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Question

One mole of ice is converted into water at 273 K and 1 atm. The entropies of H2O (s) and H2O (l) are 38.0 and 58.0 J/K-mol, respectively. The enthalpy change for the conversion is

Options

A.5460 kJ/mol
B.5460 J/mol
C.−5460 J/mol
D.20 J/mol

Solution

$\Delta S = S_{\text{water}} - S_{\text{ice}} = 58.0 - 38.0 = 20\text{ J/K.mol}$

Now, $\Delta H = T.\Delta S = 273 \times 20 = 5460\text{ J/mol}$

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