In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

To find the amount of heat absorbed by the solution, we can use the formula: \(q = mc\Delta T\) Where, q = heat absorbed by the solution (J) m = mass of the solution (g) c = specific heat capacity of the solution (J/g°C) ΔT = change in temperature (°C) Given that the initial temperature was 24.6°C and the final temperature was 31.3°C, we can calculate the change in temperature: ΔT = T_final - T_initial = 31.3°C - 24.6°C = 6.7°C Given that the density of the solutions is 1.0 g/cm3, to find out the mass (m) of the solutions, first, convert the volume of both solutions to grams. Since they have equal density, we can simply add up the volumes: m = 100.0 mL + 100.0 mL = 200.0 mL = 200.0 g (since the density is 1.0 g/cm³) Now, we can calculate the heat absorbed by the solution: \(q = (200.0 \,\text{g})(4.18\, \frac{\text{J}}{\text{g°C}})(6.7\, \text{°C}) = 5602.76\, \text{J}\)

Next, we need to determine the number of moles of both the HCl and NaOH that have reacted. As both the substances have equal concentration and volume, they will react in a 1:1 ratio. Number of moles of HCl = concentration × volume of HCl = 1.0 M × 0.100 L = 0.100 mol Similarly, the number of moles of NaOH reacted is 0.100 mol.

Now, we can calculate the enthalpy change per mole by dividing the heat absorbed by the number of moles reacted: Enthalpy change per mole = \( \frac{q_\text{absorbed} }{\text{moles of HCl or NaOH}} \) Enthalpy change per mole = \( \frac{5602.76\, \text{J}}{0.100\, \text{mol}} = 56027.6\, \frac{\text{J}}{\text{mol}}\) In this problem, the enthalpy change for the neutralization of HCl by NaOH is positive, so it is an endothermic process. The final answer can be reported as: Enthalpy change = \( +56027.6\, \frac{\text{J}}{\text{mol}} \) or \( 56.03\, \frac{\text{kJ}}{\text{mol}} \) (rounded to two decimal places).