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:
−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.
This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.
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