SolveItJEE Main
Chemistry · 2026

JEE Main · 6 April 2026, Shift 2 · Q74

In a closed flask at 600 K, one mole of X_2 Y_4( g ) attains equilibrium as given below: X_2 Y_4( g ) ⇌ 2 XY_2( g ) At equilibrium, 75 % X_2 Y_4( g )…

In a closed flask at $\displaystyle 600$ K , one mole of $\displaystyle \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ attains equilibrium as given below : $$\mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_2(\mathrm{~g}) $$ At equilibrium, $\displaystyle 75 \% \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ was dissociated and the total pressure is $\displaystyle 1$ atm. The magnitude of $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ (in $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ ) at this temperature is $\displaystyle \_\_\_\_$. (Nearest Integer) (Given : $\displaystyle \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ln 10=2.3, \log 2=0.3, \log 3=0.48, \log 5=0.69, \log 7=0.84$ )
ShareWhatsAppTelegram

More from Chemical Equilibrium

JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.