The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.5kJ ` and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ and -285.78kJ` respectively at `25^(@)C` and 1 atmospheric pressure. Calculate the `DeltaH` and `DeltaU` for the reaction,
`NH_(4)NO_(3)(s)toN_(2)O(g)+2H_(2)O(l)`

Explanation

To solve the problem, we need to calculate the change in enthalpy (ΔH) and the change in internal energy (ΔU) for the reaction:

\[ \text{NH}_4\text{NO}_3(s) \rightarrow \text{N}_2\text{O}(g) + 2\text{H}_2\text{O}(l) \]

### Step 1: Write the given data
- Molar heat of formation of \( \text{NH}_4\text{NO}_3(s) \): \( \Delta H_f = -367.5 \, \text{kJ} \)
- Molar heat of formation of \( \text{N}_2\text{O}(g) \): \( \Delta H_f = +81.46 \, \text{kJ} \)
- Molar heat of formation of \( \text{H}_2\text{O}(l) \): \( \Delta H_f = -285.78 \, \text{kJ} \)
...