Question

The enthalpy of neutralization of $NH_{4}OH$ and $CH_{3}COOH \, $ is $-10.5 \, kcal \, mol^{- 1}$ ​ and enthalpy of neutralisation of $NH_{4}OH \, $ with a strong acid is $-11.7 \, kcal \, mol^{- 1}$ . The enthalpy of ionization of $CH_{3}COOH$ will be

• A. $4.0\,kcal\,mol^{- 1}$
• B. $3.0\,kcal\,mol^{- 1}$
• C. $1.2\,kcal\,mol^{- 1}$
• D. $3.2\,kcal\,mol^{- 1}$

Answer 1

Answer: (C)

Enthalpy of neutralisation is the amount of energy released during the combination of one moe strong monobasic acid neutralised with the one-mole strong monoacidic base.
$H^{+}(a q)+(O H)^{-}(a q) \rightarrow H_{2} O(l)(\Delta)_{\text {neut. }} H=-13.7 \,kcal$
Enthalpy of ionisation of $CH _{3} COOH \& NH _{4} OH$ together
$=-10.5-(-13.7)=+3.2\, Kcal$
Enthalpy of ionisation of only $NH _{4} OH =-11.7-(-13.7)=+2\, Kcal$
Enthalpy of ionisation of only $CH _{3} COOH =3.2-2=1.2\, Kcal$