Problem 87 - Chemistry and Chemical Reactivity

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 CAN'T COPY)

Step-by-Step Solution

Answer

The value of \( \Delta U \) for the reaction is \( 35.98 \, \mathrm{kJ/mol} \).

Step 1: Calculate the heat absorbed by the water

First, determine the amount of heat absorbed by the water using the formula: \[ q = m imes c imes \Delta T \]where:- \( m \) is the mass of the water \( = 415 \, \mathrm{g} \)- \( c \) is the specific heat capacity of water \( = 4.18 \, \mathrm{J/g \, °C} \)- \( \Delta T \) is the change in temperature \( = 20.72 \, °C - 18.90 \, °C = 1.82 \, °C \)Substitute these values into the formula to find \( q_{\text{water}} \):\[ q_{\text{water}} = 415 \, \text{g} \times 4.18 \, \text{J/g°C} \times 1.82 \, °C = 3154.986 \, \text{J} \]

Step 2: Calculate the heat absorbed by the bomb

Use the formula for the bomb calorimeter to find the heat absorbed by the bomb:\[ q_{\text{bomb}} = C \times \Delta T \]where:- \( C \) is the heat capacity of the bomb \( = 155 \, \mathrm{J/K} \)- \( \Delta T \) is the change in temperature \( = 1.82 \, \mathrm{K} \)Substitute into the equation to get:\[ q_{\text{bomb}} = 155 \, \text{J/K} \times 1.82 \, \text{K} = 282.1 \, \text{J} \]

Step 3: Calculate the total heat absorbed

The total heat absorbed \( q_{\text{total}} \) is the sum of the heat absorbed by the water and the bomb:\[ q_{\text{total}} = q_{\text{water}} + q_{\text{bomb}} = 3154.986 \, \text{J} + 282.1 \, \text{J} = 3437.086 \, \text{J} = 3.437 \, \text{kJ} \]

Step 4: Calculate the moles of ammonium nitrate decomposed

Use the molar mass of ammonium nitrate \( \mathrm{NH}_{4}\mathrm{NO}_{3} \) to calculate the moles decomposed:- Molar mass of \( \mathrm{NH}_{4}\mathrm{NO}_{3} \) = 80.04 \( \mathrm{g/mol} \)- Mass given = 7.647 gThe moles of ammonium nitrate is:\[ \text{moles} = \frac{7.647 \, \text{g}}{80.04 \, \text{g/mol}} = 0.0955 \, \text{mol} \]

Step 5: Calculate the change in internal energy per mole

To find \( \Delta U \) per mole, divide the total energy change by the moles calculated:\[ \Delta U = \frac{q_{\text{total}}}{\text{moles of NH}_4\text{NO}_3} = \frac{3.437 \, \text{kJ}}{0.0955 \, \text{mol}} = 35.98 \, \text{kJ/mol} \]

Key Concepts

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It helps us determine the amount of energy exchanged in reactions, particularly in thermal processes. Typically, a calorimeter is utilized for these measurements, which isolates the reaction to minimize heat loss to the surroundings.

In this exercise, a bomb calorimeter is used to measure the energy change during the decomposition of ammonium nitrate. The bomb calorimeter allows accurate measurement of heat changes because it maintains a constant volume and insulates the reaction from external temperature changes. By knowing the specific heat capacities and measuring the temperature change, we can accurately determine the heat involved in the chemical reaction. Understanding calorimetry is crucial because it helps predict reaction outcomes based on their energy requirements.

Internal Energy Change

The internal energy change, \(\Delta U\), in a reaction represents the total change in energy, typically including heat and work. In a bomb calorimeter, because the reaction volume is constant, work done by expanding gases is minimal, and thus \(\Delta U\) primarily consists of heat transfer.

To determine \(\Delta U\) for the decomposition of ammonium nitrate, we first calculate the heat absorbed by the system. The heat interactions include energy absorbed by water and the calorimeter itself. These values are summed to find the total heat absorbed by the system, providing a clear picture of the energy change associated with the reaction. \(\Delta U\) helps interpret how energy is stored or released in chemical reactions, essential for designing industrial processes and energy-efficient reactions.

Ammonium Nitrate Decomposition

Ammonium nitrate decomposition is a chemical reaction where ammonium nitrate ( \(\mathrm{NH}_{4}\mathrm{NO}_{3}\)) breaks down into nitrous oxide ( \(\mathrm{N}_{2}\mathrm{O}\)) and water vapor ( \(\mathrm{H}_{2}\mathrm{O}\)).

This decomposition is an exothermic reaction, which means it releases energy in the form of heat. The reaction is significant for both environmental science and industrial applications, as nitrous oxide is a greenhouse gas, while also being used as a propellant or a fuel component in rockets.

• The balanced equation: $$ \mathrm{NH}_{4}\mathrm{NO}_{3}\rightarrow \mathrm{N}_{2}\mathrm{O}+2 \mathrm{H}_{2} \mathrm{O} $$ illustrates the stoichiometry of the process.
• Analyzing decomposition helps understand energy changes during such reactions.

By studying this reaction in a calorimeter, we learn more about how energy changes impact chemical processes and how to manage these changes for desired outcomes.

Molar Mass Calculation

The calculation of molar mass is a fundamental skill in chemistry, as it allows us to transition between mass and the amount of substance (moles). Molar mass is the mass of one mole of a given substance, commonly expressed in grams per mole ( \(\text{g/mol}\)).

For ammonium nitrate ( \(\mathrm{NH}_{4}\mathrm{NO}_{3}\)), the molar mass is calculated by summing the atomic masses of all the atoms in one molecule:

• Nitrogen (N): 2 atoms \(\approx 14.01 \, \text{g/mol}\) per atom
• Hydrogen (H): 4 atoms \(\approx 1.01 \, \text{g/mol}\) per atom
• Oxygen (O): 3 atoms \(\approx 16.00 \, \text{g/mol}\) per atom

This yields a total molar mass for ammonium nitrate of approximately 80.04 \(\text{g/mol}\).

Accurately determining the molar mass is crucial because it allows chemists to relate the physical mass of a substance to its chemical moles, thereby enabling precise calculations necessary for understanding and predicting chemical reaction behavior.