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 \(\mathrm{KCl}\) (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.

We are given that the balanced chemical equation is: \( 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{kJ} \) Since the reaction produces 3 moles of O₂, we can find the enthalpy change for the formation of 1 mol of O₂ by dividing -89.4 kJ by 3: \[\frac{-89.4 \mathrm{kJ}}{3}= -29.8 \frac{\mathrm{kJ}}{\mathrm{mol}}\] Now, to find \(\Delta H\) for the formation of 1.36 mol of O₂, multiply the moles of O₂, 1.36 mol, by the enthalpy change per mole of O₂, -29.8 kJ/mol: \[\Delta H = (1.36 \mathrm{mol})(-29.8 \frac{\mathrm{kJ}}{\mathrm{mol}})= -40.5 \mathrm{kJ}\] Therefore, the enthalpy change for the formation of 1.36 mol of O₂ is -40.5 kJ.

First, we need to determine the number of moles of KCl in 10.4 g. The molar mass of KCl is: \(\mathrm{K} : 39.10 \frac{\mathrm{g}}{\mathrm{mol}} \) \(\mathrm{Cl} : 35.45 \frac{\mathrm{g}}{\mathrm{mol}} \) So, the molar mass of KCl is: \(39.10 + 35.45 = 74.55 \frac{\mathrm{g}}{\mathrm{mol}}\) Now we can find the number of moles of KCl in 10.4 g: \[\frac{10.4 \mathrm{g}}{74.55 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.14 \mathrm{mol}\] Since the reaction produces 2 moles of KCl, we can find the enthalpy change per mole of KCl by dividing -89.4 kJ by 2: \[\frac{-89.4 \mathrm{kJ}}{2}= -44.7 \frac{\mathrm{kJ}}{\mathrm{mol}}\] Now, to find \(\Delta H\) for the formation of 0.14 mol of KCl, multiply the moles of KCl, 0.14 mol, by the enthalpy change per mole of KCl, -44.7 kJ/mol: \[\Delta H = (0.14 \mathrm{mol})(-44.7 \frac{\mathrm{kJ}}{\mathrm{mol}})= -6.26 \mathrm{kJ}\] Therefore, the enthalpy change for the formation of 10.4 g of KCl is -6.26 kJ.

We know that the forward reaction, the decomposition of KClO₃, is spontaneous and exothermic, as \(\Delta H\) is negative. However, the reverse reaction, the formation of KClO₃ from KCl and O₂, would be endothermic, as it would require the absorption of heat. Since the forward reaction is spontaneous and releases energy, it is likely that the reverse reaction is not feasible under ordinary conditions, as it would require the input of energy. Moreover, the creation of KClO₃ from KCl and O₂ might require specific conditions like elevated pressure or the presence of a catalyst, which are not considered ordinary conditions.