Question

Calculate the standard enthalpy change for the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Given the following information: ΔHf° [NaOH(aq)] = -469.1 kJ/mol ΔHf° [HCl(aq)] = -167.2 kJ/mol ΔHf° [NaCl(aq)] = -407.3 kJ/mol ΔHf° [H2O(l)] = -286 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf  products  -  Hf  reactants For the given reaction:H = [Hf  NaCl  + Hf  H2O ] - [Hf  NaOH  + Hf  HCl ]H = [ -407.3 kJ/mol  +  -286 kJ/mol ] - [ -469.1 kJ/mol  +  -167.2 kJ/mol ]H =  -693.3 kJ/mol  -  -636.3 kJ/mol H = -57 kJ/molThe standard enthalpy change for the reaction is -57 kJ/mol.