The enthalpy change `(DeltaH)` for the reaction
`N_2(g) + 3H_2 (g) to 2NH_3(g)`
is -92.38 kJ at 298 K. What is `DeltaU` at 298 K ? `(R = 8.314 j K^(-1) mol^(-1))`

Explanation

To solve the problem, we need to find the internal energy change (ΔU) for the given reaction using the relationship between enthalpy change (ΔH) and internal energy change (ΔU). The relationship is given by:

\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]

Where:
- ΔH is the enthalpy change
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