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Chemistry Question on Enthalpy change

Calculate the standard enthalpy of formation of CH3OH(l)CH_3OH(l) from the following data.
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_3OH (l) +\frac 32 O_2(g)→CO_2(g)+2H_2O (l) ; ∆rHΘ=–726 kJmol–1 ∆_rH^Θ= –726 \ kJ mol^{–1},
C(graphite)+O2(g)→CO2(g)C(graphite)+O_2(g)→CO_2(g); ∆cHΘ=–393 kJmol–1∆_cH^Θ = –393 \ kJ mol^{–1},
H2(g)+12O2(g)→H2O(l)H_2(g)+\frac 12 O_2(g)→H_2O (l); ∆fHΘ=–286kJmol–1∆_f H^Θ= –286 kJ mol^{–1}

Answer

The reaction that takes place during the formation of CH3OH(l)CH_3OH(l) can be written as:
C(s)+2H2O(g)+12O2(g)→CH3OH(l)C(s) + 2H_2O(g) + \frac 12 O_2(g) → CH_3OH(l) ……… (i)
The reaction (i) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)
∆fHΘ[CH3OH(l)]=∆cHΘ+2∆fHΘ[H2O(l)]–∆rHΘ∆_fH^Θ [CH_3OH(l)] = ∆_cH^Θ + 2∆_fH^Θ [H_2O(l)] – ∆_rH^Θ
= (–393 kJmol–1)+2(–286 kJmol–1)–(–726 kJmol–1)(–393\ kJ mol^{–1}) + 2(–286\ kJ mol^{–1}) – (–726 \ kJ mol^{–1})
= (–393–572+726) kJmol–1(–393 – 572 + 726) \ kJ mol^{–1}
∴ ∆fHθ[CH3OH(l)]=–239 kJmol–1∆_fH^θ [CH_3OH(l)] = –239\ kJ mol^{–1}