Answer
The standard enthalpy of formation of $CH_{3}OH_{(l)}$ = - 238.7 kJ/mol
Work Step by Step
Equation (1) $CH_{3}OH_{(l)} +(3/2) O_{2(g)} →CO_{2(g)} + 2 H_{2}O_{(l)}$
$ \Delta H^{\circ}_rxn(1) = - 726.4 kJ/mol$
Equation (2) $C_{(graphite)} + O_{2(g)} →CO_{2(g)}$
$ \Delta H^{\circ}rxn(2) = - 393.5 kJ/mol$
Equation (3) $H_{2(g}) +(1/2) O_{2(g)} → H_{2}O_{(l)}$
$ \Delta H^{\circ}_rxn(3) = - 285.8 kJ/mol$
Reverse Equation (1) to get Equation (4)
Equation (4) $ CO_{2(g)} + 2 H_{2}O_{(l)} → CH_{3}OH_{(l)} +(3/2) O_{2(g)}$
$ \Delta H^{\circ}_rxn(4) = - (- 726.4 kJ/mol) = + 726.4 kJ/mol$
Multiply Equation (3) by 2 to get Equation (5)
Equation (5) $ 2H_{2(g}) + O_{2(g)} → 2H_{2}O_{(l)}$
$ \Delta H^{\circ}_rxn(5) = - 571.6 kJ/mol$
The required equation,
$C_{(graphite)} + 2H_{2(g)} + (1/2) O_{2(g)} → CH_{3}OH_{(l)} $ is formed by adding Equation (2), Equation (4), and Equation (5).
$ \Delta H^{\circ}_{f} = \Delta H^{\circ}rxn(2) +\Delta H^{\circ}rxn(4) + \Delta H^{\circ}rxn(5) = (-393.5+ 726.4 +- 571.6) kJ/mol = - 238.7 kJ/mol$
