Standard entropies of X2,Y2and XY3, are 60, 40 and 50 K−1Jmol−1respectively. For the reaction: 12X2+32Y2⇌XY3,ΔH=−30kJto be at equilibrium, the temperature should be:
A
500 K
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B
750 K
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C
1000 K
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D
1250 K
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Solution
The correct option is B750 K 12X2+32Y2XY3 ΔS∘=∑S∘p−∑S∘R=50−(30+60) =−40JK−1mol−1 ΔG∘=ΔH∘−TΔS∘,atequilibrium,ΔG∘=0 TΔS∘=ΔH∘ T=−30×103−40=750K