A coffee-cup calorimeter initially contains 125 \(\mathrm{g}\) water at \(24.2^{\circ} \mathrm{C} .\) Potassium bromide \((10.5 \mathrm{g}),\) also at \(24.2^{\circ} \mathrm{C},\) is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\) . Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and $\mathrm{kJ} / \mathrm{mol}$ . Assume that the specific heat capacity of the solution is 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

The initial and final temperatures of the solution are given. Determine the temperature change (ΔT) by subtracting the final temperature from the initial temperature: ΔT = T_final - T_initial ΔT = \(21.1^{\circ} \mathrm{C} - 24.2^{\circ} \mathrm{C}\) ΔT = --3.1°C

Use the formula for heat transfer (q = mcΔT) to find the amount of heat absorbed or released by the solution, where m is the mass of the solution, c is the heat capacity, and ΔT is the temperature change. The mass of the solution is the combined masses of the water and potassium bromide: m = 125 g (water) + 10.5 g (KBr) = 135.5 g Using the heat capacity (c) of 4.18 J/g°C and the previously calculated temperature change (ΔT = -3.1°C), calculate q: q = mcΔT q = (135.5 g)(4.18 J/g°C)(-3.1°C) q = -1763.67 J

Divide the heat transfer (q) by the mass of potassium bromide to find the enthalpy change per gram: Enthalpy change per gram = q / mass of KBr = -1763.67J / 10.5g Enthalpy change per gram = -167.97 J/g

To find the enthalpy change per mole, first find the molar mass of potassium bromide (KBr). The molar mass of K is 39.1 g/mol, and the molar mass of Br is 79.9 g/mol. Thus, the molar mass of KBr is: Molar mass of KBr = 39.1 g/mol + 79.9 g/mol = 119 g/mol Next, divide the heat transfer (q) by the number of moles of potassium bromide: Number of moles of KBr = mass of KBr / molar mass of KBr = 10.5g / 119 g/mol = 0.0882 mol Enthalpy change per mole = q / number of moles of KBr = -1763.67J / 0.0882 mol Enthalpy change per mole = -19994.15 J/mol Convert this value to kJ/mol: Enthalpy change per mole = -19.99415 kJ/mol Finally, the enthalpy change for dissolving potassium bromide in water is approximately -167.97 J/g or -19.99415 kJ/mol.