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The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at $0^\circ C$ is
A
0.526 cal/(mol K)
B
10.52 cal/(mol K)
C
21.04 cal/(mol K)
D
5.260 cal/(mol K)
Detailed Solution
At the melting point the process is reversible, so $\Delta S_{fusion} = \frac{\Delta H_f}{T}$
$\Delta H_f = 1.435$ kcal/mol $= 1.435\times10^3$ cal/mol; T = 273 K
$\Delta S_{fusion} = \frac{1.435\times10^3}{273} = 5.26$ cal/(mol K)
$\Delta H_f = 1.435$ kcal/mol $= 1.435\times10^3$ cal/mol; T = 273 K
$\Delta S_{fusion} = \frac{1.435\times10^3}{273} = 5.26$ cal/(mol K)