The reaction of HNO3 with H2S? [closed]








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I am totally confused about the gas that would be released in this reaction. It is given NO in one book and NO2 in other. Even the internet has mixed opinions. Help me out guys.




Nitric acid reacts with plenty of substances. This reaction may produce both $\ce{NO}$ and $\ce{NO2}$. It is a question of concentration, because $\ce{NO}$ may react with $\ce{HNO3}$ according to :$$\ce{NO + 2 HNO3 -> 3 NO2 + H2O}$$ and $\ce{NO2}$ may react with water according to the same reaction, written backwards : $$\ce{3 NO2 + H2O -> NO + 2 HNO3}$$ So both gas are always produced simultaneously. And if there is a majority of $\ce{NO}$, this colorless gaz does react with oxygen from air, producing brown fumes of $\ce{NO2}$ according to $$\ce{2NO + O2 -> 2 NO2}$$


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According to Ref.1, the reaction of $\ce{HNO3}$ with $\ce{H2S}$ has been processed in multi steps, which agrees with Maurice's suggestion in the first answer. The authors found that:



> 
> Solutions containing 5-30% nitric acid show no signs of reaction for
> some time but; if nitrogen peroxide $(\ce{NO2})$ is added, a violent reaction with rise of temperature follows.   
> 
> When a 40% solution is used the reaction takes place after an interval of apparent passivity called in this paper the “induction period."
> 
> 
> 


The "induction period" became longer when, before passing hydrogen sulphide, carbon dioxide was bubbled through the solution to remove any $\ce{NO2}$ resulting from the decomposition of nitric acid (keep in mind that before passing $\ce{H2S}$,
the air in the vessels was always displaced by carbon dioxide throughout this experiment).



> 
> Milkiness due to sulphur appeared on the surface and spread rapidly through the liquid. When, after stopping the flow of hydrogen sulphide, the vessel was shaken, the atmosphere above the solution turned brown owing to the presence of nitrogen peroxide $(\ce{NO2})$:
> $$\ce{HNO3 + H2S -> HNO2 + S} \tag1$$
> $$\ce{HNO3 + HNO2 <=> H2O + 2NO2} \tag2$$
> 
> 
> 


The $\ce{HNO2}$ may also break down according to the equation:
$$\ce{3HNO2 <=> 2NO + H2O + HNO3} \tag3$$


On passing a further supply of $\ce{H2S}$, the brown color disappeared as $\ce{NO2}$ reacts with $\ce{H2S}$ giving sulphur, water, and nitric oxide $(\ce{NO})$:
$$\ce{NO2 + H2S -> NO + S + H2O} \tag4$$


Thus, either $\ce{NO2}$ or $\ce{NO}$ or both could be present in the reaction, basd on the strength of the nitric acid and how long $\ce{H2S}$ has been passed through. Also, keep in mind that solutions containing less than 5% nitric acid (dilute conditions) are not reactive with $\ce{H2S}$ even if traces of $\ce{NO2}$ are added to the reaction (Ref.1)·


**References:**


1. H. B. Dunnicliff and Sardar Mohammad, "The Action of Hydrogen Sulphide on Solutions of Nitric Acid," *J. Phys. Chem.* **1929**, *33(9)*, 1343–1362 (DOI: <https://doi.org/10.1021/j150303a005>).


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