Question

Calculate \[\;{\mathbf{\Delta }}H\] for the conversion of graphite to diamond.(in kJ/mol)
as the nearest integer.

Answer 1

Hint:
The enthalpy of a chemical process is constant, irrespective of the reaction happening in one step or many steps. Hess' Law states that: If a chemical equation is written because the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the opposite chemical equations..

Complete step by step answer:
Now we will use Hess’ Law to find out the \[\;{\mathbf{\Delta }}H\] for conversion of graphite to diamond.
\[C{\text{ }}\left( {s,{\text{ }}graphite} \right){\text{ }} - - - > {\text{ }}C{\text{ }}\left( {s,{\text{ }}diamond} \right)\]
\[\Delta H^\circ {\text{ }} = \]??? kJ
Now notice the presence of the degree sign, °, on the enthalpy. This means that the reaction has occurred under standard conditions.
If we had two (or more) reactions that could be added together, then we may add the respective enthalpies of the reactions to get what we want. Here are the two reactions that we need:
\[\begin{array}{*{20}{l}}
  {C{\text{ }}\left( {s,{\text{ }}graphite} \right){\text{ }} + {\text{ }}{O_2}\left( g \right){\text{ }} - - - > {\text{ }}C{O_2}\left( g \right)}&{\Delta H^\circ {\text{ }} = {\text{ }} - 393.5{\text{ }}kJ}
\end{array}\]


\[\begin{array}{*{20}{l}}
  {C{\text{ }}\left( {s,{\text{ }}diamond} \right){\text{ }} + {\text{ }}{O_2}\left( g \right){\text{ }} - - - > {\text{ }}C{O_2}\left( g \right)}&{\Delta H^\circ {\text{ }} = {\text{ }} - 395.4{\text{ }}kJ}
\end{array}\]


Now we will reverse the second equation. This will put the \[C{\text{ }}\left( {s,{\text{ }}diamond} \right)\] on the product side, where we need it to be. Also, on adding the two equations together, the oxygen and carbon dioxide will cancel each other out. Here are the two equations again, with the second one reversed is :
\[\begin{array}{*{20}{l}}
  {C{\text{ }}\left( {s,{\text{ }}graphite} \right){\text{ }} + {\text{ }}{O_2}\left( g \right){\text{ }} - - - > {\text{ }}C{O_2}\left( g \right)}&{\Delta H^\circ {\text{ }} = {\text{ }} - 393.5{\text{ }}kJ}
\end{array}\]


\[\begin{array}{*{20}{l}}
  {C{O_2}\left( g \right){\text{ }} - - - > {\text{ }}C{\text{ }}\left( {s,{\text{ }}diamond} \right){\text{ }} + {\text{ }}{O_2}\left( g \right)}&{\Delta H^\circ {\text{ }} = {\text{ }} + 395.4{\text{ }}kJ}
\end{array}\]


For the enthalpy for the second equation, reversing an equation means reversing the sign on the enthalpy values.
Here is the added equation without taking anything out:
\[\begin{array}{*{20}{l}}
  {C{O_2}\left( g \right){\text{ }} + C\left( {s,{\text{ }}graph} \right) + {O_2}\left( g \right) - > C{O_2}\left( g \right){\text{ }} + C{\text{ }}\left( {s,{\text{ }}di} \right) + {O_2}\left( g \right)}&{\Delta H^\circ = \left( { - 393.5{\text{ }}kJ} \right) + \left( { + 395.4{\text{ }}kJ} \right)}
\end{array}\]See the items which are the same on both sides and remove them:
\[\begin{array}{*{20}{l}}
  {C{\text{ }}\left( {s,{\text{ }}graphite} \right){\text{ }} - - - > {\text{ }}C{\text{ }}\left( {s,{\text{ }}diamond} \right)}&{\Delta H^\circ {\text{ }} = {\text{ }} + 1.9{\text{ }}kJ}
\end{array}\]

Note:Reversing an equation means reversing the sign in the enthalpy value because the unreversed equation is exothermic. As we all know this from the negative ahead of the which implies that the other, reverse equation is endothermic. Fitting enthalpy (endothermic) is the opposite of exothermic (giving off enthalpy). Therefore, we modify the sign each time we reverse an equation.