Verified Solution • 2025–26 Session moderate

QJEEM22Ch742d: JEE Main 2022 Chemistry - Class 11 JEEM22Ch742d Solution: While performing a thermodynamics experiment, a student made…

While performing a thermodynamics experiment, a student made the following observations.

HCl + NaOH →\to NaCl + H2O Δ\Delta H = −- 57.3 kJ mol −- 1

CH3COOH + NaOH →\to CH3COONa + H2O Δ\Delta H = −- 55.3 kJ mol −- 1

The enthalpy of ionization of CH3COOH as calculated by the student is _____________ kJ mol −- 1. (nearest integer)

Step-by-Step Verified Solution

(I) HCl+NaOH→NaCl+H2O\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}

ΔH1=−57.3 KJmol−1\Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1}

(II) CH3COOH+NaOH→CH3COONa+H2O\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}

ΔH2=−55.3 KJmol−1\Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1}

Reaction (I) can be written as

 (III) NaCl+H2O→HCl+NaOH\text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH}

ΔH3=57.3 KJmol−1\Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1}

By adding (II) and (III)

amp;CH3COOH+NaCl→CH3COONa+HClΔHramp;ΔHr=ΔH3+ΔH2amp;=57.3−55.3amp;=2 kJ mol−1\begin{aligned}&\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\&\begin{aligned}\Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\&=2 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\end{aligned}

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