Given $C_{(graphite)}+O_2 (g) \to CO_2(g) ; \Delta_r H^{\circ} =-393.5\;kj\;mol^{-1}\\$ $H_2(g)+\large\frac{1}{2}$$O_2(g) \to H_2O(l) ; \Delta_r H^{\circ} =-285.8\;kj\;mol^{-1}\\$ $CO_2(g)+2H_2O(l) \to CH_4(g) +2O_2(g) ; \Delta_{r}H^{\circ} =+890.3\;kJ mol^{-1}\\$ Based on the above thermochemical equations, the value of $\Delta_r H^{\circ}$ at $298\; K$ for the reaction $C_{(graphite)}+2H_2(g) \to CH_4(g)$ will be

Question

$\begin{array}{1 1} (1) −74.8\; kJ mol^{−1} \\ (2) −144.0 kJ\; mol^{−1} \\ (3) +74.8\; kJ mol^{−1} \\ (4) +144.0\; kJ mol^{−1} \end{array}$