Column-I (process)		Column-II ($∆\mathrm{G}$)
(I)	Freezing of water at 270 K	(A)	-Ve
(II)	Freezing of water at 273 K	(B)	Zero
(III)	Freezing of water at 279 K	(C)	+Ve

Correct match is

$\begin{array}{l}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right);\mathrm{\Delta H}=-\mathrm{Ve}; \mathrm{Freezing} \mathrm{is} \mathrm{exothermic};\\ \mathrm{For}\mathrm{this}\mathrm{process}\mathrm{\Delta S}=-\mathrm{Ve}; \mathrm{Freezing} \mathrm{leads} \mathrm{to} \mathrm{decrease} \mathrm{in} \mathrm{randomness};\\ \mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}\\ \mathrm{\Delta G}=\left(-\mathrm{Ve}\right)-\mathrm{T}\left(-\mathrm{Ve}\right)\\ \mathrm{Freezing}\mathrm{point}\mathrm{of}\mathrm{water}\mathrm{is}273\mathrm{K}\\ \therefore \mathrm{At}273\mathrm{K};\mathrm{\Delta G}=0\\ \mathrm{When}\mathrm{T}>273\mathrm{K},\mathrm{\Delta G}=+\mathrm{Ve}\\ \mathrm{When}\mathrm{T}<273\mathrm{K},\mathrm{\Delta G}=-\mathrm{Ve}\end{array}$