JEE Main 2022ChemistryThermodynamics (C)MediumNumerical

JEE Main 2022 — Thermodynamics (C) Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

For the reaction H2F2g→H2g+F2g

ΔU=-59.6 kJ mol-1 at 27 °C

The enthalpy change for the above reaction is -____ kJmol-1 (nearest integer)

(Given : R=8.314JK-1 mol-1) .

Enter your answer

Show full solutionCorrect answer: 57
Correct answer
57

Step-by-step explanation

For the reaction H2F2g→H2g+F2g

ΔH=ΔU+ΔngRT

Given ΔU=-59.6 kJ mol-1

Δng=2-1

=-59.6+1×8.3141000×300

=-57.11  kJ mol-1

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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.