Problem 56 - Chemistry and Chemical Reactivity

Question

Limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=3.87 .\) If pure \(\mathrm{CaCO}_{3}\) is placed in a \(5.00-\mathrm{I}\) flask and heated to \(1000^{\circ} \mathrm{C},\) what quantity of \(\mathrm{CaCO}_{3}\) must decompose to achieve the equilibrium pressure of \(\mathrm{CO}_{2} ?\)

Step-by-Step Solution

Answer

0.148 moles of \( \mathrm{CaCO}_3 \) must decompose.

Step 1: Write the equilibrium expression

The reaction is \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g})\). For gases, the equilibrium constant \(K_p\) is defined as the ratio of partial pressures of the products to the reactants raised to the power of their stoichiometric coefficients. Since \(\mathrm{CaCO}_{3}\) and \(\mathrm{CaO}\) are solids, they do not appear in the expression. Therefore, \(K_p = P_{\mathrm{CO}_2}\).

Step 2: Relate equilibrium pressure with equilibrium constant

Given \(K_p = 3.87\) at \(1000^{\circ} \mathrm{C}\), this means that the equilibrium partial pressure of \(\mathrm{CO}_2\), \(P_{\mathrm{CO}_2}\), is 3.87 atm.

Step 3: Calculate moles of CO2 at equilibrium

The relationship of pressure and moles can be expressed with the ideal gas law: \( PV = nRT \). Solve for \( n \) (moles of \( \mathrm{CO}_2 \)) as follows:\[ n = \frac{P_{\mathrm{CO}_2} \cdot V}{RT} \]Substitute the known values: \( P_{\mathrm{CO}_2} = 3.87 \text{ atm}, V = 5.00 \text{ L}, R = 0.0821 \text{ L atm/mol K}, T = 1273 \text{ K} \) (\(1000^{\circ} \text{C} = 1273 \text{ K}\)).\[ n = \frac{3.87 \cdot 5.00}{0.0821 \cdot 1273} \approx 0.148 \text{ moles}.\]

Step 4: Conclude decomposition amount

Since 1 mole of \( \mathrm{CaCO}_3 \) produces 1 mole of \( \mathrm{CO}_2 \), 0.148 moles of \( \mathrm{CaCO}_3 \) are decomposed to achieve the equilibrium pressure.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that describes the behavior of gases under various conditions. It is expressed as \( PV = nRT \), where:

• \( P \) stands for pressure, measured in atmospheres (atm).
• \( V \) is the volume in liters (L).
• \( n \) is the number of moles of gas.
• \( R \) is the universal gas constant (0.0821 L atm/mol K).
• \( T \) is the temperature in Kelvin (K).

This law helps us understand how variables such as pressure, volume, and temperature are interrelated for a gas. In our exercise, the ideal gas law is pivotal for calculating the moles of carbon dioxide at equilibrium. By rearranging the formula to \( n = \frac{PV}{RT} \), we can find how many moles are present in the 5.00-liter flask at a pressure of 3.87 atm and a temperature of 1273 K. This aids in solving the problem of determining the amount of limestone decomposition needed to achieve equilibrium.

Equilibrium Constant

The equilibrium constant \( K_p \) is crucial in understanding how reactions reach equilibrium. It is a measurement of the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients.
For gaseous reactions, \( K_p \) specifically refers to the partial pressures of gases. In this particular decomposition reaction:

• The reaction is \( \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \).
• Since \( \mathrm{CaCO}_3 \) and \( \mathrm{CaO} \) are solids, they do not affect the \( K_p \) expression, which becomes \( K_p = P_{\mathrm{CO}_2} \).
• Given that \( K_p = 3.87 \), it means the partial pressure of \( \mathrm{CO}_2 \) is also 3.87 atm at equilibrium at 1000°C.

The \( K_p \) value not only helps identify equilibrium conditions but also guides calculations for pressure and concentrations, which are pivotal in predicting reactions.”

Decomposition Reaction

Decomposition reactions are a type of chemical reaction where one compound breaks down into two or more simpler substances. These reactions are essential in various scientific and industrial processes.
The decomposition of \( \mathrm{CaCO}_3 \) is our focus here. When limestone (\( \mathrm{CaCO}_3 \)) is heated, it breaks down into calcium oxide (\( \mathrm{CaO} \)) and carbon dioxide gas (\( \mathrm{CO}_2 \)). The balanced chemical equation for this reaction illustrates:

• Limestone \( \mathrm{CaCO}_3 \) (solid) decomposes into \( \mathrm{CaO} \) (solid) and \( \mathrm{CO}_2 \) (gas).
• This type of reaction is usually endothermic, meaning it requires heat input to proceed.
• The production of gas from solids affects the system’s total volume, impacting factors like pressure.

Understanding this reaction helps predict how heat and other conditions influence the decomposition process and resultant changes in pressure within a confined system.

Partial Pressure

Partial pressure is a concept used to describe the pressure contributed by each gas in a mixture. Each gas’s partial pressure contributes to the total pressure in a system.
For example, in the reaction \( \mathrm{CaCO}_3(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) \), only \( \mathrm{CO}_2 \) is in gaseous form.

• The partial pressure of \( \mathrm{CO}_2 \) is equal to the total pressure of the gas in this system because it is the only gaseous component.
• Under equilibrium, this partial pressure matches the equilibrium constant \( K_p \), known to be 3.87 atm from the problem statement.

In practical terms, understanding partial pressure enables chemists to calculate concentrations and predict behaviors of gases in reactions. In this exercise, it simplifies the approach to finding how much of a solid needs to decompose to reach a certain gas pressure.