Question
Given :
$$\eqalign{
& {\text{(i)}}\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right); \cr
& \Delta {H^ \circ }_{298K} = - 285.9\,kJ\,mo{l^{ - 1}} \cr
& {\text{(ii)}}\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( g \right); \cr
& \Delta {H^ \circ }_{298K} = - 241.8\,kJ\,mo{l^{ - 1}} \cr} $$
The molar enthalpy of vapourisation of water will be :
A.
$$241.8\,kJ\,mo{l^{ - 1}}$$
B.
$$22.0\,kJ\,mo{l^{ - 1}}$$
C.
$$44.1\,kJ\,mo{l^{ - 1}}$$
D.
$$527.7\,kJ\,mo{l^{ - 1}}$$
Answer :
$$44.1\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Given}} \cr
& {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right); \cr
& \Delta {H^ \circ } = - 285.9\,kJ\,mo{l^{ - 1}}\,\,\,...\left( 1 \right) \cr
& {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( g \right); \cr
& \Delta {H^ \circ } = - 241.8\,kJ\,mo{l^{ - 1}}\,...\left( 2 \right) \cr
& {\text{We have to calculate}} \cr
& {H_2}O\left( l \right) \to {H_2}O\left( g \right)\,;\,\,\Delta {H^ \circ } = ? \cr
& {\text{On substracting eqn}}{\text{. (2) from eqn}}{\text{. (1) we get}} \cr
& {{\text{H}}_2}O\left( l \right) \to {H_2}O\left( g \right)\,;\Delta {H^ \circ } = - 241.8 - \left( { - 285.9} \right) \cr
& = 44.1\,kJ\,mo{l^{ - 1}} \cr} $$