Lecture 4

Lecture 4 - So far we discussed ­ ­ ­...

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Unformatted text preview: So far we discussed ­ ­ ­ • meaning of spontaneous and nonspontaneous processes • meaning of entropy • recogni2on of processes associated with increases and decreases in entropy • factors affec2ng entropy • the second and third laws of thermodynamics Important Equa6ons ΔSuniverse = ΔSsystem + ΔSsurroundings > 0; The Second Law ΔSsys = Sfinal – Sini6al ; entropy is a state func6on like enthalpy ΔSsys = k ln Wfinal  ­ k ln Wini-al ; k = Boltzmann constant = R/NA ΔSsys= qrev/T S = k ln W = k ln 1; The Third Law (entropy of a perfect crystal at the absolute temperature is zero ΔSsys==ΔSorxn = ΣmΔSoproducts  ­ ΣnΔSoreactants ΔSsurroundings =  ­ ΔHsystem T ΔHsys==ΔHorxn = ΣmΔHoproducts  ­ ΣnΔHoreactants Entropy Change •  Entropy increases when *gases are formed from solids & liquids (more freedom) *liquids/solu2ons are formed from solids (more freedom) *the number of gas molecules increases (more number of states) *the MW increases (more vibra2onal mo2on) *the volume increases (gas) (more posi2onal mo2on) *the temperature increases (more energy states) Surroundings & System Entropy & Free Energy •  Free Energy –  another thermodynamic func6on •  related to spontaneity •  G = H  ­ TS •  for a process that occurs at constant temperature (i.e. for the system): ΔG = ΔH  ­ TΔS Spontaneity, Entropy & Free energy ΤΔSuniverse =  ­ TΔSsystem + ( ­ΔHsystem) -ΤΔSuniverse =  ­ TΔSsystem + ΔHsystem -ΤΔSuniverse = ΔHsystem  ­ TΔSsystem -ΔG = TΔSuniverse  ­TΔSuniverse is defined as the Gibbs free energy, ΔG. For spontaneous processes:ΔSuniverse>0 And therefore: ΔG<0 ΔG is easier to determine than ΔSuniverse. Use ΔG to decide if a process is spontaneous. Gibbs Free Energy (G) ΔG, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it. ΔGsystem = ΔHsystem  ­ TΔSsystem ΔG < 0 for a spontaneous process ΔG > 0 for a nonspontaneous process ΔG = 0 for a process at equilibrium Standard free energies of forma6on, ΔGf° are analogous to standard enthalpies of forma6on, ΔHf°. ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants) Standard Free Energy Changes ΔG° can be looked up in tables, or calculated from S° and ΔH°. ΔG and the Work a System Can Do For a spontaneous process, ΔG is the maximum work obtainable from the system as the process takes place: ΔG = workmax For a nonspontaneous process, ΔG is the minimum work that must be done to the system as the process takes place. Sample Problem 20.4 PROBLEM: Calcula2ng ΔGorxn from Enthalpy and Entropy Values Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid ­state dispropor6ona6on reac6on when heated: 4KClO3(s) 3KClO4(s) + KCl(s) Use ΔHof and So values to calculate ΔGosys (ΔGorxn) at 25oC for this reac6on. PLAN: Use Appendix B values for thermodynamic en66es; place them into the Gibbs Free Energy equa6on and solve. SOLUTION: ΔHorxn = ΣmΔHof(products) – ΣnΔHof(reactants) ΔHorxn = (3 mol)( ­432.8 kJ/mol) + (1 mol)( ­436.7 kJ/mol)  ­ (4 mol)( ­397.7 kJ/mol) ΔHorxn =  ­144 kJ Sample Problem 20.5 PROBLEM: Calcula2ng ΔGorxn from ΔGof Values Use ΔGof values to calculate ΔGrxn for the reac6on in Sample Problem 20.4: 4KClO3(s) 3KClO4(s) + KCl(s) PLAN: Use the ΔG summa6on equa6on. SOLUTION: ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants) ΔGorxn = (3 mol)( ­303.2 kJ/mol) + (1 mol)( ­409.2 kJ/mol)  ­ (4 mol)( ­296.3 kJ/mol) ΔGorxn =  ­134 kJ Free Energy and Temperature •  There are two parts to the free energy equa6on:   ΔH°— the enthalpy term   TΔS° — the entropy term •  The temperature dependence of free energy comes from the entropy term. Temperature & Reac6on Spontaneity •  In most cases, the enthalpy contribu6on is greater than entropy— most exothermic reac6ons are spontaneous (ΔHsys <0) •  Temp of a reac6on can affect the magnitude of the TΔS term •  There are four possible combina6on of posi6ve and nega6ve ΔH and ΔS; two are temp independent and two are temp dependent T ­independent ΔS > 0 and ΔH < 0 rxn always spontaneous) ΔS < 0 and ΔH > 0 (rxn always nonspontaneous) T ­dependent ΔS > 0 and ΔH > 0 (rxn spontaneous at high temp) ΔS < 0 and ΔH < 0 (rxn spontaneous at low temp) Sample Problem 20.6 Using Molecular Scenes to Examine the Signs of ΔH, ΔS, and ΔG PROBLEM: The following scenes represent a familiar phase change for water (blue spheres): (a)  What are the signs of ΔH and ΔS for this process? Explain. (b)  Is the process spontaneous at all T, no T, low T, or high T? Explain. PLAN: The process shown is water going from gas to liquid. (a) Determine whether there is an increase or decrease in entropy based on the states. (b) Use Equa6on 20.6 and Table 20.1 to determine spontaneity by the sign of ΔG. Sample Problem 20.6 con2nued Using Molecular Scenes to Examine the Signs of ΔH, ΔS, and ΔG SOLUTION: (a) The scenes represent condensa6on of a gas, ΔS < 0 (more order) and ΔH < 0 (exothermic) (b) ΔG = ΔH  ­ TΔS With a nega6ve ΔS and nega6ve ΔH, ΔG will be nega6ve. Therefore, the process is spontaneous at low T. ΔGsystem = ΔHsystem – TΔSsystem 2SO2(g) + O2(g)  ­ ­ ­ ­ ­ ­ 2SO3(g) ΔH =  ­198.4 kJ and ΔS =  ­187.9 J/K What is the ΔG at 298 K? Is the reac2on spontaneous at 298 K? Is the reac2on spontaneous at 900 0C? The effect of temperature on reac2on spontaneity Cu2O(s) + C(s)  ­ ­ ­ ­ ­ ­ 2Cu(s) + CO(g) ΔH = 58.1 kJ and ΔS = 165 J/K The reac6on occurs at high but not at low temp What is the transi6on temperature? ΔG = ΔH – TΔS At ΔG = 0; T = ΔH/ΔS T = 58.1 x1000 J/165 J/K= 352 Below K= 352, rxn is non ­ spontaneous and above 352 the rxn becomes spontaneous Reac2on coupling and spontaneity Cu2O(s) + C(s)  ­ ­ ­ ­ ­ ­ 2Cu(s) + CO(g) ΔG375 =  ­3.8 kJ Cu2O(s) 2Cu(s) + 1/2O2(g) ΔG = 140.0 kJ C(s) + 1/2O2(g) CO(g) ΔG =  ­143.8 kJ The favorable second reac2on drives the unfavorable first reac2on Coupling of reac2ons drive nonspontaneous reac2on into a spontaneous one ...
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This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.

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