Question

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 \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

Short answer

The enthalpy change for dissolving potassium bromide is approximately 154.4 J/g and 18.375 kJ/mol.

Step-by-step solution

Calculate the heat change of water and KBr

To calculate the heat change, we will use the heat capacity formula: \[q = mc\Delta T\] where q represents heat change, m is mass, c is specific heat capacity, and ΔT is the temperature change. We will first calculate the q for water and then for KBr.

Step 1a: Calculate the heat change of water

For water, mass (m) = 125 g, specific heat capacity (c) = 4.18 J/°C·g, initial temperature (T1) = 24.2°C, and final temperature (T2) = 21.1°C. So, the temperature change, ΔT = T2 - T1 ΔT = 21.1 - 24.2 = -3.1°C Now, using the heat capacity formula for water: q_water = (125 g)(4.18 J/°C·g)(-3.1°C) = -1621.7 J

Step 1b: Calculate the heat change of KBr

Since no heat is transferred to the surroundings or the calorimeter, the heat change of KBr must be equal but opposite in sign to that of water. q_KBr = -q_water = 1621.7 J

Calculate the enthalpy change per gram of KBr

Now that we have the heat change of KBr, we can calculate the enthalpy change per gram by dividing the heat change by the mass of KBr. Enthalpy change per gram = q_KBr / mass of KBr Enthalpy change per gram = 1621.7 J / 10.5 g ≈ 154.4 J/g

Calculate the enthalpy change per mole of KBr

Finally, to calculate the enthalpy change per mole, we must first determine the molar mass of KBr. The atomic mass of potassium (K) is 39 g/mol, and the atomic mass of bromine (Br) is 80 g/mol. Therefore, the molar mass of KBr is: Molar mass of KBr = 39 g/mol + 80 g/mol = 119 g/mol Now, to find the enthalpy change per mole, we will multiply the enthalpy change per gram with the molar mass of KBr: Enthalpy change per mole = Enthalpy change per gram × Molar mass of KBr Enthalpy change per mole = 154.4 J/g × 119 g/mol ≈ 18374.6 J/mol We can convert this value to kilojoules per mole by dividing by 1000: Enthalpy change per mole = 18374.6 J/mol ÷ 1000 = 18.375 kJ/mol The enthalpy change for dissolving potassium bromide is approximately 154.4 J/g and 18.375 kJ/mol.