Question

In a coffee-cup calorimeter, 1.60 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with 75.0 \(\mathrm{g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\) . After dissolution of the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\) . a. Assuming the solution has a heat capacity of 4.18 \(\mathrm{J} / \mathrm{g}\) \(^{\circ} \mathrm{C},\) and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\mathrm{soln}}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol} .\) b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-630 . \mathrm{kJ} / \mathrm{mol}\) calculate the lattice energy of \(\mathrm{NH}_{4} \mathrm{NO}_{3} .\)

Short answer

The enthalpy of dissolution (ΔHsoln) for the dissolution of NH4NO3 is -25.995 kJ/mol, and the lattice energy of NH4NO3 is -604.005 kJ/mol.

Step-by-step solution

Calculate the heat change in the solution

To calculate the heat change (q) in the solution we use the formula: \[ q = m \times c \times \Delta T \] where m is the mass of the solution, c is the heat capacity, and ΔT is the change in temperature. In this case, we have the mass and initial temperature of water, heat capacity, and the final temperature of the calorimeter contents. First, calculate the change in temperature, then find the heat change in the solution.

Calculate the moles of NH4NO3

To determine the enthalpy of dissolution, we need the moles of NH4NO3. Use the mass of NH4NO3 given and its molar mass: Molar mass of NH4NO3 = 14.01 (N) + (4x1.01 (H)) + 14.01 (N) + 15.99 (O) x 3 = 80.04 g/mol Moles = mass / molar mass

Calculate the enthalpy of dissolution ΔHsoln

Once we have the heat change (q) and the moles of NH4NO3, we can calculate the enthalpy of dissolution: \[ \Delta H_{soln} = \frac{q}{moles} \] Convert the enthalpy of dissolution to kJ/mol, as the problem asks for the answer in these units.

Calculate the lattice energy

To calculate the lattice energy of NH4NO3, use the given enthalpy of hydration and the enthalpy of dissolution calculated in Step 3: Lattice energy = Enthalpy of hydration - Enthalpy of dissolution Now we will go through the steps with the given values.

Step 1

Given: m_water = 75 g c = 4.18 J/g°C T_initial = 25°C T_final = 23.34°C Calculate ΔT: ΔT = T_final - T_initial = 23.34°C - 25°C = -1.66°C Calculate the heat change: q = m_water × c × ΔT = 75 g × 4.18 J/g°C × (-1.66°C) = -519.87 J

Step 2

Given: mass_NH4NO3 = 1.60 g molar_mass_NH4NO3 = 80.04 g/mol Calculate moles of NH4NO3: moles = mass_NH4NO3 / molar_mass_NH4NO3 = 1.60 g / 80.04 g/mol = 0.020 mol

Step 3

Calculate the enthalpy of dissolution: ΔHsoln = q / moles = -519.87 J / 0.020 mol = -25,995 J/mol Convert to kJ/mol: ΔHsoln = -25,995 J/mol × (1 kJ / 1000 J) = -25.995 kJ/mol

Step 4

Given: Enthalpy_of_hydration = -630 kJ/mol Calculate lattice energy: Lattice_energy = Enthalpy_of_hydration - ΔHsoln = -630 kJ/mol - (-25.995 kJ/mol) = -604.005 kJ/mol Answer: a. The enthalpy of dissolution (ΔHsoln) for the dissolution of NH4NO3 is -25.995 kJ/mol. b. The lattice energy of NH4NO3 is -604.005 kJ/mol.

Key concepts

Calorimetry

Calorimetry is the branch of science that measures the heat of chemical reactions or physical changes. It is an essential technique used to determine the heat changes associated with these processes. In our exercise, a coffee-cup calorimeter helps in determining the enthalpy of solution. This type of calorimeter is quite simple; it consists of a polystyrene cup, a thermometer, and a stirrer.

When a substance such as ammonium nitrate (\(\mathrm{NH}_4\mathrm{NO}_3\)) dissolves in water, heat is either absorbed or released. In the described experiment, the dissolution process caused a decrease in temperature, indicating that the process is endothermic. This means that the system absorbs heat from the surroundings.
- The initial and final temperatures were measured to calculate the change in temperature (\(\Delta T\)). - The mass of the solution (water plus dissolved substance) and its heat capacity are used in the calculation of heat change.Understanding calorimetry helps us establish how much energy was absorbed or released during the dissolution process, vital for calculating the enthalpy of solution.

Heat Capacity

Heat capacity is a crucial property in calorimetry. It indicates how much energy is needed to raise the temperature of a substance by a degree Celsius. Water, for example, has a known heat capacity of 4.18 J/g°C. In our exercise, this value was assumed for the solution in the coffee-cup calorimeter.

To find out how much heat was involved in our calorimeter experiment, we used:\[ q = m \times c \times \Delta T \]where:

• \( m \) is the mass of the solution (75 g of water and the dissolved \( 1.60 \mathrm{g} \ \mathrm{NH}_4\mathrm{NO}_3 \))
• \( c \) is the heat capacity (4.18 J/g°C)
• \( \Delta T \) is the change in temperature (-1.66°C in this case)

Because the temperature decreased, the sign of \( q \) is negative, reflecting heat absorbed by the dissolving substance. Understanding heat capacity is vital in calculating how much energy a process requires or releases in practical applications.

Lattice Energy

Lattice energy is an essential concept in understanding ionic compounds. It represents the energy needed to separate one mole of a solid ionic compound into gaseous ions. In the dissolution process, we often need to compare lattice energy with other energies such as hydration energy to understand various reactions better.

For ammonium nitrate (\(\mathrm{NH}_4\mathrm{NO}_3\)), we used the enthalpy of dissolution (\(\Delta H_{\mathrm{soln}}\)) and the enthalpy of hydration to find the lattice energy. Knowing these values helps predict how the compound behaves when transitioning between states (solid to dissolved).
- Difference between the hydration enthalpy and dissolution enthalpy gives lattice energy.- Lattice energy impacts the stability and solubility of ions in solutions.Understanding lattice energy helps in predicting how and why certain reactions happen, especially in solutions.

Dissolution Process

The dissolution process is a fascinating chemical reaction where a solute (like \(\mathrm{NH}_4\mathrm{NO}_3\)) dissolves in a solvent (like water). This process is affected by several factors, including temperature, pressure, and the nature of the solute and solvent.

When \(\mathrm{NH}_4\mathrm{NO}_3\) dissolves, it absorbs heat, leading to an increase in entropy. The energy absorbed is what we calculated as the enthalpy of solution (\(\Delta H_{\mathrm{soln}}\)). Understanding how substances dissolve helps us in making predictions about reaction directions and in designing processes for industrial applications.
- The dissolution involves breaking the ionic lattice of the solute and forming interactions with the solvent.- This process is energetically favorable under certain conditions, governed by changes in enthalpy and entropy.Comprehending the dissolution process aids us in grasping broader chemistry concepts, including why certain salts dissolve spontaneously and how energy is exchanged during dissolution.