110c-lecture20 - Gibbs Free Energy ( r G) and K eq ( ) ( )...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Gibbs Free Energy ( r G) and K eq ( ) ( ) ( ) ( ) g Z g Y g B g A Z Y B A + = + B B A A Z Z Y Y dn dn dn dn dG + = bar P P P P rxn r r Q RT G G ln + = B B A A Z Z Y Y rxn P P P P Q = P EQ B B A A Z Z Y Y rxn K P P P P Q = = Initial state: Z Y B A 1 = = = = Final state: equilibrium ( r G = 0) and rxn r r Q RT G G ln + = P r K RT G ln + = P EQ B B A A Z Z Y Y r K RT P P P P RT G ln ln = = Rxn Coordinate Free Energy G < G r K P > 1 = G r K P = 1 > G r K P < 1 Gibbs Free Energy vs. Extent of Reaction N 2 O 4 (g) 2NO 2 (g) initial conditions: n 0 0 total moles n after reaction: n- 2 total moles + n Start with 1 mole of dinitrogen tetroxide in a cylinder with a piston. At an extent of reaction , the total Gibbs energy (G, NOT G) of the system is ( ) ( ) 2 4 2 2 1 NO O N G G G + = Substituting ln P P RT G G i i i + = ( ) ( ) ( ) 2 2 4 2 4 2 ln 2 2 ln 1 1 NO NO O N O N P RT G P RT G G + + + = Substituting + = P P O N 1 1 4 2 and + = P P NO 1 2 2 ( ) ( ) ( ) + + + + + = P RT P RT G G G NO O N 1 2 ln 2 1 1 ln 1 2 1 2 4 2 ( ) ( ) ( ) + + + + + = P RT P RT G G G NO f O N f 1 2 ln 2 1 1 ln 1 2 1 2 4 2 ( ) ( ) ( ) + + + + + = P RT P RT mol kJ mol kJ G 1 2 ln 2 1 1 ln 1 ) / 51 ( 2 ) / 97 ( 1 Gtot, P=1 bar 96 97 98 99 100 101 102 103 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 extent of reaction Gtot, P=0.1 bar 89 89.5 90 90.5 91 91.5 92 92.5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 extent of reaction Gtot, P=0.01 bar 79 80 81 82 83 84 85 86 87 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 extent of reaction N 2 O 4 (g) 2NO 2 (g) ( ) ( ) ( ) + + + + + = P RT P RT mol kJ mol kJ G 1 2 ln 2 1 1 ln 1 ) / 51 ( 2 ) / 97 ( 1 tot ( ) , = = P T EQ G G EQ P P eq K P K + = 4 The slope of ( ) G vs is G at a given (slope depends on , and goes to zero at EQ ). Equilibrium Constant and LeChateliers Principle Dissociation of dinitrogen tetroxide: N 2 O 4 (g) 2NO 2 (g) initial conditions: n 0 0 total moles n after reaction: n- 2 total moles + n At 298 K, 4 2 2 2 O N NO P P P K = = 0.148 bar for this reaction....
View Full Document

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.

Page1 / 18

110c-lecture20 - Gibbs Free Energy ( r G) and K eq ( ) ( )...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online