Question

The white pigment \({\bf{Ti}}{{\bf{O}}_{\bf{2}}}\)is prepared by the reaction of titanium tetrachloride, \({\bf{TiC}}{{\bf{l}}_{\bf{4}}}\), with water vapor in the gas phase:

\({\bf{TiC}}{{\bf{l}}_{\bf{4}}}\left( {\bf{g}} \right){\bf{ + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right) \to {\bf{Ti}}{{\bf{O}}_{\bf{2}}}\left( {\bf{s}} \right){\bf{ + 4HCl}}\left( {\bf{g}} \right)\)

How much heat is evolved in the production of exactly 1 mole of \({\bf{Ti}}{{\bf{O}}_{\bf{2}}}\)(s) under standard state conditions?

Short answer

The amount of heat evolved during the production of exactly one mole of TiO₂(s) will be 66.388 kJ.

Step-by-step solution

Enthalpy of formation of substances

To calculate the amount of heat evolved in the production of exactly one mole of TiO₂(s), we have to know the formation enthalpy of each compound involved in the reaction. The reaction is:

\({\rm{TiC}}{{\rm{l}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right) \to {\rm{Ti}}{{\rm{O}}_{\rm{2}}}\left( {\rm{s}} \right){\rm{ + 4HCl}}\left( {\rm{g}} \right)\)

\(\begin{array}{l}{\rm{The enthalpy of formation of TiC}}{{\rm{l}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ is }} - 763.2{\rm{ kJ/mol}}{\rm{.}}\\{\rm{The enthalpy of formation of }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right){\rm{ is }} - 241.82{\rm{ kJ/mol}}{\rm{.}}\\{\rm{The enthalpy of formation of Ti}}{{\rm{O}}_2}\left( {\rm{g}} \right){\rm{ is }} - 944{\rm{ kJ/mol}}{\rm{.}}\\{\rm{The enthalpy of formation of HCl}}\left( {\rm{g}} \right){\rm{ is }} - 92.307{\rm{ kJ/mol}}{\rm{.}}\end{array}\)

Enthalpy of reaction

\(\begin{array}{l}{\rm{Enthalpy of reaction, }}\\{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reaction}}}}{\bf{ = }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{products}}}}} {\bf{ - }}\sum {{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{4 \times \Delta }}{{\rm{{\rm H}}}_{{\rm{HCl}}\left( {\rm{g}} \right)}}{\rm{ + \Delta }}{{\rm{{\rm H}}}_{{\rm{Ti}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{TiC}}{{\rm{l}}_{\rm{4}}}\left( {\rm{g}} \right)}}{\rm{ + 2 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{4 \times - 92}}{\rm{.307 + - 944}}} \right){\rm{ - ( - 763}}{\rm{.2 - 2 \times 241}}{\rm{.82)}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{ - 1313}}{\rm{.228}}} \right){\rm{ + }}\left( {{\rm{1246}}{\rm{.84}}} \right){\rm{ kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 66}}{\rm{.388 kJ}}\end{array}\)

Hence, the amount of heat evolved during the production of exactly one mole of TiO₂(s) will be 66.388 kJ.