Gibbs free energy-minimum or zero?

Question

A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is minimum.

On the other hand, we have $$\Delta G=\Delta G^\circ + RT\ln Q$$, where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.

Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there was a change in $G$ and $\Delta G < 0$.

What am I missing here?

Answer 1

I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)

However, in the equation you quote:

$$\Delta G = \Delta G^0 + RT \ln Q$$

the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_\mathrm r G$, with the correct equation being:

$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$

The free energy of reaction is defined as $\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$.

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Thus, this $\Delta_\mathrm r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.

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PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!

Answer 2

Gibbs free energy is a measure of how much "potential" a reaction has left to do a net "something." So if the free energy is zero, then the reaction is at equilibrium, an no more work can be done.

It may be easier to see this using an alternative form of the the Gibbs free energy, such as $\Delta G = -T\Delta S$.

Answer 3

It might be helpful if we defined the changes in free energy in these equations in a more precise manner. Let's just consider an ideal gas reaction.

$\Delta G^\circ$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:

State 1: Pure reactants (in separate containers) in stochiometric proportions at 1 atm pressure and temperature T

State 2: Pure products (in separate containers) in corresponding stochiometric proportions at 1 atm pressure and temperature T

To directly measure $\Delta G^\circ$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G^\circ$ for that path. This path might involve the use of constant temperature reservoirs and semipermeable membranes.

$\Delta G$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:

State 1: Pure reactants (in separate containers) in stochiometric proportions at specified pressures and temperature T

State 2: Pure products (in separate containers) in corresponding stochiometric proportions at specified pressures and temperature T

If the specified pressures just happen to correspond to the partial pressures of the gases in a reaction mixture at equilibrium, then $\Delta G = 0$. For small excursions of the partial pressures from the equilibrium values, $\Delta G$ will increase as the squares of the partial pressure increments. That is what we mean when we say that it is at a minimum at equilibrium.

To directly measure $\Delta G$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G$ for that path. This path might involve the use of constant temperature reservoirs, small weights to be added or removed from a piston, and semipermeable membranes.

Answer 4

I suppose you are confused between $∆G$ and $G$. After equilibrium is achieved, $∆G$ is indeed zero, as no more reduction of free energy is possible (keeping conditions intact). What this also means is that $G$ is minimum.

Throughout the reaction, $∆G$ will have been negative until equilibrium is attained. This means the free energy will keep on reducing until no more reduction is possible. This indicates that $G$ is minimum, since $∆G$ is zero, i.e. the value of $G$ at equilibirum cannot be further diminished.