Question #7b856

1 Answer
Stefan V.
Jan 26, 2015

Since this is an equilibrium reaction, the data given point to the formation of more reactants.

The Gibbs free energy change for a system that has a reaction quotient equal to #Q_c# is

#DeltaG_("rxn") = DeltaG^@ + RT * ln(Q_c)#

You know that at equilibrium, #DeltaG = 0# and #DeltaG^@ = -RT * ln(K_c)#where #K_c# is the equilibrium constant for the reaction.

The above equation can be written as

#DeltaG_("rxn")= - RT * ln(K_c) + RT * ln(Q_c)#, or

#DeltaG_("rxn") = RT * ln(Q_c/K_c)#

This means that

#ln(Q_c/K_c) = (DeltaG_("rxn"))/(RT)#

Since #R# and #T# will always be positive, the value on the right of the equation is positive, which means that #(Q_c)/(K_c)# is greater than 1, or that #Q_c# is greater than #K_c# .

This means that the reverse reaction will be spontaneous and the formation of more reactants will be favored.

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