JEE Main 2013 — Thermodynamics (C) Question with Solution
JEE Main 2013 (09 Apr Online)
Question
Given:
$\begin{aligned}
& \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta \mathrm{H}^{\circ}{ }_{298 \mathrm{~K}}=-285.9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
$\begin{aligned}
& \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}); \\
& \Delta \mathrm{H}^{\circ}{ }_{298 \mathrm{~K}}=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
The molar enthalpy of vapourisation of water will be :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
Given
;
;
We have to calculate
On substracting eqn. from eqn. we get
;
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This is a previous-year question from JEE Main 2013, 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.