Reaction between lead dioxide and nitric acid

A [wikipedia article](http://en.wikipedia.org/wiki/Lead_dioxide) suggests that lead dioxide is produced by this reaction:
$$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$


The same article also suggests that lead dioxide reacts with nitric acid in the following way:
$$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$


I am unable to understand why then, in the first reaction, $\ce{PbO2}$ doesn't further react with the residual nitric acid. Is it because the quantity of the acid is regulated? Or is there some error in the reactions?




Exact statements are:



> 
> Because of the instability of its $\ce {Pb^4+}$ cation, lead dioxide reacts with **warm** acids, converting to the more stable $\ce {Pb^2+}$ state and liberating oxygen
>  $$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$
> 
> 
> 


and 



> 
> Lead dioxide is produced commercially by several methods, which include ... or reacting $\ce {Pb3O4}$ with **dilute** nitric acid:
>  $$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$
> 
> 
> 


Since nitric acid in warmer in first reaction, the following reaction occurs easily, which is actually a decomposition reaction:



> 
> When heated, lead(II) nitrate crystals decompose to lead(II) oxide, oxygen and nitrogen dioxide.
>  $$\ce{Pb(NO3)2->PbO +NO2 +O2}$$
> 
> 
> 


Hence the oxygen might be liberated with decomposition-cum-oxidation reaction.


 3

You need to consider the reaction conditions. 


The Wikipedia [Lead Dioxide article](http://en.wikipedia.org/wiki/Lead_dioxide) is saying that in **dilute** nitric acid the first reaction occurs, while in **warm** nitric acid the second reaction occurs. 


So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.


 3