Question

What is standard enthalpy of formation of \[S{O_2}(g)\]?
A scientist measures the standard enthalpy change for the following reaction to be \[ - 171.2\]kJ
\[2S{O_2}(g) + {O_2} \rightleftarrows 2S{O_3}(g)\]
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of \[S{O_2}(g)\]is ______ kJ/mol.

Answer 1

Hint : Whenever two reactants react to carry out a reaction and form a single product, then it is called a combination reaction. We have given that \[S{O_3}\] is formed from the reaction between \[S{O_2}\] and \[{O_2}\]. The reaction is as follows:
\[2S{O_2}(g) + {O_2} \rightleftarrows 2S{O_3}(g)\]
As in this reaction we can see that two reactants sulphur dioxide, and oxygen molecule, react to form a single product which is sulphur trioxide, so this reaction is a type of a combination reaction. Hence, formation of \[S{O_3}\] from \[S{O_2}\] and \[{O_2}\] is a combination reaction.

Complete Step By Step Answer:
Enthalpy of reaction can be calculated as follows:
\[\vartriangle {H_r} = \sum {Enthalpies{\text{ }}of{\text{ }}formation{\text{ }}of{\text{ }}products - } \sum {Enthalpies{\text{ }}of{\text{ }}formation{\text{ }}of{\text{ }}reactants} \]
For the given reaction:
\[2S{O_2}(g) + {O_2} \rightleftarrows 2S{O_3}(g)\] \[\vartriangle {H_r} = - 171.2kJ\]
Or above reaction can be written as by dividing by 2
\[S{O_2}(g) + \dfrac{1}{2}{O_2} \rightleftarrows S{O_3}(g)\] \[\vartriangle {H_r} = - 85.6\dfrac{{kJ}}{{mol}}\]
Therefore, \[\vartriangle {H_r} = \sum {\vartriangle {H_f}(S{O_3}) - } \sum {\vartriangle {H_f}(S{O_2}) + } \vartriangle {H_f}({O_2})\]
We know that, \[\vartriangle {H_f}({O_2}) = 0\dfrac{{kJ}}{{mol}}\], since it is present in its standard state which is its most stable form at STP.
\[\vartriangle {H_f}(S{O_3}) = - 395.77\dfrac{{kJ}}{{mol}}\]
Putting all the value in above equation we get,
\[\begin{gathered}
  \vartriangle {H_r} = - 395.77 - [\vartriangle {H_f}(S{O_2}) + 0] \\
   - 85.6 + 395.77 = - \vartriangle {H_f}(S{O_2}) \\
  \vartriangle {H_f}(S{O_2}) = - 310.17\dfrac{{kJ}}{{mol}} \\
\end{gathered} \]
Hence, enthalpy of formation of \[S{O_2}(g)\]is \[ - 310.17\dfrac{{kJ}}{{mol}}\]

Additional Information:
 Enthalpy of formation is the heat exchange when 1 mole of a substance is formed from its constituent elements in their most stable form or in their standard states.
Enthalpy or reaction is the heat exchange between the system and surrounding at constant pressure in a chemical reaction.

Note :
Standard enthalpy of formation of an element in its standard state is zero. Standard state is the most stable form of a substance under normal(STP) conditions of temperature and pressure.