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

Options

A. 10
B. 0.05
C. 20
D. Can't be calculated

Answer

C

Explanation

`A+B rarr C`
Given: `dx/dt = (2 xx 10^(3))[A][B]-(1 xx 10^(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)
`implies dx/dt = (2 xx 10^(3))[A][B]-(1 xx 10^(2))[C] = 0`
`implies` Equilibrium constant `(K_(eq)) = [C]/([A][B]) = 2 xx 10^(3)/1 xx 10^(2) = 20`