JEE Main 2025 Jan 24 Shift 1, Chemistry Q22: Gibbs Energy Change and Equilibrium

JEE Main2025Jan 24, Shift 1Chemistry

Standard entropies of X $_{2}$ , Y $_{2}$ and XY $_{5}$ are 70, 50 and 110 J K $^{- 1}$ mol $^{- 1}$ respectively. The temperature in Kelvin at which the reaction

$\frac{1}{2} X_{2} + \frac{5}{2} Y_{2} \to X Y_{5} Δ H^{-} = - 35$ kJ mol $^{- 1}$

Will be at equilibrium is ____ (Nearest integer)

Your answer

Solution

At equilibrium $Δ G^{0} = 0$ , hence $T = \frac{Δ H ^{0}}{Δ S ^{0}}$ .

Compute $Δ S^{0}$ for the reaction (products $-$ reactants, weighted by stoichiometry):

$Δ S^{0} = S^{0} (X Y_{5}) - [\frac{1}{2} S^{0} (X_{2}) + \frac{5}{2} S^{0} (Y_{2})]$

$= 110 - [\frac{1}{2} (70) + \frac{5}{2} (50)] = 110 - [35 + 125] = - 50 J K^{- 1} mo l^{- 1}$ .

Using $Δ H^{0} = - 35000 J mo l^{- 1}$ :

$T = \frac{- 35 000}{- 50} = 700 K$ .

$∴$ Answer: $700$ .

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JEE Main 2025 Jan 24 Shift 1, Chemistry Q22: Gibbs Energy Change and Equilibrium

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