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Calculate (1) DeltaG^(0) and (2) The equ...

Calculate (1) `DeltaG^(0)` and (2) The equilibrium constant for the formation of `NO_(2)` from NO and `O_(2)` at 298K.
`NO(g)+1//2O_(2)(g)hArr NO_(2)(g)`
where `Delta_(f)DeltaG^(0)(NO_(2))=52.0kJ//mol`,
`Delta_(f)G^(0)(NO)=87.0kJ//mol,Delta_(f)G^(0)(O_(2))=0kJ//mol`

Text Solution

Verified by Experts

(1) `DeltaG^(0)=sumDelta_(f)G^(0)("Products")-sumDelta_(f)G^(0)" (Reactants)"`
`=Delta_(f)G^(0)(NO_(2))-[Delta_(f)G^(0)(NO)+(1)/(2)Delta_(f)G^(0)(O_(2))]`
`=[52.0-(87.0+(1)/(2)xx0)]kJ*mol^(-1)=-35kJ*mol^(-1)`
(2) `DeltaG^(0)=-2.303RTlogK`
`-35xx10^(3)=-2.303xx8.314xx298logK`
`therefore K=1.36xx10^(6)`.
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