Inachemical equilibrium,the rate constant for the forward reaction is 2.5 xx 10^2 and the equilibrium constant is 50 . The rate constant for the reverse reaction is …………………… .

Inachemical equilibrium,the rate constant for the forward reaction is

1.	Inachemical equilibrium,the rate constant for the forward reaction is 2.5 xx 10^2 and the equilibrium constant is 50 . The rate constant for the reverse reaction is …………………… .
Answer» 11.5 5 `2 xx10^2` `2xx 10^(-3)` Solution :`K_f = 2.5 XX 10^(2) ,K_C = 50,K_r = ?` `K_C = K_f/K_r` `50= (2.5 xx 10^2)/K_r` `K_r = 5`