Standard entropy of X2 ,Y2and XY3 are 60, 40 and 50 JK-1mol-1 respectively, for this reaction 12X2 +32Y2⇌XY3,∆H=-30KJto be at equilibrium,the temperature should be

12X2 +32Y2⇌XY3∆S0=∑Sproducts0-∑Sreactents0=50-40×32 + 12×60=50-60+30=-40KJmole-1 ∆G0=∆H0-T∆S0 , at equilibrium ∆G0 =0 ∆H0=T∆S0 ; T =∆H0∆S0=-30×103-40=750K

Standard entropy of X2 ,Y2and XY3 are 60, 40 and 50 JK-1mol-1 respectively, for this reaction 12X2 +32Y2⇌XY3,∆H=-30KJto be at equilibrium,the temperature should be

12X2 +32Y2⇌XY3∆S0=∑Sproducts0-∑Sreactents0=50-40×32 + 12×60=50-60+30=-40KJmole-1 ∆G0=∆H0-T∆S0 , at equilibrium ∆G0 =0 ∆H0=T∆S0 ; T =∆H0∆S0=-30×103-40=750K

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Standard entropy of $X_{2}, Y_{2}$ and ${XY}_{3}$ are 60, 40 and 50 ${JK}^{- 1} {mol}^{- 1}$ respectively, for this reaction $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {XY}_{3}$ , $∆ H = - 30 KJ$ to be at equilibrium,the temperature should be

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300 K

750 K

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1250 K

answer is B.

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Detailed Solution

$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {XY}_{3}$

$∆ S^{0} = \sum S_{products}^{0} - \sum S_{reactents}^{0} = 50 - (40 \times \frac{3}{2} + \frac{1}{2} \times 60) = 50 - (60 + 30) = - 40 {KJmole}^{- 1} ∆ G^{0} = ∆ H^{0} - T ∆ S^{0}, at equilibrium ∆ G^{0} = 0 ∆ H^{0} = T ∆ S^{0}; T = \frac{∆ H^{0}}{∆ S^{0}} = \frac{- 30 \times 10^{3}}{- 40} = 750 K$