The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The solid form of $\mathrm{H}_2 \mathrm{O}$ is ice. In ice, molecules of $\mathrm{H}_2 \mathrm{O}$ are less random than in liquid water.

Thus, molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})<$ molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$. The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})$ is less than $70 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

State functions are those values which depend only on the state of the system and not on how it is reached e.g., enthalpy, entropy, temperature and free energy. Path functions are those values which depend on the path of the system. e. $g$, heat and work.

Amount of heat required to vaporise one mole of a liquid at constant temperature and under standard pressure ( 1 bar ) is called its molar enthalpy of vaporisation $\Delta_{\text {vap }} H^{\mathrm{s}}$. Molar enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in $\mathrm{H}_2 \mathrm{O}$ molecule.

Gibbs energy for a reaction in which all reactants and products are in standard state. $\Delta_r G^{\mathrm{s}}$ is related to the equilibrium constant of the reaction as follows

$$\begin{aligned} \Delta_r G & =\Delta_r G^{\mathrm{s}}+R T \ln K \\ \text{At equilibrium,}\quad 0 & =\Delta_r G^{\circ}+R T \ln K \quad \left(\because \Delta_r G=0\right)\\ \text{or}\quad \Delta_r G^{\mathrm{s}} & =-R T \ln K \\ \Delta_r G^{\mathrm{s}} & =0 \text { when } K=1 \end{aligned}$$

For all other values of $K, \Delta_r G^{\mathrm{s}}$ will be non-zero.

For isolated system there is no transfer of energy as heat, i.e., $q=0$ and there is no transfer of energy as work. i.e., $\mathrm{W}=0$. According to the first law of thermodynamics

$$\begin{aligned} \Delta U & =q+W \\ \Delta U & =0+0=0 \end{aligned}$$