Problem 44 - 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) \(0.632 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and (b) \(8.57 \mathrm{~g}\) 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

The enthalpy change (ΔH) for the formation of (a) 0.632 mol of O2 is -18.84 kJ, and (b) 8.57 g of KCl is -1.7115 kJ. (c) The reverse reaction, the formation of KClO3 from KCl and O2, is less likely to occur under ordinary conditions, as it requires an input of energy and is not spontaneous. It may become feasible if enough energy is supplied to the system.

Step 1: Mole to mole conversion

Find the moles of O2 formed by the given reaction according to the balanced chemical equations. From the balanced equation, \( 2\, KClO_{3}(s) \rightarrow 3\, O_{2}(g) \) We know that the ΔH for the balanced equation is: ΔH = -89.4 kJ. Therefore, for the formation of three moles of O2, ΔH = -89.4 kJ.

Step 2: Calculate the enthalpy change for the formation of 0.632 moles of O2

Use the ratio we found to evaluate the ΔH for the formation of the specified amount of O2. We are asked to find the ΔH for the formation of 0.632 mol of O2: For 3 moles of O2: ΔH = -89.4 kJ For 1 mole of O2: ΔH = -89.4 kJ / 3 = -29.8 kJ/mol For 0.632 moles of O2: ΔH = -29.8 kJ/mol × 0.632 mol = -18.84 kJ Thus, the enthalpy change for the formation of 0.632 mol of O2 is -18.84 kJ. (b) Calculate ΔH for the formation of 8.57 g of KCl:

Step 1: Gram to mole conversion for KCl

Convert the given mass of KCl (8.57 g) to moles by dividing the mass by the molar mass of KCl. Molar mass of KCl = 39.1 g/mol (K) + 35.45 g/mol (Cl) = 74.55 g/mol Moles of KCl = 8.57 g / 74.55 g/mol = 0.115 mol

Step 2: Mole-to-mole conversion for KCl to O2

Relate the moles of KCl formed to the moles of O2 and find the moles of O2 formed. From the balanced equation, \( 2\, KClO_{3}(s) \rightarrow 2\, KCl(s) + 3\, O_{2}(g) \) 2 moles of KCl form 3 moles of O2. So, for the formation of 0.115 moles of KCl, the moles of O2 formed are: Moles of O2 = 0.115 mol × (3 mol O2 / 2 mol KCl) = 0.0575 mol O2

Step 3: Calculate ΔH for the formation of O2.

Use the ratio of moles of O2 and the enthalpy change per mole to find the overall ΔH for the formation of 0.0575 mol of O2. ΔH for 1 mole of O2 = -29.8 kJ/mol ΔH for 0.0575 moles of O2 = -29.8 kJ/mol × 0.0575 mol = -1.7115 kJ Thus, the enthalpy change for the formation of 8.57 g of KCl is -1.7115 kJ. (c) Feasibility of the reverse reaction: The ΔH for 2 KCl(s) + 3 O2(g) → 2 KClO3(s) is positive (+89.4 kJ). The reverse reaction would require an input of energy, which is less likely to happen under ordinary conditions, as it is not spontaneous. The reverse reaction could become feasible if enough energy is supplied to the system, but under normal laboratory conditions, the decomposition of KClO3 is the more likely outcome.

Key Concepts

Enthalpy Change

Understanding enthalpy change is essential in thermochemistry, which deals with the heat changes accompanying chemical reactions. Enthalpy (\( H \)) is the total heat content of a system. When a reaction occurs, the system either absorbs or releases energy.
For the decomposition of potassium chlorate (\( KClO_3 \)), the enthalpy change (\( \Delta H \)) is negative, indicating an exothermic reaction where energy is released into the surroundings.

• If \( \Delta H \) is negative, the reaction releases heat, as observed in the decomposition where \( \Delta H = -89.4 \) kJ.
• If \( \Delta H \) is positive, energy input is required, making the reaction endothermic.

This energy exchange is crucial for determining reaction direction, feasibility, and whether external energy is required.

KClO3 Decomposition

The decomposition of potassium chlorate (\( KClO_3 \)) is a common laboratory method to produce oxygen gas. The equation for this chemical process is:\[2 \, KClO_3(s) \rightarrow 2 \, KCl(s) + 3 \, O_2(g)\]This balanced reaction reflects the relationship between the breakdown of \( KClO_3 \) and the production of oxygen.
In this process, two moles of \( KClO_3 \) decompose to form two moles of \( KCl \) and three moles of \( O_2 \).Understanding the stoichiometry helps to calculate how much oxygen is produced from a given amount of \( KClO_3 \).
The process releases energy, as evident by its negative enthalpy change. In practical applications, such as generating small quantities of gas, this exothermic reaction is very efficient. The decomposition leads to a positive gas law shift, increasing the number of moles of gas in the system, which further drives the reaction to completion.

Reaction Feasibility

The feasibility of a chemical reaction often depends on its enthalpy change and the conditions under which it occurs. While the decomposition of \( KClO_3 \) is spontaneous with heating, the reverse reaction is not spontaneous under normal conditions.To reverse the reaction, \( KCl \) and \( O_2 \) must form \( KClO_3 \), which requires absorbing energy (\( \Delta H = +89.4 \) kJ), indicating an endothermic process.
This makes the reverse reaction less feasible in standard laboratory settings since an input of energy is necessary. To achieve this, external sources of heat or energy might be needed.
The spontaneity of reactions is not solely about enthalpy. Other factors, such as entropy (\( \Delta S \)) and temperature, also play critical roles. Thus, while the reverse reaction has theoretical feasibility with sufficient energy, it is unlikely without controlled conditions that influence these factors.