Answer:
[tex]\large \boxed{\text{-237.0 kJ/mol}}[/tex]
Explanation:
2NH₄NO₃(s) ⟶ 2N₂(g) + O₂(g) + 4H₂O(g)
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
[tex]\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})[/tex]
2NH₄NO₃(s) ⟶ 2N₂(g) + O₂(g) + 4H₂O(g)
ΔH°f/kJ·mol⁻¹: -365.1 0 0 -241.8
[tex]\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & 4\times(-241.8) - 2\times(-365.1)\\& = & -967.2 + 730.2\\& = & \textbf{-237.0 kJ/mol}\\\end{array}\\\text{The enthalpy of reaction is } \large \boxed{\textbf{-237.0 kJ/mol}}[/tex]