The enthalpy of fusion of water is $1.435 ext{ kcal/mol}$. The molar entropy change for the melting of ice a

Question

The enthalpy of fusion of water is $1.435 	ext{ kcal/mol}$. The molar entropy change for the melting of ice at $0^\circ	ext{C}$ is:

Accepted Answer

For a phase transition like melting of ice, the entropy change ($\Delta_{fus} S$) is given by the formula:
$$\Delta_{fus} S = \frac{\Delta_{fus} H}{T_f}$$
where $\Delta_{fus} H$ is the enthalpy of fusion and $T_f$ is the freezing/melting point in Kelvin.
Given:
$\Delta_{fus} H = 1.435 	ext{ kcal/mol} = 1435 	ext{ cal/mol}$
$T_f = 0^\circ	ext{C} = 273 	ext{ K}$
Substituting the values:
$$\Delta_{fus} S = \frac{1435 	ext{ cal/mol}}{273 	ext{ K}} \approx 5.256 	ext{ cal/(mol K)}$$
This value is closest to $5.260 	ext{ cal/(mol K)}$.

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