For the reaction 

${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)⟶2{\mathrm{NO}}_2\left(\mathrm{g}\right)$

$\because \mathrm{\Delta H}=\mathrm{\Delta E}+{\mathrm{\Delta n}}_{\mathrm{g}}\mathrm{RT}$

and ${\mathrm{\Delta n}}_{\mathrm{g}}$ (for the given reaction) $=2-1=+1$

$\therefore \mathrm{\Delta H}>\mathrm{\Delta E}$

In exothermic reactions, heat from the system is lost to the surrounding. For such reactions, ∆H is negative.

In endothermic reactions, heat is absorbed by the system from the surroundings. For such reactions, ∆H is positive.