JEE Main 2022 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 26 July 2022, Shift 2, Chemistry Q83

Question 83 of 90 in this shift, Chemistry question 23 of 30, Section B.

For the reaction

\mathrm{H_2F_2(g) \rightarrow H_2(g) + F_2(g)}

\Delta U

= −59.6 kJ

\mathrm{mol^{-1}}

at 27°C. The enthalpy change for the above reaction is (−) ________ kJ

\mathrm{mol^{-1}}

[nearest integer] Given: R = 8.314 J

\mathrm{K^{-1}mol^{-1}}

Official answer

NTA final key.

Chapter: Thermodynamics (Chemistry)
Topic: Enthalpy and Hess's Law
Idea tested: Relation Between ΔH and ΔU

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Asked 10× in all

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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.