CHAPTER 11: Spontaneous Change and EquilibriumWhat is entropy?Entropy vs phases of matterThird Law of ThermodynamicsStandard Molar Entropies SoEntropies of ReactionSecond Law of ThermodynamicsThe second law, cont’dSecond Law, cont’dSpontaneous reactionsExampleExample, cont’dTrouton’s ruleStandard Gibbs Function of FormationEffects of Temperature on GoGibbs Function and Equilibrium ConstantAt equilibrium…Summary of spontaneityAnother relationship between G, Q, and KComputing K from G at other temperaturesvan’t Hoff EquationVariation of Vapor Pressure with TemperatureCHAPTER 11: Spontaneous Change and Equilibrium• Goal of chapter: Be able to predict which direction a reaction will go (cases where there is not necessarily an equilibrium) • At high temperatures, ice always melts instead of getting colder. (This is endothermic, but happens anyway).• Two gases in separate containers always mix if the containers are joined•Why???• Need to know about entropy(S) and Gibbs free energy (G)CHEM 1310 A/B Fall 2006What is entropy?• Example: a professor’s desk is often in a state of high entropy because it is very disorganized!• Entropy is a measure of the disorder in a system.• Increases when the number of ways of arranging the system (at constant E) increases• Each arrangement (for fixed E) of the system is a “microstate”• For W microstates, the entropy S = k ln W, where k is Boltzmann’sconstant = R / N0.CHEM 1310 A/B Fall 2006Entropy vs phases of matterhighest S lower S lowest SCHEM 1310 A/B Fall 2006For a given phase (s,l,g) how does an increase in Taffect things? As T goes up, atoms/molecules movemore, so S…For a phase transition, ∆Strans= ∆Htrans/ T.Is ∆Svappositive or negative?Third Law of Thermodynamics• Imagine a perfect crystal at 0K. Recall connection between T and kinetic energy – at 0K, no motion (well, quantum tells us this is not quite true…more later). • How many accessible states? Only one!• No disorder = no entropy!• Third Law of thermodynamics: The entropy of a perfect crystalline substance at equilibrium approaches zero as the absolute zero of temperature is reached.CHEM 1310 A/B Fall 2006Standard Molar Entropies So• Third law is helpful in establishing an absolute entropy scale• Standard molar entropy Sois the absolute entropy of one mole of a substance in a standard state at 298.15K.• Should be positive for any pure substance (third law), but can be negative for mixtures.a A + b B → c C + d D∆So= c So(C) + d So(D) – a So(A) – b So(B)• Can look up Sovalues in tables (see appendix)CHEM 1310 A/B Fall 2006Entropies of Reaction• Can usually guess the sign of ∆S.•(1) CaCO3(s) → CaO(s) + CO2(g)is ∆S + or -?•(2) 2H2O(g) → 2H2(g) + O2(g)is ∆S + or -?•(3) MgCl2(s) → Mg2+(aq) + 2Cl-(aq)is ∆S + or -?• Reaction (3) is actually tricky…turns out the ions order water around them, decreasing entropy! Dissolution can have ∆S + or -, depends on case.CHEM 1310 A/B Fall 2006Second Law of Thermodynamics• Defines “spontaneous” processes, e.g., water always freezes if below 0oC, etc) in terms of entropy changes.• Second Law of Thermodynamics: In any spontaneous process, the entropy of the universe (system + surroundings) always increases: ∆Suniverse= ∆Ssystem+ ∆Ssurroundings> 0for a spontaneous process• This has lots of deep implications. For example: The universe began to exist sometime in the finite past. It has not just been here forever.CHEM 1310 A/B Fall 2006The second law, cont’d• Spontaneous processes and ∆Suniv:∆Suniv> 0 spontaneous∆Suniv = 0 equilibrium∆Suniv < 0 non-spontaneous• Example: When water freezes, what are ∆S and ∆H for the system?• Is this a contradiction? No… the heat released causes ∆Ssurr> 0 … just need enough heat released to ensure that |∆Ssurr|> |∆Ssystem| , so that ∆Suniv= ∆Ssurr+ ∆Ssystem> 0.• For a phase transition, ∆Ssurr= -∆Hsys/T (const T,P). The more heat given off, the more the entropy of the surroundings increases.CHEM 1310 A/B Fall 2006Second Law, cont’d• Problem: we don’t normally know anything about ∆Ssurroundings!• Solution: A lot of chemistry has constant T, P. Under these conditions, use ∆Ssurr= -∆Hsys/T . This relates everything back to properties of the system!• ∆Suniv = ∆Ssys+ ∆Ssurr> 0∆Suniv = ∆Ssys+ (- ∆Hsys/T ) > 0T ∆Ssys- ∆Hsys > 0 or∆Hsys-T ∆Ssys< 0• Define Gibbs Free Energy ∆G = ∆Hsys-T ∆Ssys• ∆G < 0 spontaneous, ∆G = 0 equilibrium, ∆G > 0 non-spontaneous (reverse process is spontaneous).CHEM 1310 A/B Fall 2006Spontaneous reactions• From the last slide, ∆G < 0 for a spontaneous process. This means ∆H - T∆S < 0 for a spontaneous process• Two factors determine spontaneity: ∆H and T∆S• Reaction more likely to be spontaneous if ∆H < 0 (exothermic)T large and ∆S large and positive • However, none of these factors is required if the other factors are larger (e.g., water freezes even though ∆S < 0 … ∆H factor is dominant). Have to work out which factor is larger.CHEM 1310 A/B Fall 2006Example• Does PCl3boil at 25oC, 1 atm?• In other words, is the process PCl3(l) → PCl3(g) spontaneous under these conditions?∆Hof(kJ mol-1) So(J K-1mol-1)PCl3(g) -287.0 311.7 PCl3(l) -319.7 217.1CHEM 1310 A/B Fall 2006Example, cont’d• Actually from this data we can also estimate the boiling point, because at the boiling point, ∆G=0. Assuming P=1atm, what must be the temperature?CHEM 1310 A/B Fall 2006Trouton’s rule• Because ∆G=0 when two phases are in equilibrium, that means ∆H = T∆S. For vaporization, that means∆Svap= ∆Hvap/ Tb• The entropy of vaporization is almost the same for any liquid at its normal boiling point. ∆Svap= 88 ± 5 J K-1mol-1, Trouton’s RuleCHEM 1310 A/B Fall 2006Standard Gibbs Function of Formation∆G = ∆H - T∆S∆Gof= ∆Hof-T∆SoDefined as 0 for elements in their most stable form at 25oC, just like ∆Hof(would be forming element from itself, must be 0)a A + b B → c C + d D∆Go= c ∆Gof(C) + d ∆Gof(D) -a ∆Gof(A) – b ∆Gof(B)CHEM 1310 A/B Fall 2006Effects of Temperature on ∆Go• Tables only give ∆Go for 25oC. For other T, need to look up ∆Go and ∆So (assuming they don’t change much with T, which is usually true), and use ∆Go=