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Q.

Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 J 12X2+32Y2 → XY3, ∆H=-30  kJ, to be at equilibrium, the temperature will be

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a

1250 K

b

500 K

c

750 K

d

1000 K

answer is C.

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Detailed Solution

12X2+32Y2 → XY3 ∆Sreaction =50-32×40+12 × 60 = -40J mol-1 ∆G=∆H-T∆S at equilibrium ∆G=0 ∆H-T∆S 30 × 103=T × 40 ∴  T=750
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Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 J 12X2+32Y2 → XY3, ∆H=-30  kJ, to be at equilibrium, the temperature will be