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Entropy. The entropy of any system may be found, at least theoretically, using the relationship: S = (Boltzman Constant)*Ln(microstates)
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Entropy The entropy of any system may be found, at least theoretically, using the relationship: S = (Boltzman Constant)*Ln(microstates) Where “microstates” measure the # of possible position and energy states of the particles making up the system. This is a probability. The book shows, for a volume change, ΔS = nRLn(Vf/Vi) & earlier we saw: qrev= nRTLn(Vf/Vi), soΔS = qrev/T
Entropy • ΔS = nRLn(Vf/Vi) • qrev= nRTLn(Vf/Vi) • ΔS = qrev/T &, at ΔP = 0 & ΔT >< 0 • ΔS = nCPLn(Tf/Ti), or, if ΔV = 0 • ΔS = nCVLn(Tf/Ti),
Entropy ΔS = qrev/T To melt one mole of a solid at its melting point: Qrev = ΔHfusion and T is the MP in K. The same relationship holds for any change of state. Therefore, ΔS = ΔHstate change T IF ΔE is used only to change S
Gibbs Free Energy, G The Free Energy is the amount of energy available to/from a system. ΔG = ΔH – TΔS • ΔG = 0 for a system at equilibrium • ΔG > 0 for an endothermic process • ΔG < 0 for an exothermic process
Gibbs Free Energy, G For the reaction: C(graphite)+ O2 CO2(g) ΔHo = -393.5kJ & ΔSo = 3.05J/K ΔGo = -393.5kJ – 298K*3.05J/K ΔGo = 394.4kJ
Gibbs Free Energy Maximum useful work, w = ΔG. If we combine this with what we know about ΔE, ΔH and ΔG ΔE = ΔH – PΔV, or ΔH = ΔE + PΔV and ΔG = ΔH – TΔS we could show: qP = ΔH if wusefull = 0, or qP = ΔS if wusefull 0
Gibbs Free Energy ΔG if ΔP 0 we can show: ΔG = ΔGo + RTLn(P), or, in general, ΔG = ΔGo + RTLn(Q) where “Q” is the mass expression. Q becomes K & ΔG = 0 at equilibrium, so: ΔGo = RTLn(K) (remember, “ΔG = 0” is the definition of a system at equilibrium.)