Standard entropies of X2g- Y2g and XY3g are 60- 40 and 50 J K-1 mol-1 respectively- For the reaction 12X2g - 32Y2g - XY3g- - H - -30 kJ- to be at equilibrium- the temperature should be-

Given reaction is: 12X_2(g) + 32Y_2(g) ⇌ XY_3(g) We know, nbsp; Δ S^o=∑ S^o_products ∑ S^o_reactants =50 12(60) + nbsp;32(40) =50 (30+60)= 40 J K^ 1 mol^ 1 ...

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Standard XII

Chemistry

Effective Equilibrium Constant

Standard entr...

Question

Standard entropies of $X_{2} (g)$ , $Y_{2} (g)$ and $X Y_{3} (g)$ are $60, 40$ and $50 {J K}^{- 1} {mol}^{- 1}$ respectively. For the reaction $\frac{1}{2} X_{2} (g) + \frac{3}{2} Y_{2} (g) ⇌ X Y_{3} (g)$ , $Δ H = - 30 kJ$ , to be at equilibrium, the temperature should be:

750 K

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1000 K

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1250 K

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500 K

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Solution

The correct option is A $750 K$
Given reaction is:
$\frac{1}{2} X_{2} (g) + \frac{3}{2} Y_{2} (g) ⇌ X Y_{3} (g)$
We know,
$Δ S^{o} = \sum S_{p r o d u c t s}^{o} - \sum S_{r e a c t a n t s}^{o}$
$= 50 - \frac{1}{2} (60) + \frac{3}{2} (40)$
$= 50 - (30 + 60) = - 40 {J K}^{- 1} {mol}^{- 1}$
At equilibrium,
$Δ G^{o} = 0$
$Δ H^{o} = T Δ S^{o}$
$∴ Δ T = \frac{Δ H^{o}}{Δ S^{o}} = \frac{- 30 \times 10^{3} {J mol}^{- 1}}{- 40 {J K}^{- 1} {mol}^{- 1}} = 750 K$

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Standard entropies of X2(g), Y2(g) and XY3(g) are 60, 40 and 50 J K−1 mol–1 respectively. For the reaction 12X2(g)+32Y2(g)⇌XY3(g), ΔH=–30 kJ, to be at equilibrium, the temperature should be:

The correct option is A 750 KGiven reaction is: 12X2(g)+32Y2(g)⇌XY3(g) We know, ΔSo=∑Soproducts−∑Soreactants =50−12(60)+32(40) =50−(30+60)=−40 J K−1 mol−1 At equilibrium, ΔGo=0 ΔHo=TΔSo ∴ΔT=ΔHoΔSo=−30×103 J mol−1−40 J K−1 mol−1=750 K

Standard entropies of $X_{2} (g)$ , $Y_{2} (g)$ and $X Y_{3} (g)$ are $60, 40$ and $50 {J K}^{- 1} {mol}^{- 1}$ respectively. For the reaction $\frac{1}{2} X_{2} (g) + \frac{3}{2} Y_{2} (g) ⇌ X Y_{3} (g)$ , $Δ H = - 30 kJ$ , to be at equilibrium, the temperature should be: