Consider the following reaction 

${\mathrm{OF}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+2\mathrm{HF}\left(\mathrm{g}\right)$ if ${\mathrm{H}}_{\mathrm{f}}^0$ for ${\mathrm{OF}}_2\left(\mathrm{g}\right),{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\mathrm{and}\mathrm{HF}\left(\mathrm{g}\right)$are +20,-250,-270 ${\mathrm{KJmol}}^{-1}$at 298K then calculate the standard internal energy change for the reaction

$\begin{array}{l}\Delta H_{\text{reaction }}^0=\sum H_P^0-\sum H_R^0\\ \Delta H^0=\left[2\times H_{\mathrm{HF}}^0+H_{{\mathrm{O}}_2}^0\right]-\left[H_{{\mathrm{OF}}_2}^0+H_{{\mathrm{H}}_2\mathrm{O}}^0\right]\\ =[2\times (-270)-0]-[20-250]\left(\because H_{{\mathrm{O}}_2}^0=0\right)\\ =-310\mathrm{KJ}\\ \Delta H^0=\Delta U^0+\Delta \cap RT\\ \Delta n=3-2=1;R=8.314{\mathrm{Jk}}^{-1}{ \mathrm{mol}}^{-1},T=298 \mathrm{K}\\ \therefore -310\times {10}^3=\Delta U^0+1\times 8.314\times 298\\ \Delta Q^0=-312477.5 \mathrm{J}\\ =-312.47\mathrm{KJ}\end{array}$