If K_b and K_f for a reversible reactions are 0.8xx10^(-5) and 1.6xx10^(-4) respectively , the value of the equilibrium constant is………………. .

If K_b and K_f for a reversible reactions are 0.8xx10^(-5) and 1.6xx10

1.	If K_b and K_f for a reversible reactions are 0.8xx10^(-5) and 1.6xx10^(-4) respectively , the value of the equilibrium constant is………………. .
Answer» 20 `0.2 XX 10^(-1)` `0.05` None of these Solution :`K_b = 0.8xx10^(-5)` `K_f = 1.6xx10^(-4)` `K_(eq) = (K_f)/(k_b) =(1.6xx10^(-4))/(0.8xx10^(-5)) = 20`