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For the reaction: [Cu(NH(3))(4)]^(2+) ...

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

Text Solution

Verified by Experts

The correct Answer is:
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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