Given C_(graphite) + O_2( g ) → CO_2( g ); Δ_r H^°=-393.5 kJ mol^-1; H_2( g )+1/2 O_2( g ) → H_2 O ( l ); Δ_r H^°=-285.8 kJ mol^-1; CO_2( g )+2 H_2 O…

Given $$\begin{aligned} & \mathrm{C}_{\text {(graphite) }}+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \\ & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ $$\begin{aligned} & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \\ & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ $$\begin{aligned} & \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) ; \\ & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ Based on the above thermochemical equations, the value of $\displaystyle \Delta_{\mathrm{r}} \mathrm{H}^{\circ}$ at $\displaystyle 298$ K for the reaction $$\mathrm{C}_{\text {(graphite) }}+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g}) \text { will be : } $$