Problem 54 - Chemistry The Molecular Science

Question

When \(\mathrm{KClO}_{3}(\mathrm{~s})\), potassium chlorate, is heated, it melts and decomposes to form oxygen gas. [Molten \(\mathrm{KClO}_{3}\) was shown reacting with a Fritos chip earlier in this chapter ( \(\in\) Sec. \(4-\mid a\) ). \(]\) The thermochemical expression for decomposition of potassium chlorate is $$ 2 \mathrm{KClO}_{3}(\mathrm{~s}) \longrightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{t}} H^{\circ}=-89.4 \mathrm{~kJ} / \mathrm{mol} $$ Calculate \(q\) at constant pressure for (a) Formation of \(97.8 \mathrm{~g} \mathrm{KCl}(\mathrm{s})\). (b) Production of \(24.8 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\). (c) Decomposition of \(35.2 \mathrm{~g} \mathrm{KClO}_{3}(\mathrm{~s})\)

Step-by-Step Solution

Answer

(a) \(-58.63 \text{ kJ}\), (b) \(-738.4 \text{ kJ}\), (c) \(-12.8 \text{ kJ}\).

Step 1: Understanding the Chemical Reaction

The decomposition of potassium chlorate is described by the equation: \[ 2 \mathrm{KClO}_3(s) \rightarrow 2 \mathrm{KCl}(s) + 3 \mathrm{O}_2(g) \]\The provided enthalpy change (\(\Delta_t H^{\circ}\)) for this reaction is \(-89.4 \mathrm{~kJ/mol}\) for the production of 2 moles of \(\mathrm{KCl}\) and 3 moles of \(\mathrm{O}_2\). This means the process releases energy (exothermic) as described.

Step 2: Calculate \( q_a \) for Formation of \(97.8 \text{ g } \mathrm{KCl}\)

First, determine the number of moles of \(\mathrm{KCl}\) produced. The molar mass of \(\mathrm{KCl}\) is approximately \(74.55 \text{ g/mol}\). Calculate moles of \(\mathrm{KCl}\):\[ n = \frac{97.8 \text{ g}}{74.55 \text{ g/mol}} \approx 1.31 \text{ moles} \]Since \(-89.4 \text{ kJ/mol}\) corresponds to \(2 \text{ moles of } \mathrm{KCl}\) produced, use proportion:\[ q_a = -89.4 \text{ kJ} \times \frac{1.31}{2} \approx -58.63 \text{ kJ} \]

Step 3: Calculate \( q_b \) for Production of \(24.8 \text{ mol } \mathrm{O}_2\)

The reaction produces \(3 \text{ moles of } \mathrm{O}_2\) as \(-89.4 \text{ kJ/mol}\) of reaction. Calculate the energy for \(24.8\) moles of \(\mathrm{O}_2\):\[ q_b = -89.4 \text{ kJ/mol} \times \frac{24.8}{3} \approx -738.4 \text{ kJ} \]

Step 4: Calculate \( q_c \) for Decomposition of \(35.2 \text{ g } \mathrm{KClO}_3\)

Determine the number of moles of \(\mathrm{KClO}_3\). The molar mass of \(\mathrm{KClO}_3\) is approximately \(122.55 \text{ g/mol}\).Calculate moles of \(\mathrm{KClO}_3\):\[ n = \frac{35.2 \text{ g}}{122.55 \text{ g/mol}} \approx 0.287 \text{ moles} \]\[-89.4 \text{ kJ/mol}\] refers to \(2 \text{ moles of } \mathrm{KClO}_3\), calculate proportion:\[ q_c = -89.4 \text{ kJ} \times \frac{0.287}{2} \approx -12.8 \text{ kJ} \]

Key Concepts

Chemical Reactions

In the field of chemistry, chemical reactions are processes where one or more substances are transformed into other substances. This transformation involves the change in the chemical composition and properties of the original substance.

In the given exercise, the focus is on the decomposition reaction of potassium chlorate (\(\mathrm{KClO}_3\)). When heated, potassium chlorate breaks down into potassium chloride (\(\mathrm{KCl}\)) and oxygen gas (\(\mathrm{O}_2\)). This is a typical decomposition reaction where a single compound breaks apart into simpler substances.

• The given chemical equation is:\[ 2\mathrm{KClO}_3(s) \rightarrow 2\mathrm{KCl}(s) + 3\mathrm{O}_2(g) \]
• This reaction is central to understanding how reactants transform into products, emphasizing the conversion of solid potassium chlorate to solid potassium chloride and gaseous oxygen.

This specific reaction is significant not only for demonstrating a decomposition process but also for illustrating how energy is involved in chemical changes.

Enthalpy Change

Enthalpy change (\(\Delta H\)) is an important concept in thermochemistry that refers to the heat absorbed or released during a chemical reaction at constant pressure. This change in heat content helps determine whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat).

In the problem provided, the decomposition of potassium chlorate is described as having an enthalpy change of \(\Delta_{t} H^{\circ} = -89.4\,\mathrm{kJ/mol}\). This negative value indicates that the reaction releases heat, classifying it as exothermic.

• This means each mole of reaction liberates 89.4 kJ of energy.
• The enthalpy change is related to the stoichiometry of the reaction, with the standard notation showing the energy change for the specific number of moles provided in the balanced equation: 2 moles of \(\mathrm{KCl}\) and 3 moles of \(\mathrm{O}_2\) are produced from 2 moles of \(\mathrm{KClO}_3\).

Understanding how to compute the energy changes for given quantities, as seen in the exercise, is crucial for predicting how much energy will be absorbed or released as the reaction proceeds.

Moles Calculation

Moles calculation is a fundamental task in stoichiometry that involves converting between mass, moles, and molecules, which are core values for predicting the outcomes of chemical reactions.

For this particular exercise, the emphasis is on moles calculation to find the energy (\(q\)) associated with the formation and decomposition of substances. - **Calculating Moles of Product:**For example, when forming \(97.8\, \mathrm{g}\) of \(\mathrm{KCl}\), the molar mass of \(\mathrm{KCl}\) (\[74.55\, \mathrm{g/mol}\]), is used to find moles:\[ n = \frac{97.8\, \mathrm{g}}{74.55\, \mathrm{g/mol}} \approx 1.31\text{ moles} \]This result then aids in determining the energy released.- **Energy Calculation for Reactants and Products:**Similarly, the moles of \(\mathrm{KClO}_3\) and \(\mathrm{O}_2\) are calculated using their respective molar masses, which are integral in computing the total heat released or absorbed in their reactions.

• The rule of thumb here is to always begin by finding the number of moles, then relate it to the enthalpy change provided for the reaction.
• This approach allows for the conversion of mass measurements to understandable energy outputs, bridging the gap between theoretical chemistry and practical application.

Mastering these calculations ensures a clear grasp of how quantities of reactants and products relate to energy changes during reactions.