Derive the relationship between `Delta H and Delta U`
Given `N_(2)(g) + 3H_(2) (g) rarr 2NH_(3) (g), Delta_(r ) H^(theta) = - 92.4 kJ mol^(-1)`,
What is the standard enthalpy of formation of `NH_(3)` gas?

Explanation

To derive the relationship between ΔH (enthalpy change) and ΔU (internal energy change), we start with the equation:

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

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