A certain reaction, for which the enthalpy change is $24.85 \mathrm{kJ},$ is at equilibrium at $82°\mathrm{C}.$ Find the value of $∆\mathrm{S}$ for this reaction.

Find the $\mathrm{\Delta H}°$ for the reaction ${\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{{\mathrm{Br}}_2}_{\left(\mathrm{g}\right)}⟶{\mathrm{HBr}}_{\left(\mathrm{g}\right)}+{\mathrm{HOBr}}_{\left(\mathrm{g}\right)}$ using:
Bond energy of $\mathrm{O}-\mathrm{H}$ bond in Water $=463 \mathrm{kJ}$
Bond energy of $\mathrm{Br}-\mathrm{Br}$ bond in ${\mathrm{Br}}_2=192 \mathrm{kJ}$
Bond energy of $\mathrm{O}-\mathrm{Br}$ bond in $\mathrm{HOBr}=234 \mathrm{kJ}$
Bond energy of $\mathrm{H}-\mathrm{Br}$ bond in $\mathrm{HBr}=364 \mathrm{kJ}$

The formation of an oxide ion from Oxygen atom takes place as shown below. This two-step process is unfavourable, even though ${\mathrm{O}}^{2-}$ is iso-electronic with Neon, due to the fact that

${\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{\left(\mathrm{g}\right)}^{-}; {\mathrm{\Delta H}}_{\mathrm{f}}°=-141 \mathrm{kJ}.{\mathrm{mol}}^{-1}$

${\mathrm{O}}_{\left(\mathrm{g}\right)}^{-}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{\left(\mathrm{g}\right)}^{2-};{\mathrm{\Delta H}}_{\mathrm{f}}^0=+780 \mathrm{kJ}.{\mathrm{mol}}^{-1}$