How does nitrogen dioxide react with water in a disproportionation reaction?

Nitrogen dioxide ($N{O}_2$) reacts with water through a disproportionation reaction, a process in which the nitrogen atom is simultaneously oxidized and reduced. The specific products of this reaction depend on the temperature of the water:123

• In cold water: Nitrogen dioxide disproportionates to form a mixture of nitrous acid ($HN{O}_2$) and nitric acid ($HN{O}_3$).
• At higher temperatures: The reaction produces nitric acid ($HN{O}_3$) and nitric oxide ($NO$). In this specific reaction, the nitrogen atom in the $+4$ oxidation state is reduced to $+2$ (in $NO$) and oxidized to $+5$ (in $HN{O}_3$), as shown in the following equation:123

$$3N{O}_2+H_{2}O\to 2HN{O}_3+NO$$