Question

Calculate the standard enthalpy change (in kJ/mol) for the following reaction involving solutions: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)Given the following enthalpy values:NaOH(aq): -469.2 kJ/mol  HCl(aq): -167.2 kJ/mol  NaCl(aq): -411.2 kJ/mol  H2O(l): -285.8 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the formula:H_reaction =  H_products -  H_reactantswhere H_reaction is the standard enthalpy change of the reaction, H_products are the enthalpy values of the products, and H_reactants are the enthalpy values of the reactants.For this reaction, the reactants are NaOH aq  and HCl aq , and the products are NaCl aq  and H2O l . Using the given enthalpy values, we can calculate the standard enthalpy change as follows:H_reaction = [ -411.2 kJ/mol  +  -285.8 kJ/mol ] - [ -469.2 kJ/mol  +  -167.2 kJ/mol ]H_reaction =  -696.0 kJ/mol  -  -636.4 kJ/mol H_reaction = -59.6 kJ/molThe standard enthalpy change for the reaction is -59.6 kJ/mol.