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} .$

Step by step solution

01

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.

02

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

03

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.

04

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.

05

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

06

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

07

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

08

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.

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