The given thermochemical equations are $\mathrm{CH_3OH}_{(l)} + \frac{3}{2} \mathrm{O}_{2(g)}$ → $\mathrm{CO}_{2(g)} + 2\mathrm{H_2O}_{(l)}$ ; $\Delta_r H^\circ$ = - 726 $\mathrm{kJmol^{-1}}$ ... (1) $\mathrm{C}_{(\text{graphite})} + \mathrm{O}_{2(g)}$ → $\mathrm{CO}_{2(g)}$ ; $\Delta_c H^\circ$ = - 393 $\mathrm{kJ\,mol^{-1}}$ ... (2) $\mathrm{H}_{2(g)} + \frac{1}{2} \mathrm{O}_{2(g)}$ → $\mathrm{H_2O}_{(l)}$ ; $\Delta_f H^\circ$ = - 286 $\mathrm{kJ\,mol^{-1}}$ ... (3) The required thermochemical equation is $\mathrm{C}_{(s)} + 2\mathrm{H}_{2(g)} + \frac{1}{2} \mathrm{O}_{2(g)}$ → $\mathrm{CH_3OH}_{(l)}$ ; ΔH = ? 2 × eqn. (3) + eqn. (2) – eqn. (1) gives $\mathrm{C}_{(s)} + 2\mathrm{H}_{2(g)} + \frac{1}{2} \mathrm{O}_{2(g)}$ → $\mathrm{CH_3OH}_{(l)}$$\mathrm{C}_{(s)} + 2\mathrm{H}_{2(g)} + \frac{1}{2} \mathrm{O}_{2(g)}$ → $\mathrm{CH_3OH}_{(l)}$ ΔH = (–286 × 2) + (–393) – (–726) = –572 – 393 + 726 = –239 kJ ∴ $\Delta_f H^\circ$ for $\mathrm{CH_3OH}_{(l)}$ = –239 $\mathrm{kJ\,mol^{-1}}$