Is Gibbs free energy change applicable to forward and reverse reactions at equilibrium?

Question

Consider a reaction

$$\ce{A + B <=> C + D}. \tag{R1}$$

Now $\Delta G$ for the forward reaction is

$$\Delta G_\mathrm{fwd} = \Delta H - T\Delta S. \tag{1}$$

For the reverse reaction $\Delta H$ and $\Delta S$ are clearly the same in magnitude but opposite in sign. Therefore,

$$\Delta G_\text{rev} = -\Delta G_\text{fwd}, \tag{2}$$

which implies that if the forward reaction is feasible, then the reverse isn’t and vice versa. But both the reactions occur simultaneously in an equilibrium reaction. Can anyone please help me sort this discrepancy?

Answer 1

I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts.

$G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the equation chosen, and independent of the concentrations. $\Delta G°$ is obtained by calculating the sum of the Gibbs energies of the products in the pure state, and substracting the sum of the Gibbs energies of the reactants in the pure state, before the initial mixture is done.

During the reaction, the $G$ values are changing all the time, because they are proportional to the concentration of the reactant (or of the products) . The $G°$ values are never changing during the course of a reaction.

As soon as you mix the initial reactants, the Gibbs energy of each reactant decreases from $G°$ to a lower value $G$ < $G°$ even if no chemical reaction occurs. And the Gibbs energy of each reactant continues to decrease due to the chemical reaction. Simultaneously the Gibbs energy of the products are moving in the opposite direction. But they are is very low at the beginning of the chemical reaction. The reaction is finished when the sum of the Gibbs energies of the reactants is equal to the sum of the Gibbs energies of the products.

Answer 2

The Gibbs energy of reaction $\Delta_\mathrm{r} G$ determines in which direction equilibrium lies, i.e. in which direction there has to be a net reaction (with a change in concentrations) to reach equilibrium.

When equilibrium has been reached already, there is no net reaction (i.e. concentrations are constant). Nevertheless, at the molecular level, reactions in both directions are observed. They just are happening at the same rate, so there is no net change.

In general, thermodynamics makes statements about the bulk properties, not about individual particles. So if you look really closely, there are fluctuations in concentration, but they are negligible because an change of a single molecule in, say, a mole of molecules is a relatively small change (and not detectable by measuring bulk properties).

But both the reactions occur simultaneously in an equilibrium reaction. Can anyone please help me sort this discrepancy?

In a nutshell, reactions don't stop just because the system has reached equilibrium.

Answer 3

First you considered the reaction reversible:

$$\ce{A + B <=> C + D}\tag{R1}$$

Then you considered each reaction irreversible to determine its $\Delta_\mathrm{r}G:$

\begin{align} \ce{A + B &-> C + D} \tag{R2} \\ \ce{C + D &-> A + B} \tag{R3} \end{align}

Its not allowed at all! This way the $\Delta H$ and $\Delta S$ of all the reversible reactions would come out zero which is totally vague. In a reversible, a forward reaction may occur with a greater/lesser spontaneity than the backward reaction giving us net $\Delta H$ and $\Delta G.$ So you cannot say that both reactions occur equally.

If $\Delta G$ is more negative for a reversible reactions at a specific concentration, forward reaction proceeds more than the backward reaction and likewise we can say that reaction is less spontaneous in the backward direction. In a reversible reaction $\Delta G$ can be determined as overall net effect taking place and not for separate reactions because we cannot consider both to be equally taking place.

Answer 4

Being a professor, the most difficult thing to instill in the mind of students is that thermodynamics don't consider microscopic changes.

Koushal, you did the same thing. In calculating the change in free energy, we should only see what is visible. At a particular instant when equilibrium is not reached, both forward and backward reactions proceed with different rates and there is a net reaction in any particular direction.

Suppose the rate of forward reaction is assumed to be greater than the rate of the reverse one, so microscopically products are converting into reactants and reactants into products but macroscopically where it actually matters for thermodynamics reaction proceeds in the forward direction and hence change in free energy for the whole reversible reaction will be calculated to have a negative value only and only at this instant because rate is not constant since equilibrium is not reached.

What you did is that you split the reversible reaction into two independent irreversible reactions. Although that's true at microscopic level but not in thermodynamics.