Question
Enthalpy change for the process, $${H_2}O\left( {{\text{ice}}} \right) \rightleftharpoons {H_2}O\left( {{\text{water}}} \right)$$ is $$6.01\,kJ\,mo{l^{ - 1}}.$$ The entropy change of $$1\,mole$$ of ice at its melting point will be
A.
$$12\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
B.
$$22\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
C.
$$100\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
D.
$$30\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
Answer :
$$22\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {S_{{\text{fusion}}}} = \frac{{\Delta {H_{{\text{fusion}}}}}}{{{T_{{\text{fusion}}}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6.01 \times 1000}}{{273}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 22\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$