While performing a thermodynamics experiment, a student made the following observations, $HCl + NaOH \to NaCl + {H}_{2}O\DeltaH = - 57 . 3kJ{mol}^{- 1}$
${CH}_{3}COOH + NaOH \to {CH}_{3}COONa + {H}_{2}O$
$\Delta H = - 55 . 3kJ{mol}^{- 1}$ .
The enthalpy of ionization of ${CH}_{3}COOH$ as calculated by the student is $kJ{mol}^{- 1}$ .

Question

While performing a thermodynamics experiment, a student made the following observations, $HCl + NaOH \to NaCl + {H}_{2}O\DeltaH = - 57 . 3kJ{mol}^{- 1}$

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

$\Delta H = - 55 . 3kJ{mol}^{- 1}$ .

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

Quantrex Academy · Accepted Answer

Correct answer: 2

The enthalpy of neutralization (&Delta;H 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 Ionizationofweakacid = ∆ _ H NeutralizationofWA + SB - ∆ _ H NeutralizationofSA + SB = Δ H _ ionisation of CH _ 3 COOH = - 55 . 3 - ( - 57 . 3 ) = 2KJ / mol Hence, 2KJ / mol of heat is absorbed for the ionization of acetic acid.

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

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