Answer
$1 N_2(g) + {2}H_2(g) + \frac{3}{2} O_2(g)-- \gt NH_4NO_3(s)$ $\Delta_fH^{\circ}\{NH_4NO_3(s)\} = -365.56kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $NH_4NO_3$
$2N + 4H + 3O -- \gt NH_4NO_3$
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:
$\frac{2}{2} N_2(g) + \frac{4}{2}H_2(g) + \frac{3}{2} O_2(g)-- \gt NH_4NO_3(s)$
Just add the $\Delta_fH^{\circ}$** after the reaction.
** Appendix J.
$1 N_2(g) + {2}H_2(g) + \frac{3}{2} O_2(g)-- \gt NH_4NO_3(s)$ $\Delta_fH^{\circ}\{NH_4NO_3(s)\} = -365.56kJ/mol$
