While performing a thermodynamics experiment, a student made the following observations, $\ce{HCl + NaOH -> NaCl + H2O}$ $\Delta \mathrm{H}=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\ce{CH3COOH + NaOH -> CH3COONa + H2O}$ $\Delta \mathrm{H}=-55.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy of ionization of $\ce{CH3COOH}$ as calculated by the student is _____ $\mathrm{kJ} \mathrm{mol}^{-1}$. (nearest integer)

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Chemical Thermodynamics

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

HCl + NaOH NaCl + H X_{2} O

Δ H = - 57.3 kJ mol^{- 1}

CH X_{3} COOH + NaOH CH X_{3} COONa + H X_{2} O

Δ H = - 55.3 kJ mol^{- 1}

.

The enthalpy of ionization of

CH X_{3} COOH

as calculated by the student is _____

kJ mol^{- 1}

. (nearest integer)

Answer: 2

The first reaction involves complete ionization of strong acid HCl. The second involves weak acid

CH X_{3} COOH

which is partially ionized. The difference in enthalpy represents the ionization enthalpy of acetic acid:

Δ H_{ionization} = ∣ - 57.3 - (- 55.3) ∣ = ∣ - 2.0∣ = 2

kJ/mol.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer