09 - In this case, the entropy of the mixed gas is greater...

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10/15/09 Chapter 10 Standard Free Energies of Formation ∆G f °= the change in standard free energies when one mole of a compound is formed in its standard state from its constituent elements in their standard states ∆G f ° (element, standard state) = 0 ∆G rxn ° = Σ ∆G f °(products) – Σ ∆G f ° (reactants) Determine ∆G f ° of an irreversible reaction A + B C is spontaneous or not Chemical equilibrium - the state where the concentration of products and reactants remains constant with time *This does not mean A, B, & C are not reacting—only that their rates of formation & decomposition are balanced At equilibrium ∆G=0 ∆G = 0 = ∆H – T∆S ∆S system = ∆H/T ∆S for a reversible process may be directly measured The equilibrium point The lowest value of G for the system Why: There is another term to consider S

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Unformatted text preview: In this case, the entropy of the mixed gas is greater than either pure gas S equilbrium > S NO2 or S N2O4 The Law of Mass Action—gives a general description of the equilbriu condition j A + k B ( ) l C m D [C] l [D] m = Q [A] j [B] k Q = reaction quotient of reactants/products At equilubrium, Q = K equilibrium constant a true constant its value at a given T does not depend on the initial concentration of reactants & products K will always vary with T Q = K equilibrium Q > K reaction shifts left Q < K reaction shifts right We noted earlier that at equilibrium ,we reach a minimum on the free energy curve There must be some relationship between the equilibrium constant and the standard free 10/15/09 energy of reaction...
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This note was uploaded on 09/19/2011 for the course CHEM 43L taught by Professor Therien during the Fall '09 term at Duke.

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09 - In this case, the entropy of the mixed gas is greater...

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