Calculate the standard enthalpy change (in kJ/mol) for the reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O), given the following balanced chemical equation:HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)Assume the reaction takes place under standard conditions (298 K and 1 atm pressure) and that all solutions have a concentration of 1.0 M. The standard enthalpies of formation for HCl (aq), NaOH (aq), NaCl (aq), and H2O (l) are -92.31 kJ/mol, -469.14 kJ/mol, -407.76 kJ/mol, and -285.83 kJ/mol, respectively.

Question

Calculate the standard enthalpy change (in kJ/mol) for the reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O), given the following balanced chemical equation:HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)Assume the reaction takes place under standard conditions (298 K and 1 atm pressure) and that all solutions have a concentration of 1.0 M. The standard enthalpies of formation for HCl (aq), NaOH (aq), NaCl (aq), and H2O (l) are -92.31 kJ/mol, -469.14 kJ/

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction:HCl  aq  + NaOH  aq   NaCl  aq  + H2O  l H = [Hf NaCl  + Hf H2O ] - [Hf HCl  + Hf NaOH ]Substitute the given standard enthalpies of formation:H = [ -407.76 kJ/mol  +  -285.83 kJ/mol ] - [ -92.31 kJ/mol  +  -469.14 kJ/mol ]H =  -693.59 kJ/mol  -  -561.45 kJ/mol H = 132.14 kJ/molThe standard enthalpy change for the reaction of hydrochloric acid and sodium hydroxide to form sodium chloride and water is 132.14 kJ/mol.