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...

Question

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 J K^ -1 mol^ -1 .

Accepted Answer

57

JEE Main 2022 Chemistry question

Solution

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