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Thermodynamics of Reactions

Thermodynamics of Reactions. Entropy and Chemical Reactions Example : N 2 (g) + 3H 2 (g) 2NH 3 (g) System: positional probability 4 reactant particles 2 product particles Fewer possible configurations, D S = -

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Thermodynamics of Reactions

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  1. Thermodynamics of Reactions • Entropy and Chemical Reactions • Example: N2(g) + 3H2(g) 2NH3(g) • System: positional probability • 4 reactant particles 2 product particles • Fewer possible configurations, DS = - • An increase in number of gas particles is entropically favored • 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) • Nine particles 10 particles, DS = + • Example: Predict sign of DS • CaCO3(s) CaO(s) + CO2(g) • 2SO2(g) + O2(g) 2SO3(g)

  2. Third Law of Thermodynamics = the entropy of a perfect crystal at 0K = 0 (there is only one possible configuration) • As temperature increases, random vibrations occur; DS = + • We can calculate S for any substance if we know how S depends on T [Standard So values, Appendix IIB] • DSoreaction = SnSoprod - SnSoreactants • Example: Find DSo at 25 oC for 2NiS(s) + 3O2(g) 2SO2(g) + 2NiO(s) So(J/K mol) 53 205 248 38 DSo = 2(248) + 2(38) – 2(53) – 3(205) = -149 J/Kmol (3 gas particles 2 gas particles)

  3. Factors Affecting Standard Entropies • Phase of Matter: as discussed earlier, Sgas >> Sliquid > Ssolid • So H2O(l) = 70.0 J/Kmol • So H2O(g) = 188.8 J/Kmol • Molar Mass: heavy elements have more entropy than light (in same state) Has to do with translation energy states • Allotropes: more rigid structures have less entropy • Molecular Complexity: more complex = more entropy • So Ar(g) = 154.8 (39.948 g/mol) • So NO(g) = 210.8 (30.006 g/mol)

  4. Free Energy and Chemical Reactions • Standard Free Energy Change = DGo = reactants/products at standard states • Gases at 1 atm, solution = 1 M, element at 25 oC and 1 atm has Go = 0 • N2(g) + 3H2(g) 2NH3(g) DGo = -33.3 kJ • DGo can’t be measured directly, it must be calculated from DHo and DS • Usefulness of DGo: The more negative DGo is, the more likely reaction is • If DGo is negative, the reaction is spontaneous as written • If DGo is positive, the reaction is not spontaneous • Why use standard states? Because DG changes with P, T, concentration • Calculating DGo: DGo = DHo - TDSo • Example: C(s) + O2(g) CO2(g) • DHo = -393.5 kJ DSo = 3.05 J/K • DGo = (-3.935 x 105 J) – (298 K)(3.05 J/K) = -394.4 kJ • Example: Find DHo, DSo, DGo for 2SO2 + O2 2SO3 DHo (kJ/mol) -297 0 -396 DSo (J/Kmol) 248 205 257 T = 298 K

  5. 3) Use the DGo of known reactions to find DGo of unknown reactions • 2CO + O2 2CO2DGo = ? • Known Reactions i) 2CH4 + 3O2 2CO + 4H2O DGo = -1088 kJ/mol • CH4 + 2O2 CO2 + 2H2O DGo = -801 kJ/mol • If we reverse the first reaction and double the second… i) 2CO + 4H2O 2CH4 + 3O2DGo = +1088 kJ/mol • 2(CH4 + 2O2 CO2 + 2H2O) DGo = -1602 kJ/mol • 2CO + O2 2CO2DGo = -514 kJ/mol • Example: Find DGo for Cdiamond Cgraphite Given that Cd + O2 CO2DGo = -397 kJ/mol Cg + O2 CO2DGo = -394 kJ/mol 4) The DGfo Method • Free energy of formation DGfo is tabulated for many compounds • We can sum these for reactants and products to find DGo

  6. Example: 6C(s) + 6H2(g) + 3O2(g) C6H12O6(s) (glucose) • DGfo = free energy change of formation of one mole of the product from its elements in their standard states (-911 kJ/mol for glucose) • Example: 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g) DGfo (kJ/mol) -163 0 -394 -229 DGo = SnGfoprod - SnGforeactants DGo = 2(-394) + 4(-229) - 2(-163) - 3(0) = -1378 kJ/mol • DG and Pressure • Pressure dependencies of the state functions • H is not pressure dependent • S is pressure dependent • S(large volume) > S(small volume) • S(low pressure) > S(high pressure) • PV = nRT

  7. G must depend on pressure since G = H – TS G = Go + RTlnP Use R = 8.3145 J/Kmol • Example: N2(g) + 3H2(g) 2NH3(g) • DG = 2G(NH3) – G(N2) – 3G(H2)

  8. K = PH2O(g) = 0.0313

  9. Example: CO(g) + 2H2(g) CH3OH(l) Find DG at 25 oC, P(CO) = 5 atm, P(H2) = 3 atm • From Appendix IIB, find DGo = -166 – (-137) – 0 = -29 kJ/mol • Find RTlnQ = (8.3145)(298)ln(1/(5)(3)2) = -9.4 kJ/mol • DG = DGo + RTlnQ = -38 kJ/mol • More spontaneous at these conditions than at standard conditions • Meaning of DG for Reaction at Equilibrium • Even if DG is negative, the spontaneous reaction doesn’t necessarily go to completion • Phase changes always go to completion if spontaneous • Reactions often have a minimum G that is somewhere before completion of the reaction

  10. 4) The equilibrium mixture of reactants and products might be more stable (lower DG) than the completely formed product alone CO(g) + 2H2(g) CH3OH(l) • Equilibrium • Kinetics: forward and reverse reaction rates are the same at equilibrium • Thermodynamics: the lowest free energy state is at equilibrium • A(g) B(g) • GA = GAo + RTlnPA (this is decreasing as the reaction proceeds) • GB = GBo + RTlnPB (this is increasing as the reaction proceeds) • G = GA + GB (this is decreasing as the reaction proceeds) • At equilibrium GA = GB and we have a new PAE and PBE • G is no longer decreasing, it is at its minimum point • No further driving force for the reaction to proceed (DG = 0) Equilibrium Reaction Starts Reaction Proceeds

  11. Example: A(g) B(g) • PAE = 0.25(2 atm) = 0.5 atm • PBE = 0.75(2 atm) = 1.5 atm • K = PB/PA = 1.5/.5 = 3.0 • The same Equilibrium Position would be reached from any initial condition where A + B = 2.0 atm • At Equilibrium: Gprod = Greact (DG = Gprod – Greact = 0) 1 mol A, 2 atm 1 mol B, 2 atm 1 mol A/B, 2 atm

  12. At Equilibrium: DG = 0 = DGo + RTlnK DGo = -RTlnK • If DGo = 0, then K = 1 and we are at equilibrium • If DGo < 0, then K > 1 and the reaction proceeds forward • If DGo > 0, then K < 1 and the reaction proceeds in reverse • Example: N2(g) + 3H2(g) 2NH3(g) • Given DGo = -33.3 kJ/mol at 25 oC • Predict the direction when P(NH3) = 1 atm, P(N2) = 1.47 atm and P(H2) = 0.01 atm • Predict direction when P(NH3) = P(N2) = P(H2) = 1 atm • Example: 4Fe(s) + 3O2(g) 2Fe2O3(s) at 25 oC DHfo 0 0 -826 kJ/mol So 27 205 90 J/Kmol Find K 11) Dependence of K on T Slope = -DH/R Intercept = DS/R lnK 1/T

  13. DG and Work • Wmax = DG All the free energy produced from a spontaneous reaction could be used to do work (Reversible Process only) • For a non-spontaneous reaction, DG tells us how much work we would have to do on the system to get the reaction to occur • Wactual < Wmax We always lose some energy to heat in any process (Irreversible Processes) • Theoretically, Reversible Process utilize all energy for work, Unfortunately, all real processes are Irreversible DG = -50.5 kJ DS = -80.8 J/K Irreversible (real) Process Increases Ssurr to make DSuniv = +

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