If the standard enthalpy of neutralization reaction of HCl and NaOH is −57.3 kJ mol⁻¹, then find out the enthalpy of neutralization of 0.25 mol of HCl by 0.25 mol of NaOH:
−57.3 kJ
−28.3 kJ
−14.32 kJ
+57.3 kJ
The standard enthalpy of neutralization is the heat evolved when 1 gram equivalent (or 1 mole for monobasic acids/bases) of a strong acid is completely neutralized by a strong base. For the reaction between HCl and NaOH, the enthalpy change is given as −57.3 kJ mol⁻¹.
Since 0.25 mol of HCl is neutralized by 0.25 mol of NaOH, exactly 0.25 mol of H⁺ reacts with 0.25 mol of OH⁻ to form 0.25 mol of H₂O.
The enthalpy change for this specific amount will be: \Delta H = 0.25 mol × (−57.3 kJ mol⁻¹) \Delta H = −14.325 kJ
Rounding off, we get −14.32 kJ.
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