Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:

(a) Si(s) + 2F₂(g)⟶SiF₄(g)

(b) 2C(s) + 2H₂(g) + O₂(g)⟶CH₃CO₂H(l)

(c) CH₄(g) + N₂(g)⟶HCN(g) + NH₃(g)

(d) CS₂(g) + 3Cl₂(g)⟶CCl₄(g) + S₂Cl₂(g)

We have to calculate the standard enthalpy change for each reaction. For that, we have to know the enthalpy of formation for each compound in its respective states.

For the first reaction, we have to know the enthalpy of formation of the silicon, the fluorine, and the SiF₄(g).

\(\begin{array}{l}{\bf{The enthalpy of formation of Si(s) is 0 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of }}{{\bf{F}}_{\bf{2}}}{\bf{(g) is 0 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of Si}}{{\bf{F}}_{\bf{4}}}{\bf{(g) is - 1615}}{\bf{.0 kJ/mol}}{\bf{.}}\end{array}\)

The reaction is:

\(\begin{array}{l}{\rm{Si(s) + 2}}{{\rm{F}}_{\rm{2}}}{\rm{(g) }} \to {\rm{Si}}{{\rm{F}}_4}{\rm{(g)}}\\\\{\rm{The change in enthalpy of the reaction is calculated below}}{\rm{. }}\\\Delta {\rm H}{\rm{ = }}\sum {\Delta {{\rm H}_{products}} - \sum {\Delta {{\rm H}_{reac\tan ts}}} } \\\Delta {\rm H} = \Delta {{\rm H}_{Si{F_4}(g)}} - (\Delta {{\rm H}_{{F_{\rm{2}}}{\rm{(g)}}}} + \Delta {{\rm H}_{{\rm{Si(s)}}}})\\\Delta {\rm H} = - 1615.0 - (0 + 0){\rm{ kJ}}\\\Delta {\rm H} = - 1615.0{\rm{ kJ}}\end{array}\)

Hence, the change in enthalpy of Si(s) + 2F₂(g)⟶SiF₄(g) is -1615.0 kJ.

We have to calculate the standard enthalpy change for each reaction. For that, we have to know the enthalpy of formation for each compound in their respective states.

For the second reaction, we have to know the enthalpy of formation of the carbon, the hydrogen, the oxygen, and the acetic acid.

\(\begin{array}{l}{\bf{The enthalpy of formation of C(s) is 0 kJ/mol}}.\\{\bf{The enthalpy of formation of }}{{\bf{O}}_{\bf{2}}}{\bf{(g) is 0 kJ/mol}}.\\{\bf{The enthalpy of formation of }}{{\bf{H}}_{\bf{2}}}{\bf{(g) is 0 kJ/mol}}.\\{\bf{The enthalpy of formation of C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{H(g) is - 484}}{\bf{.2 kJ/mol}}.\end{array}\)

The reaction is:

\(\begin{array}{l}{\rm{2C(s) + }}{\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) + }}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H(g)}}\\\\{\rm{The change in enthalpy of the reaction is calculated below}}{\rm{. }}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H(g)}}}}} \right){\rm{ - }}\left( {{\rm{2 \times \Delta }}{{\rm{{\rm H}}}_{{\rm{C(s)}}}}{\rm{ + 2 \times \Delta }}{{\rm{{\rm H}}}_{{{\rm{H}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + \Delta }}{{\rm{{\rm H}}}_{{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 484}}{\rm{.2 - (0 + 0 + 0)}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 484}}{\rm{.2 kJ}}{\rm{.}}\end{array}\)

Hence, the change in enthalpy of \(2{\rm{C(s) + }}{\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}} + {{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H(g)}}\) is -484.2 kJ.

We have to calculate the standard enthalpy change for each reaction. For that, we have to know the enthalpy of formation for each compound in their respective states.

For the third reaction, we have to know the enthalpy of formation of the methane, the nitrogen, the HCN(g), and the ammonia.

\(\begin{array}{l}{\bf{The enthalpy of formation of C}}{{\bf{H}}_{\bf{4}}}{\bf{(g) is - 74}}{\bf{.6 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of }}{{\bf{N}}_{\bf{2}}}{\bf{(g) is 0 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of HCN(g) is 135}}{\bf{.5 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of N}}{{\bf{H}}_{\bf{3}}}{\bf{(g) is - 45}}{\bf{.9 kJ/mol}}{\bf{.}}\end{array}\)

The reaction is:

\(\begin{array}{l}{\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ + }}{{\rm{N}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{HCN}}\left( {\rm{g}} \right){\rm{ + N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\\\\{\rm{The change in enthalpy of the reaction is calculated below}}{\rm{. }}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{(g)}}}}{\rm{ + \Delta }}{{\rm{{\rm H}}}_{{\rm{HCN(g)}}}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}}}{\rm{ + \Delta }}{{\rm{{\rm H}}}_{{{\rm{N}}_{\rm{2}}}{\rm{(g)}}}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = ( - 45}}{\rm{.9 + 135}}{\rm{.5) - ( - 74}}{\rm{.6 + 0)}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = (89}}{\rm{.6 + 74}}{\rm{.6) kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = 164}}{\rm{.2 kJ}}\end{array}\)

Hence, the change in enthalpy of \({\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ + }}{{\rm{N}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{HCN}}\left( {\rm{g}} \right){\rm{ + N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\) is 164.2 kJ.

We have to calculate the standard enthalpy change for each reaction. For that, we have to know the enthalpy of formation for each compound in their respective states.

For the fourthreaction, we have to know the enthalpy of formation of \({\rm{C}}{{\rm{S}}_2}{\rm{(g),}}\) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g),}}\) \({\rm{CC}}{{\rm{l}}_4}{\rm{(g),}}\) and \({{\rm{S}}_2}{\rm{C}}{{\rm{l}}_2}{\rm{(g)}}{\rm{.}}\)

\(\begin{array}{l}{\bf{The enthalpy of formation of C}}{{\bf{S}}_{\bf{2}}}{\bf{(g) is 116}}{\bf{. kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of C}}{{\bf{l}}_{\bf{2}}}{\bf{(g) is 0 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of CC}}{{\bf{l}}_{\bf{4}}}{\bf{(g) is - 95}}{\bf{.7 kJ/mol}}{\bf{.}}\\{\bf{The enthalpy of formation of }}{{\bf{S}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g) is - 19}}{\bf{.50 kJ/mol}}{\bf{.}}\end{array}\)

The reaction is:

\(\begin{array}{l}{\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g) + 3C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{(g) + }}{{\rm{S}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\\\\{\rm{The change in enthalpy of the reaction is calculated below}}{\rm{.}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{products}}}}} {\rm{ - }}\sum {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reactants}}}}} \\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{{\rm{S}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + \Delta }}{{\rm{{\rm H}}}_{{\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{(g)}}}}} \right){\rm{ - }}\left( {{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g)}}}}{\rm{ + 3 \times \Delta }}{{\rm{{\rm H}}}_{{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}}}} \right)\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = ( - 19}}{\rm{.50 - 95}}{\rm{.7) - (116}}{\rm{.9 + 0)}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = }}\left( {{\rm{ - 115}}{\rm{.2 - 116}}{\rm{.9}}} \right){\rm{ kJ}}\\{\rm{\Delta }}{{\rm{{\rm H}}}_{{\rm{reaction}}}}{\rm{ = - 232}}{\rm{.1 kJ}}\end{array}\)

Hence, the change in enthalpy of \({\rm{C}}{{\rm{S}}_2}{\rm{(g)}} + 3{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{CC}}{{\rm{l}}_4}{\rm{(g)}} + {{\rm{S}}_2}{\rm{C}}{{\rm{l}}_2}{\rm{(g)}}\) is -232.1 kJ.