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

The correct option is A 750 K 12X_2+32Y_2⟶ XY_3, Δ H= 30 kJ Δ S_reaction=∑Δ S_product ∑Δ S_reactant → X_2+3Y_2⟶ 2XY_3 Δ H= 60 k ...

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Standard XII

Chemistry

Entropy

Standard entr...

Question

Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60, 40 and 50 $Z J K^{- 1} m o l^{- 1}$ ,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 A 750 K
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⟶ X Y_{3}, Δ H = - 30$ $k J$
$Δ S_{r e a c t i o n} = \sum Δ S_{p r o d u c t} - \sum Δ S_{r e a c t a n t}$
$\to X_{2} + 3 Y_{2} ⟶ 2 X Y_{3}$
$Δ H = - 60$ $k J$
$Δ S_{r e a c t i o n} = 2 \times 50 - 3 \times 40 - 1 \times 60 = 100 - 120 - 60 = - 80$ $J K^{- 1} m o l^{-}$
$Δ G = Δ H - T Δ S = 0$
$Δ H = T Δ S$
$1000 \times (- 60) = - 80 \times T$
$T = 750$ $K$

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

The correct option is A 750 K12X2+32Y2⟶XY3,ΔH=−30 kJΔSreaction=∑ΔSproduct−∑ΔSreactant→X2+3Y2⟶2XY3ΔH=−60 kJΔSreaction=2×50−3×40−1×60=100−120−60=−80 JK−1mol−ΔG=ΔH−TΔS=0ΔH=TΔS1000×(−60)=−80×TT=750 K

Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60, 40 and 50 $Z J K^{- 1} m o l^{- 1}$ ,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 :