Q14 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Chemistry
For strong acid and strong base neutralisation net chemical change is
$\begin{aligned} \mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow & \mathrm{H}_2 \mathrm{O}(l) ; \\ \Delta_r H^{\circ} & =-55.84 \mathrm{kJmol}^{-1} .\end{aligned}$
If enthalpy of neutralisation of $\mathrm{CH}_3 \mathrm{COOH}$ by $\mathrm{NaOH}$ is $49.86 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then enthalpy of ionisation of $\mathrm{CH}_3 \mathrm{COOH}$ is
- Correct answerA$5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$-5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$105.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$-59.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\because \mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O} ; \Delta_r H^{\circ}=-55.84 \mathrm{~kJ} / \mathrm{mol}$
$\therefore \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} ; \Delta_r H^{\circ}=55.84 \mathrm{~kJ} / \mathrm{mol}$ ...(i)
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}$
$\Delta_r H=49.86$ ...(ii)
$\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \Delta H_{\mathrm{i}}=?$
Subtracting Eq. (ii) from Eq. (i)
$\Delta H_{\mathrm{i}}=55.84-49.86$
$=5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Hence, the correct answer is option: 'A'.