## Chemistry: The Molecular Science (5th Edition)

$C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) -- \gt CH_3OH(l)$ $\Delta_fH^{\circ}\{CH_3OH(l)\} = -238.66kJ/mol$
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. $1 C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) -- \gt CH_3OH(l)$ $\Delta_fH^{\circ}\{CH_3OH(l)\} = -238.66kJ/mol$