$$\ce{HNO3 + P2O5 -> 2HPO3 + N2O5}$$

In this reaction, both nitrogen and phosphorus have the same oxidation number before and after the product is formed. Since the $+5$ oxidation of nitrogen is unstable, it has to undergo reduction, but here it remains the same after the product is formed. So what makes the reaction go forward?

Is it because of the dehydrating property of $\ce{P2O5}$?

Why does this compound have this property and why doesn't $\ce{N2O5}$ have this dehydrating property?

  • $\begingroup$ Somewhat related: chemistry.stackexchange.com/questions/33742/… $\endgroup$ Jul 26, 2020 at 6:43
  • $\begingroup$ and: chemistry.stackexchange.com/questions/851/… $\endgroup$ Jul 26, 2020 at 6:44
  • $\begingroup$ But they all state this just as a fact, What about N2O5? It is unstable than P2O5 because of the oxidation no it has, but in the reaction I mentioned, the product has the same oxidation number as before. What gives P2O5 this property? $\endgroup$
    – R. Anusha
    Jul 26, 2020 at 6:50
  • $\begingroup$ N2O5 does react with water but not as exothermically compared to P2O5 who is just looking for water. So, it does not take role as dehydrating agent. $\endgroup$ Jul 26, 2020 at 6:56
  • 1
    $\begingroup$ The true molecule is P4O10,not P2O5. $\endgroup$
    – Poutnik
    Jul 26, 2020 at 8:05

1 Answer 1


First of all, your reaction is wrong. $\ce{HPO3}$ does not exist as a single entity. It is actually a cylic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid. See this answer for more details.

So, what happens is that phosphorus pentoxide is very powerful dehydrating agent and can rip off water molecules from compounds. Phosphorus pentoxide can dehydrate nitric acid to form its anhydride, dinitrogen pentoxide.

$$\ce{2HNO3 ->[P2O5][- H2O] N2O5}$$

Why dinitrogen pentoxide is not used as a dehydrating agent?

My take on this is that hydrolysis of dinitrogen pentoxide is not that exothermic as compared to hydroysis of phosphorus pentoxide. Hydrolysis of phosphorus pentoxide is very exothermic, releasing so much energy to dehydrate compounds to form corresponding anhydrides and phosphoric acid.

$$\ce{P4O10 + 6 H2O → 4 H3PO4 (–177 kJ)}$$

Phosphoric acid is hygroscopic which further sucks remaining water from compound. If we compare all the dehydrating agents, phosphorus pentoxide will be on a different league. Dinitrogen pentoxide is nothing.

Nitric acid also hydrolyses to form nitric acid but the reaction is not that exothermic. Also, nitric acid is not that hygroscopic as compared to phosphoric acid for which it is not that great of a dehydrating agent. But hydrolysis of dinitrogen pentoxide has its importance. A solution of dinitrogen pentoxide and nitric acid is a good nitrating agent and hence $\ce{N2O5}$ is known as a nitrating agent than a dehydrating agent. Also, this hydrolysis reaction is a basis of forming atmospheric aerosols ($\ce{HNO3}$ is a good $\ce{NO_x}$ reservoir).

  • 1
    $\begingroup$ "phosphorus pentoxide is a water-loving compound" is descriptive, but doesn't answer the core question of why the reaction with water is so favorable, especially as compared to that of N2O5. $\endgroup$
    – Andrew
    Jul 26, 2020 at 11:49
  • $\begingroup$ @Andrew removed that statement and added an explanation of how P2O5 removes water and how it can be a good dehydrating agent based on that and also why N2O5 cannot achieve that. $\endgroup$ Jul 26, 2020 at 12:53
  • 1
    $\begingroup$ An important experimental consideration is that $\ce{N2O5}$ is a volatile solid that melts at 41 degrees C and boils at 47 degrees C. If this reaction is run at a moderately elevated temperature (but below the boiling points or decomposition temperature of the other components), $\ce{N2O5}$ is readily removed, and Le Chatelier's principle drives any equilibrium that might be established. $\endgroup$
    – Ben Norris
    Jul 26, 2020 at 17:03

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