Question

Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide, given the balanced chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) and the enthalpy change of formation values: ∆Hf°(NaCl(aq)) = -407.3 kJ/mol  ∆Hf°(H2O(l)) = -285.83 kJ/mol ∆Hf°(HCl(aq)) = -167.2 kJ/mol  ∆Hf°(NaOH(aq)) = -469.11 kJ/mol

Answer 1

To calculate the standard enthalpy change for the neutralization reaction, we can use the following equation:H reaction  =  [Hf products ] -  [Hf reactants ]For the given reaction:HCl aq  + NaOH aq   NaCl aq  + H2O l The enthalpy change of the reaction is:H reaction  = [Hf NaCl aq   + Hf H2O l  ] - [Hf HCl aq   + Hf NaOH aq  ]Substitute the given values:H reaction  = [ -407.3 kJ/mol  +  -285.83 kJ/mol ] - [ -167.2 kJ/mol  +  -469.11 kJ/mol ]H reaction  =  -693.13 kJ/mol  -  -636.31 kJ/mol H reaction  = -56.82 kJ/molThe standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -56.82 kJ/mol.