First, I should mention that the question was already asked here: Why is active mass of a pure solid or liquid always taken as unity? However, the treatment there is qualitative and I do not really understand the reasoning for the disclusion of solids and liquids from the reaction qoutient that is presented there. I have attempted to derive this fact mathematically and failed, which why I ask this question again.

Let us look at an example reaction which involves solids and gases. The treatment for other reactions or reactions with liquids shouldn't differ significantly. A very simple reaction for this matter will be the combustion of graphite

$$\ce{C(s) + O2(g) -> CO2(g)}$$

Reaction quotients are derived from the equation for change in Gibbs free energy at nonstandard pressures, so let's find that. Suppose the reaction is carried out in nonstandard pressure, and $\Delta G^o_r$ is known. Let's (I) compress the system isothermally to standard pressure, (II) carry out the reaction and then (III) return it isothermally to the original pressure.

I. $\Delta G$ for the oxygen here is the well known $RT\ln (1/(P/P_o))$. For the carbon: In any isothermal process $dG=VdP$, let's assume for our ideal solid that the volume is constant (I have also tried this in a model where the volume changes with pressure with bulk moduli, the result does not differ significantly), so $\Delta G=V(P-P_o)$.

II. Here simply $\Delta G=\Delta G^o_r$.

III. Here it's just gas compressing so again $\Delta G=RT\ln (P/P_o)$

Adding the contributions from all three stages we have $$ \Delta G=RT\ln (1/(P/P_o))+V(P-P_o)+\Delta G^o_r+RT\ln (P/P_o)=\Delta G^o_r+V(P-P_o) $$

This result is different from what a model of an activity of unity for the carbon would predict. What went wrong? And why are activities of solids and liquids taken to be unity? Any help will be appreciated.


2 Answers 2


It is not that case that the activity of solids and liquids is always taken as unity.

Instead, activity is unity for a pure solid or pure liquid at standard pressure.

This is a matter of definition of "activity" and "standard state".

According to the IUPAC definitions:

activity (relative activity), a

Defined by the equation
$a=e^\frac{\mu-\mu ^0}{RT}$
where $R$ is the gas constant, $T$ the thermodynamic temperature, $\mu$ the chemical potential and $\mu ^0$ the standard chemical potential the definition of which depends on the choice of the standard state.

and standard state

State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p °, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p ° ...

So activity is 1 almost by definition when a substance is in its standard state, because $\mu - \mu ^0 = 0$ and $e^0 = 1$


Nothing "went wrong". Hf for CO2 is ~393 kJ/mole. 1 latm = 9.87 kJ, so lets see what a mole of C does Gibbs energy wise going from 1 atm to 11. OK 1 mole is 0.0053 l and multiplying it by 10 atm we get 0.053 latm ~ 0.15 kJ or 0.15% of the heat of formation for the CO2. If you want to get down to this level of accuracy, you gotta throw out the Ideal Gas Law (the simple approximations it is based on aren't that accurate) and all sorts of consequences follow. Then, we have to consider how you actually change pressure "reversibly" (hint:you don't). The point of this level of treatment of thermodynamics is to show you how useful it can be to get an accurate idea of what is going on. It is accurate, unless your idea of "accurate" is errors in the parts per thousand range. (Ideal Gas Law is generally good to between 1 and 3% (at near STP), but it can be off over 10% for some not-so-uncommon gasses (and let's not even consider about gasses near their boiling points, or at cryo-temperatures or at mega-pressures...). A quick search for activity of graphite vs pressure indicates that at 100 bar, its activity coefficient is 1.01. That is, assuming an activity of 1.00 is a great approximation near STP. Liquids (but NOT solutions!) will be similar (but not quite as well 'behaved'). So, pragmatically, the answer to your question is:"Because it gives us an accurate estimate of the system's behavior." Another way to put it is: "Because it is useful." One other factor to consider is what the alternatives are, that is, if we wanted to get more accurate predictions, could we formulate a generally applicable means to do so? The answer is, as far as I know:no. There is no generally applicable (call this a "powerful") method; to get more accurate requires case-by-case considerations: how is heat gained or lost by system, what kind of container walls is it in contact with, how was complete combustion insured, what were the losses due to friction during compression and expansion, etc. etc.

  • $\begingroup$ Thank you very much. This is in general what I was suspecting would be the explanation, but I really wasn't sure so thank you. $\endgroup$
    – roymend
    Commented Sep 19, 2016 at 19:13

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