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 \((\mathbf{b}) 10.4 \mathrm{g}\) 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.

Short answer

The enthalpy change for the formation of 1.36 mol of O₂ is -40.5 kJ, and for the formation of 10.4 g of KCl is -6.26 kJ. The reverse reaction, the formation of KClO₃ from KCl and O₂, is not likely to be feasible under ordinary conditions because it would require the input of energy and might need specific conditions like elevated pressure or a catalyst.

Step-by-step solution

Part (a): Find \(\Delta H\) for the formation of 1.36 mol of O₂.

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.

Part (b): Find \(\Delta H\) for the formation of 10.4 g of KCl.

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.

Part (c): Determine the feasibility of the reverse reaction under ordinary conditions.

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.

Key concepts

KClO3 decomposition

Potassium chlorate (\(\text{KClO}_3\)) is a chemical compound that is widely used in laboratories for oxygen production. The process of decomposing \(\text{KClO}_3\) involves breaking it down into potassium chloride (\(\text{KCl}\)) and oxygen gas (\(\text{O}_2\)). This transformation is represented by the balanced chemical equation:

• \(2\,\text{KClO}_3(s) \longrightarrow 2\,\text{KCl}(s) + 3\,\text{O}_2(g)\)

In this reaction, energy is released, making it an exothermic reaction with a negative change in enthalpy (\(\Delta H = -89.4\,\text{kJ}\)).

The decomposition of \(\text{KClO}_3\) is often achieved by heating it, which provides the necessary energy to overcome the activation barrier. The products, \(\text{KCl}\) and \(\text{O}_2\), are more stable than the reactants, which is why energy is liberated in the form of heat during the reaction.
This characteristic makes the process particularly useful when a ready supply of pure oxygen is needed with minimal equipment.

Oxygen gas formation

The formation of oxygen gas (\(\text{O}_2\)) is one of the key outcomes of the \(\text{KClO}_3\) decomposition reaction. During the reaction, three moles of oxygen gas are produced for every two moles of \(\text{KClO}_3\) decomposed. This means that oxygen formation is efficient and occurs in large quantities relative to the reactants.

Such production is crucial for various applications, from laboratory experiments to industrial processes. The oxygen produced is pure, making it suitable for reactions where the presence of other gases could be problematic.

• Oxygen is essential for combustion processes.
• It supports respiration and other chemical reactions requiring oxygen.

Laboratories take advantage of this reaction by heating \(\text{KClO}_3\) in a controlled environment, thereby obtaining \(\text{O}_2\) on demand. This process, however, must be conducted with caution due to the highly reactive nature of oxygen, which can widely accelerate other reactions.

Reaction spontaneity

Spontaneity refers to the natural tendency of a process to occur without external intervention over time. For the decomposition of \(\text{KClO}_3\) into \(\text{KCl}\) and \(\text{O}_2\), spontaneity is aided by the application of heat. The exothermic nature of the reaction demonstrates a spontaneous process, highlighted by the negative enthalpy change, \(\Delta H = -89.4\,\text{kJ}\).

Conversely, the reverse reaction—synthesizing \(\text{KClO}_3\) from \(\text{KCl}\) and \(\text{O}_2\)—is not ordinarily spontaneous. This is because the reverse process would be endothermic, requiring an input of energy to proceed, making it energetically unfavorable under normal conditions.

• The breakdown process is spontaneous with heat.
• Reverse synthesis is non-spontaneous without additional energy input.

Generally, reactions that are spontaneous tend to increase the entropy of the system—a principle occurring during the breakdown of complex molecules into simpler constituents. While the decomposition reaction is feasible once initiated, practical synthesis of \(\text{KClO}_3\) may demand catalysts and special conditions to proceed.