1.

Standard entropy of X_(2), Y_(2) and XY_(3) are 60, 40 and 50" J K"^(-1) mol^(-1), respectively. For the reaction, 1/2 X_(2)+3/2 Y_(2) rarr XY_(3) Delta H=-30 kJ, to be at equilibrium, the temperature will be

Answer»

500 K
750 K
1000 K
1250 K

Solution :`DeltaS_("REACTION")=DeltaS_("Product")-DeltaS_("Reactant")`
`=50-[3//2xx40+1//2xx60]=50-[60+30]=-4 J//K`
We know
`DeltaG=DELTAH-T DeltaS`
at equilibrium `DeltaG=0`
`0=DeltaH-T DeltaS`
`DeltaH=T DeltaS`
`T=(DeltaH)/(DeltaS)=((-) 30xx1000)/((-)40)=750 K`


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