91Ó°ÊÓ

Here, to calculate the amount of energy released during the combustion reaction, we should know the exact combustion reaction of manganese.

\(\begin{array}{l}{\rm{Combustion reaction of manganese (Mn):}}\\{\rm{3Mn(s) + 2}}{{\rm{O}}_{\rm{2}}}{\rm{(g) }} \to {\rm{M}}{{\rm{n}}_{\rm{3}}}{{\rm{O}}_{\rm{4}}}{\rm{(s)}}\\{\rm{Hence, the enthalpy of reaction, }}{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{reaction}}}{\bf{ = }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{product}}}{\bf{ - }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}^{\bf{^\circ }}}_{{\bf{reactant}}}} } \\{\rm{ \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{reaction}}}{\rm{ = \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{M}}{{\rm{n}}_{\rm{3}}}{{\rm{O}}_{\rm{4}}}{\rm{(s)}}}{\rm{ - (3 \times \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{Mn(s)}}}{\rm{ + 2 \times \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}{\rm{)}}\\{\rm{ \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{reaction}}}{\rm{ = ( - 1378}}{\rm{.83 kJ) - (0 + 0)}}\\{\rm{ \Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{reaction}}}{\rm{ = - 1378}}{\rm{.83 kJ}}\\{\rm{[Enthalpy of formation of M}}{{\rm{n}}_{\rm{3}}}{{\rm{O}}_{\rm{4}}}{\rm{(s),(\Delta }}{{\rm{{\rm H}}}^{\rm{^\circ }}}_{{\rm{formation}}}{{\rm{)}}_{{\rm{M}}{{\rm{n}}_{\rm{3}}}{{\rm{O}}_{\rm{4}}}{\rm{(s)}}}}{\rm{ = - 1378}}{\rm{.83 kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{]}}\end{array}\)

Previously, we have calculated the amount of heat released when the combustion of 3 mole manganese takes place.

\(\begin{array}{l}{\rm{When 3 mole of Mn, takes part in combustion reaction, }}1378.83{\rm{ KJ of energy is released}}\\{\rm{When 1 mole of Mn will take part in combustion reaction, }}\\{\rm{The amount of energy released = }}\frac{{1378.83{\rm{ kJ}}}}{3}\\{\rm{ }} = 459.61{\rm{ kJ}}\end{array}\)

Hence, the amount of energy released after the combustion of one mole of manganese is 459.61 kJ.