Standard entropy of X2Y2 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 B 750 K Solution: (C) 750 K 12X_2 + 32Y_2⟶ XY_3 ΔH = 30 kJ For the above reaction ΔS = ∑ΔS_( Product ) ∑ΔS_( ...

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Standard XII

Chemistry

Entropy

Standard entr...

Question

Standard entropy of $X_{2} Y_{2} a n d X Y_{3}$ 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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750 K

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1000 K

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Solution

The correct option is B $750 K$
Solution:- (C) $750 K$
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⟶ X Y_{3} Δ H = - 30 k J$
For the above reaction-
$Δ S = {\sum Δ S}_{(P r o d u c t)} - {\sum Δ S}_{(r e a c t a n t)}$
$Δ S = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40)$
$\Rightarrow Δ H = 50 - 90 = - 40$
As we know that,
$Δ G = Δ H - T Δ S$
But at equilibrium,
$Δ G = 0$
$Δ H - T Δ S = 0$
$\Rightarrow T = \frac{Δ H}{Δ S}$
Given:- $Δ H = - 30 k J = - 30000 J$
$∴ T = \frac{- 30000}{- 40} = 750 K$
Hence the temperature of the given reaction is $750 K$ .

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Standard entropy of X2Y2 and XY3 are 60,40,and 50 JK−1mol−1 respectively. For the reaction, 12X2+32Y2→XY3,ΔH=−30 kJ to be at equilibrium, the temperature will be

Standard entropy of $X_{2} Y_{2} a n d X Y_{3}$ 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