Standard entropy of X2- Y2 and XY3 are 60- 40 and 50 JK-1mol-1 respectively- For the reaction- 12X2-32Y2- XY3- - H-30 kJ to be at equilibrium- the temperature will be -

The correct option is D 750 KAt equilibrium, Δ G=0=Δ H TΔ S #160;Δ H=TΔ S Now, Δ S=50 602 402× 3 Δ S= 40 JK^ 1 Again, 30× 1000=T× ( 40) ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Standard entr...

Question

Standard entropy 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: $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}, Δ H = - 30 k J$ to be at equilibrium, the temperature will be :

1250 K

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500 K

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1000 K

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750 K

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Solution

The correct option is D 750 K
At equilibrium,
$Δ G = 0 = Δ H - T Δ S$
$⟹ Δ H = T Δ S$
Now,
$Δ S = 50 - \frac{60}{2} - \frac{40}{2} \times 3$
$Δ S = - 40 J K^{- 1}$
Again,
$⟹ - 30 \times 1000 = T \times (- 40)$
$∴ T = 750 K$

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Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK−1mol−1 respectively. For the reaction: 12X2+32Y2→XY3,ΔH=−30kJ to be at equilibrium, the temperature will be :

The correct option is D 750 KAt equilibrium,ΔG=0=ΔH−TΔS⟹ΔH=TΔSNow,ΔS=50−602−402×3ΔS=−40JK−1Again,⟹−30×1000=T×(−40)∴T=750K

Standard entropy 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: $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}, Δ H = - 30 k J$ to be at equilibrium, the temperature will be :

The correct option is D 750 K
At equilibrium,
$Δ G = 0 = Δ H - T Δ S$
$⟹ Δ H = T Δ S$
Now,
$Δ S = 50 - \frac{60}{2} - \frac{40}{2} \times 3$
$Δ S = - 40 J K^{- 1}$
Again,
$⟹ - 30 \times 1000 = T \times (- 40)$
$∴ T = 750 K$