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Standard entropt of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-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:A. `1250K`B. `500K`C. `750K`D. `1000K` |
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Answer» Correct Answer - A `(1)/(2)X_(2)+(3)/(2)Y_(2)rarr XY_(3)` `DeltaS_("reaction")=50-((3)/(2)xx40+(1)/(2)xx60)=-40Jmol^(-1)` `DeltaG=DeltaH-TDeltaS` At equilibrium `DeltaG=0` `DeltaH=TDeltaS` `30xx10^(3)=Txx40` `:. T=750` |
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