Standard entropies of X2,Y2 and XY3 are 60, 40 and 50JK-1 mol-1 respectively. For the reaction 12X2+32Y2⇋XY3, ΔH=−30 kJ/mole, to be at equilibrium, the temperature should be:

12X2+32Y2⇌XY3Δr0S=∑ΔP0S−∑ΔR0SΔr0S=(50)−(12×60+32×40)= 50 – (30 + 60) = 50 – 90 = –40Apply, ΔG = ΔH = TΔS(At equilibrium., ΔG = 0)ΔH=TΔS; T=ΔHΔS−30×1000−40=T30004=T=750 K

Standard entropies of X2,Y2 and XY3 are 60, 40 and 50JK-1 mol-1 respectively. For the reaction 12X2+32Y2⇋XY3, ΔH=−30 kJ/mole, to be at equilibrium, the temperature should be:

12X2+32Y2⇌XY3Δr0S=∑ΔP0S−∑ΔR0SΔr0S=(50)−(12×60+32×40)= 50 – (30 + 60) = 50 – 90 = –40Apply, ΔG = ΔH = TΔS(At equilibrium., ΔG = 0)ΔH=TΔS; T=ΔHΔS−30×1000−40=T30004=T=750 K

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Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60, 40 and $50 J K^{- 1} m o l^{- 1}$ respectively. For the reaction $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇋ X Y_{3}, Δ H = - 30$ kJ/mole, to be at equilibrium, the temperature should be:

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500 K

750 K

1000 K

1250 K

answer is B.

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Detailed Solution

$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}$
$Δ_{r}^{0} S = \sum Δ_{P}^{0} S - \sum Δ_{R}^{0} S$
$Δ_{r}^{0} S = (50) - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40)$
$= 50 - (30 + 60) = 50 - 90 = - 40$
$A p p l y, Δ G = Δ H = T Δ S$
$(A t e q u i l i b r i u m ., Δ G = 0)$
$Δ H = T Δ S; T = \frac{Δ H}{Δ S}$