Question

Ammonium nitrate can be used as an effective explosive because it decomposes into a large number of gaseous products. At a sufficiently high temperature, ammonium nitrate decomposes into nitrogen, oxygen, and steam. (a) Write a balanced net ionic equation for the decomposition of ammonium nitrate. (b) If 2.00 kg of ammonium nitrate are sealed in a 50.0-L steel drum and heated to 745°C, what is the resulting pressure in the drum after decomposition? (Assume 100% decomposition.) 40\. Acetone peroxide, C9H

Short answer

Answer: The balanced net ionic equation for the decomposition of ammonium nitrate is 2 NH4NO3 (s) -> 2 N2 (g) + O2 (g) + 4 H2O (g). The resulting pressure in the steel drum after decomposition is approximately 183,816 Pa (or 183.8 kPa).

Step-by-step solution

Part (a): Balancing the net ionic equation

To balance the net ionic equation, we need to consider the decomposition of ammonium nitrate into nitrogen, oxygen, and steam (water vapor). The formula of ammonium nitrate is NH4NO3. Decomposition produces N2 (nitrogen gas), O2 (oxygen gas), and H2O (water vapor). The balanced net ionic equation is: 2 NH4NO3 (s) -> 2 N2 (g) + O2 (g) + 4 H2O (g)

Part (b): Calculating the resulting pressure

We are given 2.00 kg of ammonium nitrate sealed in a 50.0 L steel drum and heated to 745°C. We must first convert the mass to moles: Molar mass of NH4NO3 = (14.01 * 1 + 1.01 * 4) + (14.01 * 1 + 16.00 * 3) Molar mass of NH4NO3 = 80.05 g/mol 2.00 kg = 2000 g n (moles) = 2000 g / 80.05 g/mol ≈ 25.0 mol Now we need to determine the moles of gas produced from decomposition: From the balanced equation, 2 mol NH4NO3 -> 2 mol N2, 1 mol O2, and 4 mol H2O For 25.0 mol NH4NO3, we get 25.0 mol N2, 12.5 mol O2, and 50.0 mol H2O Total moles of gas = 25.0 + 12.5 + 50.0 = 87.5 mol Now we convert the temperature from Celsius to Kelvin by adding 273.15: 745°C + 273.15 = 1018.15 K Finally, we apply the ideal gas law to calculate the pressure: PV = nRT P = nRT / V R (ideal gas constant) = 8.314 J/(mol K) P = (87.5 mol * 8.314 J/(mol K) * 1018.15 K) / 50.0 L P = 183816.09 Pa So, the resulting pressure in the drum after decomposition is approximately 183,816 Pa (or 183.8 kPa).

Key concepts

Balanced Net Ionic Equation

Understanding the balanced net ionic equation is pivotal in chemistry as it helps reveal the substances that are actively participating in a chemical reaction. In the decomposition of ammonium nitrate (NH_4NO_3), the balanced net ionic equation is a straightforward representation of what happens at the molecular level. Through this equation, students can see that two molecules of ammonium nitrate decompose to form two molecules of nitrogen gas (N_2), one molecule of oxygen gas (O_2), and four molecules of steam or water vapor (H_2O). It's essential to note that the balancing of molecules on both sides of the equation follows the law of conservation of mass, meaning every atom on the reactant side appears on the product side. In simple terms, what goes into the reaction must come out, but in different forms. This equation simplifies complex processes to help students visualize the change and reinforces the understanding of stoichiometric coefficients in chemical reactions.

For students looking to grasp the concept better, breaking down the original compound into its ionic constituents and then demonstrating how these recombine into new products can be instructive. It emphasizes the fundamental rule that the charge must also be balanced, just like the atoms. Remember, while writing net ionic equations, ions or molecules that do not participate directly in the reaction, known as spectator ions, are omitted for clarity and simplicity.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the case of ammonium nitrate (NH_4NO_3), the decomposition process is a key focus, particularly due to its implications in industrial and safety applications. When heated to a sufficiently high temperature, ammonium nitrate breaks down into nitrogen, oxygen, and water vapor. This specific type of decomposition is known as thermal decomposition, a reaction that is initiated by heat. Learning about this process gives students insight into the behavior of compounds under temperature changes and the energy required to break chemical bonds.

Thermal decomposition reactions are intriguing due to the variety of products they can produce and their real-world applications. For instance, ammonium nitrate's decomposition is exothermic and releases a large volume of gases, a principle utilized in explosives and fertilizers. When students study these reactions, they should pay attention to the states of matter—for example, recognizing that water is in its gaseous form as steam in this scenario. The theoretical understanding gained from these concepts enables students to anticipate the outcomes of reactions and apply this knowledge to predict the behavior of compounds in various conditions.

Ideal Gas Law Calculations

The ideal gas law, represented as (PV = nRT), is an equation of state that provides a close approximation to the behavior of many gases under various conditions. It is a cornerstone of gas law calculations, helping students relate the quantities of pressure (P), volume (V), number of moles (n), and temperature (T) for an ideal gas with the universal gas constant (R). By rearranging the formula, students can compute any one of the variables as long as the others are known. This proves incredibly valuable when dealing with problems involving changes in gas conditions, such as in the given exercise with ammonium nitrate decomposing in a sealed drum.

It is vital to ensure that the units used match the units for the gas constant (R = 8.314 J/(mol·K)), which means converting temperature to Kelvin and mass to moles. Understanding these conversions is crucial for students, as missteps here are common sources of errors. Another key point is recognizing that the ideal gas law assumes gas molecules do not interact with each other and occupy no space—assumptions that simplify calculations but can stray from reality at high pressures or low temperatures. Therefore, although this law is a powerful tool in making predictions about gaseous behavior, students should learn about its limitations and the conditions under which it best applies.