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 the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\). a. Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\text {soln }}\right)\) 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 solution (ΔHsoln) for the dissolution of NH4NO3 is -26.04 kJ/mol, and the lattice energy of NH4NO3 is 603.96 kJ/mol.

Step-by-step solution

Calculate heat change of the solution

First, we need to determine the heat absorbed or released by the solution during the dissolution process. We'll use the formula: \(q = mc\Delta{ T }\), where - q is the heat change, - m is the mass of the solution, - c is the specific heat capacity of the solution, - \(\Delta{ T }\) is the temperature change. The given values are: - m = 75.0 g (mass of water), - c = 4.18 J/g°C (specific heat capacity of the solution), - \(\Delta{ T } = T_{final} - T_{initial} = 23.34^{\circ} \mathrm{C} - 25.00^{\circ} \mathrm{C} = -1.66^{\circ} \mathrm{C}\). Now, calculate q: \(q = (75.0 \mathrm{~g}) (4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot ^\circ \mathrm{C}})(-1.66 ^\circ \mathrm{C}) = -5.20724\times10^2 \mathrm{~J}\)

Calculate moles of NH4NO3

We need to determine the moles of NH4NO3 dissolved in water. The molar mass of NH4NO3 is \(M = 1(14.01) + 4(1.01) + 1(14.01) + 3(16.00) = 80.05 \mathrm{~g} / \mathrm{mol} \). Now, calculate the number of moles: \(n = \frac{m}{M} = \frac{1.60 \mathrm{~g}}{80.05 \mathrm{~g} / \mathrm{mol}} = 2.00\times10^{-2} \mathrm{~mol} \)

Calculate enthalpy of solution

Now, we can determine the enthalpy of solution (ΔHsoln) using the heat change (q) and the number of moles (n). To find the enthalpy of solution per mole of NH4NO3, we'll use the following relationship: \(\Delta{H_{\text {soln }}}=\frac{q}{n}\). Calculate ΔHsoln: \(\Delta{H_{\text {soln }}}= \frac{-5.20724\times10^2 \mathrm{~J}}{2.00\times10^{-2} \mathrm{~mol}} = -2.604\times10^4 \mathrm{~J/mol} = -26.04 \mathrm{~kJ/mol}\)

Calculate lattice energy

Finally, we can determine the lattice energy using the enthalpy of hydration (ΔHhydration) and the enthalpy of solution (ΔHsoln). The relationship between these quantities is as follows: \(Lattice \thinspace energy = \Delta{H_{\text {soln }}} - \Delta{H_{\text {hydration}}}\). Using the given enthalpy of hydration (-630 kJ/mol), we can compute the lattice energy: \(Lattice \thinspace energy = -26.04 \mathrm{~kJ/mol} - (-630 \mathrm{~kJ/mol}) = 603.96 \mathrm{~kJ/mol}\). Therefore, the lattice energy of NH4NO3 is 603.96 kJ/mol.

Key concepts

Calorimetry

Calorimetry is a technique used to measure the amount of heat absorbed or released during a chemical or physical process. In a coffee-cup calorimeter, this process is used to observe dissolutions, reactions, or heat changes in a liquid solution. Here, the dissolution of ammonium nitrate (\(\text{NH}_4\text{NO}_3\)) was studied. Calorimetry ensures that we can track how the temperature changes when a substance like \(\text{NH}_4\text{NO}_3\) dissolves in water. This is crucial for understanding energy changes and is widely used in chemistry for calculating enthalpies.

By knowing the heat change \(q\), using the formula \(q = mc\Delta T\) (where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is temperature change), we can gain insights into the enthalpies involved. Apply this concept to find the heat either absorbed or released during such reactions, helping to describe the energy flow within the system.

Heat Capacity

Heat capacity helps us understand how much heat energy a substance can absorb before its temperature changes. It's a crucial concept in calorimetry. In this exercise, the solution had a specific heat capacity of \(4.18 \text{ J/g°C}\), which is the same as water. This value indicates how resistant the solution is to changing temperature as it absorbs heat.

When \(1.60 \text{ g}\) of \(\text{NH}_4\text{NO}_3\) was dissolved, the heat capacity of the solution played a role in determining the final temperature change. Heat capacity is crucial in energy calculations and helps us make sense of temperature changes resulting from heat exchange in chemical processes.

Always ensure to take into account the specific heat capacity of your solution when calculating temperature-based energy changes. It provides a way to quantify the heat required to change the temperature and is pivotal for precise enthalpy measurements.

Lattice Energy

Lattice energy refers to the energy required to break apart an ionic compound into its individual ions. It's also the energy released when gaseous ions form a solid crystal lattice. This concept is vital for understanding ionic compounds' stability and formation.

In this exercise, the lattice energy of \(\text{NH}_4\text{NO}_3\) was calculated using the relationship: Lattice energy = \(\Delta H_{\text{soln}} - \Delta H_{\text{hydration}}\). Using the solution's enthalpy and the hydration enthalpy, you find the lattice energy, giving insight into the compound's inherent energy when forming its solid structure.

Lattice energy is not something you can observe directly but is inferred from other thermodynamic measurements. It plays a key role in predicting the properties of ionic compounds and understanding their energy changes during dissolution.

Enthalpy of Hydration

Enthalpy of hydration deals with the energy changes when ions become surrounded by water molecules upon dissolution. It's a measure of how much energy is released when ions engage with the solvent. In our example, the enthalpy of hydration for \(\text{NH}_4\text{NO}_3\) was given as \(-630 \text{ kJ/mol}\). This negative sign suggests that energy is released when the compound dissolves and ions interact with water.

Understanding the enthalpy of hydration is crucial when examining how substances dissolve, as it helps determine the overall enthalpy change in these processes. It is one part of the solution enthalpy equation, alongside lattice energy, to explore the complete energetics of dissolution in ionic compounds.

By combining these thermodynamic pieces, you can analyze the net effect of dissolution, showing whether the entire process is endothermic or exothermic, enlightening the full energy profile of such a chemical event.