Question

Calculate the standard enthalpy change (in kJ/mol) for the neutralization reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH) using the following information: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O(l)ΔHf° (kJ/mol): HCl (aq) = -167.2 NaOH (aq) = -469.2 NaCl (aq) = -407.3 H2O(l) = -285.8

Answer 1

To calculate the standard enthalpy change  H  for the neutralization reaction, we can use the following equation:H =  Hf products  -  Hf reactants For the given reaction:HCl  aq  + NaOH  aq   NaCl  aq  + H2O l H = [Hf NaCl  + Hf H2O ] - [Hf HCl  + Hf NaOH ]Using the given Hf values:H = [ -407.3  +  -285.8 ] - [ -167.2  +  -469.2 ]H =  -693.1  -  -636.4 H = -56.7 kJ/molThe standard enthalpy change for the neutralization reaction of hydrochloric acid and sodium hydroxide is -56.7 kJ/mol.