Lecture 6 - K 2 2 2 3 = = = 8 V = 10.0 L After reaching...

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1 Ice Cream Social Date: Tuesday September 11 Time: 4:30 p.m. Undergraduate Lounge, G-29 Baker Today: Ch. 6.1- 6.3 PS2 due Friday
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4 Comparing w and q for A C B vs. A D B We see that w and q are path dependent, Δ E is path independent
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5 Magnitude of work = w = P Δ V = shaded area
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6 Chapter 6: Equilibrium Phase Equilibria : H 2 O (l) H 2 O (g) P H2O = 24 Torr = 0.0316 atm at 25 o C P H2O =150 Torr = 0.197 atm at 60 o C P H2O =760 Torr = 1.00 atm at 100 o C Solvation : AgCl (s) Ag + (aq) + Cl - (aq) Sat’d solution: 1.82 mg/liter @25 o C Chemical Reaction : NO 2 (g) + NO 2 (g) N 2 O 4 (g) Δ H = - 57.2 kJ/mol K P = P H2O K = [Ag + ][Cl - ] = 1.6 x 10 -10 [] [] [] [ ] [] 2 2 4 2 2 2 4 2 NO O N NO NO O N K = = () 2 NO O N NO NO O N P 2 4 2 2 2 4 2 P P P P P K = = OR K P = 0.0316 K P = 0.197 K p = 1.00
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7 V = 10.0 L CO H 2 CH 3 OH moles After reaching equilibrium at t e : [ ] [] [] 5 . 14 ) 0822 . 0 )( 0911 . 0 ( 0.00892 H CO OH H C
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Unformatted text preview: K 2 2 2 3 = = = 8 V = 10.0 L After reaching equilibrium at t e : [ ] [ ] [ ] 4 . 14 ) 151 . )( 0753 . ( 0.0247 H CO OH H C K 2 2 2 3 = = = CO H 2 CH 3 OH 9 V = 10.0 L After reaching equilibrium at t e : [ ] [ ] [ ] 5 . 14 ) 176 . )( 138 . ( 0.0620 H CO OH H C K 2 2 2 3 = = = H 2 CO CH 3 OH 10 CO(g) + 2H 2 (g) CH 3 OH (g) For ANY initial conditions, after reaching equilibrium at t e : [ ] [ ] [ ] 5 . 14 H CO OH H C K 2 2 3 = CO H 2 CH 3 OH CO H 2 CH 3 OH CO H 2 CH 3 OH 11 For the generalized reaction: aA + bB +. . gG +hH+. The equilibrium constant in terms of concentration (K) is: [ ] [ ] [ ] [ ] ..... B A .... H G K b a h g =...
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This note was uploaded on 04/16/2009 for the course CHEM 2150 taught by Professor Chirik,p during the Fall '08 term at Cornell University (Engineering School).

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Lecture 6 - K 2 2 2 3 = = = 8 V = 10.0 L After reaching...

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