Question

The decomposition of hydrogen peroxide, \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}\), has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}\)under standard conditions.

\({\bf{2}}{{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}\left( {\bf{l}} \right) \to {\bf{2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)\)

Short answer

The decomposition of exactly one mole of H₂O₂will be -54.04 kJ.

Step-by-step solution

Enthalpy of formation

The given reaction is:

\({\rm{2}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\left( {\rm{l}} \right) \to {\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right){\rm{ + }}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\)

Firstly, we have to know the enthalpy of formation of each compound participating in the chemical reaction.

\(\begin{array}{l}{\rm{Enthalpy of formation of }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\left( {\rm{l}} \right){\rm{ is - 187}}{\rm{.78}}\,{\rm{kJ/mol}}{\rm{.}}\\{\rm{Enthalpy of formation of }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right){\rm{ is - 241}}{\rm{.82}}\,{\rm{kJ/mol}}{\rm{.}}\\{\rm{Enthalpy of formation of }}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{ is 0}}\,{\rm{kJ/mol}}{\rm{.}}\end{array}\)

Change in enthalpy for the reaction

\(\begin{array}{l}{\rm{Hence, the change in enthalpy for the reaction will be, }}\\{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reaction}}}}{\bf{ = }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{products}}}}} {\bf{ - }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)}}{\rm{ + 2 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)}}} \right){\rm{ - (2 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\left( {\rm{l}} \right)}}{\rm{)}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{0 - 2 \times 241}}{\rm{.82}}} \right){\rm{ - ( - 2 \times 187}}{\rm{.78) kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 108}}{\rm{.08 kJ}}\\\\{\rm{Hence, the change in enthalpy during the composition will be - 108}}{\rm{.08 kJ}}{\rm{.}}\end{array}\)

Change in enthalpy for 1 mole

If we observe the balanced chemical reaction, we will see that the calculated amount of heat was released for 2 moles of hydrogen peroxide decomposition. But, we have to calculate for 1 mole of hydrogen peroxide.

\(\begin{array}{l}{\rm{The change in enthalpy during the composition will be - 108}}{\rm{.08 kJ}}{\rm{.}}\\{\rm{For 2 mole of }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{(l) enthalpy change equal to - 108}}{\rm{.08 kJ}}{\rm{.}}\\{\rm{Thus, for 1 mole of }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{(l) enthalpy change equal to - 54}}{\rm{.04 kJ}}{\rm{.}}\\\\{\rm{Hence, the change in enthalpy for one mole of }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{(l) will be - 54}}{\rm{.04 kJ}}{\rm{.}}\end{array}\)