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Thermodynamics question

2022 · 26 Jul · Shift 2 · Q14
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Thermodynamics question

2022 · 26 Jul · Shift 2 · Q14

JEE MainChemistryThermodynamicsNumerical+4 / −1
For the reaction H2F2( g)→H2( g)+F2( g)\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})H2​F2​( g)→H2​( g)+F2​( g)ΔU=−59.6 kJ mol−1\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}ΔU=−59.6 kJ mol−1 at 27∘C27^{\circ} \mathrm{C}27∘C. The enthalpy change for the above reaction is (−-−) ‾kJ mol−1\underline{\hspace{2cm}}\mathrm{kJ} \,\mathrm{mol}^{-1}​kJmol−1 [nearest integer] Given: R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}R=8.314 J K−1 mol−1.
Numerical answer
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