Question

Calculate the enthalpy change for the dissolution of 5.00 g of potassium chloride (KCl) in 100 mL of water. Given that the specific heat capacity of the resulting solution is 4.18 J/g°C, and its temperature increases from 25.0°C to 30.0°C. The molar mass of KCl is 74.55 g/mol, and the enthalpy of hydration of K+ and Cl- ions is -716 and -364 kJ/mol, respectively.

Answer 1

First, we need to calculate the moles of KCl dissolved:moles of KCl = mass / molar massmoles of KCl = 5.00 g / 74.55 g/mol = 0.0671 molNext, we need to calculate the enthalpy change for the dissolution of KCl:H_dissolution = moles of KCl *  enthalpy of hydration of K+ + enthalpy of hydration of Cl- H_dissolution = 0.0671 mol *  -716 kJ/mol +  -364 kJ/mol  H_dissolution = 0.0671 mol *  -1080 kJ/mol H_dissolution = -72.5 kJNow, we need to calculate the heat absorbed by the solution:q_solution = mass of solution * specific heat capacity * Tq_solution =  100 g + 5 g  * 4.18 J/gC *  30.0C - 25.0C q_solution = 105 g * 4.18 J/gC * 5.0Cq_solution = 2194.5 J = 2.1945 kJSince the heat absorbed by the solution is equal to the heat released by the dissolution process, we can write:H_dissolution = -q_solutionH_dissolution = - -2.1945 kJ H_dissolution = 2.1945 kJTherefore, the enthalpy change for the dissolution of 5.00 g of potassium chloride in 100 mL of water is 2.1945 kJ.