Given are the following data at 298 K.${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }\left({\mathrm{H}}_2\mathrm{O},\mathrm{l}\right)=-286\mathrm{kJ}{\mathrm{mol}}^{-1}$ ${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }\left({\mathrm{HNO}}_3,1\right)=-174\mathrm{kJ}{\mathrm{mol}}^{-1};{\mathrm{\Delta }}_f{H}^{\circ }\left({\mathrm{N}}_2{\mathrm{O}}_5,\mathrm{g}\right)=14.5\mathrm{kJ}{\mathrm{mol}}^{-1}$, The ${\mathrm{\Delta }}_{\mathrm{r}}{H}^{\circ }$ for the reaction ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶2{\mathrm{HNO}}_3\left(\mathrm{l}\right)$ is

$$\begin{gathered} \text{ (i) }{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);{\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }=-286\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \text{ (ii) }\frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)⟶{\mathrm{HNO}}_3\left(\mathrm{l}\right)\text{; }{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\circ }=-174\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \text{ (iii) }{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{5}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)\text{; }{\mathrm{\Delta H}}^{\circ }=-14.5{\mathrm{kJmol}}^{-1} \end{gathered}$$

The required reaction ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶2{\mathrm{NHO}}_3\left(\mathrm{l}\right)$ is obtained as -Eq.(iii) - Eq. (i) + 2 Eq. (ii) Hence, $\left.{\mathrm{\Delta }}_1{H}^{\circ }=[-14.5-(-286\left)\right)+2(-174)\right]\mathrm{kJ}{\mathrm{mol}}^{-1}=-76.5\mathrm{kJ}{\mathrm{mol}}^{-1}$