Assume each reaction is carried out in an open container. For which reaction will $\mathrm{\Delta H}=\mathrm{\Delta E}$ ?

We know that

$\mathrm{\Delta H}=\mathrm{\Delta E}+\mathrm{P\Delta V}$

In the reactions, ${\mathrm{H}}_2+{\mathrm{Br}}_2\to 2\mathrm{HBr}$ there is no change in volume or $\mathrm{\Delta V}=0$;$∆{\mathrm{n}}_{\mathrm{g}}= 0$

So, $\mathrm{\Delta H}=\mathrm{\Delta E}$ for this reaction