For the reaction H₂ F₂(~g) → H₂(~g)+F₂(~g) Δ 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.

Chemical Thermodynamics: JEE Main 2022 Chemistry question

JEE Main 2022 · 26 Jul, Evening shift · Q23

JEE Main 202226 Jul, Evening shiftChemistryHardChemical Thermodynamics84× since 2002

For the reaction

\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})

\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}

27^{\circ} \mathrm{C}

. The enthalpy change for the above reaction is (

-

) __________

\mathrm{kJ} \,\mathrm{mol}^{-1}

[nearest integer] Given:

\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Chemical Thermodynamics Question with Answer (26 Jul Evening shift, Q23) | JEE Arc