Given are the following data at 298 K

${\mathrm{\Delta }}_{\mathrm{c}}{H}^{\circ }\left({\mathrm{H}}_2,\mathrm{g}\right)=-286.1\mathrm{kJ}{\mathrm{mol}}^{-1};$ ${\mathrm{\Delta }}_{\mathrm{c}}{H}^{\circ }\left(\mathrm{C}\text{, graphite }\right)=-394.9\mathrm{kJ}{\mathrm{mol}}^{-1}$ ; ${\mathrm{\Delta }}_{\mathrm{c}}{H}^{\circ }\left({\mathrm{CH}}_4,\mathrm{g}\right)=-882.0\mathrm{kJ}{\mathrm{mol}}^{-1}$

The value of ${\mathrm{\Delta }}_4{H}^{\circ }\left({\mathrm{CH}}_4,\mathrm{g}\right)$ will be 

Solution:

$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);{\mathrm{\Delta }}_{\mathrm{c}}{H}^{\circ }=-286.1\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \text{ (ii) }\mathrm{C}\text{ (graphite) }+{\mathrm{O}}_2\text{ (g) }⟶{\mathrm{CO}}_2\left(\mathrm{g}\right)\text{; }{\mathrm{\Delta }}_{\mathrm{c}}{H}^{\circ }=-394.9\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \text{ (iii) }{\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2⟶{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);{\mathrm{\Delta cH}}^{\circ }=-882.0\mathrm{kJ}{\mathrm{mol}}^{-1} \end{gathered}$$

The required chemical equation for which ${\mathrm{\Delta }}_f{H}^{\circ }$ is required is C(graphite) + 2H₂ (g) ~ CH₄(g)

This equation is obtained by the manipulations Eq.(ii) + 2 Eq. (i) - Eq. (iii)

$\mathrm{Hence} {\mathrm{\Delta }}_f{H}^{\circ }\left({\mathrm{CH}}_4,\mathrm{g}\right)=[-394.9+2(-286.1)-(-882.0\left)\right]\mathrm{kJ}{\mathrm{mol}}^{-1}=-85.1\mathrm{kJ}{\mathrm{mol}}^{-1}$