Chem 170 1st Edition Lecture 20Outline of Last Lecture I. 2nd Law of Thermodynamics:∆ Suniverse=∆ Ssystem+∆ SsurroundingsII.∆ Ssurroundings=−∆ HsystemTIII.∆ G=∆ Hsystem−T ∆ SsystemOutline of Current Lecture IV. Chapter 14 – Entropy and Free EnergyV. Ch. 15 – Chemical equilibriuma. Characteristics of equilibrium stateb. Empirical law of mass actionCurrent LectureChapter 14 – Entropy and Free Energy∆ Suniverse=∆ Ssystem+∆ Ssurroundings∆ Ssurroundings=−∆ HsystemT∆ S0system=∑(n Sproducts0)−∑(n Sreactants0)∆ G=∆ Hsystem−T ∆ Ssystem∆ G0system=∑(n Gproducts0)−∑(nGreactants0)ex. ∆ Gf0C O2(g)at 25 ℃ ;C(s)+O2(g)→C O2(g)∆ G=∆ HC O2−T ∆ Ssys¿−393.36KJmol−(25+273 K)(213.63JK−(205.03JK+5.74JK))¿−393.36−(298)(0.00286KJK)¿−394.36 KJOR look up ∆ G0 in the tableex. 3 NO(g)→ N20(g)+N O2(g); What is ∆ Gf0?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.3 mol(86.3KJmol)∆ G=∆ HC O2−T ∆ Ssys=(104.18 KJ +51.29 KJ)−¿¿−104.18 KJ <0therefore spontaneousnote that ΔH and ΔS are not very temperature dependent, therefore it is OK to assume ΔH and ΔS remain constant over a large range of temperatures to calculate ΔG at varioustemperatures. ΔG is sensitive to temperature.Ch. 15 – Chemical equilibriumLet’s take a look at ¿6H2O¿¿Cl4Co ¿¿2−¿+6 H2O−¿ ↔¿2+¿+4 Cl¿Co ¿¿the equilibrium constant (K) characterizes the approach to a state of equilibrium.Characteristics of equilibrium state:1. it deisplays no macroscopic evidence of change2. it is reached through spontaneous processes3. it shows dynamic behavior between forward and reverse processesThis means that the rate for the reaction forward equals the rate of the reaction going backwards when the reaction is at equilibrium.This makes sense because when the amount of reactants is high then the rate of collisions between reactions will be high and the production rate of products will increase as well.The reverse is true as well. When the amount of products is high then therate of collisions between products will be high and the production rate of reactants will increase as well4. they are the same regardless of direction of approachEmpirical law of mass action (from 1864 Guldberg and Waage)when there is the reaction aA+bB ↔cC +dD, thenK=[C ]eqc[D]eqd[ A ]eqa[B]eqb for solutionsPPPP(¿¿ A)a(¿ ¿B)b¿(¿¿C)c(¿¿ D)d¿¿K =¿ for gases (use partial pressures)note that pure liquids do not appear in the
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