Q16.4-36E
Question
Is the formation of ozone \(\left( {{{\bf{O}}_{\bf{3}}}{\bf{(g)}}} \right)\) from oxygen \(\left( {{{\bf{O}}_{\bf{2}}}{\bf{(g)}}} \right)\) spontaneous at room temperature under standard state conditions?
Step-by-Step Solution
VerifiedThe formation of ozone from oxygen is not spontaneous under standard state conditions.
- The definition of spontaneous reaction in science is a reaction that occurs in a specified set of conditions without interference. A spontaneous reaction is one that occurs in the absence of intervention in a given set of conditions. A spontaneous reaction is completed without the assistance of anyone else.
- A reaction is spontaneous if the overall entropy, or disorder, increases, according to the Second Law of Thermodynamics.
The equation for the formation of ozone from oxygen:
\(3{{\rm{O}}_2}(g) \to 2{{\rm{O}}_3}(g)\)
In order to determine whether this reaction is spontaneous under standard state conditions, the easiest way would be to calculate the Gibbs energy change\((\Delta G)\). If the \(\Delta G\)is less than zero, it means that the reaction Is spontaneous under standard state conditions. Otherwise, it is not spontaneous under these conditions. To determine the \(\Delta G\)for this reaction, we can use the formula
\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)
\(\Delta G = 2{G_f}\left( {{O_3},g} \right) - 3{G_f}\left( {{O_2},g} \right)\)
\(\Delta G = (2 \cdot 163.2 - 3 \cdot 0)\frac{{kJ}}{{mol}}\)
\(\Delta G = 326.4\frac{{kJ}}{{mol}}\)
Since the \(\Delta G\)is positive, this reaction is not spontaneous under standard state conditions.
Therefore the formation of ozone from oxygen is not spontaneous under standard state conditions.