I was trying to find relation between Gibb's free energy change and equilibrium constant for the following homogeneous gaseous phase reaction at constant temperature $$\ce{aA +bB }\rightleftharpoons \ce{cC + dD}$$

Let the initial partial pressures are $P_{1,i}$ and final partial pressures are $P_{2,i}$ for $i=A,B,C,D$

Now Gibb's free energy change of reaction is given by

$$\Delta_rG=\Delta G_{reac} + \Delta G_{prod}$$

However when I tried to find Gibb's free energy change for $\ce{A}$, I got stuck at first step

Using the relation $dG=VdP-SdT$

$$\Delta G_{A}=\int_1^2VdP=\int_1^2\frac{n_ART}{P_A}dP$$

However I cannot perform this integration as $n_A$ is not constant.

I saw other answers relating to same question but cannot gather much from them

Is my method correct ? If yes, then how can I continue ?

  • $\begingroup$ Are you familiar with the effect of concentration of a species on the free energy of a mixture of gases? $\endgroup$ Apr 28, 2022 at 9:03
  • $\begingroup$ The chemical potential (in this case free energy of the gas divided by amount of gas in moles) remains constant when you evaluate the Gibbs free energy of reaction. It is the amount of each gas that changes. $\endgroup$
    – Buck Thorn
    Apr 28, 2022 at 14:35
  • $\begingroup$ @BuckThorn $\mu_i=\mu^o_i+RT\ln P_i$ , as the reaction proceeds in forward direction the partial pressure of $i$ decreases and hence chemical potential should decrease , shouldn't? $\endgroup$
    – Lalit
    Apr 28, 2022 at 18:07
  • $\begingroup$ Please see my answer. By the way, it's Gibbs' (although the apostrophe is usually dropped), not Gibb's. $\endgroup$
    – Buck Thorn
    Apr 28, 2022 at 22:59

1 Answer 1


It's sometimes not obvious because of the way the subject may be taught, but $\Delta_rG$ is what's called a differential form, so it actually describes how the free energy changes as the reaction progresses by an infinitely small change in the amount of reactants and products at a given composition, $T$ and $p$.

The equation presented for the chemical potential $\mu_i=\mu^o_i+RT\ln p_i$ is correct, but the partial pressures are constant when evaluating $\Delta_rG$. In fact, that equation can be used to derive the relation between the reaction quotient and the Gibbs free energy of reaction:

$$\Delta_rG = \sum_i c_i \mu_i= \sum_i c_i \left( \mu^o_i+RT\ln p_i \right)\\ = \sum_i c_i \mu^o_i + RT \sum_i c_i\ln p_i \\ = \Delta_rG ^o + RT \ln Q $$

where $Q = \sum_i c_i\ln p_i$ is the reaction quotient and the $c_i$ are the coefficients of the reaction (not to be confused with the amount $n_i$ of each reactant or product, which remains constant).

This idea can be conceptually difficult at first. Another way to think of it is to imagine a very (=infinitely) large reaction vessel for which you measure the change in free energy upon changing the amount of reactants and products by $c_i$ (see for instance the textbook by Klotz and Rosenberg). Since the $c_i$ are finite and the vessel infinite, the composition remains constant. It's a mathematical sleight of hand equivalent to adding an infinitely small amount of substance to a finite system.

  • $\begingroup$ so basically $dG=\Delta_rG d \xi$ and change in $\Delta G \neq \Delta_rG$ ($\xi$ is extent of reaction and $\Delta G$ is overall change in Gibbs' energy) , am I right? $\endgroup$
    – Lalit
    Apr 29, 2022 at 6:04
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Buck Thorn
    Apr 29, 2022 at 8:10

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