CHEM 170 1st Edition Lecture 19Outline of Last Lecture I. EntropyII. Entropy for Ideal GasesIII. Standard EntropyIV. Trends of EntropyOutline of Current Lecture V. 2nd Law of Thermodynamics:∆ Suniverse=∆ Ssystem+∆ SsurroundingsVI.∆ Ssurroundings=−∆ HsystemTVII.∆ G=∆ Hsystem−T ∆ SsystemCurrent LectureChapter 14 – Entropy and Free Energy2nd Law of Thermodynamics:∆ Suniverse=∆ Ssystem+∆ Ssurroundingswhen ∆ Suniverse>0 then the reaction is spontaneousAll spontaneous processes produce an increase in the entropy of the universeex. What is the ΔS of N2(g)+3 H2(g)→2 NH3(g) at 25°C, given that SN H30=193.0Jmol × K, SN20=191.5Jmol × K, and SH20=131Jmol × K.∆ Sreaction0=∑(n Sproducts0)−∑(n Sreactants0)¿2(193.0)−(191.5Jmol × K+3(131Jmol × K))¿−198.5JKif ∆ Suniverse=∆ Ssystem+∆ Ssurroundings>0,t hen the reaction is spontaneousif ∆ Suniverse=∆ Ssyste m+∆ Ssurroundings<0,then the reaction is not spontaneous, but the reverse process is spontaneousif ∆ Suniverse=∆ Ssystem+∆ Ssurroundings=0,then the reaction is at an equilibrium.ex. If the reaction H2O(l )→ H2O(s) (∆ Hfreezing=−6.0KJmol) takes place in a room at -10°C then the reaction is spontaneous. However looking at the reaction itself the liquid is become a solid, which means the order is increasing (ΔS<0), which is not spontaneous.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.Therefore the ∆ Ssurroundings>∆ Ssystem so that the ∆ Suniverse>0 and the reaction will be spontaneous.∆ Ssurroundings=−∆ HsystemT, where T is assumed to stay constantex. Is the reaction H2O(l)→ H2O(s) spontaneous or non-spontaneous at -10°C? S0 of H2O(s)=43.2 J/mol*K and S0 of H2O(l )=65.2 J/mol*K.∆ Ssystem=∑(n Sproducts0)−∑(n Sreactants0)¿(1)(43.2Jmol × K)−(1)(65.2Jmol × K)¿−22 J / K∆ Ssurroundings=−−6000Jmo l263 K=23JK∆ Suniverse=∆ Ssystem+∆ Ssurroundings=−22+23=1>0therefore the reaction is spontaneousChange in Enthalpy of the Universe (∆ Suniverse) to Gibbs Free Energy (∆ G)∆ Suniverse=∆ Ssystem+∆ Ssurroundings∆ Suniverse=∆ Ssystem+−∆ HsystemTT ∆ Suniverse=T ∆ Ssystem+−∆ HsystemTT−T ∆ Suniverse=∆ Hsystem−T ∆ Ssystem∆ G=∆ Hsystem−T ∆ SsystemGibbs Free Energy (∆ G)The Gibbs free energy predicts the spontaneity of a reaction using only the parameters of the system.if ∆ G<0, then the reaction is spontaneous.if ∆ G>0, then the reaction is not spontaneous, but the reverse process is spontaneous.if ∆ G=0, then the reaction will be at equilibrium.ex. If the reaction H2O(l )→ H2O(s)occurs at the following temperatures will the reaction be spontaneous? ΔH=-6007 J/mol and ΔS=-21.99 J/K.If the temperature was 0°C, then∆ G= ΔH −T ΔS=(−6007Jmol)(273 K)(−21.99JK)=0 J therefore the reaction is at equilibrium.If the temperature was -10°C, then∆ G= ΔH −T ΔS=(−6007Jmol)(263 K)(−21.99JK)=−220 J <0therefore the reaction is spontaneousIf the temperature was 10°C, then∆ G= ΔH −T ΔS=(−6007Jmol)(283 K)(−21.99JK)=502 J >0therefore the reaction is not spontaneousStandard free energy (∆ G0) can be found whengas→1atmliquid→puresolvent→pureelement→in its most stable state at 25°Csolutions→1M∆ G0system=∑(n
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