Does a high +∆G mean that the reaction is irreversible (and equilibrium cannot be attained) or that the reaction lies to the right?
2 Answers
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.
Explanation:
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.
I think you've got it mixed up.
Large positive
Equilibrium reactions necessarily have
[You may want to read this for further discussion on irreversibility.](https://socratic.org/questions/for-a-spontaneous-process-in-an-isolated-system-the-change-in-entropy-is-positiv#532832)
Consider the deviation of the change in Gibbs' free energy
DeltaG = DeltaG^@ + RTlnQ
- First of all, reactions that have positive
DeltaG are necessarily nonspontaneous, and therefore, the forward direction is easily reversed.
Thus, the reactants have a really hard time reaching a dynamic equilibrium (where
r_(fwd) = r_(rev) ), whatever the size ofbbQ happens to be, since the reverse rate is much larger than the forward rate, i.e.r_(fwd) "<<" r_(rev) .
- Second, all spontaneous reactions are irreversible... in the forward direction. Therefore,
DeltaG must be negative for irreversible forward reactions. In that case,r_(fwd) ">>" r_(rev) .