JEE Main 2018ChemistryThermodynamics (C)MediumMCQ

JEE Main 2018 — Thermodynamics (C) Question with Solution

JEE Main 2018 (08 Apr)

Question

The combustion of benzene l gives CO2g and H2Ol. Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25°C; the heat of combustion in kJ mol-1 of benzene at constant pressure will be

R=8.314 JK-1 mol-1

Choose an option

Show full solutionCorrect option: A
Correct answer
A-3267.6

Step-by-step explanation

C6H6l+152O2g→6CO2g+3H2O(I)    ∴  ∆ng= -32

Use, ΔH=ΔU+ΔngRT

=-3263.5×103-32×8.314×298×10-3 

 =-3267.6 kJ mol-1

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About this question

This is a previous-year question from JEE Main 2018, 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.