Reaction between sulfur and sodium hydroxide

Sulfur reacts with sodium hydroxide in the following way:

$\ce{3S + 6NaOH->2Na_2S + Na_2SO_3 + 3H_2O}$

It is a disproportionation reaction: sulfur's oxidation changes from zero to -2 in sodium sulfide and from zero to +4 in sodium sulfite.

But how to remember the products of this reaction? By rote memorization?

The preparation of sodium sulfide is apparent. Is there a way to understand why sodium sulfite forms? We have the hydroxide ion, and it somehow reacts with sulfur to yield the $\ce{SO3^{2-}}$ ion - but how?

• Why, that's simple. Sulfide (a reduced form of $\ce{S}$) is apparent, you say. OK, then some $\ce{S}$ is going to be oxidized as well. What is the next stable oxidation state? It is +4. So sulfite it is. Then both react further with neutral $\ce{S}$, and things start to get hairy, with all them polysulfides and thiosulfates. – Ivan Neretin May 10 '16 at 18:20
• – Nilay Ghosh Nov 15 '17 at 12:56

The reaction products depends on the conditions(temperature, pressure etc.) and stoichiometric ratio of reactants taken. Depending on that, you will get various products in each case if you slightly change the conditions or the ratio of sodium hydroxide to sulfur taken.

1. $$\ce{4S + 8NaOH ->[600 C] Na2SO4 + 3Na2S + 4H2O}$$

Sulfur react with sodium hydroxide to produce sodium sulfate, sodium sulfide and water. This reaction takes place at a temperature of over 600°C.(source)

1. $$\ce{4S + 6NaOH → Na2S2O3 + 2Na2S + 3H2O}$$

Sulfur react with sodium hydroxide to produce sodium thiosulfate, sodium sulfide and water. Sodium hydroxide - concentrated solution. The reaction takes place in a boiling solution.(source)

When you compare the three equation, you can see that though stoichiometric ratio of the reactants in the 1st and 3rd equation is same, the moles of sulfur and sodium hydroxide is different and thus gives different products in each case.

Also, from reaction 2 we can say that sodium thiosulfate may have disproportionate into sodium sulfite and sulfur.

$$\ce{Na2S2O3 ->[220-300 C] Na2SO3 + S}$$

and thus one mole sulfur cancel from each side and gave reaction 3 (your reaction). This site explains the disproportionation reaction.

Chemical reactions of sulfur compounds have been studied in equimolar $\ce{NaOH-H2O}$ melt at 100°C by voltammetry and UV spectrophotometry. Disproportionation of sulfur is fast and quantitative according to:

$$\ce{S8 + 6OH- -> 2S3^2- + S2O3^2- + 3H2O}$$

On the contrary, addition of sulfur to sulfide is not quantitative and gives $\ce{S2^2-}$ besides prevailing sulfur disproportionation.$\ce{SO3^2-}$ reacts with $\ce{S2^2-}$ and $\ce{S3^2-}$ , while $\ce{S2O3^2-}$ decomposes $\ce{S^2−}$. A pseudoequilibrium is observed:

$$\ce{S^2- + S2O3^2- <=> S2^2- + SO3^2-}$$

with concentration quotient Formula Q = $\ce{10^{-0.8}}$.