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Saturday, 31 October

JEE Main 2022 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 25 July 2022, Shift 2, Chemistry Q84

Question 84 of 90 in this shift, Chemistry question 24 of 30, Section B.

While performing a thermodynamics experiment, a student made the following observations. HCl+NaOH→NaCl+H2O\mathrm{HCl + NaOH \rightarrow NaCl + H_2O} ΔH=−57.3\Delta H = -57.3 kJ mol−1^{-1} CH3COOH+NaOH→CH3COONa+H2O\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O} ΔH=−55.3\Delta H = -55.3 kJ mol−1^{-1} The enthalpy of ionization of CH3COOH\mathrm{CH_3COOH} as calculated by the student is __________ kJ mol−1^{-1}. (nearest integer)

Official answer

2

NTA final key.

Chapter
Thermodynamics (Chemistry)
Topic
Enthalpy and Hess's Law
Idea tested
Enthalpy of Ionization of Weak Acids/Bases

Same topic in other shifts

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.