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.