Thermo Notes #3Entropy and 2nd Law ofThermodynamicsMonday, January 30CHEM 102HT. HughbanksReading You should reading Chapter 7. Some of this material is quitechallenging, be sure to read thismaterial carefully.SpontaneitySpontaneous: “Occurring withoutoutside intervention.” A reaction or change of state is said tobe spontaneous if it isthermodynamically allowed. For a chemist, prediction of spontaneityis a major goal of thermodynamics.Factors Affecting Spontaneity “Spontaneity” is a somewhat misleadingterm referring to whether a chance canoccur (even in principle). Energy: ∆U or ∆H• Not good predictors. Both endothermic andexothermic reactions can occur. Temperature• Some processes are spontaneous only atcertain temperatures. Concentrations, pressures, etc.Entropy Entropy (S) is a thermodynamic statefunction which can be describedqualitatively as a measure of theamount of disorder present in asystem. From a chemical perspective, weusually mean molecular disorder.Entropy and Disorder Entropy is a measure of disorder.more disorder → greater entropy Entropy of a substance depends onphysical state. Sgas >> Sliquid > Ssolid Entropy depends on temperature.Increasing T will increase entropy dueto increase in molecular motion.Entropy & SpontaneityIn many spontaneous processes, entropyof a system increases. (∆S > 0) Examples:• expansion of a gas into vacuum• mixing of gases or generation of a gas fromsolid or liquid reactantsExpansion of a Gas This process obviously has a preferreddirection. Once the valve is open, the probabilitythat all molecules will return to one sideis astronomically small. ∆S > 0 for this spontaneous changeProbabilities and EntropyProbability thatboth moleculeswill be on lefthand side: (1/2)2= 1/4.Probabilities and Entropy= A very small number when N isAvogadro’s number!Probability thatboth moleculeswill be on lefthand side: (1/2)NExtra DetailsNumber of ways of having N 2 molecules on each side: NN 2!"#$%&=N !(N 2)![ ]2Number of ways of having N 2 ! N molecules on one side: NN 2 ! N / 2"#$%&'=N !(N 2 ! N / 2)!(N 2 + N / 2)! NN 2!"#$%&NN 2 ' N / 2!"#$%&( e, as N ( ); 2.709 for N = 200)use Stirling's Approximation: N! ! 2"NNe#$%&'(Nand definition of e = limx)*1 +1x#$%&'(xDistribution of ProbabilitiesN/2!NP(N)PmaxPmax1eGas Expansion & Probabilities Once the valve is open, the probability that we’llfind N molecules on one side is (1/2)N. If wehad started with 2N molecules, the probabilitywould have been (1/2)2N.Gas Expansion & Probabilities Once the valve is open, the probabilitythat N molecules will be found on theleft side is (VA/(VA+ VB))N = (Vi/Vf)NFor 2N molecules, the probability wouldbe (Vi/Vf)2N.VAVBSome notes on Nature and Probability For a system with a given amount ofenergy, the system will usually be foundin its most probable state(s). The probabilities we’ve computed in gasexpansion example do not scale withthe size of the system, but theirlogarithms do scale with the size of thesystem.Gas Expansion & Probabilities Entropy: S = k ln W (“W” is the number ofsystem microstates of the same energy) ΔS = k {ln(Wfinal / Winitial} Wfinal / Winitial = (Vf/Vi)N ΔS = Nk ln (Vf/Vi) > 0 (Nk = R) (!)VAVBEntropy, Probabilties, Disorder For systems with equal energycontent, those that are mostdisordered also turn out to be moststatistically likely (most probable). Entropy increases as the statisticallikelihood increases. The “disorder” of the universe tendsto increase: (∆S)universe > 0 in allchangesExample: Boiling and Entropy For H2O(l), S˚ = 69.91 J• K–1• mol–1 For H2O(g), S˚ = 188.3 J• K–1• mol–1 Therefore ∆S˚vap = 118.4 J• K–1• mol–1(at 298 K) Problem: At 373 K, ∆Svap = 109 J• K–1•mol–1 (why different?). What part of thisis due to the greater volume occupiedby the vapor phase?Generation of a Gas “Frozen Beaker” reaction :Ba(OH)2• 8H2O(s) + 2NH4Cl(s) →BaCl2• 2H2O(s) + 2NH3(g) + 8H2O(l) Solid reactants → solid, liquid, & gas inproducts ∆Ssystem >> 0Entropy & Spontaneity In some spontaneous processes, theentropy of a system decreases. (∆S < 0) Examples:• freezing of a liquid• condensation of a gas• formation of a solid product from gas orliquid reactants These are common events, and theyclearly have ∆S < 0.Example: 2 NO2 → N2O4, ∆S < 0 The conversion of 2 moles of gasinto one mole of gas decreases theentropy of the system.Second Law of Thermodynamics Entropy change of the system does notcorrectly predict spontaneity. One correct statement of the secondlaw of thermodynamics is:“The entropy of the universe isalways increasing.” In equation form: ΔSuniverse > 0.∆Suniverse = ∆Ssystem + ∆SsurroundingsEntropy and Heat Simplest case is a process whichoccurs at constant T. Phase changesare good examples. For the case of constant T, ∆S = qrev/T qrev is heat transferred in a reversibleprocess, at constant T (signs as usual) From this equation, ∆S has units of J/KSome Subtleties We’ve said that, for constant T, ∆S = qrev/T This is a way of calculating ∆S (∆Ssysrecall) even if we don’t actually transferthe heat reversibly as long as in theirreversible process the state of thesystem is the same as it would havebeen in the reversible process. (!)∆S for Phase Changes Since, for constant T, ∆S = qrev/Tand, qrev = ∆H˚ (at constant P), we canget ∆S for a phase change at thenormal melting or boiling points.Ex: For ethanol, compute ∆Sfus & ∆Svap∆Hfus˚ = 4.6 kJ/mol; Tm = 158.7 K∆Hvap˚ = 43.5 kJ/mol; Tb = 351.5 KIsothermal Ideal Gas Expansion Suppose we carry out the expansion of anideal gas at constant temperature. Sincethe energy of an ideal gas depends only ontemperature ∆U = 0 !U = q + w " q = #w for each infinitesimal step of the expansion, dw = !PdV q = ! dw =ViVf"PdV =ViVf"nRTVdVViVf"= nRTdVVViVf"q = nRT lnVfVi # $S =qT= nRlnVfViExamples Suppose we carry out a reversible,isothermal expansion of 1.0 mole of anideal gas such that we double itsvolume. What is ∆S? Suppose we carry out a free expansionof 1.0 mole of an ideal gas, such thatthe volume the gas occupies after theexpansion has doubled. What is ∆S?Illustrative Phase Change Example Dry ice is solid CO2. At T = 195 K, dryice sublimes (is converted directly from asolid into a