$\mathrm{CH_4}+2\mathrm{O_2}$ → $\mathrm{CO_2}+2\mathrm{H_2O}$ ; ΔH = - 890.3 kJ ... (1) C + $\mathrm{O_2}$ → $\mathrm{CO_2}$ ; ΔH = - 393.5 kJ ... (2) $\mathrm{H_2}+\frac{1}{2}\mathrm{O_2}$ → $\mathrm{H_2O}$ ; ΔH = - 285.8 kJ or $2\mathrm{H_2}+\mathrm{O_2}$ → $2\mathrm{H_2O}$ ; ΔH = - 571.6 kJ ... (3) Add equations (2) and (3) and subtract equation (1) we get, C + $2\mathrm{H_2}$ → $\mathrm{CH_4}$ ΔH = –393.5 – 571.6 + 890.3 = –74.8 kJ/mol ∴ Enthalpy of formation of $\mathrm{CH_4}$ is –74.8 kJ/mol.