The enthalpy change `(DeltaH)` for the reaction, `N_(2(g))+3H_(2(g)) rarr 2NH_(3g)` is `-92.38kJ` at `298K` What is `DeltaU` at `298K ?`

Explanation

To find the internal energy change (ΔU) for the reaction \( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \) given that the enthalpy change (ΔH) is \(-92.38 \, \text{kJ}\) at \(298 \, \text{K}\), we can use the relationship between enthalpy change and internal energy change:

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where:
- \(\Delta H\) is the enthalpy change.
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