g) • Where Go = -33.3 kJ per mole of N2 consumed at 25 oC. For each • of the following mixtures of reactants and products at 25 oC, predict • the direction in which the system will shift to reach equilibrium • PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00  10-2 atm • 2. PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm

  31. Sample 16.15: Consider the overall reaction of the corrosion of iron (rusting): 4Fe (s) + 3O2 (g)  2 Fe2O3 (s) Using the following data, calculate the equilibrium constant for This reaction at 25 oC Substance Hof(kJ/mol)So (J/K.mol) Fe2O3 (s) -826 90 Fe (s) 0 27 O2 (g) 0 205

  32. Temperature dependence of K G = RT ln(K) = H -T S • y = mx + b • (H and S independent of temperature over a small temperature range) ln(K) = -H/RT + S/R Plot of ln K vs. 1/T linear; note sign of slope and dependence on exo- or endo- thermicity Exothermic, K decreases with increasing T Endothermic, K increases with increasing T

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