Question

Calculate the enthalpy of combustion of butane, C₄H₁₀(g) for the formation of H₂O(g) and CO₂(g). The enthalpy of formation of butane is -126 kJ/mol.

Short answer

The change in enthalpy for the combustion reaction of butane will be -2657.14 kJ.

Step-by-step solution

Formation enthalpy of substances

Here, we need to calculate the combustion enthalpy of propane. For doing so, we have to know the formation enthalpy of H₂O(g) and CO₂(g) and the balanced chemical reaction of combustion of propane.

\(\begin{array}{l}{\rm{The formation enthalpy of }}{{\rm{H}}_{\rm{2}}}{\rm{O(g) is - 241}}{\rm{.82 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{O}}_{\rm{2}}}{\rm{(g) is 0 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) is - 393}}{\rm{.51 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{(g) is - 126 kJ/mol}}{\rm{.}}\\\\{\rm{The balanced chemical equation for combustion of propane,}}\\{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{(g) + }}\frac{{{\rm{13}}}}{{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ 4C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + 5}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\end{array}\)

Change in enthalpy

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

Hence, the change in enthalpy for the combustion reaction of butane will be -2657.14 kJ.