The standard enthalpy of fusion of crystalline ice is $6.0 ext{ kJ/mol}$. What is the precise mola

Question

The standard enthalpy of fusion of crystalline ice is $6.0 	ext{ kJ/mol}$. What is the precise molar entropy change ($\Delta S_{	ext{fus}}$) characterizing the melting of ice specifically at exactly $273 	ext{ K}$?

Professor PC Thomas Classes · Answer

(b) : Melting (fusion) at the normal melting point is an equilibrium process, meaning $\Delta G = 0$. Therefore, the entropy of fusion is given by $\Delta S_{	ext{fus}} = \frac{\Delta H_{	ext{fus}}}{T_{	ext{fus}}}$. Substituting the given values (ensuring to convert kJ to J): $\Delta S_{	ext{fus}} = \frac{6000 	ext{ J/mol}}{273 	ext{ K}} \approx 21.978 	ext{ J/(mol K)}$, which rounds neatly to $22.0 	ext{ J/(mol K)}$.