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My textbook outlines reductive behaviour of $\ce{H2S}$ with the reaction

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> 3S + Cr2(SO4)3 + K2SO4 + 7H2O}.\label{rxn:QR1}\tag{R1}$$

This reaction oxidizes $\ce{S^2-}$ to $\ce{S}$. But why sulfur is not oxidized further to $\ce{SO2}$ according to

$$\ce{H2S + K2Cr2O7 + 4H2SO4 -> SO2 + Cr2(SO4)3 + K2SO4 + 5H2O}?\label{rxn:QR2}\tag{R2}$$

The reaction seems stoichiometrically plausible, yet likely requires highly concentrated oxidants. I know \eqref{rxn:QR1} happens at low concentration of $\ce{H2SO4}$. Maybe using highly concentrated $\ce{H2SO4}$ would do the work for \eqref{rxn:QR2}. What are the prerequisites for \eqref{rxn:QR2} to occur?

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    $\begingroup$ All reactions are possible. H2S can be oxidized to sulfur, sulfite, or sulfate. $\endgroup$ Apr 28, 2017 at 8:32
  • $\begingroup$ Well, permeakra's answer suggests me it's quite hard to produce SO<sub>2</sub> in this case! $\endgroup$ Apr 28, 2017 at 8:58
  • $\begingroup$ I never said it's easy. In fact, I think, permeakra is right. SO2 will be produced as a minor product, if at all. Still, it will be produced at least in some cases, hence the reaction is possible, like I said. $\endgroup$ Apr 28, 2017 at 9:03

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But I thought why not $\ce{SO2}$

While the reaction is stechiometrically correct, it is hard to perform.

  • In excess of $\ce{H2S}$ sulfur will be produced.
  • In excess of oxidant sulfate will be produced.

And exact balance is impossible to achieve.

That said, $\ce{SO2}$ reacts with $\ce{H2S}$ finally producing sulfur. This reaction, just like many inorganic reactions, is an unholy mess of many different processes with many products obtained simulationaly and the only way to produce understandable and reproduceable results is to simplify the situation. One of the way to simplify the situation is to use a vast excess of one of the reagents.

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  • $\begingroup$ Have you seen the second answer? It says H2S will also be produced in this reaction. will it? $\endgroup$ Apr 28, 2017 at 22:46
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Actually the first reaction is possible at normal conditions.

$$ \begin{align} \ce{3H2S + K2Cr2O7 + 4H2SO4 → 3S + Cr2(SO4)3 + K2SO4 + 7H2O} & \tag{R1}\\ \end{align} $$

Now, if you observe the two reactions, you see that there is 7 moles of water molecule in 1st reaction and 5 moles of water molecule in 2nd reaction. So, the 2 moles of water must somehow reacts with sulfur to produce sulfur dioxide.

$$ \begin{align} \ce{3S + 2H2O ->[400 °C] 2H2S + SO2} & \tag{R2}\\ \end{align} $$

You observe that 2 moles of hydrogen sulfide gas is produced which get cancelled from each side and give the 2nd reaction.

$$ \begin{align} \ce{3H2S + K2Cr2O7 + 4H2SO4 &-> (3S + 2H2O) + Cr2(SO4)3 + K2SO4 + 5H2O} \tag{R1}\\ \ce{3H2S + K2Cr2O7 + 4H2SO4 &-> (SO2 + 2H2S) + Cr2(SO4)3 + K2SO4 + 5H2O} \tag{From R2}\\ \ce{H2S + K2Cr2O7 + 4H2SO4 &-> (SO2) + Cr2(SO4)3 + K2SO4 + 5H2O} & \tag{R3}\\ \end{align} $$

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  • $\begingroup$ I changed the alignment of the bottom equations to make them more legible. Hope you agree with the change. $\endgroup$
    – Buck Thorn
    Feb 22, 2023 at 16:25
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    $\begingroup$ @BuckThorn I'd probably go even further and retag '(From R2)' to something shorter of a label, and also replace full reactions with the net ionic equations. That way it's arguably easier to track the difference and there will be no horizontal scroll bar. But this is probably for the author to decide. $\endgroup$
    – andselisk
    Feb 22, 2023 at 16:37
  • $\begingroup$ Three critics : 1)The equations are not readable. 2) Potassium is missing. 3) H2S appears on both sides in equation (2). $\endgroup$
    – Maurice
    Feb 22, 2023 at 16:52

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