For the reaction H_2F_2(g) → H_2(g) + F_2(g) ∆U = -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 JK^-1 mol^-1

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Accepted Answer

Correct answer: 57. (57) ∆H = ∆U + ∆n_gRT ∆H = -59.6 + 1 x 8.314 x 300 x 10^-3 = -57.10

$For the reaction H_{2} F_{2} (g) \to H_{2} (g) + F_{2} (g) ∆ U = -59.6 kJ m o l^{- 1} at 27°C. The enthalpy change for the above reaction is (-) ___kJ m o l^{- 1} [nearest integer] Given : R = 8.314 J K^{- 1} m o l^{- 1}$

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