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For a reversible reaction A+BhArrC (...

For a reversible reaction
`A+BhArrC`
`((dx)/(dt))=2.0xx10^(3) L mol^(-1) s^(-1) [A][B]-1.0xx10^(2) s^(-1) [C]` where `x` is the amount of 'A' dissociated. The value of equilibrium constant `(K_(eq))` is

A

`10`

B

`0.05`

C

`20`

D

Cannot be calculated

Text Solution

Verified by Experts

The correct Answer is:
C
`A+BhArrC`
Given: `(dx)/(dt)=(2xx10^(3))[A][B]-(1xx10^(2))[C]`
where x is the amount of 'A' dissociated.
At equilibrium: `(dx)/(dt)=0` (since no change in the concentration of any reactant or product with respect to time)
`rArr (dx)/(dt)=(2xx10^(3))[A][B]-(1xx10^(2))[C]=0`
rArr Equilibrium constant `(K_(eq))=([C])/([A][B])=(2xx10^(3))/(1xx10^(2))=20`
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