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I know that you can have 2 acids reacting with 1 base. These 4 reactions are an example of this:

$\ce{3 HCl + H2SO4 + NaOH -> 3 H2O + SCl2 + O2 + NaCl}$

$\ce{6 HCl + 4 H2SO4 + 2 NaOH -> 8 H2O + 2 S2Cl2 + 5 O2 + 2 NaCl}$

$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2S + 2 O2 + NaCl}$

And

$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2SO4 + NaCl}$

I checked and all these equations are balanced. But will these reactions forming sulfur chlorides and ionic compounds with sodium and sulfur as well as $\ce{H2O}$ and for 3 of them, $\ce{O2}$, actually occur without any added heat or are they endothermic acid base reactions and thus require heat or will the stronger acid(in this case $\ce{HCl}$) be the one that reacts and the weaker acid (in this case $\ce{H2SO4}$) just make the pH of the solution lower?

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    $\begingroup$ Those first three equations might be stoichiometrically valid, but they look extremely weird from a chemical perspective, and I would very surprised if they happened. I suspect if you calculate the reaction free energy change using the free energy of formation for each compound, you will find that the last reaction is far more favourable than the rest. $\endgroup$
    – Nicolau Saker Neto
    Commented May 19, 2015 at 12:32

2 Answers 2

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As Klaus and I mentioned, the first reactions seem very strange from a chemical perspective, even if they are correctly balanced. This is where a chemist needs to exercise their intuition and experience.

Balancing reactions is nothing but solving a linear system of equations, so it is entirely possible to propose reactions which balance perfectly, but which are based on questionable chemistry. For example, in all but the last reaction, the sulphur atoms in $\ce{SO_4^2-}$ anions are being reduced to lower oxidation states. While this is possible in some conditions, it takes quite strong reducing agents, and none of the reagents shown are known to possess significant reducing character. Reactions which release gaseous oxygen are also not particularly common. Furthermore, two reactions propose the formation of sulphur chlorides, which would actually hydrolyse in contact with water.

Disregarding intuition, it should be possible to pick out which reaction is the most favourable by simply comparing the Gibbs free energy of reaction in each case; the most exergonic reaction, with the most negative value of $\boldsymbol{\Delta_\mathbf{r}G}$, should dominate (assuming no kinetic factors bar it from happening, which is unlikely in this situation). I did a very quick and dirty calculation, out of curiosity, by taking the standard Gibbs free energy of formation $\Delta_\mathrm{f}G^\circ$ for each compound and summing them together to obtain the standard Gibbs free energy of reaction:

$$\Delta_\mathrm{r}G^\circ = \sum\limits_\text{products}\Delta_\mathrm{f}G^\circ - \sum\limits_\text{reagents}\Delta_\text{f}G^\circ$$

This is only meant as a qualitative way to compare the reactions, as I am assuming standard state conditions for the reactions, which are not necessarily true. Furthermore, for a few species, I didn’t find the correct values in these two sources [1] [2]. $\ce{SCl2}$ and $\ce{S2Cl2}$ probably can’t be found tabulated for the aqueous phase because of their propensity to hydrolysis, so I used the free energies of formation for the gas and solid phases, respectively. I also didn’t find a source for aqueous $\ce{Na2SO4}$, so instead I used the data for the solid decahydrate $\ce{Na2SO4 . 10 H2O (s)}$. The results I got were:

$$ \small \begin{array}{lcc} \hline \text{Reaction} & \Delta_\mathrm{r}G^\circ / \mathrm{kJ\ mol^{-1}}\\ \hline \text{A} & +440 \\ \text{B} & +1940 \\ \text{C} & +590 \\ \text{D} & \mathbf{-2610} \\ \hline \end{array} $$

As you can see, even though the calculation was very rough, the huge disparity in free energy values suggests that the first three reactions are completely overwhelmed by the last one, so that of the four possibilities listed, only the last has any chance of occurring. Strictly speaking this doesn’t mean the last reaction happens, only that it’s much more favourable than the others presented; there could be an unlisted reaction which is even more favourable than all four mentioned in the question. However, empirical data confirms that the last reaction, a simple neutralisation reaction, is the major event when mixing $\ce{HCl}$, $\ce{H2SO4}$ and $\ce{NaOH}$ in practically any conditions.

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    $\begingroup$ Very good answer. This is a very good example of the value of teaching thermodynamics in chemistry. $\endgroup$
    – Leeser
    Commented May 19, 2015 at 16:44
  • $\begingroup$ So does that mean that these redox reactions are endothermic and thus require heat energy to be added in order for it to happen? Or would I need to use a different sulfur acid where the sulfur is either positively charged or negatively charged depending on whether the atoms it is bonded to other than hydrogen break off as cations or as anions(Which I think requires that it is a carbide since carbon doesn't mind if it is positive or negative when it is charged)? $\endgroup$
    – Caters
    Commented May 19, 2015 at 18:42
  • $\begingroup$ If you want to make one of the first three reactions more likely, the problem is that you have to selectively favour them, while not accidentally favouring the last one too. Just increasing the temperature is a rather blunt tool, and could well make a favourable reaction even more favourable. One way to force the first three reactions is to consume a product which is only present there. For example, in this case, you could add a reagent to consume the oxygen gas. This doesn't necessarily mean one of the first three reactions would then take place; there are other reaction possibilities. $\endgroup$
    – Nicolau Saker Neto
    Commented May 19, 2015 at 23:28
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Fortunately, sulfurdichloride and disulfurdichloride (= sulfurmonochloride), which is the decomposition product of the former,
$$\ce{2SCl2 <=>> S2Cl2 + Cl2}$$ cannot be prepared that way.

I suggest to balance the neutralisation reaction without proposing redox reactions that lead to these compounds and I would stay with your fourth equation.

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