Standard entropy of X 2- Y 2 and X Y 3 are 60-40 and 50 JK -1 mol -1 respectively- For the reaction 1-2 X 2-3-2 Y 2- XY 3 - - H -30 kJ - mol to be at equilibrium- the temperature will be-A- 1250 KB- 500 KC- 750 KD- 100 K

The correct option is C 750 KFor the reaction 1/2X_2+3/2Y_2 → XY_3; Δ H= 30 kJ/mol Calculating Δ S for the reaction, we get Δ S=S_product S_reactant Δ S = 50 ...

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Standard XII

Chemistry

Heat Capacity

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

1250 K

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500 K

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750 K

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100 K

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Solution

The correct option is C 750 K
For the reaction
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}; Δ H = - 30 k J / m o l$
Calculating $Δ S$ for the reaction, we get
$Δ S = S_{p r o d u c t} - S_{r e a c t a n t}$
$Δ S = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40] J K^{- 1} m o l^{- 1}$
$= 50 - (30 + 60) J K^{- 1} = - 40 J K^{- 1} m o l^{- 1}$
At equilibrium $Δ H = T Δ S$ $(∵ Δ G = 0)$
$∴ T \times (- 40) = - 30 \times 1000$
$∴ T = \frac{- 30 \times 1000}{- 40} = 750 K$

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