Question

For the reaction H₂F₂(g) → H₂(g) + F2(g), ΔU = –59.6 kJ mol^–1 at 27°C. The enthalpy change for the above reaction is (–) _______ kJ mol–1 [nearest integer] Given: R = 8.314 J K–1 mol^–1.

Answer 1

Correct Answer: 57

H₂F₂(g) → H₂(g) + F₂(g)
ΔU = -59.6 kJ mol^-1 at \(27 \degree\) C
ΔH = ΔU +Δn_gRT

=-59.6+ \(\frac{1 \times 8.314 \times 300 }{ 1000}\)
\(= -57.10 kJ mol^-1\)