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 \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) 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 kJ/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

a. The enthalpy of solution for the dissolution of NH4NO3 is approximately -20.78 kJ/mol. b. The lattice energy of NH4NO3 is approximately -609.22 kJ/mol.

Step-by-step solution

(Step 1: Calculate the energy released/absorbed)

Formula to calculate energy is: \(q = mc\Delta T\), where 'm' is the mass (in grams) of the water, 'c' is the specific heat capacity (in J/g°C) of the solution, and \(\Delta T\) is the change in temperature. Given, Initial temperature, \(T_{initial} = 25.00\)°C Final temperature, \(T_{final} = 23.34\)°C Mass of water, m = 75.0 g Specific heat capacity of the solution, c = 4.18 J/g°C First, we will calculate the change in temperature: \(\Delta T = T_{final} - T_{initial} = 23.34°C - 25.00°C = -1.66°C\) Now, let's calculate the energy released/absorbed: \(q = (75.0\,\text{g})(4.18\, \frac{\text{J}}{\text{g}\cdot°C})(-1.66°C) = -519.54\,\text{J}\)

(Step 2: Convert mass to moles)

Given, mass of NH4NO3 = 1.60g. Now, we will find moles of NH4NO3. Molar mass of NH4NO3 = (14.01 + 4 x 1.01) + (14.01 + 3 x 16.00) = 64.06 g/mol Moles of NH4NO3 = mass / molar mass = 1.60 g / 64.06 g/mol = 0.0250 mol

(Step 3: Calculate the enthalpy of solution)

Enthalpy of solution, \(\Delta H_{soln}\) = Energy absorbed / moles of solute \(\Delta H_{soln}\) = -519.54 J / 0.0250 mol = -20781.6 J/mol Converting to kJ/mol, we get: \(\Delta H_{soln}\) = -20.78 kJ/mol (approx.)

(Step 4: Calculate the lattice energy)

Now, we will use the enthalpy of hydration and enthalpy of solution to find the lattice energy of NH4NO3. Enthalpy of Hydration, \(\Delta H_{hydration}\) = -630.0 kJ/mol Enthalpy of Solution, \(\Delta H_{soln}\) = -20.78 kJ/mol Lattice energy, \(\Delta H_{lattice}\) = \(\Delta H_{hydration}\) - \(\Delta H_{soln}\) \(\Delta H_{lattice}\) = -630.0 kJ/mol - (-20.78 kJ/mol) = -609.22 kJ/mol So, the lattice energy of NH4NO3 is approximately -609.22 kJ/mol.

Key concepts

Enthalpy of Solution

Understanding the enthalpy of solution is crucial in calorimetry. This term refers to the heat change that occurs when a substance dissolves in a solvent, in this case, water. In this exercise, \(\Delta H_{\text{soln}}\) was calculated for ammonium nitrate (NHeaten{4}NOeaten{3}).

Here is what happened:

• The salt was mixed with water in a calorimeter, leading to a temperature change of the water from 25.00°C to 23.34°C.
• Using the formula \(q = mc\Delta T\), we calculated the amount of energy absorbed, as the solution process was endothermic (absorbing heat). \(q\), being negative, indicates this absorption.
• The calculated \(\Delta H_{\text{soln}}\) was -20.78 kJ/mol, exemplifying the energy absorbed per mole of NHeaten{4}NOeaten{3} dissolved.

Notice how using mass and temperature changes can give us insights into reaction energetics!

Lattice Energy

Lattice energy can be thought of as the energy you would need to execute a reversal of forming a solid lattice of ions into its separate gaseous ions. It is a key concept in predicting a substance's solubility as well as its stability.

In our exercise:

• The enthalpy of solution (\(\Delta H_{\text{soln}}\)) and the enthalpy of hydration (\(\Delta H_{\mathrm{hydration}}\)) together contribute to finding the lattice energy.
• The lattice energy (\(\Delta H_{\mathrm{lattice}}\)) was deduced using the relationship: \(\Delta H_{\mathrm{lattice}} = \Delta H_{\mathrm{hydration}} - \Delta H_{\mathrm{soln}}\).
• For \(\mathrm{NH}_{4}\mathrm{NO}_{3}\), this came out to be approximately -609.22 kJ/mol, indicating a considerable energy release when the compound forms from its ions.

It's vital for diagnosing the energy needs of reactions and predicting whether a compound will readily dissolve!

Enthalpy of Hydration

The term "enthalpy of hydration" involves the heat change when ions from a solid become surrounded by water molecules in a solution. This is particularly relevant for ionic compounds.

In the problem at hand:

• Ammonium nitrate's heat of hydration was given as -630.0 kJ/mol.
• This negative enthalpy indicates the process is exothermic, meaning energy is released as bonds form between water molecules and the ions.
• Enthalpy of hydration forms part of the considerations for calculating lattice energy.

Such calculations help detail how solutions behave energetically, describing how compounds break down and interface with a solvent.

Specific Heat Capacity

Specific heat capacity (\(c\)) is an important concept in calorimetry and thermodynamics. It refers to the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. It influences how much temperature changes when energy is added or removed from a system.

What we have seen in our exercise:

• The water/solution had a specific heat capacity of 4.18 J/g°C.
• This value was pivotal in calculating the heat change using the formula \(q = mc\Delta T\).
• It helped us determine the energy absorbed when the ammonium nitrate dissolved, highlighting the efficiency of calorimeters in energy studies.

Emphasizing specific heat capacity sheds light on a substance's heat absorption or emission tendencies, playing a big role in thermal calculations.