Question

When NH₄NO₂(s), decomposes at 373 K, it forms N₂ (g) and H₂O (g). The ∆H for the reaction at one atmospheric pressure and 373 K is −223. 6 kJ mol ^-1 . What is the value of ∆E for the reaction under the same conditions? (Given R = 8.31 JK^-1 mol^-1)

Answer 1

NH₄NO₂(S) → N₂ (g) + 2H₂O (g),

∆_ng= 3 − 0 = 3

∆U = ∆H − ∆_ng RT

= −223.6 kJ mol⁻¹ − (3 mol) x (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × (373 K)

= −232.9 kJ mol⁻¹ .