The equilibrium constant `(K_(C))` for the reaction `HA +B hArr BH^(+) + A` is 100. If the rate constant for the forward reaction is `10^(5)`, rate constant for the reverse reaction is

The equilibrium constant (K_(c)) for the reaction HA + B hArrBH^(+) + A^(-) is 100. If the rate constant for the forward reaction is 10^(5) , then constant for the backward reaction is

The equilibrium constant ( K ) of the reaction A + B rarr C + D at 297 K is 100. If the rate constant of the forward reaction is 4 xx 10^(5) , the rate constant of the reverse reaction is x xx 10^(3) . Find the value of x.

The unit of equilibrium constant K_(c) for the reaction A+B hArr C would be

The unit of equilibrium constant K_(c) for the reaction A+B hArr C would be

The equilibrium constant Kc for A_((g))hArrB_((g)) is 2.5xx10^(-2) . The rate constant of the forward reaction is 0.05" sec"^(-1) . Calculate the rate constant of the reverse reaction.

The equilibrium constant Kc for A_((g))hArrB_((g)) is 2.5xx10^(-2) . The rate constant of the forward reaction is 0.05" sec"^(-1) . Calculate the rate constant of the reverse reaction.