Problem 46 - Chemistry: The Central Science

Question

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 (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.

Step-by-Step Solution

Answer

For the formation of (a) 1.36 mol of O₂, the enthalpy change (ΔH) is -40.6 kJ, and for the formation of (b) 10.4 g of KCl, the enthalpy change is -6.24 kJ. (c) Under ordinary conditions, the reverse reaction of KClO₃ formation is not feasible because the forward reaction is spontaneous and exothermic, and it does not favor the formation of KClO₃ from KCl and O₂.

Step 1: (a) Calculating ΔH for 1.36 mol of O₂

To find the enthalpy change (ΔH) for producing 1.36 mol of O₂, first, we must determine the stoichiometric relationship between the reactants and products in the given reaction. The balanced reaction provided is \[2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \] For every 3 moles of O₂ produced, 2 moles of KClO₃ are consumed. Now, we proceed to calculate the enthalpy change for the formation of 1.36 mol of O₂: 1. Calculate the moles of KClO₃ required for 1.36 mol of O₂: \(\frac{1.36\;\text{mol}\;\text{O}_{2}}{3\;\text{mol}\;\text{O}_{2}} \times 2\;\text{mol}\;\text{KClO}_{3} = 0.9067\;\text{mol}\;\text{KClO}_{3} \) 2. Calculate the ΔH for 0.9067 mol of KClO₃: \(\Delta H = 0.9067\;\text{mol}\;\text{KClO}_{3} \times \frac{-89.4 \;\text{kJ}}{2\;\text{mol}\;\text{KClO}_{3}} = -40.6\;\text{kJ} \) For the formation of 1.36 mol of O₂, the enthalpy change is -40.6 kJ.

Step 2: (b) Calculating ΔH for 10.4 g of KCl

To find the enthalpy change (ΔH) for the formation of 10.4 g of KCl, we first convert the mass of KCl to moles using its molar mass: The molar mass of KCl = 39.10 (K) + 35.45 (Cl) = 74.55 g/mol Now, convert 10.4 g of KCl to moles: \(\frac{10.4\;\text{g}\;\text{KCl}}{74.55\;\text{g/mol}} = 0.1394\;\text{mol}\;\text{KCl} \) Next, we calculate the enthalpy change for 0.1394 mol of KCl: 1. Calculate the moles of KClO₃ required for 0.1394 mol of KCl: \(\frac{0.1394\;\text{mol}\;\text{KCl}}{2\;\text{mol}\;\text{KCl}} \times 2\;\text{mol}\;\text{KClO}_{3} = 0.1394\;\text{mol}\;\text{KClO}_{3} \) 2. Calculate the ΔH for 0.1394 mol of KClO₃: \(\Delta H = 0.1394\;\text{mol}\;\text{KClO}_{3} \times \frac{-89.4 \;\text{kJ}}{2\;\text{mol}\;\text{KClO}_{3}} = -6.24\;\text{kJ} \) For the formation of 10.4 g of KCl, the enthalpy change is -6.24 kJ.

Step 3: (c) Assessing the feasibility of the reverse reaction

To assess the feasibility of the reverse reaction, we can analyze the enthalpy change (ΔH) and the conditions in which the decomposition of KClO₃ occurs. The given reaction has a negative ΔH (-89.4 kJ), indicating that it is exothermic, releasing heat to the surroundings. The decomposition of KClO₃ proceeds spontaneously when heated, which further supports the exothermic nature of the reaction. In the reverse reaction, the formation of KClO₃ would require taking in the same amount of heat (ΔH=+89.4 kJ) to form 2 moles of KClO₃ and absorb 3 moles of O₂ gas. This process would be endothermic. Under ordinary conditions, the reverse reaction of KClO₃ formation is not feasible because the forward reaction is spontaneous and exothermic. Thus, it does not favor the formation of KClO₃ from KCl and O₂ under normal circumstances.