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 + NaOH → CH 3 COONa + H 2 O ΔH = - 55 . 3 kJ mol

Question

While performing a thermodynamics experiment, a student made the following observations, $HCl + NaOH \to NaCl + H_{2} O Δ H = - 57.3 kJ {mol}^{- 1}$

${CH}_{3} COOH + NaOH \to {CH}_{3} COONa + H_{2} O$

$ΔH = - 55.3 kJ {mol}^{- 1}$ .

The enthalpy of ionization of ${CH}_{3} COOH$ as calculated by the student is $kJ {mol}^{- 1}$ .

TestHub · Accepted Answer

The enthalpy of neutralization (ΔH n ) is the change in enthalpy that occurs when one equivalent of an acid and a base undergo a neutralization reaction to form water and a salt. It is a special case of the enthalpy of reaction. It is defined as the energy released with the formation of 1 mole of water. △ H Ionization of weak acid = ∆ H Neutralization of W A + S B − ∆ H Neutralization of S A + S B = ΔH ionisation of CH 3 COOH = - 55 . 3 - ( - 57 . 3 ) = 2 KJ / mol Hence, 2 KJ / mol of heat is absorbed for the ionization of acetic acid.

Chemistry - Thermochemistry Question with Solution | TestHub

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