If K_b and K_f for a reversible reactions are 0.8 xx 10^(-5) and 1.6 xx 10^(-4) respectively, the value of the equilibrium constant is,

If K_b and K_f for a reversible reactions are 0.8 xx 10^(-5) and 1.6 x

1.	If K_b and K_f for a reversible reactions are 0.8 xx 10^(-5) and 1.6 xx 10^(-4) respectively, the value of the equilibrium constant is,
Answer» 20 `0.2 xx 10^(-1)` 0.05 None of these Solution :`K_b = 0.8 xx 10^(-5)` `K_f = 1.6 xx 10^(-4)` `k_(AQ)= K_f/K_b = (1.6 xx 10^(-4))/(0.8 xx 10^(-5))= 20`