Answer
$\frac{1}{2} N_2(g) + \frac{3}{2}H_2(g) -- \gt NH_3(g)$; $\Delta_fH^{\circ}\{NH_3\} = -46.11kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $NH_3$
$N + 3H -- \gt NH_3(g)$
2. Analyze the appendix J, and find the standart state for nitrogen and hydrogen.
- Divide the coefficient by the number of atoms in each compound in standard state:
$\frac{1}{2} N_2(g) + \frac{3}{2}H_2(g) -- \gt NH_3(g)$
Just add the $\Delta_fH^{\circ}$ after the reaction.
$\frac{1}{2} N_2(g) + \frac{3}{2}H_2(g) -- \gt NH_3(g)$; $\Delta_fH^{\circ}\{NH_3\} = -46.11kJ/mol$
