For the reaction H 2 F 2 g → H 2 g + F 2 g ΔU = - 59 . 6 kJ mol - 1 at 27 ° C The enthalpy change for the above reaction is - ____ kJmol - 1 (nearest integer) (Given : R = 8 . 314 JK - 1 mol - 1 ) .

Step-by-step solution

For the reaction H 2 F 2 g → H 2 g + F 2 g ΔH = ΔU + Δn g RT Given ΔU = - 59 . 6   kJ   mol - 1 Δn g = 2 - 1 = - 59 . 6 + 1 × 8 . 314 1000 × 300 = - 57 . 11     kJ   mol - 1