The entropy of fusion of water is +22.0 J.K^-¹.mol^-¹ and the enthalpy of fusion of water is +6.01 J.K^-¹.mol^-¹at 0^$\omicron$C.At 0^$\omicron$C, $\Delta$S_total for the melting of ice is

Question

The entropy of fusion of water is +22.0 J.K^-¹.mol^-¹> and the enthalpy of fusion of water is +6.01 J.K^-¹.mol^-¹at 0^$\omicron$C.At 0^$\omicron$C, $\Delta$S_total for the melting of ice is

Quizplus · Accepted Answer

The Gibbs free energy change ($\Delta G$) for a process at constant temperature and pressure can be calculated using the equation $\Delta G = \Delta H - T\Delta S$, where $\Delta H$ is the enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S$ is the entropy change. For the melting of ice at 0°C (273.15 K), $\Delta H = +6.01 \times 10^3 \, \text{J/mol}$ and $\Delta S = +22.0 \, \text{J/mol}\cdot K$. Converting 0°C to Kelvin gives 273.15 K. Plugging these values into the equation gives $\Delta G = 6010 \, \text{J/mol} - 273.15 \, K \times 22.0 \, \text{J/mol}\cdot K = 6010 \, \text{J/mol} - 6010 \, \text{J/mol} = 0$. Therefore, the Gibbs free energy change for the melting of ice at 0°C is 0, indicating the process is at equilibrium.

The Entropy of Fusion of Water Is +22 ο\omicronο CAt 0 ο\omicronο

The Entropy of Fusion of Water Is +22 $\omicron$ CAt 0^$\omicron$