Problem 46 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\mathrm{KCl} .(\mathbf{c})\) 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.

Step-by-Step Solution

Answer

The enthalpy change \(\Delta H\) for the formation of 1.36 mol of O₂ is -81.1 kJ, and for the formation of 10.4 g of KCl is -6.21 kJ. Since the decomposition of KClO₃ to KCl and O₂ is exothermic, the reverse reaction, the formation of KClO₃ from KCl and O₂, is endothermic and is unlikely to be feasible under ordinary conditions without external energy input.

Step 1: Calculate \(\Delta H\) for the formation of 1.36 mol of O₂

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: $$ \Delta H = -59.6 \mathrm{~kJ/mol} \times 1.36 \mathrm{~mol} = -81.1 \mathrm{~kJ} $$

Step 2: Calculate \(\Delta H\) for the formation of 10.4 g of KCl

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: $$ \Delta H = -44.7 \mathrm{~kJ/mol} \times 0.139 \mathrm{~mol} = -6.21 \mathrm{~kJ} $$

Step 3: Discuss the feasibility of the reverse reaction

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.

Key Concepts

Enthalpy Change

Enthalpy change, denoted by the symbol \( \Delta H \), represents the heat absorbed or released in a chemical reaction at constant pressure. It's an important aspect of thermodynamics as it provides insight into the energy dynamics within a reaction. When \( \Delta H \) is negative, the reaction releases energy to its surroundings, classifying it as an exothermic reaction. Conversely, a positive \( \Delta H \) implies the reaction absorbs energy, making it endothermic. Understanding enthalpy change allows us to predict the energy requirements or releases associated with chemical processes.

In the example provided, we calculate the enthalpy change for forming specific amounts of products. By knowing how much energy is released or absorbed per mole of a substance, students can determine the overall energy change for varying quantities of reactants or products in a reaction.

Spontaneity

Spontaneity in thermodynamics refers to a reaction's ability to occur naturally without external assistance. It is influenced by factors such as temperature, pressure, and energy changes, including enthalpy and entropy. A spontaneous reaction is generally irreversible under given conditions and tends to achieve an equilibrium state.

For the decomposition of \( \mathrm{KClO}_3 \) to \( \mathrm{KCl} \) and \( \mathrm{O}_2 \), the process occurs spontaneously with the application of heat. Spontaneity is often associated with negative enthalpy changes (exothermic reactions). However, it's important to note that not all exothermic reactions are spontaneous; the reaction's entropy change also plays a crucial role. In our context, the reverse reaction (formation of \( \mathrm{KClO}_3 \)) is likely non-spontaneous due to its positive enthalpy change, requiring continuous energy input.

Exothermic Reaction

Exothermic reactions are those that release energy, usually in the form of heat, to their surroundings. These reactions are characterized by a negative enthalpy change (\( \Delta H < 0 \)). As the system releases heat, the products formed are generally more stable and have lower energy than the reactants. This energy release often results in an increase in temperature of the surroundings.

A practical example is the decomposition of \( \mathrm{KClO}_3 \) to produce \( \mathrm{KCl} \) and \( \mathrm{O}_2 \), where \( \Delta H = -89.4 \mathrm{~kJ} \). The negative \( \Delta H \) indicates the energy released during the reaction. Such processes are typically favorable and present as spontaneous under ordinary conditions, largely due to the energy surplus given off, which propels the reaction forward.

Endothermic Reaction

Endothermic reactions, in contrast to exothermic reactions, absorb energy from their surroundings, indicated by a positive enthalpy change (\( \Delta H > 0 \)). These reactions often require continuous energy input to proceed as the reactants are of higher energy than the products. Because these processes absorb heat, they are usually accompanied by a reduction in the temperature of the surroundings.

An example related to the provided exercise is the hypothetical reverse reaction—the formation of \( \mathrm{KClO}_3 \) from \( \mathrm{KCl} \) and \( \mathrm{O}_2 \). For this process, energy must be supplied to offset the energy required for the reaction to proceed, meaning it's endothermic and not feasible under spontaneous or ordinary conditions without external assistance. Understanding endothermic reactions helps students grasp why certain processes require energy investment to drive them forward.