I am trying to understand the following equation:

$$\Delta G = \Delta G^{\circ} + RT \ln\left(\frac{[C][D]}{[A][B]}\right)$$

for a reversible reaction with reactants A and B and products C and D.

The way I understand it is that the Gibbs free energy change $\Delta G$ is built up by two components. One is the standard Gibbs free energy change $\Delta G^{\circ}$ which is something we can measure in the lab so to say and use as a reference point. The second term accounts for the given conditions such as actual temperature and reactant and product concentrations which deviate from those used to calculate/measure $\Delta G^{\circ}$.

My confusion stems from the following. In the textbook I am reading (Biochemistry by Berg et al) they say that:

"A simple way to determine $\Delta G^{\circ}$ is to measure the concentrations of reactants and products when the reaction has reached equilibrium. At equilibrium, there is no net change in reactants and products; in essence, the reaction has stopped and $\Delta G$ = 0." They then set $\Delta G = 0 $ and solve for $\Delta G^{\circ}$.

So they appear to be doing the backwards process of what I described above. Also why is this allowed at all? Isn't setting $\Delta G = 0 $ akin to enforcing a statement about the spontaneity of the reaction? The way I understood $\Delta G$ was that if it is positive the reaction is not spontaneous and if $\Delta G$ is negative the reaction is spontaneous. By extension if $\Delta G=0$ the reaction is at equilibrium, so I guess this means that $K=1$? Those should be things intrinsic to a given reaction, so how can we just set $\Delta G$=0? Also does $\Delta G$ change throughout a reaction at all? I mean, it clearly does, because it is a process that evolves in time but I thought that when we say $\Delta G$ we always take initial and the final free energy, so in that sense it is a constant and not an evolving quantity.

As you have probably gathered from the question I have no chemistry background so any help would be greatly appreciated.


1 Answer 1


Isn't setting $\Delta G = 0$ akin to enforcing a statement about the spontaneity of the reaction?

Yes, it is. But they stipulate that the reaction is at equilibrium, which tells you that neither the forward nor backward reaction is spontaneous, which is precisely what allows you to assert that $\Delta G = 0$.* In other words, when a system is at equilibrium, we have $\Delta G = 0$.

By extension if $\Delta G = 0$ the reaction is at equilibrium, so I guess this means that $K = 1$?

No, that's not right, because we have $\Delta G^\circ = -RT \ln K$ (note the standard sign). $\Delta G$ has no direct relation to $K$.

One way of putting it is that $\Delta G^\circ$ and $K$ are properties of the reaction. That is, once you write down a chemical equation, these two are fixed, immutable constants.† However, $\Delta G$ (and $Q$) depend on the exact composition of the reacting system, i.e. they depend on the concentrations of reactants and products at any given point in time.

Also does $\Delta G$ change throughout a reaction at all?

Yes, it does. As the reaction proceeds towards equilibrium, it goes from whatever value it is initially at, to zero.

I thought that when we say $\Delta G$ we always take initial and the final free energy

... Kind of. You need to be careful here because the difference of a free energy should have units of free energy, traditionally kJ. However, $\Delta G$ is usually given in units of molar free energy, i.e. kJ/mol.

The reason is because what you're calling $\Delta G$ is actually not the difference between the free energy of products and free energy of reactants. It's actually a lazy shorthand for $\Delta_\mathrm{r}G$, the reaction Gibbs free energy. Actually, $\Delta_\mathrm{r}G$ is the difference between the chemical potentials of reactants and products in the reaction, multiplied by their stoichiometric coefficients.

I wrote about this difference before, see What is the difference between ΔG and ΔrG?, although I must apologise as it's a bit convoluted. (In fairness to myself, the topic in general is pretty convoluted.)

* A quick and dirty definition of spontaneity is that if you set something up and the reaction naturally goes in the net forward direction, the forward reaction is spontaneous. A formal definition would involve $\Delta G$, but this is not suitable in the present case, as it would be a circular argument.

† To be precise, they're constant for a given set of reaction conditions. $K$, and $\Delta G^\circ$, will change with temperature or pressure, for example. But that's not under consideration here.

  • $\begingroup$ Thank you for the quick response. I will also check the other question you linked out. That clears it up I guess, I just found it a bit strange for some reason that we calculate the standard free energy change from an assumption on the free energy change. If it is not too much to ask, could you give an example of how this equation is used in the lab so to say? It obviously can be used in many ways but what is the prototypical research application from your experience? $\endgroup$ Jan 22, 2022 at 21:59
  • $\begingroup$ @terraregina Er, that's kind of hard to discuss! It really depends on what kind of chemistry you're doing. Often, determining $\Delta_\mathrm{r}G^\circ$ (note the standard symbol again) is only a means to an end, in the sense that you use it to in turn equilibrium concentrations / binding constants / etc which are more interesting. It also depends on what kind of values you are expected to output - if you're studying drug binding then you want binding constants, but if you're just benchmarking some computational method, you could just report $\Delta G$'s and be done with it. $\endgroup$ Jan 22, 2022 at 22:09
  • $\begingroup$ @terraregina You can imagine how to do this yourself. Suppose you have a reaction corresponding to the expression you wrote above ($\ce{A + B->C + D}$) and measure that, at equilibrium, [A] = 2 M, [B]= 7 M, [C] = 3 M, and [D] = 5M. Assuming $T=298 \pu{K}$, what would be $\Delta G^{\circ}$? I'd highly enourage you to try this calculation. I think after you do it, things will become much clearer to you. $\endgroup$
    – theorist
    Jan 22, 2022 at 22:33
  • $\begingroup$ That is, once you write down a chemical equation, these two are fixed, immutable constants. $K$ does change with temperature though! $\endgroup$
    – AVS
    Feb 28, 2023 at 7:48
  • 1
    $\begingroup$ @AVS That's fair, thanks. $\endgroup$ Feb 28, 2023 at 17:41

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