I wanted to ask clarifications about a passage in Atkin's physical chemistry book, chapter 9, in the paragraph 9.2 description of equilibrium.

For studying equilibrium of a reaction, where $\nu_i$ are the stechiometric coefficients it imposes $\sum_i \nu_i \mu_i=0$ with $\mu$ the chemical coefficients. At this point it writes the chemical coefficients as a function of the activities $\mu_i=\mu_i^\circ+RT\ln(a_i)$ and identifies the term $\sum_i \nu_i \mu_i^\circ$ with the standard reaction Gibbs energy. I do not understand this identification. I thought that the standard state for the definition of activities (even dependent on wether the substance is a solute or a solvent) was not the same reference state for computing the reaction Gibbs energy (and does not depend on the role played by the substance in the reaction). I'm surely getting something wrong...


2 Answers 2


$\mu_i=\mu_i^o+RT ln(a_i)$ is the definition of activity ($a_i$).

Therefore, the standard state for standard chemical potential ($\mu_i^o$) is necessarily part of the definition of activity.

For a pure liquid or solid, the standard state is the pure substance at 1 bar.

For a liquid solvent, the standard state is also the pure liquid at 1 bar.

For a gas, the standard state is the fictitious state of an ideal gas at 1 bar.

For solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution.

$\Delta G^o = \sum_i \nu_i \mu_i^o$ by definition.

So there is no one standard state for $\Delta G^o$, because the various substances may have different standard states.

  • $\begingroup$ Actually Atkins says that for a solute the $\mu_i^0$ is your equation is defined as $\mu_i^0=\mu_i^*+RT ln(K_B/p^*_B)$, where $\mu_i^*$ is the chemical potential of the pure solute at the same temperature, $p^*_B$ the vapour pressure of the pure solute ad that pressure, and $K_B$ the constant given bu Henry's rule. I think this is the chemical potential of the pure substance when the vapour pressure is $K_B$ (not sure) but anyway I do not see here a link with the standard conditions (1 bar) that are used to define the standard Gibs free energy of reaction... $\endgroup$
    – tirrel
    Jan 23, 2015 at 21:59
  • $\begingroup$ I wanted to say that $p_B$ is the vapour pressure at that temperature (I do noto think the external pressure plays a role).. I also used i and B ad the same symbol I'm sorry bit I'm noto able to edit the message... $\endgroup$
    – tirrel
    Jan 23, 2015 at 22:14
  • $\begingroup$ @tirrel The way you and/or Atkins have $\mu_i^o =$ (function of temperature) doesn't make sense. Should $\mu_i^o$ just be $\mu_i$? The closest equation that I know of to what you have written is $\mu_i = \mu_i^* +RTln(x_i)$ for an ideal solution and $P_i = K_ix_i$ for Henry's law and $P_i = x_iP_i^*$ for Raoult's law. $\endgroup$
    – DavePhD
    Jan 24, 2015 at 19:31
  • $\begingroup$ DavePhD: this is what I understand from Atkins, which is a well respected book, so maybe I do not understand it. The section is 7.2, if you want to have a look. By the way in your equation $μ_i=μ^∗_i+RTln(x_i)$, which also appears in Atkin's book how do you interpret the symbols? I though that $μ^∗_i$ was the chemical potential of the pure substance at the temperature given (so it is temperature dependent). And by the way if $x_i=1$ we get $μ_i=μ^∗_i$, which suggests to me that the reference state must indeed be temperature dependent. $\endgroup$
    – tirrel
    Jan 24, 2015 at 20:39
  • $\begingroup$ @tirrel I agree Atkins is a great book, but I don't have it at home, only Levine. The superscript in $\mu_i^o$ means "standard state" and the superscript in $\mu_i^*$ means "pure state". The equation $\mu_i=\mu_i^*+RTln(x_i)$ doesn't say anything about the standard state, and isn't generally true, it is an ideal-law type of equation. Make sure you check if the equations are true generally, or just in some ideal condition, and whether Atkins says $\mu_i^o$ or $\mu_i$ $\endgroup$
    – DavePhD
    Jan 24, 2015 at 21:32

So is the point of contention here "how does the standard state for the activities relate to the standard state for the Gibbs free energy change"?

To see how choice of standard state for an activity relates to the standard state chosen for the chemical potential, write the equation as

$$\mu_i = \mu_i^\circ + R T \ln a_i = \mu_i^\circ + R T \ln \frac{\gamma_i c_i}{c_i^\circ}$$

where the standard states are consistent.The standard state is being defined so that when $\gamma_i c_i = c_i^\circ$, $\mu_i = \mu_i^\circ$. If we're using molarities, we can choose $c_i^\circ = 1\ M$.

Now suppose you want to define a standard state that's different from 1 in the units you're using for concentration; e. g, suppose we have a reaction that includes $\rm H^+$ and we want to define the biological standard state ($^\oplus$) as $c_{\rm H^+}^\oplus = 10^{-7}\ M = 10^{-7} c_{\rm H^+}^\circ$, while using molarities as our concentration unit.

We can compute a new chemical potential:

$$\begin{array}{rcl}\mu_{\rm H^+} &=& \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\oplus}\\ &=& \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{10^{-7} c_{\rm H^+}^\circ}\\ &=& (\mu_{\rm H^+}^\circ + 7 R T\ln 10) + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\ &=& \mu_{\rm H^+}^\oplus + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\ &=& \mu_{\rm H^+}^\oplus + R T \ln a_{\rm H^+} \end{array}$$

Now at unit activity $\gamma_i c_i = c_i^\circ$ and $\mu_i = \mu_i^\oplus$. Your chemical potential (and your Gibbs free energy changes) are now defined in terms of this new standard state. And note that the chemical potential at this new standard state ($\mu_{\rm H^+}^\oplus$) is indeed temperature dependent. To convert between the two standard state Gibbs free energies in this particular case,

$$\Delta G^\oplus = \Delta G^\circ + 7 \nu R T \ln 10$$

where $\nu$ is the stoichiometric coefficient of $\rm H^+$ in the reaction.

TL;DR: The standard state you choose for the concentration at unit activity becomes the standard state for the standard state Gibbs free energy change.


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