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 \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

Step by step solution

01

(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: \(q = (75.0\,\text{g})(4.18\, \frac{\text{J}}{\text{g}\cdot°C})(-1.66°C) = -519.54\,\text{J}\)

02

(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

03

(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.)

04

(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.

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