JEE Main 2022ChemistryThermodynamics (C)MediumMCQ

JEE Main 2022 — Thermodynamics (C) Question with Solution

JEE Main 2022 (24 Jun Shift 2)

Question

At 25 °C and 1 atm pressure, the enthalpies of combustion are as given below:

Substance H2 C (graphite) C2H6g
ΔcHΘkJmol-1 -286.0 -394.0 -1560.0

The enthalpy of formation of ethane is

Choose an option

Show full solutionCorrect option: C
Correct answer
C-86.0 kJ mol-1

Step-by-step explanation

Given

(i) C2H6g+72O2g⟶2CO2g+3H2OlΔHcomb°=-1560.0 kJ/mole

(ii) Cs+O2g⟶CO2gΔHcomb°=-394.0 kJ/mole

(ii) H2g+12O2g⟶H2OgΔHcomb°=-286.0 kJ/mole

Target 2Cs+3H2g⟶C2H6gΔHrx°=ΔHf°C2H6, g

ΔHr°=ΔHc°reactant-ΔHc°Product

=2×-394+3-286--1560

=-788-858+1560

=-86.0KJ/mole

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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.