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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 beA. 500 KB. 750 KC. 1000 KD. 1250 K |
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Answer» Correct Answer - B `(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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