Calculate the enthalpy of combustion of propane, C3H8(g), for the formation of H2O(g) and CO2(g). The enthalpy of formation of propane is -104 kJ/mol.

Short Answer

Expert verified

The amount of heat released in the combustion reaction of propane will be 2043.81 kJ.

Step by step solution

01

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.

\(\begin{array}{l}{\rm{The formation enthalpy of }}{{\rm{H}}_{\rm{2}}}{\rm{O(g) is - 241}}{\rm{.82 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{O}}_{\rm{2}}}{\rm{(g) is 0 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) is - 393}}{\rm{.51 kJ/mol}}{\rm{.}}\\{\rm{The formation enthalpy of }}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{(g) is - 104 kJ/mol}}{\rm{.}}\\\\{\rm{The balanced chemical equation for combustion of propane,}}\\{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{(g) + 5}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{3C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + 4}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\end{array}\)

02

Change in enthalpy

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

Hence, the change in enthalpy for the combustion reaction of propane will be -2043.81 kJ.

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from the following information:

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