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For the reaction: `[Cu(NH_(3))_(4)]^(2+) + H_(2)O hArr [Cu(NH_(3))_(3)H_(2)O]^(2+)+NH_(3)` The net rate of reaction at any time is given net by: rate `=2.0xx10^(-4) [[Cu(NH_(3))_(4)]^(2+)][H_(2)O]-3.0xx10^(5) [[Cu(NH_(3))_(3)H_(2)O]^(2+][NH_(3)]` Then correct statement is (are) :A. Rate constant for forward reaction `=2xx10^(-4)`B. Rate constant for backward reaction= `3xx10^(5)`C. Equilibrium constant for the reaction= `6.6xx10^(-10)`D. All of these |
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Answer» Correct Answer - d Net rate of reaction = rate of forward recation - rate of backward reaction `=K_(f) ["reactant"]-K_(b)`[product] Also `K_(c )=K_(f)/K_(b)` (at equilibrium) |
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