Ch.10 - Gases: Their Properties & Behavior

Problem 143a

Chapter 10, Problem 143a

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 K.

Verified step by step guidance

Identify the chemical reaction: \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \). Note that only \( \text{CO}_2 \) is a gas and will contribute to the pressure.

Calculate the number of moles of \( \text{CaCO}_3 \) using its molar mass: \( n = \frac{\text{mass}}{\text{molar mass}} \).

Since \( \text{CaCO}_3 \) decomposes completely, the moles of \( \text{CO}_2 \) produced will be equal to the moles of \( \text{CaCO}_3 \) initially present.

Use the ideal gas law \( PV = nRT \) to find the pressure \( P \). Rearrange to \( P = \frac{nRT}{V} \), where \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), \( T \) is the temperature in Kelvin, and \( V \) is the volume in liters.

Substitute the values for \( n \), \( R \), \( T \), and \( V \) into the equation to calculate the pressure inside the container.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law allows us to predict the behavior of gases under various conditions, making it essential for calculations involving gas reactions.

Stoichiometry of Decomposition Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. In the decomposition of calcium carbonate (CaCO3), it breaks down into calcium oxide (CaO) and carbon dioxide (CO2). Understanding the stoichiometry of this reaction is crucial for determining the amount of gas produced, which directly influences the pressure in the container after heating.

Gas Behavior at High Temperatures

At high temperatures, gases tend to behave more ideally, meaning they follow the Ideal Gas Law more closely. The kinetic energy of gas molecules increases with temperature, leading to greater molecular motion and, consequently, higher pressure when confined in a fixed volume. This concept is important for understanding how the pressure inside the container will change as the temperature rises to 1500 K.

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Related Practice

Textbook Question

When 10.0 g of a mixture of Ca1ClO322 and Ca1ClO22 is heated to 700 °C in a 10.0-L vessel, both compounds decompose, forming O21g2 and CaCl21s2. The final pressure inside the vessel is 1.00 atm. (b) What is the mass of each compound in the original mixture?

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Textbook Question

When 10.0 g of a mixture of Ca(ClO₃)₂ and Ca(ClO)₂ is heated to 700 °C in a 10.0-L vessel, both compounds decompose, forming O₂(g) and CaCl₂(s). The final pressure inside the vessel is 1.00 atm. (a) Write balanced equations for the decomposition reactions.

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Textbook Question

A 5.00-L vessel contains 25.0 g of PCl3 and 3.00 g of O2 at 15 °C. The vessel is heated to 200.0 °C, and the contents react to give POCl3. What is the final pressure in the vessel, assuming that the reaction goes to completion and that all reactants and products are in the gas phase?

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Textbook Question

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (b) Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density = 3.34 g/mL) in the container, and use the van der Waals equation to calculate the pressure. The van der Waals constants for CO₂(g) are a = 3.59 (L²-atm)/mol² and b = 0.0427 L/mol.

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Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ

(a) What mass of CO₂(g) is produced in the reaction?

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