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

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

Q. Standard entropies 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} ⇌ X Y_{3}; △ H = - 30 k J / m o l

to be at spontaneous, the temperature should be:

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

The correct option is D 750K12X2+32Y2→XY3;ΔH=−30KJΔS(Reaction)=∑ΔS(product)−∑ΔS(reactants) =ΔS(XY3)−32ΔS(Y2)−12ΔS(X2) =50−32×40−12×60ΔS=−40JK−1mol−1Now, ΔG=ΔH−TΔSFor equilibrium, ΔG=0∴ΔH=TΔS∴T=ΔHΔS=−30×103J−40JK−1=750K∴T=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:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to {X Y}_{3}$ . $Δ H = - 30 k J$ . to be equilibrium, the temperature will be: