Calculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$

(N/A) The reaction for the formation of $CH_{3}OH_{(l)}$ is:
$C_{(graphite)} + 2 H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow CH_{3}OH_{(l)}$
Using Hess's Law,we manipulate the given equations:
$(i) CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$(ii) C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$(iii) H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$
Target reaction = $(ii) + 2 \times (iii) - (i)$
$\Delta_{f} H^{\ominus} = (-393) + 2(-286) - (-726)$
$\Delta_{f} H^{\ominus} = -393 - 572 + 726$
$\Delta_{f} H^{\ominus} = -239 \ kJ \ mol^{-1}$