Standard entropies of X 2- Y 2 and XY 3- are 60-40 and 50 K -1 J mol -1 respectively- For the reaction-1-2 X 2-3-2 Y 2- XY 3- - H -30 kJ to be at equilibrium- the temperature should be-A- 500 KB- 750 KC- 1000 KD- 1250 K

The correct option is B 750 K1/2X_2+3/2Y_2 XY_3 Δ S^∘ = ∑ S_p^∘ ∑ S_R^∘=50 (30+60) = 40 JK^ 1mol^ 1 Δ G^∘=Δ H^∘ T Δ S^∘, at equilibrium, Δ G^∘ = 0 T Δ S^∘ = Δ ...

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Standard XII

Chemistry

Entropy

Standard entr...

Question

Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ , are 60, 40 and 50 $K^{- 1} J m o l^{- 1}$ 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:

500 K

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750 K

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1000 K

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1250 K

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Solution

The correct option is B 750 K
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} X Y_{3}$
$Δ S^{\circ} = \sum S_{p}^{\circ} - \sum S_{R}^{\circ} = 50 - (30 + 60)$
$= - 40 J K^{- 1} m o l^{- 1}$
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}, a t e q u i l i b r i u m, Δ G^{\circ} = 0$
$T Δ S^{\circ} = Δ H^{\circ}$
$T = \frac{- 30 \times 10^{3}}{- 40} = 750 K$

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