1.

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

Answer»

500 K
750 K
1000 K
1250 K

Solution :`(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3)`
`DeltaS=50-((60)/(2)+(3)/(2)xx40)=50-(30+60)=-40J//k, mol`
at equilibrium `DeltaG=0`
`DeltaH=TDeltaS, T=(DeltaH)/(DeltaS)=(-30xx10^(3))/(-40)=750 K`


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