The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$

(A) The combustion reactions are:
$(i) CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_{2}O_{(l)}; \Delta_cH = -890.3 \, kJ \, mol^{-1}$
$(ii) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta_cH = -393.5 \, kJ \, mol^{-1}$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(l)}; \Delta_cH = -285.8 \, kJ \, mol^{-1}$
The formation reaction of $CH_{4(g)}$ is:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
Using Hess's Law:
$\Delta_fH_{CH_4} = \Delta_cH_{C(s)} + 2(\Delta_cH_{H_2(g)}) - \Delta_cH_{CH_4(g)}$
$\Delta_fH_{CH_4} = [-393.5 + 2(-285.8) - (-890.3)] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = [-393.5 - 571.6 + 890.3] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = -74.8 \, kJ \, mol^{-1}$
Thus,the enthalpy of formation is $-74.8 \, kJ \, mol^{-1}$,which corresponds to option $(i)$.