For H₂O(l) → H₂(g) + (1/2)O₂(g) , Δ H = 285.80 kJ/mol at 298 K. What is Δ U ? ( R = 8.314 J/mol·K )
For Δ ng = 1.5 - 0 = 1.5 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = 285.80 - 1.5 × 2.4776 = 285.80 - 3.7164 = 282.08 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
