If the standard enthalpy of neutralization reaction of HCl and NaOH is −57.3 kJ mol⁻¹, then find out the entha

Question

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:

Accepted Answer

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:
ΔH = 0.25 mol × (−57.3 kJ mol⁻¹)
ΔH = −14.325 kJ

Rounding off, we get −14.32 kJ.

Step-by-Step Solution

Practice All CHEMISTRY Questions