Chemical Thermodynamics Question 352
Question: Given:
(I) $ H_2(g)+\frac{1}{2}O_2(g)\to H_2O(l); $
$ \Delta H^{o} _{298k}=-285.9,kJ,mol^{-1} $
(II) $ H_2(g)+\frac{1}{2}O_2(g)\to H_2O(g); $
$ \Delta H^{o} _{298k}=-241.8,kJ,mol^{-1} $
The molar enthalpy of vapourisation of water will be:
Options:
A) $ 241.8kJmo{{l}^{-1}} $
B) $ ~22.0kJmo{{l}^{-1}} $
C) $ 44.1kJ,mo{{l}^{-1}} $
D) $ 527.7kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given $ H_2(g)+\frac{1}{2}O_2(g)\xrightarrow{{}}H_2O(l); $
$ \Delta H{}^\circ =-285.9kJ,mo{{l}^{-1}} $ ….. (1) $ H_2(g)+\frac{1}{2}O_2\xrightarrow{{}}H_2O(g); $
$ \Delta {{H}^{{}^\circ }}=-241.8kJ,mo{{l}^{-1}} $ ….. (2) We have to calculate $ H_2O( l )\xrightarrow{{}}H_2O( g );\Delta {{H}^{{}^\circ }}=- $ On substracting eqn. (2) from eqn. (1) we get $ H_2O( l )\xrightarrow{{}}H_2O( g ); $
$ \Delta {{H}^{{}^\circ }}=-241.8-( -285.9 ) $
$ =44.1kJ,mo{{l}^{-1}} $
