Answer
$C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) -- \gt CH_3OH(l)$ $\Delta_fH^{\circ}\{CH_3OH(l)\} = -238.66kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $CH_3OH$
$C + 4H + O -- \gt CH_3OH$
2. Analyze the appendix J, and find the standard state for each element.
- Hint: It is the compound with $\Delta_fH^{\circ}$ = 0.
- Divide the coefficient by the number of atoms in each compound in standard state:
$\frac{1}{1} C(graphite) + \frac{4}{2}H_2(g) + \frac{1}{2}O_2(g) -- \gt CH_3OH(l)$
Just add the $\Delta_fH^{\circ}$** after the reaction.
** Appendix J.
$1 C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) -- \gt CH_3OH(l)$ $\Delta_fH^{\circ}\{CH_3OH(l)\} = -238.66kJ/mol$
