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Chapter 6: Problem 46

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
Therefore, the more stable form of oxygen at this temperature is \(O_{2}\).

Step by step solution

01

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

You are given that the \(\Delta H_{\mathrm{f}}^{\circ}\) of \(O_{2}\) is 0 kJ/mol while that of \(O_{3}\) is 142.2 kJ/mol. Now, compare these two values.
02

Identify the more stable form

The allotrope of oxygen having the lower \(\Delta H_{\mathrm{f}}^{\circ}\) value will be more stable because it requires less energy to form. From step 1, we find that \(O_{2}\) has a lower \(\Delta H_{\mathrm{f}}^{\circ}\) value than \(O_{3}\).

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Most popular questions from this chapter

(a) For most efficient use, refrigerator freezer compartments should be fully packed with food. What is the thermochemical basis for this recommendation? (b) Starting at the same temperature, tea and coffee remain hot longer in a thermal flask than chicken noodle soup. Explain.

Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(A\) is larger than that of \(B\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$ \begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

These are various forms of energy: chemical, heat, light, mechanical, and electrical. Suggest ways of interconverting these forms of energy.

The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state at \(25^{\circ} \mathrm{C}\) is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions.
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