Question

Question #7b856

Answer 1

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.