The average $\mathrm{Xe}-\mathrm{F}$ bond energy is $34 \mathrm{kcal}/\mathrm{mol}$, first I.E. of $\mathrm{Xe}$ is $279 \mathrm{kcal}/\mathrm{mol}$, electron affinity of $\mathrm{F}$ is $85 \mathrm{kcal}/\mathrm{mol}$ and bond dissociation energy of ${\mathrm{F}}_2$ is $38 \mathrm{kcal}/\mathrm{mol}$. Then, the enthalpy change for the reaction

${\mathrm{XeF}}_4\to {\mathrm{Xe}}^{+}+{\mathrm{F}}^{-}+{\mathrm{F}}_2+\mathrm{F}$ will be

Select the correct statement for the equilibrium under standard conditions:

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); {\mathrm{\Delta S}}_1^{\circ }$

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right); {\mathrm{\Delta S}}_2^{\mathrm{o}}$

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right); {\mathrm{\Delta S}}_3^{\mathrm{o}}$

For an ideal gas, consider only $\mathrm{P}-\mathrm{V}$ work in going from an initial state $\mathrm{X}$ to the final state $\mathrm{Z}$. The final state $\mathrm{Z}$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct? [Take $\mathrm{\Delta S}$ as change in entropy and $\mathrm{w}$ as work done.]