Question

When 1.00 L of 1.00 M Ba(NO₃)₂ solution at 25.0°C is mixed with 1.00 L of 1.00 M Na₂SO₄ solution at 25.0°C in a calorimeter, the white solid BaSO₄ forms, and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/°C g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO₄ formed.

Answer 1

To calculate the enthalpy change per mole of BaSO4, the heat absorbed by the solution from the temperature change is determined and divided by the moles of BaSO4 formed.

Step-by-step explanation:

To calculate the enthalpy change per mole of BaSO4 formed when 1.00 L of 1.00 M Ba(NO3)2 solution is mixed with 1.00 L of 1.00 M Na2SO4 solution, first determine the amount of heat absorbed by the solution. Using q = mcΔT, where m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature. The total mass of the solution is the combined volume of the solutions times the density, which is 2.00 L × 1.00 g/mL = 2000 g. The specific heat capacity c is given as 4.18 J/°C g.

Because the temperature increased by 28.1°C - 25.0°C = 3.1°C, we use q = (2000 g)(4.18 J/°C g)(3.1°C) to find the total heat absorbed. Once the heat q has been calculated, the moles of BaSO4 can be determined. Because the starting solutions were of equal molarity and volume, and barium sulfate precipitates in a 1:1 mole ratio from solutions of barium nitrate and sodium sulfate, the number of moles of BaSO4 will equal the original moles of either reactant in solution, which is 1.00 mole. Finally, the enthalpy change per mole of BaSO4 is obtained by dividing the total heat by the number of moles of BaSO4 formed.