Standard entropies of X 2- Y 2 and XY 3 are 60- 40 and 50JK -1 mol -1 respectively- For the reaction 1 - 2 X 2 - 3 - 2 Y 2 - XY 3 - H-30kJ at equilibrium- the temperature should be-

The correct option is B 750K 12× 2+32y_2⇌ xy_3 Δ H= 30kJ S_x_2=60 S_y_2=40 S_xy_3=50 ∴Δ S_rkn=50 32× 40 12× 60 =50 60 30 40J/k A ...

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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 ${50 J K}^{- 1} {m o l}^{- 1}$ respectively. For the reaction
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {X Y}_{3}, △ H = - 30 k J$ at equilibrium, the temperature should be:

500K

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750K

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1000K

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1250K

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Solution

The correct option is B 750K
$\frac{1}{2} \times 2 + \frac{3}{2} y_{2} ⇌ x y_{3} Δ H = - 30 k J$
$S_{x_{2}} = 60 S_{y_{2}} = 40 S_{x y_{3}} = 50$
$∴ Δ S_{r k n} = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60$
$= 50 - 60 - 30$
$- 40 J / k$
At equation , $Δ G = 0$
$Δ H - T Δ S = 0$
$\Rightarrow T = \frac{Δ H}{Δ S} = \frac{30 \times 10^{3}}{40}$
$= 750 k$

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Standard entropies of X2, Y2 and XY3 are 60, 40 and 50JK−1mol−1 respectively. For the reaction 12X2+32Y2⇌XY3,△H=−30kJ at equilibrium, the temperature should be:

The correct option is B 750K12×2+32y2⇌xy3ΔH=−30kJSx2=60Sy2=40Sxy3=50∴ΔSrkn=50−32×40−12×60=50−60−30−40J/kAt equation , ΔG=0ΔH−TΔS=0⇒T=ΔHΔS=30×10340=750k

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
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {X Y}_{3}, △ H = - 30 k J$ at equilibrium, the temperature should be: