Question
If the enthalpy change for the transition of liquid water to steam is $$30\,kJ\,mo{l^{ - 1}}$$ at $${27^ \circ }C,$$ the entropy change for the process would be
A.
$$1.0\,J\,mo{l^{ - 1}}{K^{ - 1}}$$
B.
$$0.1\,J\,mo{l^{ - 1}}{K^{ - 1}}$$
C.
$$100\,J\,mo{l^{ - 1}}{K^{ - 1}}$$
D.
$$10\,J\,mo{l^{ - 1}}{K^{ - 1}}$$
Answer :
$$100\,J\,mo{l^{ - 1}}{K^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ } \cr
& {\text{Given,}}\,\,\Delta {H_{vap.}} = 30\,kJ\,mo{l^{ - 1}} \cr
& T = 27 + 273 \cr
& \,\,\,\,\, = 300\,k \cr
& \Delta {G^ \circ } = 0\,\,{\text{at equilibrium,}} \cr
& \Delta {S_{vap}} = \frac{{\Delta {H_{vap}}}}{T} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{30 \times {{10}^3}J\,mo{l^{ - 1}}}}{{300\,K}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100\,J\,mo{l^{ - 1}}{K^{ - 1}} \cr} $$