Q.

Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 J K–1 mol–1 respectively. The temperature in Kelvin at which the reaction
12X2+52Y2XY5ΔH=35kJmol1
Will be at equilibrium is ____ (Nearest integer)

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answer is 700.

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Detailed Solution

12X2+52Y2XY5ΔSRxn0=11012×70+52×50=110160=50JK1mol1ΔG0=0 at eqb ΔG0=ΔH0TΔS00=35000T(50)T=700 Kelvin 

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