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The equilibrium constant for the reactio...

The equilibrium constant for the reaction :
`Fe^(3) + (aq) +SCN^(-)(aq) hArr Fe SCN^(2+) (aq)`
at 298 K is 138. What is the value of the equilibrium for the reaction?
`2Fe^(3+) (aq) + 2SCN^(-) (aq) hArr 2Fe SCN^(2+) (aq)`

Text Solution

Verified by Experts

The expressions for the equilibrium constants of two reactions may be written as :
`K_(1) = ([FeSCN^(2+) (aq)])/([Fe^(3+) (aq)][SCN^(-) (aq)]) = 138`
`and " "K_(2) = ([FeSCN^(2+) (aq)]^(2))/([Fe^(3+) (aq)]^(2)[SCN^(-)(aq)]^(2))=[(FeSCN^(2+)(aq))/[[Fe^(3+) (aq)][SCN^(-)(aq)))]^(2)`
`= (K_(1))^(2) = (138)^(2) = 19044`
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