91Ó°ÊÓ

We have to determine who is going to produce more heat per gram when burned among propane and butane. We will follow the below steps to find that out.

First, we will calculate the amount of heat produced during the combustion of propane.

\(\begin{array}{l}{\rm{The formation enthalpy of }}{{\rm{H}}_{\rm{2}}}{\rm{O(g) is - 241}}{\rm{.82 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{O}}_{\rm{2}}}{\rm{(g) is 0 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) is - 393}}{\rm{.51 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{(g) is - 104 kJ/mol}}{\rm{.}}\end{array}\)

\(\begin{array}{l}{\bf{The balanced chemical equation for combustion of propane,}}\\{{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{8}}}{\bf{(g) + 5}}{{\bf{O}}_{\bf{2}}}{\bf{(g) }} \to {\bf{ 3C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 4}}{{\bf{H}}_{\bf{2}}}{\bf{O(g)}}\\\\{\rm{Change in enthalpy of the reaction, }}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{3 \times \Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + 4 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{(g)}}}}{\rm{ + 5 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g) }}}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{3 \times - 393}}{\rm{.51 + 4 \times - 241}}{\rm{.82}}} \right){\rm{ - }}\left( {{\rm{ - 104 + 0}}} \right){\rm{ kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 2147}}{\rm{.81 + 104 kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 2043}}{\rm{.81 kJ}}\end{array}\)

Now, we will calculate the amount of heat produced during the combustion of butane.

\(\begin{array}{l}{\rm{The formation enthalpy of }}{{\rm{H}}_{\rm{2}}}{\rm{O(g) is - 241}}{\rm{.82 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{O}}_{\rm{2}}}{\rm{(g) is 0 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) is - 393}}{\rm{.51 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{(g) is - 126 kJ/mol}}{\rm{.}}\end{array}\)

\(\begin{array}{l}{\bf{The balanced chemical equation for combustion of propane,}}\\{{\bf{C}}_{\bf{4}}}{{\bf{H}}_{{\bf{10}}}}{\bf{(g) + }}\frac{{{\bf{13}}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g) }} \to {\bf{ 4C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 5}}{{\bf{H}}_{\bf{2}}}{\bf{O(g)}}\\\\{\rm{Change in enthalpy of the reaction, }}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{4 \times \Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + 5 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{(g)}}}}{\rm{ + }}\frac{{{\rm{13}}}}{{\rm{2}}}{\rm{ \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g) }}}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{4 \times - 393}}{\rm{.51 + 5 \times - 241}}{\rm{.82}}} \right){\rm{ - }}\left( {{\rm{ - 126 + 0}}} \right){\rm{ kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 2783}}{\rm{.14 + 126 kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 2657}}{\rm{.14 kJ}}\end{array}\)

After observing the combustion reaction, we can say that the amount of heat released during the combustion of one mole of propane is 2043.81 kJ, and the amount of heat released during the combustion of one mole of butane is 2657.14 kJ.

Hence, the amount of heat released during the combustion of propane per gram will be:

\(1{\rm{ g }}{{\rm{C}}_3}{{\rm{H}}_8} \times \frac{{1{\rm{ mole of }}{{\rm{C}}_3}{{\rm{H}}_8}{\rm{ }}}}{{44.084{\rm{ g }}{{\rm{C}}_3}{{\rm{H}}_8}}} \times \frac{{ - 2043.81{\rm{ kJ}}}}{{1{\rm{ mole }}{{\rm{C}}_3}{{\rm{H}}_8}}} = - 46.35{\rm{ kJ/gram}}\)

Hence, the amount of heat released during the combustion of butane per gram will be:

\(1{\rm{ g }}{{\rm{C}}_4}{{\rm{H}}_{10}} \times \frac{{1{\rm{ mole of }}{{\rm{C}}_4}{{\rm{H}}_{10}}{\rm{ }}}}{{{\rm{58}}{\rm{.12 g }}{{\rm{C}}_4}{{\rm{H}}_{10}}}} \times \frac{{ - 2657.14{\rm{ kJ}}}}{{1{\rm{ mole }}{{\rm{C}}_4}{{\rm{H}}_{10}}}} = - 45.61{\rm{ kJ/gram}}\)

Hence, the amount of heat released is more in case of propane compared to butane during the combustion reaction.