Calculate the standard enthalpy change for the following neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given the following tabulated standard enthalpies of formation values: ΔHf°(HCl) = -167 kJ/mol, ΔHf°(NaOH) = -469 kJ/mol, ΔHf°(NaCl) = -411 kJ/mol, ΔHf°(H2O) = -286 kJ/mol.

Question

Calculate the standard enthalpy change for the following neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given the following tabulated standard enthalpies of formation values: ΔHf°(HCl) = -167 kJ/mol, ΔHf°(NaOH) = -469 kJ/

Answer 1

To calculate the standard enthalpy change  H  for the given reaction, we can use the following formula:H =  Hf products  -  Hf reactants For the given reaction:HCl aq  + NaOH aq   NaCl aq  + H2O l The standard enthalpies of formation  Hf  for the products are:Hf NaCl  = -411 kJ/molHf H2O  = -286 kJ/molThe standard enthalpies of formation  Hf  for the reactants are:Hf HCl  = -167 kJ/molHf NaOH  = -469 kJ/molNow, we can plug these values into the formula:H = [ -411  +  -286 ] - [ -167  +  -469 ]H =  -697  -  -636 H = -61 kJ/molSo, the standard enthalpy change for the given neutralization reaction is -61 kJ/mol.