Question

The enthalpy change for the reaction, 

N₂ (g) + 3H₂ (g) → 2NH₃ (g) is -92.38 kJ at 298 K. 

What is ∆U at 298 K?

Answer 1

∆U = ∆H — ∆n_gRT 

= -92.38 × 10³ J – [-2 × 8.314J K mol × 293 K)] 

= -92.38 × 10³J +4.872 × 10³J 

= -87.51 × 10³J 

= – 87.51 kJ