| (I) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l);\]\[\Delta {{H}^{o}}_{298k}=-285.9\,kJ\,mo{{l}^{-1}}\] |
| (II) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(g);\]\[\Delta {{H}^{o}}_{298k}=-241.8\,kJ\,mo{{l}^{-1}}\] |
| The molar enthalpy of vapourisation of water will be: |
A) \[241.8\text{ }kJ\text{ }mo{{l}^{-1}}\]
B) \[~22.0\text{ }kJ\text{ }mo{{l}^{-1}}\]
C) \[44.1kJ\,mo{{l}^{-1}}\]
D) \[527.7\text{ }kJ\text{ }mo{{l}^{-1}}\]
Correct Answer: C
Solution :
[c] Given \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l);\] \[\Delta H{}^\circ =-285.9\text{ }kJ\,mo{{l}^{-1}}\] ..... (1) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}{{H}_{2}}O(g);\] \[\Delta {{H}^{{}^\circ }}=-241.8\text{ }kJ\,mo{{l}^{-1}}\] ..... (2) We have to calculate \[{{H}_{2}}O\left( l \right)\xrightarrow{{}}{{H}_{2}}O\left( g \right);\Delta {{H}^{{}^\circ }}=?\] On substracting eqn. (2) from eqn. (1) we get \[{{H}_{2}}O\left( l \right)\xrightarrow{{}}{{H}_{2}}O\left( g \right);\] \[\Delta {{H}^{{}^\circ }}=-241.8-\left( -285.9 \right)\] \[=44.1\text{ }kJ\,mo{{l}^{-1}}\]
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