Problem 99 - Chemistry The Molecular Science

Question

Limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) $$ At \(1000 .{ }^{\circ} \mathrm{C}, K_{\mathrm{P}}=3.87\). Pure \(\mathrm{CaCO}_{3}\) is placed into an empty \(5.00-\mathrm{L}\) flask. The flask is sealed and heated to \(1000 .{ }^{\circ} \mathrm{C}\). Calculate the mass of \(\mathrm{CaCO}_{3}\) that must decompose to achieve the equilibrium pressure of \(\mathrm{CO}_{2}\).

Step-by-Step Solution

Answer

18.52 g of \(\mathrm{CaCO}_3\) must decompose.

Step 1: Write the equilibrium expression

The problem provides the value of \(K_{P}\) at \(1000^{\circ}C\). For the reaction: \[\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{~g})\] Since \(K_{P}\) is given, the equilibrium expression is: \[K_{P} = P_{\mathrm{CO}_2}\]. Given \(K_{P} = 3.87\), the equilibrium pressure of \(\mathrm{CO}_2\) is \(P_{\mathrm{CO}_2} = 3.87\; \text{atm}\).

Step 2: Use the ideal gas law to find moles of CO₂

Once we know the pressure of \(\mathrm{CO}_2\), we apply the ideal gas law: \[PV = nRT\] where \(P\) is the pressure, \(V\) is the volume of the flask, \(n\) is the number of moles, and \(R = 0.0821 \text{ L atm/mol K}\) is the ideal gas constant. We need to convert the temperature to Kelvin: \[T = 1000 + 273.15 = 1273.15\;\text{K}\]. Then, using the ideal gas law: \[3.87 \times 5.00 = n \times 0.0821 \times 1273.15\]. Solving for \(n\): \[n = \frac{3.87 \times 5.00}{0.0821 \times 1273.15}\].

Step 3: Calculate moles of CO₂

Calculate the number of moles \(n\) of \(\mathrm{CO}_2\): \[n = \frac{3.87 \times 5.00}{0.0821 \times 1273.15} \approx 0.185\;\text{mol}\].

Step 4: Relate moles of CaCO₃ to moles of CO₂

From the chemical equation, we observe that the decomposition of \(1\) mole of \(\mathrm{CaCO}_3\) produces \(1\) mole of \(\mathrm{CO}_2\). Therefore, \(0.185\) moles of \(\mathrm{CO}_2\) comes from \(0.185\) moles of \(\mathrm{CaCO}_3\).

Step 5: Calculate mass of decomposed CaCO₃

The molar mass of \(\mathrm{CaCO}_3\) is \(40.08\;\text{g/mol (Ca)} + 12.01\;\text{g/mol (C)} + 3 \times 16.00\;\text{g/mol (O)} = 100.09\;\text{g/mol}\). So, the mass of \(\mathrm{CaCO}_3\) decomposed is \[\text{mass} = 0.185 \times 100.09 \approx 18.52\;\text{g}\].

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used in chemistry and physics to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as:

• \( PV = nRT \)

where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume of the container holding the gas.
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant \( (0.0821 \text{ L atm/mol K}) \).
• \( T \) is the temperature in Kelvin.

The Ideal Gas Law helps calculate the unknown variable among the four parameters when the other three are known. To convert temperature into Kelvin, a crucial step in using the Ideal Gas Law, add 273.15 to the Celsius temperature. This equation assumes the gas behaves "ideally," typically valid under high temperature and low pressure conditions.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. This can be represented in a general form as:

• \( AB \rightarrow A + B \)

In the given exercise, limestone \( (\text{CaCO}_3) \) undergoes a decomposition reaction:

• \( \text{CaCO}_{3}(\text{s}) \rightleftharpoons \text{CaO}(\text{s}) + \text{CO}_{2}(\text{~g}) \)

Most decomposition reactions require an external source of energy like heat or electricity to initiate the reaction, which is why the limestone is heated in this problem. Decomposition reactions are crucial in various industrial processes, including cement production and metal extraction, making understanding of this concept fundamentally important in chemistry.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), quantifies the concentration ratio of products and reactants in a reversible chemical reaction at equilibrium. Different forms include \( K_c \) for concentrations and \( K_p \) for partial pressures. For the provided reaction at 1000 °C:

• \( K_{P} = 3.87 \)

This value indicates the pressure of \( \text{CO}_{2} \) at equilibrium, since solids like \( \text{CaCO}_{3} \) and \( \text{CaO} \) do not appear in the expression. Using \( K_p \):

• \( K_{P} = \text{P}_{\text{CO}_2} \)

Calculating equilibrium conditions helps predict the direction of reaction tendencies; a high \( K_p \) suggests product formation is favored. Students should understand how equilibrium constants tie into Le Chatelier's principle and the behaviors of dynamic equilibria.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in a chemical reaction. It is essential for quantitative chemistry analysis, involving balanced chemical equations to derive the relationships between reactants and products. In the decomposition reaction:

• \( \text{CaCO}_{3} \rightarrow \text{CaO} + \text{CO}_{2} \)

The balanced equation indicates that one mole of \( \text{CaCO}_{3} \) yields one mole of \( \text{CO}_{2} \). Thus, the decomposition of 0.185 moles of \( \text{CO}_{2} \) implies 0.185 moles of \( \text{CaCO}_{3} \) decomposed, shown by:

• Moles of \( \text{CaCO}_{3} = \text{Moles of }\text{CO}_{2} = 0.185 \)

To find mass, multiply the moles by the molar mass of \( \text{CaCO}_{3} \) (100.09 g/mol). This calculation demonstrates the practical use of stoichiometry in determining the necessary amount of a substance in a given reaction, essential for laboratory work and industrial applications.