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

ΔS for the reaction 12X2+32Y2⇌XY3 is:ΔS=50−(12×60+32×30)=−25 JFor equilibrium ∆G=0ΔG=ΔH−TΔS 0=ΔH−TΔS T=ΔHΔS =−30000−25=1200 K.

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

ΔS for the reaction 12X2+32Y2⇌XY3 is:ΔS=50−(12×60+32×30)=−25 JFor equilibrium ∆G=0ΔG=ΔH−TΔS 0=ΔH−TΔS T=ΔHΔS =−30000−25=1200 K.

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

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$1200 K$

$1000 K$

$750 K$

$500 K$

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$Δ S$ for the reaction $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}$ is:

$Δ S = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 30) = - 25 J$

For equilibrium $(∆ G = 0)$

$Δ G = Δ H - T Δ S 0 = Δ H - T Δ S T = \frac{Δ H}{Δ S} = \frac{- 30000}{- 25} = 1200 K .$

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

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