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Thermodynamics question

2025 · 24 Jan · Shift 1 · Q25
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Thermodynamics question

2025 · 24 Jan · Shift 1 · Q25

JEE MainChemistryThermodynamicsNumerical+4 / −1
Standard entropies of X2,Y2\mathrm{X}_2, \mathrm{Y}_2X2​,Y2​ and XY5\mathrm{XY}_5XY5​ are 70, 50 and 110 J K−1 mol−1110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}110 J K−1 mol−1 respectively. The temperature in Kelvin at which the reaction 12X2+52Y2⇌XY5ΔH⊖=−35 kJ mol−1\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}21​X2​+25​Y2​⇌XY5​ΔH⊖=−35 kJ mol−1 will be at equilibrium is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
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