Question

calculate the entropy change for 2 mole of ice.Freezing point of water is 273 k and molar enthalpy for fashion is 6 KJ Mole^-1.

Answer 1

Calculate the entropy change for 2 mole of ice. Freezing point of water is $273\,\mathrm{k}$ and molar enthalpy for fashion is 6 KJ Mole$^{\wedge}$-1.

Entropy is a mathematically-defined thermodynamic quantity that helps to account for the flow of energy through a thermodynamic process. Entropy was originally defined for a thermodynamically reversible process as:

$$\Delta S = \int \frac{dQ_{rev}}{T}$$

where the uniform temperature $(T)$ of a closed system is divided into an incremental reversible transfer of heat energy into that system $(dQ_{rev})$.

Therefore:

$$\Delta S = \int \frac{dQ}{273\,K} = \frac{Lv}{273\,K}$$

L - enthalpy for fashion

$v$ - amount of ice

$$\Delta S = \frac{2\,\mathrm{mole} \cdot \frac{6\,KJ}{\mathrm{mole}}}{273\,K} = \frac{12\,KJ}{273\,K} = 44\,\frac{J}{K}$$

Answer: $\Delta S = 44\,\frac{J}{K}$