Chem 175 1st Edition Lecture 18Outline of Last Lecture I. Spontaneous and Non-spontaneousII. Spontaneity and Exothermic ReactionsIII. EntropyOutline of Current Lecture IV. EntropyV. Entropy for Ideal GasesVI. Standard EntropyVII. Trends of EntropyCurrent LectureChapter 14 – Entropy and Free EnergyEntropyThe change in entropy can be found using Boltzmann’s equation (S=kln(W)).∆ Ssystem=Sfinal−Sinitial=kln(Wfinal)−kln(Winitial)¿kln(WfinalWinitial)given that W= XN, where W is the number of configurations, N is the number of particles and X is the number of cells, there is a new equation.∆ Ssystem=kln(XfinalXinitial)N=(kN )ln (XfinalXinitial)As the volume increases the entropy also increases, because the particles have more positions they can potentially fill.X =Vν , where V is the total volume and ν is the volume of each individual cell.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Entropy for ideal gases∆ Ssystem=(kN)ln(Vfinalν÷Vinitialν)=(kN)ln(VfinalVinitial)given that N is the number of particles and NA is Avogadro’s constant, we can find n (the number of moles).N=n× NAand given that k is Boltzmann’s constant equals the ideal gas constant (8.314 J/mol×K) over NA (Avogadro’s constant).k =RNAWe get the new equation:∆ Ssystem=(RNA)(n × NA)ln(VfinalVinitial)=Rn ln(VfinalVinitial)ex. What is the change in entropy when 0.321 mol of O2 experiences a pressure change form 0.30 atm to 12 atm while temperature is constant? (Note that the volume must be decreasing, therefore the entropy should decrease)V1P1n1T1=V2P2n2T2→V1(0.30 atm)=12 V20.3012=V1V2∆ Ssystem=(0.321mol)(8.314Jmol × K)(0.312)=−9.85JKStandard Entropy: S0 is entropy at 1 atmTo find the absolute entropy of a substance at 1 atm the temperature must be defined.3rd Law of thermodynamicsThe entropy of a pure crystal at 0 K is zero. A pure crystal has every atom at a perfect orientation, which is the most ordered state tha could ever bepossible.∆ Ssystem=qreversibleT, where qreversible is heat along a special path∆ V =q+ww=−P ∆VTrends of entropy:Phases: S(s)<S(l)<S(g)As the mass increase the entropy also increases. (ex. SXe>SNe)For 2 substances (usually molecules) with about the same molecular weight the substance with the more complex structure has a greater entropy than the less complex structure.ex. molecule y-y-y has a greater entropy than molecule z-z even if they have the same molecular weight.Entropy (S) is a state function, therefore∆ Sreaction0=∑(n Sproducts0)−∑(n
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