Use the following themochemical equations to calculate the standard enthalpy change of formation of $\mathrm{HNO}_{3}(0)$
\[
\begin{aligned}
\mathrm{H}_{2} \mathrm{O}_{2}\left(\mathrm{D}+2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HNO}_{2}()\right.& \Delta_{4} H^{\circ} &=-226.8 \mathrm{kJ} \mathrm{mol}^{-1} \\
\mathrm{N}_{2}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) & \Delta_{\mathrm{r}} H^{\circ} &=+66.4 \mathrm{kJ} \mathrm{mol}^{-1} \\
\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{o}) & \Delta_{\mathrm{r}} \mathrm{H}^{\mathrm{e}} &=-187.8 \mathrm{kJ} \mathrm{mol}^{-1}
\end{aligned}
\]
(Section $13.1)$