At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) $1.36 \mathrm{~mol}\( of \)\mathrm{O}_{2}\( and \)(\mathbf{b}) 10.4 \mathrm{~g}$ of \(\mathrm{KCl} .(\mathbf{c})\) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

In the given reaction, 2 moles of KClO₃ decompose to form 3 moles of O₂ with \(\Delta H = -89.4 \mathrm{~kJ}\). To find the heat change for the formation of 1.36 moles of O₂, we need to determine the heat change per mole of O₂ first. From the balanced reaction equation: This can be determined by dividing the heat change for the reaction by the ratio of moles of O₂ formed to moles of KClO₃ decomposed, which is \(\frac{3}{2}\). $$ \Delta H_{O₂} = \frac{-89.4 \mathrm{~kJ}}{\frac{3}{2}} = -59.6 \mathrm{~kJ/mol} $$ Now, we can find the heat change for the formation of 1.36 moles of O₂ by multiplying the heat change per mole of O₂ by the number of moles of O₂: $$ \Delta H = -59.6 \mathrm{~kJ/mol} \times 1.36 \mathrm{~mol} = -81.1 \mathrm{~kJ} $$

First, we need to convert the mass of KCl to moles. The molar mass of KCl is approximately 39.1 (K) + 35.5 (Cl) = 74.6 g/mol. So, $$ \text{moles of KCl} = \frac{10.4 \mathrm{~g}}{74.6 \mathrm{~g/mol}} = 0.139 \mathrm{~mol} $$ Now, we can calculate the enthalpy change \(\Delta H\) for the formation of 0.139 mol of KCl. From the balanced reaction equation: One mole of KCl is formed for each mole of KClO₃ decomposed. So, the heat change for the formation of 1 mole of KCl from KClO₃ is half of \(\Delta H = -89.4 \mathrm{~kJ}\): $$ \Delta H_{\mathrm{KCl}} = \frac{-89.4 \mathrm{~kJ}}{2} = -44.7 \mathrm{~kJ/mol} $$ Now, we can find the heat change for the formation of 0.139 moles of KCl by multiplying the heat change per mole of KCl by the number of moles of KCl: $$ \Delta H = -44.7 \mathrm{~kJ/mol} \times 0.139 \mathrm{~mol} = -6.21 \mathrm{~kJ} $$

The decomposition of KClO₃ to KCl and O₂ has a negative enthalpy change (\(\Delta H = -89.4 \mathrm{~kJ}\)), which means this reaction is exothermic. Under ordinary conditions, exothermic reactions are generally spontaneous and favorable. However, the reverse reaction, the formation of KClO₃ from KCl and O₂, would have a positive enthalpy change (opposite sign), meaning that it would be endothermic. Endothermic reactions generally require continuous energy input to proceed. As a result, the reverse reaction is unlikely to be feasible under ordinary conditions without external energy input.