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

Step by step solution

01

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: \(q = (75.0 \mathrm{~g}) (4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot ^\circ \mathrm{C}})(-1.66 ^\circ \mathrm{C}) = -5.20724\times10^2 \mathrm{~J}\)

02

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: \(n = \frac{m}{M} = \frac{1.60 \mathrm{~g}}{80.05 \mathrm{~g} / \mathrm{mol}} = 2.00\times10^{-2} \mathrm{~mol} \)

03

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: \(\Delta{H_{\text {soln }}}= \frac{-5.20724\times10^2 \mathrm{~J}}{2.00\times10^{-2} \mathrm{~mol}} = -2.604\times10^4 \mathrm{~J/mol} = -26.04 \mathrm{~kJ/mol}\)

04

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.

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