Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) in aqueous solutions. The balanced chemical equation for the reaction is as follows:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Given the enthalpy of formation for HCl(aq) is -167.2 kJ/mol, NaOH(aq) is -469.14 kJ/mol, NaCl(aq) is -407.3 kJ/mol, and H2O(l) is -285.83 kJ/mol. The reaction is carried out at standard temperature and pressure.

Question

Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) in aqueous solutions. The balanced chemical equation for the reaction is as follows:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Given the enthalpy of formation for HCl(aq) is -167.2 kJ/mol, NaOH(aq) is -469.14 kJ/mol, NaCl(aq) is -407.3 kJ/mol, and H2O

Answer 1

To calculate the standard enthalpy change for the neutralization reaction, we can use the following formula:H_reaction =  H_f products  -  H_f reactants where H_reaction is the standard enthalpy change for the reaction, H_f products  is the sum of the enthalpies of formation of the products, and H_f reactants  is the sum of the enthalpies of formation of the reactants.For the given reaction:HCl aq  + NaOH aq   NaCl aq  + H2O l H_f HCl  = -167.2 kJ/molH_f NaOH  = -469.14 kJ/molH_f NaCl  = -407.3 kJ/molH_f H2O  = -285.83 kJ/molNow, we can plug these values into the formula:H_reaction = [ -407.3 kJ/mol  +  -285.83 kJ/mol ] - [ -167.2 kJ/mol  +  -469.14 kJ/mol ]H_reaction =  -693.13 kJ/mol  -  -636.34 kJ/mol H_reaction = -56.79 kJ/molTherefore, the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide in aqueous solutions is -56.79 kJ/mol.