Question
At $$373\,K,$$ steam and water are in equilibrium and $$\Delta H = 40.98\,kJ\,mo{l^{ - 1}}.$$ What will be $$\Delta S$$ for conversion of water into steam ?
$${H_2}{O_{\left( l \right)}} \to {H_2}{O_{\left( g \right)}}$$
A.
$$109.8\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
B.
$$31\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
C.
$$21.98\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
D.
$$326\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
Answer :
$$109.8\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {S_{{\text{vap}}}} = \frac{{\Delta {H_{{\text{vap}}}}}}{{{T_b}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40.98 \times 1000}}{{373}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 109.8\,J\,{K^{ - 1}}\,mo{l^{ - 1}} \cr} $$