Enthalpy of neutralization is defined as the enthalpy change when 1 mole of acid/base is completely neutralized by base/acid in dilute solution.
For strong acid and strong base neutralization net chemical change is
H + (aq) + OH – (aq) ⎯→ H 2 O( λ ); Δ r Hº = – 55.84 kJ/mol
Δ Hº ionization of aqueous solution of strong acid and strong base is zero.
When a dilute solution of a weak acid or base is neutralized, the enthalpy of neutralization is some what less because of the absorption of heat in the ionization of the weak acid or base, for weak acid/base
Δ Hº neutrilization = Δ Hº ionization + Δ r Hº (H + + OH – ⎯→ H 2 O)
(i) If enthalpy of neutralization of CH 3 COOH by NaOH is –49.86 kJ/mol then enthalpy of ionzation of CH 3 COOH is:
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) Δ H ionization = Δ H° neutrilization – Δ H° (H + OH – ⎯→ H 2 O)
= – 49.86 – (–55.84) kJ/mole
= 5.98 kJ/mole
(ii) Δ H° = 2 × (–55.84) kJ/mole = – 111.68 kJ
(iii) For max. rise in temp.; max. neutralization of H + and OH – required.
If we take equal volume, all H + (5 m-mole) will react with all OH – (5 m-mole).
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems