Chemistry - Thermochemistry Question with Solution | TestHub

Enthalpies of formation of elementary substances Xe and ${\mathrm{F}}_2$ are taken as zero. For (c) $\Delta H_{\text{reaction }}^{\circ }$ $=\Delta H_f^{\circ }\left({\mathrm{XeF}}_4\right)-\left[\Delta H_f^{\circ }(\mathrm{Xe})+2\Delta H_f^{\circ }\left({\mathrm{F}}_2\right)\right]$ Thus, $\Delta {\mathrm{H}}^{\circ }$ reaction $=\Delta {\mathrm{H}}^{\circ }{}_{\mathrm{f}}\left({\mathrm{XeF}}_4\right)$