The \(\Delta H_{\mathrm{f}}^{\circ}\) values of the two allotropes of oxygen, \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\), are 0 and \(142.2 \mathrm{~kJ} / \mathrm{mol}\), respectively, at \(25^{\circ} \mathrm{C}\). Which is the more stable form at this temperature?

Short Answer

Expert verified
At \(25^{\circ} \mathrm{C}\), \(\mathrm{O}_{2}\) is the more stable form of oxygen because of its 0 enthalpy of formation compared to \(142.2 \mathrm{~kJ}\)/ \(\mathrm{mol}\) for \(\mathrm{O}_{3}\).

Step by step solution

01

Analyze the provided data

The question provides the \(\Delta H_{\mathrm{f}}^{\circ}\) values for two allotropes of oxygen at \(25^{\circ} \mathrm{C}\): for \(\mathrm{O}_{2}\) it's 0 and for \(\mathrm{O}_{3}\), it's \(142.2 \mathrm{~kJ}\)/ \(\mathrm{mol}\). Remember, the \(\Delta H_{\mathrm{f}}^{\circ}\) value represents the stability of the molecule.
02

Compare the \(\Delta H_{\mathrm{f}}^{\circ}\) values

Comparing the \(\Delta H_{\mathrm{f}}^{\circ}\) values, one can see \(\mathrm{O}_{2}\) has a \(\Delta H_{\mathrm{f}}^{\circ}\) value of 0 which is lower than the \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{O}_{3}\). This indicates that the molecule of \(\mathrm{O}_{2}\) is in a more stable state than \(\mathrm{O}_{3}\).

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