Question

Determine the standard enthalpy change of the dissolution reaction of 2.50 g of NaOH in water, given that the molar enthalpy of hydration of Na+ and OH- ions are -406.8 kJ/mol and -230.0 kJ/mol respectively. The heat capacity of the calorimeter is 25.2 J/°C and the initial and final temperatures were 24.0°C and 29.2°C respectively.

Answer 1

First, we need to calculate the moles of NaOH dissolved:Molar mass of NaOH = 22.99 g/mol  Na  + 15.999 g/mol  O  + 1.007 g/mol  H  = 39.996 g/molmoles of NaOH =  2.50 g  /  39.996 g/mol  = 0.0625 molNext, we need to calculate the heat absorbed by the calorimeter:T = T_final - T_initial = 29.2C - 24.0C = 5.2Cq_calorimeter = C_calorimeter * T = 25.2 J/C * 5.2C = 131.04 JSince the heat absorbed by the calorimeter is equal to the heat released by the dissolution reaction, we have:q_reaction = -131.04 J = -0.13104 kJNow, we can calculate the enthalpy change of the dissolution reaction per mole of NaOH:H_reaction = q_reaction / moles of NaOH =  -0.13104 kJ  /  0.0625 mol  = -2.09664 kJ/molFinally, we can calculate the standard enthalpy change of the dissolution reaction using the molar enthalpy of hydration of Na+ and OH- ions:H_dissolution = H_reaction - H_hydration Na+  - H_hydration OH- H_dissolution = -2.09664 kJ/mol -  -406.8 kJ/mol  -  -230.0 kJ/mol H_dissolution = 634.70336 kJ/molThe standard enthalpy change of the dissolution reaction of NaOH in water is approximately 634.7 kJ/mol.