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In a reaction A+BhArrC+D the rate consta...

In a reaction `A+BhArrC+D` the rate constant of forward reaction & backward reaction is `K_(1) and K_(2)` respectively then the equilibrium constant `(K_(C))` for reaction is expressed as

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The correct Answer is:
`K_(c)=(K_(1))/(K_(2))`
Forward reaction rate `(r_(f))=K_(1)[A][B]`
Bacward reaction rate `(r_(f))=K_(2)[C][D]`
At equilibrium, `R_(f)=R_(b)`
`therefore K_(1)[A][B]=K_(2)[C][D]`
The concentration of reactents & products at equlibrium are related by
`K=(K_(1))/(K_(2))=([C][D])/([A][B]) therefore K=(K_(1))/(K_(2))`
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