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Chemical Equilibrium question

2025 · 24 Jan · Shift 1 · Q21
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Chemical Equilibrium question

2025 · 24 Jan · Shift 1 · Q21

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
37.8 g N2O537.8 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_537.8 g N2​O5​ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K 2 N2O5( g)⇌2 N2O4( g)+O2( g)2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}2 N2​O5( g)​⇌2 N2​O4( g)​+O2( g)​ The total pressure at equilibrium was found to be 18.65 bar. Then, Kp=\mathrm{Kp}=Kp=‾\underline{\hspace{2cm}}​×10−2\times 10^{-2}×10−2[nearest integer] Assume N2O5\mathrm{N}_2 \mathrm{O}_5N2​O5​ to behave ideally under these conditions. Given: R=0.082\mathrm{R}=0.082R=0.082 bar Lmol−1 K−1\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}Lmol−1 K−1
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