Question

The value of \(\Delta U\) for the decomposition of \(7.647 \mathrm{g}\) of ammonium nitrate can be measured in a bomb calorimeter. The reaction that occurs is \\[ \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\] The temperature of the calorimeter, which contains \(415 \mathrm{g}\) of water, increases from \(18.90^{\circ} \mathrm{C}\) to \(20.72^{\circ} \mathrm{C}\) The heat capacity of the bomb is \(155 \mathrm{J} / \mathrm{K}\). What is the value of \(\Delta U\) for this reaction, in \(\mathrm{kJ} / \mathrm{mol}\) ? (IMAGE CANNOT COPY)

Short answer

The value of \(\Delta U\) for the reaction is approximately \(36.17\,\mathrm{kJ/mol}\).

Step-by-step solution

Calculating Total Heat Absorbed

First, we calculate the total heat absorbed using the change in temperature. The total heat absorbed by the calorimeter is the sum of the heat absorbed by the water and the bomb. For water: \[ q_{ ext{water}} = m imes c imes \Delta T \]where \( m = 415 \text{ g} \), \( c = 4.184 \text{ J/g}\cdot\text{K} \), and \( \Delta T = 20.72 - 18.90 \text{ C} = 1.82 \text{ K} \).For the bomb:\[ q_{ ext{bomb}} = C_b \times \Delta T \]where \( C_b = 155 \text{ J/K} \) and \( \Delta T = 1.82 \text{ K} \).Calculating both:\[ q_{ ext{water}} = 415 \times 4.184 \times 1.82 = 3173.232 \text{ J} \]\[ q_{ ext{bomb}} = 155 \times 1.82 = 282.1 \text{ J} \]Total heat, \( q_{total} = q_{ ext{water}} + q_{ ext{bomb}} = 3173.232 + 282.1 = 3455.332 \text{ J} \).

Calculating Moles of Ammonium Nitrate

The chemical reaction involves the decomposition of ammonium nitrate, \(\text{NH}_4\text{NO}_3\). We need to determine the number of moles of ammonium nitrate decomposed.The molar mass of \(\text{NH}_4\text{NO}_3\) is calculated as:- Nitrogen (N): 14.01 g/mol \( \times 2 \)- Hydrogen (H): 1.01 g/mol \( \times 4 \)- Oxygen (O): 16.00 g/mol \( \times 3 \)\[ Molar \space mass = 2(14.01) + 4(1.01) + 3(16.00) = 80.04 \text{ g/mol} \]Calculate moles:\[ n = \frac{7.647}{80.04} \approx 0.0955 \text{ moles} \]

Calculating Internal Energy Change per Mole

The change in internal energy \(\Delta U\) per mole of reaction can be calculated as follows:Convert the total heat from joules to kilojoules:\[ q_{total} = 3455.332 \text{ J} = 3.455332 \text{ kJ} \]Change in internal energy per mole is computed by dividing total heat absorbed by moles of ammonium nitrate:\[ \Delta U = \frac{q_{total}}{n} \]\[ \Delta U = \frac{3.455332}{0.0955} = 36.17 \text{ kJ/mol} \].

Key concepts

Calorimetry

Calorimetry is a technique used to measure the heat involved in chemical reactions or physical changes. In the context of this exercise, we use it to find out how much heat is absorbed during the decomposition of ammonium nitrate in a bomb calorimeter.
Calorimeters are insulated devices that capture all the energy exchanged during a reaction, ensuring no heat is lost to the surroundings. Here's how it works in our example:

• The bomb calorimeter contains the ammonium nitrate and is surrounded by water.
• When the reaction takes place, heat is absorbed by both the water and the calorimeter's bomb.
• This heat causes a temperature change, which can be measured easily.

By knowing the heat capacity of the bomb and the specific heat of water, we can easily calculate the total heat absorbed. This total heat lets us dive deeper into understanding the reaction's energetics.

Internal Energy

Internal energy, \(\Delta U\), represents the total energy change within a system during a chemical reaction. It considers all forms of energy transfer, including heat and work. In a bomb calorimeter, the primary focus is on heat transfer, as the system is mostly isolated from doing work.
In this problem, the internal energy change is determined through the heat absorbed by the calorimeter:

• We calculate the total heat absorbed using the formula: \( q_{total} = q_{\text{water}} + q_{\text{bomb}} \).
• The units of heat obtained in joules are converted to kilojoules.
• The internal energy change per mole is found by dividing the total heat by the moles of ammonium nitrate involved.

By understanding \(\Delta U\), you gain insight into how much energy is consumed or released in the reaction, providing a crucial understanding of the reaction's behavior and viability.

Ammonium Nitrate Decomposition

Ammonium nitrate decomposition is an interesting chemical reaction where solid ammonium nitrate breaks down into nitrous oxide and water vapor. This reaction is significant due to its thermal properties, often releasing a considerable amount of energy.

• The balanced chemical equation given is: \(\text{NH}_4\text{NO}_3 (s) \rightarrow \text{N}_2\text{O} (g) + 2\text{H}_2\text{O} (g)\).
• This equation illustrates the conversion of one mole of ammonium nitrate into gas products.
• Understanding the molar mass helps calculate how much ammonium nitrate is used in the reaction.

Knowing the stoichiometry and changes in products lets us accurately measure the energy changes. This information is essential for applications like designing safe storage and handling processes for ammonium nitrate. By studying the decomposition closely, chemists can also find ways to harness or control the energy released.