I'm calculating molar changes in thermodynamic properties due to reactions between gasses (assumed to be ideal gases). I can calculate $\Delta H$ easily enough, because it's just $\sum_i \nu_i\Delta_f H^\circ_i$, with $\nu_i$ the stoichiometric coefficients. $\Delta G$ (at standard pressure) can be calculated from $\sum_i \nu_i\mu_i$, with $\mu_i = \Delta_f G^\circ_i + RT\log[i]$. Since $\Delta G = \Delta H - T\Delta S$, I can calculate $\Delta S$ as $\frac{1}{T}(\Delta H - \Delta G)$.

So far so good. But I'm inexperienced in working with tabulated quantities (I'm a more of a theoretical physicist than a chemist) and keep making mistakes with signs and units. So I figured that as a sanity check I would calculate $\Delta S$ directly, using tabulated values of $S^\circ$. But then I realised I don't know how to do this, and it seems that none of the textbooks on my desk explain it either.

Based mostly on intuition, it seems like it should be $\Delta S = \sum_i\nu_i(S^\circ_i + R\log[i])$, with the $R\log[i]$ term having something to do with the entropy of mixing. For the example reaction I chose, this gave something with the approximately correct magnitude but the wrong sign (-15.6 instead of 14.2, which is what I get by calculating it from the Gibbs energy and the enthalpy).

So my question is, what is the correct way to calculate the entropy change due to an ideal gas reaction if, for some reason, you only have access to the concentrations and the standard entropies of the reactants?


1 Answer 1


Just to make sure that I get your question correctly; the process in question is a chemical reaction that takes place at a temperature and pressure close enough to the reference temperature and pressure (those used for the tabulated values) that any changes to the tabulated values are neglible? (I assume this, as this is what you have done in the enthalpy calculation)

In that case, it is probably easier that you would expect - you can use (almost) the same approach as you used to calculate $\Delta H$ as both entropy and enthalpy are state functions. That is:

$\Delta_{rx} S = \sum_i \nu_i S^\circ_i$

where $S^\circ_i$ is the tabulated value for each of the components. This is described on Wikipedia.

If the temperature deviates sufficiently from the reference temperature used then you should adjust the entropy change accordingly

$\Delta S_{T_1 \rightarrow T_2} = \int_{T_1}^{T_2}\frac{C_V}{T}dT=C_V\ln(T_2/T_1)$

where the last equality holds if $C_V$ can be assumed to stay constant in the interval $T_1 \rightarrow T_2$. A similar adjustment can be made if the pressure, volume etc. changes during the course of the reaction. Again, as the entropy is a state function, it is enough to adjust the total change in entropy for the reaction.

Hope this helps! Please let me know if I am not being sufficiently clear, and I'll try to elaborate things a bit.

  • $\begingroup$ Hmm, but that's just the calculation for $\Delta_r S^\circ$, not $\Delta_r S$. I.e. it doesn't take account of the entropy of mixing. For a reaction at equilibrium, we should have $\Delta S = \Delta H/T$ (because $\Delta G =0$ at equilibrium) and that won't happen unless $\Delta S$ is concentration-dependent. $\endgroup$
    – N. Virgo
    Commented Aug 10, 2012 at 14:27
  • $\begingroup$ Thanks for the comment about the temperature change though. These are reactions in the atmosphere near the surface, so they're at 1 atm but at around 288K instead of 273K or 298K. $\endgroup$
    – N. Virgo
    Commented Aug 10, 2012 at 14:29
  • $\begingroup$ Ahh, sorry - in that case I think I misunderstood you a bit. The entropy change of mixing two ideal gases should be $\Delta_{mix}S = -nR(x_1\ln(x_1) + x_2\ln(x_2))$. This article on Wikipedia is a bit more thorough. In my experience, the entropy change associated with the reaction itself is (usually) quite a bit larger than the entropy of mixing. $\endgroup$ Commented Aug 10, 2012 at 15:39

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