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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)` |
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Answer» 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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