Problem 133 - Chemistry: An Atoms-Focused Approach

Question

Denitrification in the Environment In some aquatic ecosystems, nitrate \(\left(\mathrm{NO}_{3}^{-}\right)\) is converted to nitrite \(\left(\mathrm{NO}_{2}^{-}\right),\) which then decomposes to nitrogen and water. As an example of this second reaction, consider the decomposition of ammonium nitrite: $$\mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ What is the change in pressure in a sealed 10.0 L. vessel due to the formation of \(\mathrm{N}_{2}\) gas when the ammonium nitrite in \(1.00 \mathrm{L}\) of \(1.0 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{2}\) decomposes at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The change in pressure in the sealed vessel is 2.45 atm.

Step 1: Convert the given concentration and volume to moles of ammonium nitrite

We are given 1.00 L of 1.0 M ammonium nitrite. To find the moles of ammonium nitrite, we simply multiply the volume by its concentration: Moles of \(\mathrm{NH}_{4} \mathrm{NO}_{2} = 1.0 \mathrm{M} \times 1.00 \mathrm{L} = 1.00 \, \text{moles}\)

Step 2: Use stoichiometry to determine moles of \(\mathrm{N}_{2}\) formed

Now, we use stoichiometry to determine the moles of nitrogen gas, \(\mathrm{N}_{2}\), formed during the decomposition. The balanced equation shows that one mole of ammonium nitrite forms one mole of nitrogen gas: \(\mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) So the moles of nitrogen gas formed are the same as the moles of ammonium nitrite: Moles of \(\mathrm{N}_{2} = 1.00 \, \text{moles}\)

Step 3: Calculate the pressure of nitrogen gas using the ideal gas law

We use the ideal gas law, \(PV = nRT\), to calculate the pressure of nitrogen gas in the vessel. First, convert the temperature to Kelvin: \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \, \mathrm{K}\) Rearrange the ideal gas law to solve for pressure: \(P = \frac{nRT}{V}\) Substitute the values for n (1.00 mole), R (\(0.0821 \, \mathrm{L \cdot atm} / \mathrm{mol \cdot K})\), T (298.15 K), and V (10.0 L): \(P_\mathrm{N_2} = \frac{(1.00\, \text{moles})(0.0821\, \mathrm{L \cdot atm} / \mathrm{mol \cdot K})(298.15\, \mathrm{K})}{10.0\, \mathrm{L}} = 2.45\, \mathrm{atm}\)

Step 4: Calculate the change in pressure due to the formation of nitrogen gas

Since the system is sealed, the only pressure change comes from the formation of nitrogen gas. Therefore, the change in pressure is equal to the pressure of nitrogen gas: \(\Delta P = P_\mathrm{N_2} = 2.45 \, \mathrm{atm}\) So, the change in pressure in the sealed 10.0 L vessel due to the formation of \(\mathrm{N}_{2}\) gas when the ammonium nitrite decomposes at \(25^{\circ} \mathrm{C}\) is 2.45 atm.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced. In our exercise, stoichiometry helps us determine how much nitrogen gas \(\mathrm{N}_{2}\) is formed when ammonium nitrite decomposes.To use stoichiometry effectively, follow these steps:

• Start by identifying the balanced chemical equation for the reaction. For ammonium nitrite, it decomposes as \(\mathrm{NH}_{4} \mathrm{NO}_{2}(aq) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\).
• Determine the ratios between the moles of each substance. Here, the equation tells us that one mole of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) produces one mole of \(\mathrm{N}_{2}\).
• Calculate the amount of product (\(\mathrm{N}_{2}\)) based on the initial amount of reactant (ammonium nitrite). Since we have 1.00 mole of ammonium nitrite, we will also have 1.00 mole of nitrogen gas produced.

By using stoichiometry, we ensure the quantities are correct and consistent with the conservation of mass in chemical reactions.
Understanding the stoichiometric relationships provides clarity on how substances transform during reactions.

Chemical Decomposition

Chemical decomposition is the process where a single compound breaks down into simpler substances. In the context of the exercise, the decomposition of ammonium nitrite (\(\mathrm{NH}_{4}\mathrm{NO}_{2}\)) serves as a classic example.Here’s what you need to know about chemical decomposition:

• Decomposition reactions involve the breaking of chemical bonds within a compound. Energy is often released or absorbed during this process.
• In our case, ammonium nitrite decomposes to form nitrogen gas \(\mathrm{N}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\). This transformation is reflected in the balanced equation: \(\mathrm{NH}_{4} \mathrm{NO}_{2}(aq) \rightarrow \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\).
• This reaction is crucial in demonstrating the conversion of a compound into gaseous and liquid products, providing insight into the behavior of similar reactions in ecological and environmental contexts.

Recognizing decomposition reactions helps you understand how complex materials break down in nature and industrial processes.

Gas Pressure Calculation

Gas pressure calculation is an essential aspect of studying gases, especially when dealing with the products of reactions like the decomposition of ammonium nitrite.Here’s how you calculate gas pressure using the Ideal Gas Law:

• The Ideal Gas Law is expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the universal gas constant \(0.0821 \, \mathrm{L \cdot atm} / \mathrm{mol \cdot K}\), and \(T\) is temperature in Kelvin.
• To find the pressure \(P\) of \(\mathrm{N}_{2}\) formed, rearrange the formula to \(P = \frac{nRT}{V}\).
• In this example, calculate the pressure formed by substituting the values you already have: \(1.00\) mole of \(\mathrm{N}_{2}\), temperature \(298.15 \, \mathrm{K}\), volume \(10.0 \, \mathrm{L}\), and \(R\).
• This gives a calculated pressure of \(2.45 \, \mathrm{atm}\).

Understanding gas pressure calculations allows you to predict how gases will behave under different conditions, a crucial skill in both laboratory settings and real-world applications.
These calculations provide insights into how various environmental and chemical processes occur.