Standard entropies of X 2 , Y 2 and X Y 5 are 70, 50 and 110 J K -1 mol -1 respectively. The temperature in Kelvin at which the reaction 1/2 X 2 +5/2 Y 2 rightarrow XY 5 Δ H - =-35 kJ mol -1 Will be at equilibrium is (Nearest integer)

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 24 January, Morning Shift — Question 43

Standard entropies of $X_{2}, Y_{2}$ and $X Y_{5}$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The

temperature in Kelvin at which the reaction

$\frac{1}{2} \mathrm{X}_{2}+\frac{5}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{5} \Delta \mathrm{H}^{-}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Will be at equilibrium is____(Nearest integer)

Numerical answer — enter this value.

$\frac{1}{2} X_{2}+\frac{5}{2} Y_{2} \rightleftharpoons X Y_{5}$

$\Delta \mathrm{S}_{\mathrm{Rxn}}^{0}=110-\left[\left(\frac{1}{2} \times 70\right)+\left(\frac{5}{2} \times 50\right)\right]$

$=110-160=-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\Delta \mathrm{G}^{0}=0$ at eqb $\Delta \mathrm{G}^{0}=\Delta \mathrm{H}^{0}-\mathrm{T} \Delta \mathrm{S}^{0}$

$0=-35000-\mathrm{T}(-50)$

$\mathrm{T}=700$ Kelvin

Answer key and solution verified before publishing.

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