1. Tardigrade
  2. Question
  3. Chemistry
  4. Given: (I) H2(g) + (1/2)O2(g) → H2O(l); Δ H°298K = -285.9kJ mol-1 (II) H2(g) + (1/2)O2(g) → H2O(g); Δ H° 298K = -244.8kJ mol-1 The molar enthalpy of vapourisation of water will be :

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Q. Given :
$\left(I\right) \quad H_{2}\left(g\right) + \frac{1}{2}O_{2}\left(g\right) \to H_{2}O\left(l\right);$
$\Delta H^{\circ}_{298K} = -285.9kJ\,mol^{-1}$
$\left(II\right) \quad H_{2}\left(g\right) + \frac{1}{2}O_{2}\left(g\right) \to H_{2}O\left(g\right);$
$\Delta H^{\circ }_{298K} = -244.8kJ\,mol^{-1}$
The molar enthalpy of vapourisation of water will be :

JEE MainJEE Main 2013Thermodynamics

Solution:

Given
$H_{2}\left(g\right) + \frac{1}{2}O_{2}\left(g\right) \to H_{2}O\left(l\right);$
$\Delta H^{\circ} = -285.9kJ\,mol^{-1}\quad\quad...\left(1\right)$
$H_{2}\left(g\right) + \frac{1}{2}O_{2}\left(g\right) \to H_{2}O\left(g\right);$
$\Delta H^{\circ } = -244.8kJ\,mol^{-1}\quad\quad...\left(2\right)$
We have to calculate
$H_{2}O\left(l\right)\to H_{2}O\left(g\right) ; \Delta H^{\circ} = ?$
On substracting eqn. $\left(2\right)$ from eqn. $\left(1\right)$ we get
$H_{2}O\left(l\right)\to H_{2}O\left(g\right) ;$
$\Delta H^{\circ } = -241.8-\left(-285.9\right)$
$= 44.1\,kJ\,mol^{-1}$