Chemistry · 2013
JEE Main · 9 April 2013, Shift 1 · Q43
Given: (a) H_2( g )+1 / 2 O_2( g ) → H_2 O ( l ); Δ H^°_298 K =-285.9 kJ mol^-1 (b) H_2( g )+1/2 O_2( g ) → H_2 O ( g ); Δ H^°_298 K =-241.8 kJ…
Given :
(a)
$$\begin{aligned}
& \mathrm{H}_2(\mathrm{~g})+$\displaystyle 1$ / $\displaystyle 2$ \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ;
& \Delta \mathrm{H}^{\circ}{ }_{$\displaystyle 298$ \mathrm{~K}}=-$\displaystyle 285.9$ \mathrm{~kJ} \mathrm{~mol}^{-$\displaystyle 1$}
\end{aligned}
$$(b) $\displaystyle \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$;
$\displaystyle \Delta \mathrm{H}^{\circ}{ }_{298 \mathrm{~K}}=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The molar enthalpy of vapourisation of water will be :
Official answer
(3)
$\displaystyle 44.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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JEE Main 2013 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.