Given : (I) $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) ightarrow \mathrm{H}_{2}…

Given :
(I) $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) ightarrow \mathrm{H}_{2} \mathrm{O}(l)$
$\Delta \mathrm{H}^{\circ}{ }_{298 \mathrm{~K}}=-285.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(II) $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) ightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\Delta \mathrm{H}^{\circ}{ }_{298 \mathrm{~K}}=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The molar enthalpy of vapourisation of water will be :
  1. $241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $22.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $44.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $527.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Given
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$
$\Delta \mathrm{H}^{\circ}=-285.9 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad \ldots$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)$
$\Delta \mathrm{H}^{\circ}=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad \ldots$
We have to calculate
$\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=?$
On substracting eqn. (2) from eqn. (1) we get $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\Delta \mathrm{H}^{\circ}=-241.8-(-285.9)$
$=44.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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