Question

If 5.15 gof \(\mathrm{Ag}_{2} \mathrm{O}\) 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 the decomposition of \({Ag_2O}\) is 8.127 atm.

Step-by-step solution

Calculate moles of \({Ag_2O}\)

Given mass of \({Ag_2O} = 5.15 g\) Molar mass of \({Ag_2O} = 2(107.87) + 16 = 231.74g/mol\) To find moles of \({Ag_2O}\), use the following formula: \(n(Ag_2O) = \frac{mass}{molar mass}\) \(n(Ag_2O) = \frac{5.15 g}{231.74 g/mol} = 0.0222 mol\)

Calculate moles of \({O_2}\) produced

The balanced chemical equation of the decomposition of \({Ag_2O}\) is \({2Ag_2O} \rightarrow {4Ag} + {O_2}\) According to this equation, each mole of \({O_2}\) is produced from 2 moles of \({Ag_2O}\). To find moles of \({O_2}\) produced, use the following formula: \(n(O_2) = \frac{n(Ag_2O)} {2}\) \(n(O_2) = \frac{0.0222 mol}{2} = 0.0111 mol\)

Calculate partial pressure of \({O_2}\) produced

Since we know the initial pressure and temperature, we can use the ideal gas law to find the initial number of moles of \({N_2}\) in the tube. The ideal gas law is given by: \(PV = nRT\) Where: \(P = pressure\) \(V = volume\) \(n = moles\) \(R = ideal gas constant = 0.0821 \frac{L \cdot atm}{mol \cdot K}\) \(T = temperature\) Solving for initial moles of \({N_2}\): \(n(N_2) = \frac{PV} {RT}\) Before we substitute the values, we should convert torr to atm and Celsius to Kelvin: \(32^\circ C = 32 + 273.15 = 305.15K\) \(760\,torr = \frac{760}{760}\,atm = 1\,atm\) Now substitute the values: \(n(N_2) = \frac{1\,atm \cdot 75.0\,mL} {0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 305.15K}\) Since mL and L are not the same unit, we should convert 75.0 mL to L: \(75.0 mL = 0.075 L\) \(n(N_2) = \frac{1\,atm \cdot 0.075LD}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 305.15K} = 0.00306\,mol\) Now, let's find the partial pressure of \({O_2}\) using the ideal gas law at the final temperature of \(320^\circ C\): \(320^\circ C = 320 + 273.15 = 593.15K\) \(P(O_2) = \frac{n(O_2)RT}{V}\) \(P(O_2) = \frac{0.0111\,mol \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 593.15K} {0.075 L} = 6.182\,atm\)

Calculate final pressure of \({N_2}\)

The final pressure of \({N_2}\) can be found with the combined gas law: \(P_{1}V_{1}{/}T_{1} = P_{2}V_{2}{/}T_{2}\) Since the volume remains constant: \(P_{1}{/}T_{1} = P_{2}{/}T_{2}\) Solve for final pressure of \({N_2}\) : \(P_{2}(N_2) = P_{1}(N_2) \frac{T_{2}} {T_{1}}\) \(P_{2}(N_2) = 1\,atm \cdot \frac{593.15K}{305.15K} = 1.945\,atm\)

Calculate total pressure

The total pressure inside the tube is the sum of the final pressures of \({O_2}\) and \({N_2}\). \(P_{total} = P(O_2) + P_{2}(N_2)\) \(P_{total} = 6.182\,atm + 1.945\,atm = 8.127\,atm\) So, the total pressure inside the tube after the decomposition of \({Ag_2O}\) is 8.127 atm.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that connects four essential properties of gases: pressure (P), volume (V), the number of moles (n), and temperature (T). This relationship is expressed by the equation \( PV = nRT \), where \( R \) is the ideal gas constant. This law assumes that gas molecules do not attract or repel each other and occupy no space—ideal conditions for gases.
Let's break it down:

• **Pressure (P):** The force exerted by gas molecules colliding with the walls of their container, usually measured in atm, torr, or pascal.
• **Volume (V):** The space occupied by the gas, typically in liters (L) or milliliters (mL).
• **Moles (n):** The quantity of gas present, denoted in moles.
• **Temperature (T):** A measure of the thermal energy of gas particles, always in Kelvin for gas law calculations.
• **R (Ideal Gas Constant):** With a value of 0.0821 L·atm/mol·K, \( R \) acts as a proportionality constant.

The Ideal Gas Law is a powerful tool used to solve for any one of these parameters when the others are known. It's crucial in our exercise to determine partial pressures and final total pressures.

Partial Pressure

Partial Pressure refers to the pressure exerted by a single type of gas in a mixture of gases. When gases are mixed, each one contributes to the total pressure based on its proportion in the mixture. This concept is essential when analyzing gas reactions or decompositions in closed systems.
In Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of each component gas. Mathematically, it is represented as:\[ P_{total} = P_1 + P_2 + P_3 + ... + P_n \]where each \( P_n \) is the partial pressure of a component gas.
In this exercise, we are interested in finding the partial pressures of oxygen \((O_2)\) and nitrogen \((N_2)\). First, we calculated the pressure exerted solely by the oxygen produced during the decomposition using the known moles of \( O_2 \) and the volume of the container. Then, we combined this information with the known behavior of \( N_2 \) to arrive at the total pressure.

Chemical Decomposition

Chemical Decomposition is a reaction in which a single compound breaks down into two or more simpler substances. In the exercise, we observe the decomposition of silver oxide \((Ag_2O)\). This is represented by the equation:\[ 2Ag_2O \rightarrow 4Ag + O_2 \]Key points about this process include:

• **Stoichiometry:** Each mole of \( Ag_2O \) produces a specific amount of \( O_2 \) and \( Ag \) according to the balanced reaction.
• **Products:** The reaction results in the formation of metallic silver \((Ag)\) and oxygen gas \((O_2)\).
• **Conditions:** Decomposition often needs specific conditions like heat; in this problem, the reaction occurs when heating the substance to \(320^{\circ}C\).

This decomposition is crucial because it produces \( O_2 \), which contributes to the change in pressure inside the tube, making it essential to calculate the total final pressure.

Moles Calculation

In chemistry, calculating the number of moles of a substance is foundational in understanding reaction quantities. Moles provide a way to quantify the number of particles in a given mass of a substance, using Avogadro's number. The formula is:\[ n = \frac{mass}{molar\ mass} \]Here's how it's applied in the context of the exercise:

• **Given Mass:** We have a known mass of \( Ag_2O \), which is 5.15 grams.
• **Molar Mass:** The molar mass is calculated by summing the atomic masses of silver and oxygen in \( Ag_2O \), resulting in 231.74 g/mol.
• **Calculation:** Moles of \( Ag_2O \) are found by dividing its mass by the molar mass. This gives us the starting point for further calculations of gas production.

Calculating moles is essential for predicting the outcomes of reactions like the decomposition we observe here, allowing us to determine how much \( O_2 \) is produced and its effect on pressure.