While performing a thermodynamics experiment, a student made the following observations. HCl + NaOH → NaCl + H_2 O Δ H =-57.3 kJ mol^-1; CH_3 COOH +…

While performing a thermodynamics experiment, a student made the following observations. $$\begin{aligned} & \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-55.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ The enthalpy of ionization of $\displaystyle \mathrm{CH}_3 \mathrm{COOH}$ as calculated by the student is \_\_\_\_ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$. (nearest integer)