A chemistry student needs to calculate the heat of hydration for the reaction between anhydrous magnesium chloride (MgCl2) and water (H2O) using the following data: the enthalpy change for the dissolution of anhydrous magnesium chloride in water is -63.3 kJ/mol, and the enthalpy change for the hydration of magnesium ions (Mg2+) and chloride ions (Cl-) is -192 kJ/mol and -131 kJ/mol, respectively. Calculate the overall heat of hydration (in kJ/mol) for the reaction: MgCl2(s) + 6H2O(l) → MgCl2.6H2O(s)

Question

A chemistry student needs to calculate the heat of hydration for the reaction between anhydrous magnesium chloride (MgCl2) and water (H2O) using the following data: the enthalpy change for the dissolution of anhydrous magnesium chloride in water is -63.3 kJ/mol, and the enthalpy change for the hydration of magnesium ions (Mg2+) and chloride ions

Answer 1

To calculate the overall heat of hydration for the reaction, we need to consider the enthalpy changes for the dissolution of anhydrous magnesium chloride and the hydration of magnesium and chloride ions.The reaction can be broken down into three steps:1. Dissolution of anhydrous magnesium chloride in water:MgCl2 s   Mg2+ aq  + 2Cl- aq ; H1 = -63.3 kJ/mol2. Hydration of magnesium ions:Mg2+ aq  + 6H2O l   Mg H2O 6^2+ aq ; H2 = -192 kJ/mol3. Hydration of chloride ions  2 moles of Cl- ions per mole of MgCl2 :2[Cl- aq  + H2O l   Cl- H2O  aq ]; H3 = 2 * -131 kJ/mol = -262 kJ/molNow, we can calculate the overall heat of hydration by adding the enthalpy changes for each step:H_total = H1 + H2 + H3H_total = -63.3 kJ/mol +  -192 kJ/mol  +  -262 kJ/mol H_total = -517.3 kJ/molSo, the overall heat of hydration for the reaction is -517.3 kJ/mol.