Standard entropies of X2 ,Y2 and XY3 are 60 , 40 and 50 JK-1 mol-1 respectively. For the reaction (1/2) X2 + (3/2)Y2 leftharpoons XY3, Δ H = -30 kJ to be at equilibrium, the temperature should be

Question

Standard entropies of X2 ,Y2 and XY3 are 60 , 40 and 50 JK-1 mol-1 respectively. For the reaction (1/2) X2 + (3/2)Y2 leftharpoons XY3, Δ H = -30 kJ to be at equilibrium, the temperature should be (A) 500 K (B) 750 K (C) 1000 K (D) 1250 K

Tardigrade Admin · Accepted Answer

(1/2) X2 + (3/2) Y2 leftharpoons XY3 Δ S° = ∑ S°P - ∑ SR° = 50 - (30 + 60) Δ S° = -40 JK-1 mol-1 T = (Δ H°/Δ S°) = (-30 × 103 J mol-1/-40 JK-1 mol-1) = 750 K

Q. 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 $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}, Δ H = - 30 k J$ to be at equilibrium, the temperature should be

500 K

750 K

1000 K

1250 K

$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}$
$Δ S^{\circ} = \sum S_{P}^{\circ} - \sum S_{R}^{\circ}$
$= 50 - (30 + 60)$
$Δ S^{\circ} = - 40 J K^{- 1} m o l^{- 1}$
$T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{- 30 \times 1 0 ^{3} J m o l ^{- 1}}{- 40 J K ^{- 1} m o l ^{- 1}} = 750 K$

Q. Standard entropies of X2​,Y2​ and XY3​ are 60,40 and 50JK−1mol−1 respectively. For the reaction 21​X2​+23​Y2​⇌XY3​,ΔH=−30kJ to be at equilibrium, the temperature should be