The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.54kJ` and those of `N_(2)O(g)`and `H_(2)O(l)` are `+81.46kJ` and `-285.80kJ`, respectively, at `25^(@)C` and `1atm`. Calculate `DeltaH` and `DeltaU` for the reaction.
`NH_(4)NO_(3)(s) rarr N_(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 down the given data
- Molar heat of formation of \( \text{NH}_4\text{NO}_3(s) = -367.54 \, \text{kJ/mol} \)
- Molar heat of formation of \( \text{N}_2\text{O}(g) = +81.46 \, \text{kJ/mol} \)
- Molar heat of formation of \( \text{H}_2\text{O}(l) = -285.80 \, \text{kJ/mol} \)
...