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

Options

A. `K_c=(K_(2))/(K_(1))`
B. `K_c=(K_(1))/(K_(2))`
C. `K_c=K_(1) xx K_(2)`
D. `K_c=K_(1) + K_(2)`

Answer

B

Explanation

Forward reaction rate `(r_f)[A][B]=K_(1)[A][B]`
Backward reaction rate `(r_b) =K_2[C][D]=K_2[C][D]`
At equilibrium, `r_f=r_b`
`:. K_(1)[A][B]=K_2[C][D]`
The concentration of reactents & products of equilibrium are related by
`K=(K_(1))/(K_2)=([C][D])/([A][B])`
`:. (K_c)=(K_(1))/(K_2)`