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

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

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