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The net rate of reaction of the change : [Cu(NH_(3))_(4)]^(2+)+H_(2)O ?" "[Cu(NH_(3))_(3)H_(2)P]^(2+)+NH_(3) is, (dx)/(dt) =2.0xx10^(-4)[Cu(NH_(3))_(4)]^(2+)-3.0xx10^(5)[Cu(NH_(3))_(3)H_(2)O]^(2+)[NH_(3)] Calculate : the ratio of rate constant for forward and backward reactions. |
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Answer» Solution :Also, `" "K_(F) =2.0xx10^(-4)` `K_(b) =3.0xx10^(5)` `THEREFORE" "(K_(f))/(K_(b))=(2.0xx10^(-4))/(3.0xx10^(5))=6.6xx10^(-10)` |
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