Question

If \(5.15 \mathrm{~g}\) of \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a 75.0-mL tube filled with 760 torr of \(\mathrm{N}_{2}\) gas at \(32^{\circ} \mathrm{C}\), and the tube is heated to \(320^{\circ} \mathrm{C}\), the \(\mathrm{Ag}_{2} \mathrm{O}\) decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

Short answer

The total pressure inside the tube after heating to \(320^{\circ} \mathrm{C}\) and the reaction takes place is approximately 6888.26 torr.

Step-by-step solution

Find the number of moles of N₂ and Ag₂O

We are given that the tube contains 5.15 g of Ag₂O and is filled with 760 torr of N₂ gas at 32 °C. At these conditions, let's first use the Ideal Gas Law to find the number of moles of N₂ gas: PV = nRT Rearranging the equation to solve for n: n = PV/RT We are given P = 760 torr, V = 75.0 mL (convert to L by dividing by 1000), T = 32 °C (convert to Kelvin by adding 273.15). R (the ideal gas constant) = 62.36 (L·torr)/(K·mol) Substitute the values and solve for n: n(N₂) = (760 torr × 0.075 L) / (62.36 (L·torr)/(K·mol) × (32 + 273.15) K) n(N₂) ≈ 0.00303 mol Now, let's find the number of moles of Ag₂O: Use the molar mass of Ag₂O: 1 mol of Ag₂O = 231.74 g/mol n(Ag₂O) = mass / molar mass n(Ag₂O) = 5.15 g / 231.74 g/mol n(Ag₂O) ≈ 0.0222 mol

Find the number of moles of O₂ produced

When Ag₂O decomposes, it forms oxygen gas and silver according to the balanced chemical equation: 2 Ag₂O → 4 Ag + O₂ From the stoichiometry of the reaction, 2 moles of Ag₂O produces 1 mole of O₂: n(O₂) = n(Ag₂O) * (1 mol O₂ / 2 mol Ag₂O) n(O₂) = 0.0222 mol * (1 mol O₂ / 2 mol Ag₂O) n(O₂) ≈ 0.0111 mol

Find the partial pressures of N₂ and O₂

Using the Ideal Gas Law (PV = nRT), we can find the partial pressures of N₂ and O₂ at the final temperature (320 °C): The volume of the tube remains constant at 0.075 L. Convert 320 °C to Kelvin by adding 273.15, T = 593.15 K. For N₂: P(N₂) = n(N₂)RT / V P(N₂) = (0.00303 mol)(62.36 (L·torr)/(K·mol))(593.15 K) / 0.075 L P(N₂) ≈ 1461.27 torr For O₂: P(O₂) = n(O₂)RT / V P(O₂) = (0.0111 mol)(62.36 (L·torr)/(K·mol))(593.15 K) / 0.075 L P(O₂) ≈ 5426.99 torr

Find the total pressure inside the tube

In order to calculate the total pressure inside the tube, we add the partial pressures of N₂ and O₂: Total pressure = P(N₂) + P(O₂) Total pressure = 1461.27 torr + 5426.99 torr Total pressure ≈ 6888.26 torr Hence, the total pressure inside the tube after heating to 320 °C and the reaction takes place is approximately 6888.26 torr.

Key concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the quantitative relationships in a chemical reaction. It is the method we use to calculate the amounts of reactants and products in a chemical reaction. In the given exercise, stoichiometry helps us find out how much oxygen (\(\text{O}_2\)) is produced from the decomposition of silver oxide (\(\text{Ag}_2\text{O}\)).

The balanced chemical equation is crucial here: 2 Ag₂O → 4 Ag + O₂. This equation tells us that 2 moles of \(\text{Ag}_2\text{O}\) produce 1 mole of \(\text{O}_2\). By using this relationship, we can convert the moles of \(\text{Ag}_2\text{O}\) to moles of \(\text{O}_2\). This step is essential in determining how much oxygen gas is formed in the reaction.

In this problem, stoichiometry underscores the importance of the mole ratio. It helps us link the initial amount of silver oxide to the resulting amount of oxygen gas. Understanding stoichiometry makes it possible to predict how much product will form, which is ideal for computations involving gas laws and pressure changes.

Partial Pressure

The concept of partial pressure is central to solving this problem. Partial pressure refers to the pressure that each gas in a mixture would exert if it were alone in a container.

In our exercise, we have two gases after the decomposition reaction: nitrogen (\(\text{N}_2\)) and oxygen (\(\text{O}_2\)) formed during the reaction. Each of these gases contributes to the total pressure inside the tube.

To find the partial pressure of a gas, the Ideal Gas Law is used: \( PV = nRT \). By solving this for pressure (\(P\)), we get \( P = \frac{nRT}{V} \). This formula helps us calculate how much pressure each gas is exerting individually based on its mole quantity, temperature, and volume of the container. Once we have the partial pressures of \(\text{N}_2\) and \(\text{O}_2\), we simply add them together to get the total pressure inside the tube.

Recognizing how to calculate partial pressures using the Ideal Gas Law is a useful skill. It helps understand the contribution of each gas in a mixture to the total pressure.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more elements or simpler compounds. In the context of this exercise, the decomposition of silver oxide (\(\text{Ag}_2\text{O}\)) into silver (\(\text{Ag}\)) and oxygen gas (\(\text{O}_2\)) is examined.

The reaction is represented as follows: 2 Ag₂O → 4 Ag + O₂. This shows that silver oxide breaks down when heated, producing silver and releasing oxygen gas.

Decomposition reactions are important because they provide products, sometimes in a gaseous state, that can change the pressure inside a container like in this problem. Understanding the nature of the decomposition helps predict the products and subsequent changes in other factors like pressure or temperature in a closed system.

These reactions are common in chemistry, and knowing how to identify and analyze them is essential for predicting changes in physical and chemical properties.

Gas Constant

The gas constant (\(R\)) is a crucial part of the Ideal Gas Law, \( PV = nRT \). It serves as a bridge between different units and values when dealing with gases. In our problem, we use \(R = 62.36 \text{ L}\cdot\text{torr}/(\text{K}\cdot\text{mol}) \) to match the given pressure in torr.

The gas constant allows us to relate pressure, volume, and temperature to the number of moles of gas in a system. Since \(R\) is a universal constant, it provides a consistent value to use in calculations regardless of the specific gas involved.

The choice of \(R\) is important. It should align with the units of the other variables: if the pressure is in torr and the volume in liters, the \(R\) should reflect those units to maintain consistency. A correct understanding of the gas constant and its use is fundamental when solving problems using the Ideal Gas Law, ensuring that all units fit together like pieces of a puzzle.