The enthalpy of neutralisation of $\mathrm{NH}_{4} \mathrm{OH}$ and $\mathrm{CH}_{3} \mathrm{COOH}$ is $-10…

The enthalpy of neutralisation of $\mathrm{NH}_{4} \mathrm{OH}$ and $\mathrm{CH}_{3} \mathrm{COOH}$ is $-10.5 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy
of neutralisation of $\mathrm{CH}_{3} \mathrm{COOH}$ with strong base is $-12.5 \mathrm{kcal} \mathrm{mol}^{-1}$. The enthalpy of ionisation of $\mathrm{NH}_{4} \mathrm{OH}$ will be
  1. 3.2 kcal mol $^{-1}$
  2. $2.0 \mathrm{kcal} \mathrm{mol}^{-1}$
  3. $3.0 \mathrm{kcal} \mathrm{mol}^{-1}$
  4. $4.0 \mathrm{kcal} \mathrm{mol}^{-1}$

Solution

$\Delta_{\text {neu }} \mathrm{H}$ for strong base and strong acid
$=-13.7 \mathrm{kcal} \mathrm{eq}^{-1}$
$\Delta \mathrm{H}_{\text {ion }}\left(\mathrm{CH}_{3} \mathrm{COOH}ight)$
$=-12.5-(-13.7)=1.2$ kcal mol $^{-1}$
$\Delta \mathrm{H}_{\mathrm{ion}}\left(\mathrm{NH}_{4} \mathrm{OH}ight)$
$=-10.5-(-13.7)-\Delta \mathrm{H}_{\text {ion }}\left(\mathrm{CH}_{3} \mathrm{COOH}ight)$
$=13.7-10.5-1.2$
$=2 \mathrm{kcal} \mathrm{mol}^{-1}$ ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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