Solution Found!
The values of the two allotropes of oxygen, O2 and O3, are
Chapter 6, Problem 46P(choose chapter or problem)
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?
Questions & Answers
QUESTION:
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?
ANSWER:Here, we are going to find out the most stable form of oxygen at \(25^{\circ} \mathrm{C}\).
Step 1 of 2
The standard enthalpy change \(\Delta H_{\text {reaction, }}\) is defined as the difference between the enthalpies of the products and the enthalpies of the reactants.
\(\Delta H_{\text {reaction }}^{o}=\Delta H_{\text {product }}^{o}-\Delta H_{\text {reactant }}^{o} \ldots(1)\)
\(=\Delta H_{\text {product }}^{o}-\Delta H_{\text {reactant }}^{o} \ldots(1)\)
But the standard enthalpy change for the stable element is considered as zero(0).