Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(l) and O₂(g) under standard conditions?

We have to calculate the amount of heat required to decompose exactly 1 mole of red \(HgO(s)\) to \(Hg(l)\)and \({O_2}(g)\). To evaluate, we have to know the enthalpy of formation of \({\rm{HgO(s)}}\).

\(\begin{array}{l}{\rm{Decomposition reaction of HgO(s) is: }}\\HgO(s){\rm{ }} \to {\rm{ }}Hg(l){\rm{ }} + {\rm{ }}\frac{1}{2}{O_2}(g)\end{array}\)

\(\begin{array}{l}{\rm{Useful information, (\Delta }}{{\rm{{\rm H}}}_{{\rm{formation}}}}{{\rm{)}}_{{\rm{HgO}}}}{\rm{ = - 90}}{\rm{.83 kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\\{\rm{We are going to use the formula, }}{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{reaction}}}{\bf{ = }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{product}}}{\bf{ - }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{reactant}}}} } \\{\rm{ \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{reaction}}}{\rm{ = (\Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{Hg(l)}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{\Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}{\rm{) - ( - 90}}{\rm{.83)kJ}}\\{\rm{ \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{reaction}}}{\rm{ = 90}}{\rm{.83 kJ}}\end{array}\)

Hence, the amount of heat required is equal to 90.83 kJ.