110c-lecture15

# 110c-lecture15 - Problem Set #3 Due Wed May 5 th . Problem...

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Unformatted text preview: Problem Set #3 Due Wed May 5 th . Problem Set #4 Due Wed May 12th. Midterm 2 Wed May 19 th . Material covered through end of Fri May 14 th . Chapters 23, 24, 25, and 26. Activity Coefficient ( ) Corrects for Non-Ideal Vapor Phase j j sln j a RT liq ln ) ( * + = j j j sln j x RT liq ln ) ( * + = a j x j as j 1 = = j j j x a is a measure of deviation from ideality. In dilute solution limit: 1 1 for solvent if x 1 &gt; 0.95 and 2 1 for dilute solutes as x 2 0 1 1 * 1 1 ln ) ( x RT liq sln + = 2 2 * 2 2 , * 2 2 ln ln ) ( x RT P k RT liq H sln + + = * j j j j P x P = P 2 k H,2 x 2 P 1 x 1 P* 1 1.0 0.8 0.6 0.4 0.2 0.0 x 2 * 1 1 1 1 P x P = * 1 1 1 * 1 1 ln ) ( P P RT liq sln + = * 2 2 2 * 2 2 , * 2 2 ln ln ) ( P P RT P k RT liq H sln + + = * 2 2 2 2 P x P = Practice Problem: Activity Coefficient ( ) Ethanol is used as an additive to gasoline: vapor pressure of pure ethanol (P * EtOH ): 0.5768 atm at 338.15 K vapor pressure of pure octane (P * octane ): 0.1279 atm Partial pressures of ethanol and octane in equilibrium with the solution: Solution X(ethanol) p(ethanol) [atm] p(octane) [atm] I 0.9833 0.5694 0.0249 II 0.5734 0.4976 0.1130 [a] Calculate the activity coefficient of the ethanol ( EtOH ) in solution I and in solution II, using a Raoults law standard state (P EtOH = EtOH x EtOH P * EtOH ). [b] Calculate the change in the Gibbs free energy per mole for the transfer of ethanol from solution I to solution II ( I EtOH II EtOH tr G = ). 01 . 1 5768 . 9833 . 5694 . * = = = atm atm P X P EtOH I EtOH I EtOH I EtOH 5 . 1 * = = EtOH II EtOH II EtOH II EtOH P X P = = = = = 5694 . 4976 . ln 15 . 338 314 . 8 ln ln ln ln ln G P P RT X X RT a a RT a RT a RT G tr I EtOH II EtOH I EtOH I EtOH II EtOH II EtOH I EtOH II EtOH I EtOH II EtOH tr * j j j j j P x P a = = Review of Chapter 24: Binary solution made by mixing 2 volatile liquids (solvent (1) + solute (2)) The solvent (major component, x 1 &gt; 0.95) obeys Raoults Law ( * 1 1 1 * 1 1 1 P x P a P = = ) : * 1 1 * 1 1 ln 1 ln ) ( P P RT l vap s + = = (if vapor behaves as an ideal gas) 1 1 * 1 1 * 1 ln 1 ln ) ( ln ) ( x RT l a RT l s + = + = The solute (minor component, x 2 &lt; 0.05) obeys Henrys law ( 2 , 2 2 H k x P = ) : * 2 2 , 2 * 2 * 2 2 , 2 * 2 ln 2 ln ln ) ( ln ) ( P k RT x RT l P k x RT l H H s + + = + = (ideal vapor phase) Non-ideal vapor phase: 2 * 2 2 , * 2 ln 2 ln ln ) ( a RT P k RT l H s + + = where a 2 = x 2 2 2 2 * 2 2 , * 2 ln 2 ln ln ln ) ( RT x RT P k RT l H s + + + = ( 2 1 as x 2 0) The j RT ln term takes into account deviations from Henrys law....
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## This note was uploaded on 06/22/2010 for the course CHEM 21360 taught by Professor Ame during the Spring '09 term at East Los Angeles College.

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110c-lecture15 - Problem Set #3 Due Wed May 5 th . Problem...

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