Why following two reaction proceed differently?
`Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O`
and
`Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O`

Explanation

`Pb_(3)O_(4)` is actually a stoichiometric mixture of 2 mol of PbO and 1 mol of `PbO_(2)`. In `PbO_(2)`, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. `PbO_(2)` thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl– ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction
`Pb_(3)O_(4) + 8HCl to 3 PbCl_(2) + Cl_(2) + 4 H_(2)O` can be splitted into two reactions namely :
`2PbO + 4 HCl to 2 PbCl_(2) + 2H_(2)O` (acid -base reaction)
`overset(4+)(Pb)O_(2) + 4 Hoverset(-1)(Cl) to overset(+2)(Pb)Cl_(2) + overset(0)(Cl_(2)) underset("redox reaction")(2H_(2)O)`
Since `HNO_(3)` itself is an oxidising agent therefore, it is unlikely that the reaction may occur between `PbO_(2)` and `HNO_(3)`. However, the acid-base reaction occurs between PbO and `HNO_(3)` as :
`2PbO + 4 HNO_(3) to 2 Pb (NO_(3))_(2) + 2H_(2)O`
It is the passive nature of `PbO_(2)` against `HNO_(3)` that makes the reaction different from the one that follows with HCl.