For strong acid and strong base neutralisation net chemical change is $\begin{aligned}…

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
  1. $5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $105.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-59.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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) $\Delta H_{\mathrm{i}}=55.84-49.86$ $=5.98 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya