Question

The enthalpy change for the reaction of 50 ml of ethylene with 50.0 ml of H₂ at 1.5 atm pressure is ∆H = -0.31 KJ. What is the ∆U?

Answer 1

W = –P∆V

 = –1.5(50–50–50)×10^–3

= –1.5 × –50 ×10^–3 × 10⁵ × 10^–3

W = 15 = 0.0075

H = –0.31

∆V = ∆H -∆(PV)

∆V = –0.31 kJ + 0.15 kJ = 0.3025