Standard entropy of X2,Y2 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 :

12X2+32Y2→X3ΔSreaction =−50−32×40+12×60=−40 Jmol−1ΔG=ΔH−TΔSat equilibrium ΔG=0ΔH=TΔS30×103=T×40⇒T=750K

Standard entropy of X2,Y2 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 :

12X2+32Y2→X3ΔSreaction =−50−32×40+12×60=−40 Jmol−1ΔG=ΔH−TΔSat equilibrium ΔG=0ΔH=TΔS30×103=T×40⇒T=750K

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Standard entropy of $X_{2}, Y_{2}$ and ${XY}_{3}$ are 60, 40 and $50 {JK}^{- 1} {mol}^{- 1}$ , respectively. For the reaction, $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⟶ {XY}_{3}, ΔH = - 30 kJ$ , to be at equilibrium the temperature will be :

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1000 K

500 K

750 K

1250 K

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X_{3} \\ {ΔS}_{reaction} = - 50 - (\frac{3}{2} \times 40 + \frac{1}{2} \times 60) = - 40 {Jmol}^{- 1} \\ ΔG = ΔH - TΔS \end{array}$
at equilibrium $ΔG = 0$
$\begin{array}{l} ΔH = TΔS \\ 30 \times 10^{3} = T \times 40 \\ \Rightarrow T = 750 K \end{array}$