Question

Why do the following reactions proceed differently? $P{b}_3{O}_4+8HCl⟶3PbC{l}_2+C{l}_2+4H_{2}O$ and $P{b}_3{O}_4+4HN{O}_3⟶2Pb(N{O}_3{)}_2+Pb{O}_2+2H_{2}O$

Answer 1

$P{b}_3{O}_4$ is actually a stoichiometric mixture of $2mol$ of $PbO$ and $1mol$ of $Pb{O}_2.$ In $Pb{O}_2,$ lead is present in $+4$ oxidation state whereas the stable oxidation state of lead in $PbO$ is $+2.$ $Pb{O}_2$ thus can act as an oxidant (oxidizing agent and, therefore, can oxidise $C{l}^{-}$ ion of $HCl$ into chlorine. We may also keep in mind that $PbO$ is a basic oxide.

Therefore, the reaction

$P{b}_3{O}_4+8HCl⟶3PbC{l}_2+C{l}_2+4H_{2}O$
Can be splitted into two reactions namely:

$2PbO+4HCl⟶2PbC{l}_2+2H_{2}O$ (Acid-base reaction)

$+4$ $-1$ $+2$ $0$
$Pb{O}_2+4HCl⟶PbC{l}_2+C{l}_2+2H_{2}O$ (Redox reaction)

Since $HN{O}_3$ itself is an oxidizing agent therefore, it is unlikely that the reaction may occur between $Pb{O}_2$ and $HN{O}_3$.
However, the acid-base reaction occurs between $PbO$ and $HN{O}_3$ as:

$2PbO+4HN{O}_3⟶2Pb(N{O}_3{)}_2+2H_{2}O$

It is the passive nature of $Pb{O}_2$ against $HN{O}_3$ that makes the reaction different from the one that follows with $HCl$.