Ch1b2010_week6_ab

Ch1b2010_week6_ab - Wk 5 3 4 JH 5 Spontaneous Processes &...

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1 1 19 18 17 16 JH Kinetics OGC Chptr 18 15 Pres Day holiday Wk 7 12 PS5 due 11 JH equilibrium 10 9 JH Chemical equilib OGC: 14.1-14.7 8 no lecture No lecture Midterm due 8 PM Wk 6 5 4 JH Spontaneous Processes & Thermo Equilib OGC: 13.5-13.7 midterm out 10PM 3 Wk 5 Prof Reisman takes over 2 Entropy (reversible, non-isothermal process) Reversible processes Δ S of system + surroundings = 0 Example (ideal gas): n = 2; V=10 Liters; T=300K For a reversible, isobaric compression to 5 Liters Δ S gas and Δ S surrounding = ? Δ P = 0 V 2 /V 1 = 0.5 T 2 /T 1 = 0.5 So T 2 =150K = = = Δ 300 150 ln ) 314 . 8 2 5 ( 2 ln 1 1 1 2 2 1 K mol J T T nc dq S T T P rev gas = -29 J K -1 1 1 . 29 = Δ K J S gs surroundin
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2 3 Entropy Take same system (V 2 = 5 liters; T = 150 K; n=2) and do a reversible, isothermal expansion back to 10 Liters = = = 2 1 1 2 ln 1 V V q V V nRT dV V nRT w Δ T = 0 so q=-w w =-1728 J; q = 1728 J Δ S system = 1728J/150K = 11.5 J K -1 Δ S surroundings = -11.5 J K -1 4 Entropy Repeat for the same system (V 2 = 5 liters; T = 150 K; n=2) but this time do an ir reversible , isothermal expansion back to 10 Liters Δ S system = 1728J/150K = 11.5 J K -1 Δ S surroundings = -11.5 J K -1 Calculate out the P (w/ system at 10 L) and equate to P ext P ext = nRT/V = 2(0.0821)(150K)(10L) -1 = 2.46 atm J atm L L L V P w ext 1230 ~ 3 . 12 ) 5 10 ( 46 . 2 = = = Δ = Since Δ T = 0 so q = -w = 1230 J Δ S system = +11.5 J K -1 still!! Reversible, isothermal expansion results State function!!
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3 5 Entropy Continuing with the same system (V 2 = 5 liters; T = 150 K; n=2) and an ir reversible , isothermal expansion back to 10 Liters J atm L L L V P w ext 1230 ~ 3 . 12 ) 5 10 ( 46 . 2 = = = Δ = Since Δ T = 0 so q = -w = 1230 J Δ S system = +11.5 J K -1 still!! State function!! Δ S surroundings = -1230J/150K= -8 J K -1 Δ S total = 11.5J K -1 -8 J K -1 = 3.5J K -1 Δ S total > 0 6 Hey, wait a second!! Can you go over that point again??
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4 7 END OF Entropy 8 Gibbs Free Energy Three state functions: E, H, S H (enthalpy) and S (entropy) are most important A fourth state function, G, connects both H and S together G = H – TS G is the Gibbs Free Energy
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5 9 Gibbs Free Energy G is the Gibbs Free Energy Δ G sys > 0 Δ G sys = 0 Δ G sys < 0 Non-spontaneous process Reversible process Spontaneous process Δ G = Δ H -T Δ S For chemical reactions at a constant temperature 10 Gibbs Free Energy This is just like the enthalpy ( Δ H o ) discussion we had last time.
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6 11 Gibbs Free Energy Δ G = Δ H -T Δ S Entropy determines the temperature-dependence of Δ G Gases exhibit the largest T dependence Followed by solutions Followed by solids Why is this? Think about the statistical definition of entropy. 12 Gibbs Free Energy & Chemical Equilibrium [][] ln cd ab aA bB cC dD GR T K CD K AB ++ −Δ = = ± This is the equation that we want to get to
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7 13 Gibbs Free Energy & Chemical Equilibrium Characteristics of Equilibrium States They display no macroscopic evidence of change They are reached through spontaneous processes They show a dynamic balance of forward and reverse processes They are the same regardless of direction of approach 14 Empirical Law of Mass Action aA + bB Ù cC + dD
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This note was uploaded on 10/27/2010 for the course CH 1b taught by Professor Natelewis during the Winter '09 term at Caltech.

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Ch1b2010_week6_ab - Wk 5 3 4 JH 5 Spontaneous Processes &...

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