Lecture26A - Lecture 26 1 Gibbs Energy: Dependence on and...

Info iconThis preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 26 1 Gibbs Energy: Dependence on and T(26.4 7) - r As mentioned in the last lecture, while can be used to determine the equilibrium constant, it does not determine whether the reaction will occur. It only determines the extent of the reactions, o G r i.e. until 0. G = Consider the classic (at least for physical chemistry) example: ( 29 ( 29 2 4 2 2 N O g NO g G For this reaction: ( 29 2 51.258 97.787 4.729 / at 298K ln so ln 1.908 o r P P G kJ mol RT K K =- = = - = - 2 2 4 2 0.148 at equilibrium NO P N O P K P = = 2 4 2 2 So , but note that 0. It is present. N O NO NO P P P Lecture 26 2 ( 29 2 4 2 Hypothetically, let's consider starting with one mole of N O . If moles are converted to NO , then g 2 4 2 1 , 2 , 1 2 1 N O NO n n n = - = = - + = + The Gibbs energy of the system is: ( 29 ( 29 2 4 2 2 4 2 1 2 N O NO N O NO G G G G G G = = + =- + Since ln o i i i G G RT P = + ( 29 ( 29 ( 29 2 4 2 2 4 2 1 2 1 ln 2 ln o o N O NO N O NO G G G RT P RT P =- + +- + ( 29 2 4 2 T 1 2 Let's keep the math simple: ; 1 1 Since for ideal gases, let's keep P at 1 bar so 1 . N O NO i i T i i x x P x P P x - = = + + = = ( 29 ( 29 ( 29 ( 29 ( 29 2 4 2 1 2 1 2 1 ln 2 ln 1 1 o o f f G G N O G NO RT RT - =- +...
View Full Document

Page1 / 8

Lecture26A - Lecture 26 1 Gibbs Energy: Dependence on and...

This preview shows document pages 1 - 3. Sign up to view the full document.

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