Question

Calculate the standard enthalpy of hydrolysis for the following reaction: AlCl3 + 3H2O → Al(OH)3 + 3HCl Given that the standard enthalpies of formation for AlCl3(-696.5 kJ/mol), Al(OH)3(-1277.1 kJ/mol), and HCl(-92.3 kJ/mol) are known.

Answer 1

To calculate the standard enthalpy of hydrolysis for the given reaction, we can use the following equation:H_reaction =  H_f products  -  H_f reactants where H_reaction is the standard enthalpy of hydrolysis, H_f products  are the standard enthalpies of formation for the products, and H_f reactants  are the standard enthalpies of formation for the reactants.For the given reaction:AlCl3 + 3H2O  Al OH 3 + 3HClThe standard enthalpies of formation are:H_f AlCl3  = -696.5 kJ/molH_f Al OH 3  = -1277.1 kJ/molH_f HCl  = -92.3 kJ/molNow, we can calculate the standard enthalpy of hydrolysis:H_reaction = [1  H_f Al OH 3  + 3  H_f HCl ] - [1  H_f AlCl3 ]H_reaction = [1   -1277.1 kJ/mol  + 3   -92.3 kJ/mol ] - [1   -696.5 kJ/mol ]H_reaction =  -1277.1 kJ/mol - 276.9 kJ/mol  + 696.5 kJ/molH_reaction = -857.5 kJ/molTherefore, the standard enthalpy of hydrolysis for the given reaction is -857.5 kJ/mol.