What is the free energy change $\left(\mathrm{\Delta G}\right)$ when $1.0$ mole of water at $100°\mathrm{C}$ and $1 \mathrm{atm}$ pressure is converted into steam at $100°\mathrm{C}$ and $1 \mathrm{atm}$ pressure ?

For the process: ${\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right)\left(1 \mathrm{bar}, 373 \mathrm{K}\right)\to {\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right)\left(1 \mathrm{bar}, 373 \mathrm{K}\right)$, the correct set of the thermodynamic parameters is

The value of ${\log }_{10}\mathrm{K}$ for a reaction $\mathrm{A}\rightleftharpoons \mathrm{B}$ is

Given:

The standard enthalpy of the formation of ${\mathrm{NH}}_3$ is $-46.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$. If the enthalpy of the formation of ${\mathrm{H}}_2$ from its atoms is $-436 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and that of ${\mathrm{N}}_2$ is $-712 \mathrm{kJ} {\mathrm{mol}}^{-1}$, the average bond enthalpy of $\mathrm{N}-\mathrm{H}$ bond in ${\mathrm{NH}}_3$ is:

On the basis of the following thermochemical data:

${\mathrm{\Delta H}}_{\mathrm{f}}°\left({\mathrm{H}}_{\mathrm{aq}}^{+}\right)=0$

$\begin{array}{l}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right); \mathrm{\Delta H}°=57.32\mathrm{kJ}\\ {\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); \mathrm{\Delta H}°=-286.20\mathrm{kJ}\end{array}$

The value of the enthalpy of formation of ${\mathrm{OH}}^{-}$ ions at $25°\mathrm{C}$ is: