Answer
$C(graphite) + \frac{1}{2}O_2(g) -- \gt CO(g)$ $\Delta_fH^{\circ}\{CO\} = -110.525kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $CO$
$1C + 1O -- \gt CO$
2. Analyze appendix J, and find the standard state for carbon and hydrogen.
- Divide the coefficient by the number of atoms in each compound in standard state:
$\frac{1}{1} C(graphite) + \frac{1}{2}O_2(g) -- \gt CO(g)$
Just add the $\Delta_fH^{\circ}$ after the reaction.
$C(graphite) + \frac{1}{2}O_2(g) -- \gt CO(g)$; $\Delta_fH^{\circ}\{CO\} = -110.525kJ/mol$
