Q5.3-60E - Chemistry (OpenStax)

Question

Both graphite and diamond burn.

\({\bf{C}}\left( {{\bf{s, diamond}}} \right){\bf{  +  }}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{C}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)\)

For the conversion of graphite to diamond:

\({\bf{C}}\left( {{\bf{s, graphite}}} \right) \to {\bf{C}}\left( {{\bf{s, diamond}}} \right){\bf{                 \Delta {\rm H}}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = 1}}{\bf{.90 kJ}}\)

Which produces more heat, the combustion of graphite or the combustion of a diamond?

Step-by-Step Solution

Answer

On the combustion of diamond, more heat will be produced than graphite. 

Step 1: Combustion of graphite

To solve the problem, we will follow some steps. First, we will calculate the enthalpy of formation of carbon dioxide from C(s, graphite). 

\(\begin{array}{l}{\rm{C}}\left( {{\rm{s, graphite}}} \right){\rm{  +  }}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\\{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reaction}}}}{\bf{ = (\Delta {\rm H}}}_{{\bf{formation}}}^{\bf{^\circ }}{{\bf{)}}_{{\bf{C}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)}}{\bf{ - }}\left( {{\bf{\Delta {\rm H}}}_{{\bf{C}}\left( {{\bf{s, graphite}}} \right)}^{\bf{^\circ }}{\bf{ + \Delta {\rm H}}}_{{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)}^{\bf{^\circ }}} \right)\\\Delta {{\rm H}_{reaction}} =  - 393.51{\rm{ kJ}}\end{array}\)

Step 2: Combustion of diamond

Now, we have to calculate the enthalpy of formation of C(s, diamond)

C(s,graphite) → C(s, diamond)               ΔH⁰₂₉₈ =1.90kJ

ΔH⁰₂₉₈ =1.90kJ = ΔH⁰_C(s,diamond) -ΔH⁰_{C(s,graphite)}

→ ΔH⁰_C(s,diamond) =0+1.90kJ

→ ΔH⁰_C(s,diamond) =1.90kJ

Step 3: Enthalpy change of the reaction

Now, we will calculate the enthalpy of combustion of C(s, diamond)

\(\begin{array}{l}{\rm{C}}\left( {{\rm{s, diamond}}} \right){\rm{  +  }}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\\{\bf{\Delta }}{{\bf{{\rm H}}}_{{\bf{reaction}}}}{\bf{ = (\Delta {\rm H}}}_{{\bf{formation}}}^{\bf{^\circ }}{{\bf{)}}_{{\bf{C}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)}}{\bf{ - }}\left( {{\bf{\Delta {\rm H}}}_{{\bf{C}}\left( {{\bf{s, diamond}}} \right)}^{\bf{^\circ }}{\bf{ + \Delta {\rm H}}}_{{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)}^{\bf{^\circ }}} \right)\\\Delta {{\rm H}_{reaction}} = \left( { - 393.51{\rm{ }} - 1.90} \right){\rm{ kJ}}\\{\rm{               =  }} - {\rm{395}}{\rm{.41 kJ}}\end{array}\)

 

Hence, we can easily comment that on combustion of diamond, more heat will be produced than graphite.