Question

In the early days of automobiles, illumination at night was provided by burning acetylene, C₂H₂. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC₂:

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

Calculate the standard enthalpy of the reaction. The \({\bf{\Delta H}}_{\bf{f}}^{\bf{o}}\)of CaC₂is -15.14 kcal/mol.

Short answer

The standard enthalpy of the reaction is -122.795 kJ.

Step-by-step solution

Data required

To evaluate the given problem, we have to follow the following steps. Initially, we have to know the enthalpy of formation of each compound involved in the reaction and then calculate the enthalpy change of the reaction.

Enthalpy of formation

The reaction is given below:

\(\begin{array}{l}{\rm{Ca}}{{\rm{C}}_{\rm{2}}}\left( {\rm{s}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right) \to {\rm{Ca}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\left( {\rm{s}} \right){\rm{ + }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right)\\{\rm{The enthalpy of formation of Ca}}{{\rm{C}}_{\rm{2}}}\left( {\rm{s}} \right){\rm{ is - 15}}{\rm{.14 Kcal/mol}}{\rm{.}}\\{\rm{[1 Kcal/mol = 4}}{\rm{.184 kJ/mol] }}\\{\rm{ = - 15}}{\rm{.14 \times 4}}{\rm{.184 kJ/mol}}\\{\rm{ = - 63}}{\rm{.345 kJ/mol}}\\{\rm{The enthalpy of formation of }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right){\rm{ is - 285}}{\rm{.83 kJ/mol}}{\rm{.}}\\{\rm{The enthalpy of formation of Ca}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\left( {\rm{s}} \right){\rm{ is - 985}}{\rm{.2 kJ/mol}}{\rm{.}}\\{\rm{The enthalpy of formation of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{ is 227}}{\rm{.4 kJ/mol}}{\rm{.}}\end{array}\)

Standard enthalpy of reaction

The calculation is shown below:

\(\begin{array}{l}{\bf{\Delta }}{{\bf{H}}_{{\bf{reaction}}}}{\bf{ = }}\sum {{\bf{\Delta }}{{\bf{H}}_{{\bf{products}}}}} {\bf{ - }}\sum {{\bf{\Delta }}{{\bf{H}}_{{\bf{reactants}}}}} \\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{\Delta }}{{\rm{H}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right)}}{\rm{ + \Delta }}{{\rm{H}}_{{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}\left( {\rm{s}} \right)}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{H}}_{{\rm{Ca}}{{\rm{C}}_{\rm{2}}}\left( {\rm{s}} \right)}}{\rm{ + 2 \times \Delta }}{{\rm{H}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)}}} \right)\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{227}}{\rm{.4 - 985}}{\rm{.2}}} \right){\rm{ - ( - 63}}{\rm{.345 - 2 \times 285}}{\rm{.83) kJ}}\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = - 757}}{\rm{.8 + 635}}{\rm{.005 kJ}}\\{\rm{\Delta }}{{\rm{H}}_{{\rm{reaction}}}}{\rm{ = - 122}}{\rm{.795 kJ}}\end{array}\)

Hence, the standard enthalpy of the reaction is -122.795 kJ.