For the reaction H 2 F 2 (g) H 2 (g) + F 2 (g) U = -59.6 …

Chemistry Thermodynamics and Thermochemistry JEE Main 2022 Numeric Response

Published by: CGP EDU Academic Team

Published on: August 13, 2026

For the reaction H 2 F 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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The correct answer is:

(57)

H = U + n g RT

H = -59.6 + 1 x 8.314 x 300 x 10 -3 = -57.10

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