Horvath Practice Problems 106

# Horvath Practice Problems 106 - Q = RT ln Q K |{z} < 1 |...

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Turning to 4 rxn G, this tells us how the reaction will go under the current conditions of the reaction. So if we consider the case where we mix reactants to perform areact ion . Initially there are no products and therefore Q =0 and we have 4 rxn G = 4 rxn G ° + RT −∞ z}|{ ln 0 = −∞ (25.4) So regardless of what 4 rxn G ° , even if very reactant favored, initially there is a tremendous amount of thermodynamic force for the reaction to proceed. Then some products are formed but before we reach equilibrium Q<K .Thu s 4 rxn G = RT ln K z }| { 4 rxn G ° + RT ln <K Q = RT ln K +
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Unformatted text preview: Q = RT ln Q K |{z} < 1 | {z } negative = negative and the reaction will continue to proceed towards products. At equilibrium Q = K and 4 rxn G = − RT ln K z }| { 4 rxn G ° + RT ln = K Q = − RT ln K + RT ln K = 0 On the other side of equilibrium Q > K. Thus 4 rxn G = − RT ln K z }| { 4 rxn G ° + RT ln >K Q = − RT ln K + RT ln >K Q = RT ln Q K |{z} > 1 | {z } positive = positive and the reaction will reverse to form reactants. GIBBS FREE ENERGY AND EQUILIBRIUM 105...
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## This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.

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