In a chemical equilibrium, the rate constant for the forward reaction is 2.5xx10^(2) and the equilibrium constant is 50. The rate constant for the reverse reaction is.....

In a chemical equilibrium, the rate constant for the forward reaction

1.	In a chemical equilibrium, the rate constant for the forward reaction is 2.5xx10^(2) and the equilibrium constant is 50. The rate constant for the reverse reaction is.....
Answer» 11.5 5 `2XX10^(2)` `2xx10^(-3)` Solution :`K_(f)=2.5xx10^(2)`,`K_(C)=50` K=? `K_(C)=(K_(f))/(K_(R))rArr50=(2.5xx10^(2))/(K_(r))rArrK_(r)=5`