The enthalpy of neutralization of monobasic acid in decinormal solution by a dilute solution of KOH — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of neutralization of monobasic acid in decinormal solution by a dilute solution of KOH is –12,200 cal. The enthalpy of neutralization of strong acid by strong base is –13,700 cal. Assuming that the acid is 25% dissociated in decinormal solution, the enthalpy of dissociation of the acid (in kcal/mole) is
Answer: 2
💡 Solution & Explanation
The missing heat is due to the endothermic dissociation of the undissociated portion of the weak acid. Since 25% is already dissociated, 75% remains to be dissociated during neutralization. Heat consumed = $13,700 - 12,200 = 1500\text{ cal}$. Thus, $0.75 \times \Delta H_{\text{diss}} = 1500\text{ cal} \implies \Delta H_{\text{diss}} = 1500 / 0.75 = 2000\text{ cal/mol} = 2\text{ kcal/mol}$.
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