$$\ce{Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq) + Heat}$$

  1. From a conceptual standpoint, shouldn't the change in entropy be
    positive for this reaction because there are more moles of gas in the products vs the reactants? Why is this not the case? Can this be justified through a conceptual view of the problem rather than a pure mathematical approach(simply calculating delta s as negative)?
  2. From a conceptual standpoint, why is the change in entropy negative for this reaction?

Screenshot of reference data (source):

entropy data

  • $\begingroup$ What's your source for the entropy data? @ShuryuKisuke $\endgroup$
    – Zhe
    Mar 6 '19 at 2:12
  • $\begingroup$ chemistry-reference.com/reaction.asp?rxnnum=697 $\endgroup$ Mar 6 '19 at 2:15
  • $\begingroup$ Can you show me the work for the entropy calculations because when I asked my Chemistry teacher about it she also said it was negative? $\endgroup$ Mar 6 '19 at 2:26

For starters, you should be suspicious when the reference page is formatted as it is.

The real issue I see is that the page lists $\Delta S_{f}$. This is where you should immediately call BS.

We don't use this quantity because entropy is not an energy, therefore, you cannot set the standard state entropy to 0. That means that entropy of formation is pretty useless because you need to know the actual entropy of the standard state of components to know what the entropy of the species is. For elements, the standard molar entropy is positive. However, for solutions (and this is what I got wrong in my now-deleted comment), the standard molar entropy can be negative thanks to solvation (specifically, solvation shells of water can be highly "ordered").

In order to compute the entropy of this reaction, you need to know the standard molar entropy of each species. The easiest thing to do is to look it up in a trustworthy reference source, for example a NIST database or CRC Handbook of Chemistry and Physics.

$$ \begin{array}{|c||c|} \hline\hline \text{substance} & S^{\circ} (\mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) \\ \hline \ce{H2(g)} & 130.68 \\ \ce{Mg(s)} & 32.67 \\ \ce{HCl(aq, 1 M)} & 56.6 \\ \ce{MgCl2(aq, 1 M)} & -25.1 \\ \hline \end{array} $$

Sources: hydrogen, magnesium, hydrochloric acid CRC handbook under "Thermodynamic Properties of Aqueous Ions", magensium chloride CRC handbook under "Thermodynamic Properties of Aqueous Ions"

These are the same values in your reference, but I had trouble believing those data. So, basically, the answer is correct. This reaction has negative entropy change.

Why? Because solutions are kind of evil. Solubility and other solution phenomena can be really hard to predict because they depend so much on how the solvent interacts with the solute. In this case, it's probably driven by some idiosyncratic size of the chloride anion, which somehow allows water molecules to form very nice solvation spheres around the anion. I'm rationalizing this because the standard molar entropy for solutions of magnesium iodide (84.5) and magnesium bromide (26.8) are all positive. These ions are probably of a size that prevents the water molecular from fitting together nicely into a single shell. The imperfections in the solvation shell increase entropy.

That means that the large entropy of hydrogen gas is balanced by some stupid solvent effect with negative entropy, so the overall change in entropy is negative.

  • $\begingroup$ Still is difficult to relate the effect to solvation of chloride ions as for formally nothing change between the two sides of the reaction. It should be a fine balance involving Mg++ too. Puzzling. Isn't? And, again formally, we attain two "entities" starting from three, which can also contribute at least in the specific case. $\endgroup$
    – Alchimista
    Mar 6 '19 at 9:15
  • $\begingroup$ @Alchimista I did actually contemplate this last night when I wrote out the answer. My guess is that hydration of the proton is very favorable from an entropic standpoint. I probably shouldn't have put away my CRC handbook in the most hard to reach place in the basement after using it last night... Anyway, without looking it up, I would guess that the ability to easily shift a proton around in a hydrogen bonded network of water might actually provide the high entropy I'm looking for. Let me know if you have additional thoughts. This is the most interesting question I've answered in a while. $\endgroup$
    – Zhe
    Mar 6 '19 at 21:15

Zhe's answer is good and I could have added a comment but think it is worth a separate answer to ellaborate on his. It's true that it is difficult to develop an intuition around entropy changes associated with solutions and gases, because their entropy is a strong function of concentration.

The solutions in this case contain dissociated ionic species, respectively $\ce{2HCl}$ (dissociated into $\ce{2H+}$ and $\ce{2Cl-}$) and $\ce{MgCl2}$ (dissociated into $\ce{Mg^{2+}}$ and $\ce{2 Cl-}$), so the effective change in moles of ions per mole of reaction is -1. This counters the entropically favorable release of $\ce{H2}$ gas.

In addition, the data in the table refers to formation of 1 M solutions or gas at STP (1 bar, 298.15 K). But what if we'd worked with different concentrations or pressures?

Say the concentrations are 0.1 M. Then the entropy of reagents would be increased more than that of the products, by $$ \mathrm{\Delta S = Rlog(10)} = \pu{19.14 J/molK}$$ so that the total change in entropy change would be $ \pu{-59.31 J/molK}$.

Instead if the concentrations were 5.5 M (saturated $\ce{MgCl2}$), then assuming again ideality $$ \mathrm{\Delta S = Rlog\left(\frac{1}{5.5}\right)} = \pu{-14.17 J/molK}$$ so that the total change in entropy change would be $ \pu{-26.00 J/molK}$.

If the product $\ce{H2}$ gas had been very dilute, say 0.01 bar, then the entropy of the gas would have been greater by $$ \mathrm{\Delta S = Rlog(100)} = \pu{38.29 J/molK}$$ and the reaction entropy would be $ \pu{-1.88 J/molK}$, a much smaller change. Another order of magnitude reduction in the pressure of the gas and the entropy change would be favorable by $ \pu{17.26 J/molK}$.

  • $\begingroup$ A possibly useful analysis is to compute the enthalpy change and think about how much the entropic change actually even matters at ambient temperatures. Put it another way, what's the dominate contribution to the favorable free energy change, entropic or enthalpic? $\endgroup$
    – Zhe
    Mar 6 '19 at 15:26

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