Problem 56 - Chemistry and Chemical Reactivity

Question

limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=3.87 .\) If pure \(\mathrm{CaCO}_{3}\) is placed in a \(5.00-\mathrm{L}\). 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

18.62 grams of \( \mathrm{CaCO}_3 \) must decompose.

Step 1: Understand the Reaction

The given chemical equation is one of decomposition, where calcium carbonate (\( \CaCO_3 \)), when heated, breaks down into calcium oxide (\( \CaO \)) and carbon dioxide gas (\( \mathrm{CO}_2 \)). We are asked to find the amount of \( \CaCO_3 \) that must decompose to reach equilibrium under the given conditions.

Step 2: Identify Known Values

We are given that the equilibrium constant \( K_p \) is 3.87 at \(1000^{\circ}C\). We also know the volume of the flask is \(5.00\, \text{L}\). Since \( \CaCO_3 \) and \( \CaO \) are solids, they do not appear in the expression for \( K_p \). Only the pressure of \( \mathrm{CO}_2 \) is relevant for the equilibrium expression.

Step 3: Establish the Equilibrium Expression

For the decomposition reaction, at equilibrium, the expression for \( K_p \) is given by:\[K_p = P_{CO_2}\]where \(P_{CO_2}\) is the partial pressure of carbon dioxide in the flask.

Step 4: Calculate the Partial Pressure

Since \( K_p = 3.87 \) refers only to \( P_{CO_2} \), upon reaching equilibrium, the pressure of the carbon dioxide is: \\[P_{CO_2} = 3.87\, \text{atm}\]

Step 5: Use Gas Laws to Find Moles

We use the ideal gas law to find the moles of \( \mathrm{CO}_2 \) at equilibrium. The law is given by:\[ PV = nRT \]where \( P = 3.87\, \text{atm} \), \( V = 5.00\, \text{L} \), \( R = 0.0821\, \text{L}\, \text{atm}/(\text{mol}\, \text{K}) \), and \( T = 1273\, \text{K}(1000^{\circ}C).\) Solving for \(n\), the number of moles of \( \mathrm{CO_2} \): \[ n = \frac{PV}{RT} = \frac{3.87 \times 5.00}{0.0821 \times 1273} \approx 0.186 \text{mol}\]

Step 6: Determine Mass from Moles

Since each mole of \( \mathrm{CaCO}_3 \) produces one mole of \( \mathrm{CO}_2 \), 0.186 moles of \( \mathrm{CaCO}_3 \) must decompose. To find the mass of \( \mathrm{CaCO}_3 \), we multiply by its molar mass \((100.09\, \text{g/mol})\):\[ \text{Mass of } \mathrm{CaCO}_3 = 0.186 \times 100.09 \approx 18.62\, \text{g}\]

Key Concepts

Decomposition Reactions

Decomposition reactions involve breaking down a compound into simpler substances when energy is applied, such as heat. In the context of chemical equilibrium, the decomposition of calcium carbonate (\( \mathrm{CaCO}_3 \)) is an example. When heated, \( \mathrm{CaCO}_3 \) decomposes into calcium oxide (\( \mathrm{CaO} \)) and carbon dioxide gas (\( \mathrm{CO}_2 \)).In such reactions:

• The original substance breaks down into different substances.
• Energy, often in the form of heat, is crucial to initiate the reaction.

Here, the decomposition of calcium carbonate is essential in understanding how these types of reactions can be effectively utilized to achieve a desired state, such as reaching equilibrium in this example. The nature of the substances involved, especially in their solid and gaseous states, significantly impacts the equilibrium attained.

Equilibrium Constant

The equilibrium constant (\( K_p \)) is a crucial factor in chemical reactions, indicating the ratio of the concentration of products to reactants at equilibrium. In our decomposition reaction, only the gaseous product, carbon dioxide, affects the equilibrium constant because solids like \( \mathrm{CaCO}_3 \) and \( \mathrm{CaO} \) do not factor into the calculation.Equilibrium constants allow us to:

• Predict the degree to which a reaction will proceed.
• Understand whether the reactants or products are favored at equilibrium.

For our given decomposition, \( K_p \) is given as 3.87 at \( 1000^{\circ}C \), meaning that at equilibrium, the partial pressure of \( \mathrm{CO}_2 \) is significant, affecting how much of the \( \mathrm{CaCO}_3 \) must decompose to reach this state. This constant is fundamental for calculations involving gaseous equilibria.

Gas Laws

Gas laws are integral to understanding and calculating the behavior of gases, particularly when they are involved in chemical reactions and equilibrium. In this problem, the ideal gas law helps us determine the number of moles of \( \mathrm{CO}_2 \) produced at equilibrium.The ideal gas law is expressed as:\[ PV = nRT \]where \( P \) is the pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.By rearranging the formula, we find the relationship between these variables. Knowing three of them lets us solve for the fourth. Here, we calculate the moles of \( \mathrm{CO}_2 \) that develop from the decomposition reaction. By understanding gas laws, we can bridge the measurements that are often observed in lab conditions to theoretical predictions.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It is a fundamental aspect when solving for how much \( \mathrm{CaCO}_3 \) is needed to decompose into a specific amount of \( \mathrm{CO}_2 \).In our example, each mole of \( \mathrm{CaCO}_3 \) decomposes to produce exactly one mole of \( \mathrm{CO}_2 \). Therefore:

• Total moles of \( \mathrm{CO}_2 \) produced equals the moles of \( \mathrm{CaCO}_3 \) decomposed.
• The reaction involves a 1:1 mole ratio for \( \mathrm{CaCO}_3 \) to \( \mathrm{CO}_2 \).
• Stoichiometry allows us to use mole relationships to solve for the mass of reactants needed based on product demands.

This concept aids in systematically determining the amount of reactants that participate in reactions to form desired products.