Standard entropies of X2, Y2 and XY3 are given below the reaction1/2X2+3/2Y2⇌XY3; ΔH∘=−30kJmol−1S∘ 60 40 50JK−1mol−1At what temperature, reaction would be in equilibrium?

Reaction will be in equilibrium whenΔG∘=0 But ΔG∘=ΔH∘−TΔS∘ Then, T=ΔH∘ΔS∘ΔH∘=−30kJmol−1ΔS∘=S∘XY3−12S∘X2+32S∘Y2=50−12×60+32×40=50−[30+60]=−40JK−1mol−1T=−30×1000Jmol−140KK−1mol−1=750KHence, at 750 K, reaction would be in equilibrium.

Standard entropies of X2, Y2 and XY3 are given below the reaction1/2X2+3/2Y2⇌XY3; ΔH∘=−30kJmol−1S∘ 60 40 50JK−1mol−1At what temperature, reaction would be in equilibrium?

Reaction will be in equilibrium whenΔG∘=0 But ΔG∘=ΔH∘−TΔS∘ Then, T=ΔH∘ΔS∘ΔH∘=−30kJmol−1ΔS∘=S∘XY3−12S∘X2+32S∘Y2=50−12×60+32×40=50−[30+60]=−40JK−1mol−1T=−30×1000Jmol−140KK−1mol−1=750KHence, at 750 K, reaction would be in equilibrium.

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Standard entropies of X₂, Y₂ and XY₃ are given below the reaction
$\begin{array}{l} 1 / 2 X_{2} + 3 / 2 Y_{2} ⇌ {XY}_{3}; {ΔH}^{\circ} = - 30 kJ {mol}^{- 1} \\ S^{\circ} 60 40 50 {JK}^{- 1} {mol}^{- 1} \end{array}$
At what temperature, reaction would be in equilibrium?

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500 K

750 K

1000 K

1250 K

answer is B.

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Detailed Solution

Reaction will be in equilibrium when

$\begin{array}{l} {ΔG}^{\circ} = 0 \\ But {ΔG}^{\circ} = {ΔH}^{\circ} - {TΔS}^{\circ} \\ Then, T = \frac{Δ H^{\circ}}{Δ S^{\circ}} \\ {ΔH}^{\circ} = - 30 kJ {mol}^{- 1} \\ {ΔS}^{\circ} = S^{\circ} ({XY}_{3}) - [\frac{1}{2} S^{\circ} (X_{2}) + \frac{3}{2} S^{\circ} (Y_{2})] \\ = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40] = 50 - [30 + 60] \\ = - 40 {JK}^{- 1} {mol}^{- 1} \\ T = \frac{- 30 \times 1000 J {mol}^{- 1}}{40 {KK}^{- 1} {mol}^{- 1}} = 750 K \end{array}$