Does a high +∆G mean that the reaction is irreversible (and equilibrium cannot be attained) or that the reaction lies to the right?

Question

Jeremiah Adams · Accepted Answer

A positive value of ∆G indicates that the reaction is not spontaneous in the direction written, A high value of Keq indicates the equilibrium lies to the right.

Reactions are spontaneous when the change in free energy (∆G)is negative. (spontaneous does not imply rate, only favored direction). Since ∆G = ∆H- T∆S, reactions tend to be favored if the change in enthalpy (∆H) is negative (exothermic reaction, goes to lower potential energy), and the change in entropy (∆S) is positive (system increases in disorder,) If ∆H is negative and ∆S is positive, the reaction is spontaneous at all temperatures. If ∆H is positive and ∆S is negative, the reaction is never spontaneous. If both ∆H and ∆S are negative the reaction may be spontaneous at low temperatures, but when T∆S becomes greater than ∆H, reaction will no longer be spontaneous. If both ∆H and ∆S are positive, the reaction will become spontaneous when T∆S exceeds ∆H.

Mateo Aguilar · Answer

I think you've got it mixed up.

Large positive #DeltaG# indicates a nonspontaneous reaction (i.e. the reverse direction is very fast), which is irreversible in the REVERSE direction, not forward direction.

Equilibrium reactions necessarily have #DeltaG = 0#, so equilibria cannot lie to either side AND be irreversible at the same time.

You may want to read this for further discussion on irreversibility.

Consider the deviation of the change in Gibbs' free energy #DeltaG# from standard state:

#DeltaG = DeltaG^@ + RTlnQ#

Thus, the reactants have a really hard time reaching a dynamic equilibrium (where #r_(fwd) = r_(rev)#), whatever the size of #bbQ# happens to be, since the reverse rate is much larger than the forward rate, i.e. #r_(fwd)# #"<<"# #r_(rev)#.

Serenity Allen · Answer

undefined A high positive ΔG indicates that the reaction is thermodynamically unfavorable and proceeds spontaneously in the reverse direction. This suggests that the reaction is unlikely to reach equilibrium and tends to favor the formation of reactants over products.