Question

If \(5.15 \mathrm{~g}\) of \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a \(75.0-\mathrm{mL}\) tube filled with 101.3 \(\mathrm{kPa}\) 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 is 719.6 kPa.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the decomposition of silver oxide is: \[2 \mathrm{Ag}_2\mathrm{O} \rightarrow 4\mathrm{Ag} + \mathrm{O}_{2}\]

Calculate the number of moles of silver oxide and nitrogen gas

To calculate the number of moles, we need to use the given mass and molar mass for silver oxide and the Ideal Gas Law for nitrogen gas. The molar mass of silver oxide is: \( 2 * 107.87 \) g/mol (for silver) + 16.00 g/mol (for oxygen) = \( 231.74 \) g/mol Now let's calculate the moles of silver oxide: \[\text{moles of Ag}_2\text{O} = \frac{5.15 \;\text{g}}{231.74 \;\text{g/mol}} = 0.0222 \;\text{mol}\] For nitrogen gas, let's convert the temperature from Celsius to Kelvin: \(32{ }^{\circ}\text{C} = 273.15 + 32 = 305.15 \;\text{K}\) Now we can apply the Ideal Gas Law: PV = nRT Solving for n (moles) for nitrogen gas: \[ n = \frac{PV}{RT} = \frac{101.3 \; \text{kPa} * 75.0 \; \text{mL}}{8.314 \;\text{J} \cdot \text{mol}^{-1}\text{K}^{-1} * 305.15 \;\text{K}} \times \frac{1000 \;\text{mL}}{1 \; \text{L}} \times \frac{1 \; \text{Pa}}{10^3 \;\text{kPa}}\] \[ n \approx = 0.00296 \; \text{mol}\]

Calculate the number of moles of oxygen gas produced

From the balanced chemical equation, we see that 1 mole of oxygen gas is produced per 2 moles of silver oxide decomposed. So, \[\text{moles of O}_{2}\text{ produced} = 0.5 * \text{moles of Ag}_2\text{O} = 0.5 * 0.0222 \;\text{mol} = 0.0111 \; \text{mol}\]

Apply the Ideal Gas Law to calculate the final pressure of nitrogen and oxygen gas

Now, let's apply the Ideal Gas Law for both gases at the final temperature (320 °C = 593.15 K), considering the partial pressures: \[P_{\text{N}_2} V = n_{\text{N}_2} R T_{\text{final}}\] \[P_{\text{O}_2} V = n_{\text{O}_2} R T_{\text{final}}\] Solving for the final pressures of each gas: \[P_{\text{N}_2} = \frac{0.00296 \;\text{mol} * 8.314 \;\text{J/mol K} * 593.15 \;\text{K}}{75.0 \;\text{mL} * \frac{1000 \;\text{mL}}{1 \;\text{L}}} = \frac{1.463\times10^2 \; \text{Pa}}{10^3} = 146.3 \;\text{kPa}\] \[P_{\text{O}_2} = \frac{0.0111 \;\text{mol} * 8.314 \;\text{J/mol K} * 593.15 \;\text{K}}{75.0 \;\text{mL} * \frac{1000 \;\text{mL}}{1 \;\text{L}}} = \frac{5.733\times10^2 \; \text{Pa}}{10^3} = 573.3 \;\text{kPa}\]

Calculate the total pressure inside the tube

The total pressure is the sum of the partial pressures of the two gases (nitrogen and oxygen): \[P_{\text{total}} = P_{\text{N}_2} + P_{\text{O}_2} = 146.3 \;\text{kPa} + 573.3 \;\text{kPa} = 719.6 \;\text{kPa}\] So, the total pressure inside the tube after heating is 719.6 kPa.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and amount of moles of a gas. It is expressed as \(PV = nRT\), where:

• \(P\) is the pressure in units of Pa (Pascals) or kPa (kilopascals).
• \(V\) is the volume of the gas in liters.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.

The law assumes that the gas particles are ideal, meaning they do not interact with each other except through elastic collisions and occupy a negligible volume compared to the container they are in.
By applying this law, one can determine the unknown quantity if the other three are known. For instance, in our nitrogen gas example, we used the given pressure, volume, and temperature to find the moles of nitrogen gas present initially inside the tube.

Chemical Decomposition

Chemical decomposition, or analysis, refers to the breakdown of a single compound into two or more components or elements.
In the given exercise, decomposition occurs when silver oxide (\(\mathrm{Ag}_2\mathrm{O}\)) is heated and breaks apart to form silver and oxygen gas:

• Original compound: \(\mathrm{Ag}_2\mathrm{O}\)
• Products: \(\mathrm{Ag}\) (Silver) and \(\mathrm{O}_2\) (Oxygen gas)

The balanced chemical equation for this process is \(2 \mathrm{Ag}_2\mathrm{O} \rightarrow 4\mathrm{Ag} + \mathrm{O}_2\).
Decomposition reactions are crucial in understanding how compounds transform, especially under heat or light. This kind of chemical reaction is common in areas such as metallurgy, where ores are broken down into metal components.

Pressure Calculation

Pressure calculation involves finding out how much force gas molecules exert per unit area of the container they occupy. In the scenario provided, we needed to calculate the pressure for both nitrogen and oxygen gases after heating.
The Ideal Gas Law \(PV=nRT\) is applied to find partial pressures. As temperature increases, according to the law, pressure also increases if volume remains constant.
To calculate the total pressure within the tube:

• The partial pressure of nitrogen \(P_{\text{N}_2}\) was calculated based on its original amount of moles and increased temperature.
• The partial pressure of oxygen \(P_{\text{O}_2}\) was determined from the moles of oxygen produced during decomposition.
• Both partial pressures were then summed to find the total pressure: \(P_{\text{total}} = P_{\text{N}_2} + P_{\text{O}_2}\).

This systematic approach allows us to understand how individual gas components influence the overall pressure in any gaseous system.

Mole Calculations

Moles are a measure of the quantity of a substance and are crucial in stoichiometry and understanding chemical reactions. The concept of moles helps us relate mass to the amount of substance.
In the exercise, we calculated moles for both silver oxide and nitrogen to define how many particles are present.

• To find moles of silver oxide, its mass was divided by its molar mass \(\frac{5.15 \, \text{g}}{231.74 \, \text{g/mol}}\), yielding \(0.0222\) moles.
• The moles of nitrogen were calculated using the Ideal Gas Law, considering the provided conditions.
• Moles of oxygen produced were derived from the balanced chemical equation; for every 2 moles of \(\mathrm{Ag}_2\mathrm{O}\), \(1\) mole of \(\mathrm{O}_2\) is liberated."\(0.5 \times 0.0222 = 0.0111\) mole of \(\mathrm{O}_2\) was formed.

Knowing the moles allows us to better manage and foresee the amounts of both reactants and products in any chemical reaction scenario.