JEE Main 2022ChemistryThermodynamics (C)MediumNumerical

JEE Main 2022Thermodynamics (C) Question with Solution

JEE Main 2022 (25 Jul Shift 2)

Question

While performing a thermodynamics experiment, a student made the following observations, HCl+NaOHNaCl+H2OΔH=-57.3 kJ mol-1

CH3COOH+NaOHCH3COONa+H2O

ΔH=-55.3 kJ mol-1.

The enthalpy of ionization of CH3COOH as calculated by the student is kJ mol-1

Enter your answer

Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

The enthalpy of neutralization (ΔHn) 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 =

ΔHionisation of CH3COOH=-55.3-(-57.3)                   

=2KJ/mol

Hence, 2KJ/mol of heat is absorbed for the ionization of acetic acid.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermodynamics (C) chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, covering the Thermodynamics (C) chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.