Question

Calculate the standard enthalpy change (ΔH°) for the following reaction at 25°C, given that the enthalpies of formation for the reactants and products are:HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)ΔH°f[HCl (aq)] = -92.31 kJ/molΔH°f[NaOH (aq)] = -469.49 kJ/molΔH°f[NaCl (aq)] = -407.29 kJ/molΔH°f[H2O (l)] = -285.83 kJ/mol

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants In this case, the products are NaCl  aq  and H2O  l , and the reactants are HCl  aq  and NaOH  aq . Plugging in the given values for the enthalpies of formation, we get:H = [ -407.29 kJ/mol  +  -285.83 kJ/mol ] - [ -92.31 kJ/mol  +  -469.49 kJ/mol ]H =  -693.12 kJ/mol  -  -561.80 kJ/mol H = -131.32 kJ/molSo, the standard enthalpy change for the reaction is -131.32 kJ/mol.