The atomic radius of halogens increases as we go down the group due to the addition of new shells. As a result, the bond length of halogen $\ce{X-X}$ increases down the group. So, less energy is required to break the bond (bond dissociation enthalpy decreases). So, as per this assumption, iodine should be the most reactive halogen. But this is not true. Rather, fluorine is the most reactive halogen. It reacts violently with almost all chemicals. Bond cleavage of fluorine is much easier that that of other halogens, which may be due to the repulsive force of the lone pairs of electrons.
$$\ce{2F2(g) + 2H2O(g) → O2(g) + 4HF(g) + heat}$$
So, what about iodine? This source gives the values of bond dissociation enthalpy. It shows that the B.D.E value of iodine ($\pu{151 kJ/mol}$) is less than that of fluorine ($\pu{158 kJ/mol}$). So, iodine should be the most reactive halogen and not halogen. But this does not happen. In most of the reactions of iodine, equilibrium is maintained. To push the reaction forward, a catalyst is used.
$$\ce{I2(l) + H2O(l) <=> OI-(aq) + 2H+(aq) + I-(aq)}$$
Another similar phenomenon is that acidity of hydrohalic acid increases on going the group. $\ce{H-X}$ bond length increases on going down the group. Bond dissociation enthalpy decreases. Bond breaks into corresponding ions. So, $\ce{HI}$ is able to furnish $\ce{H+}$ ion faster than $\ce{HF}$. So, $\ce{HI}$ is more acidic than $\ce{HF}$. But, the reverse is observed. $\ce{HF}$ is so reactive that it reacts with glass and has to be stored in wax bottles.
$$\ce{SiO2 + 4 HF -> SiF4(g) + 2 H2O}$$
$$\ce{SiO2 + 6 HF -> H2SiF6 + 2 H2O}$$
In most of the reactions of hydroiodic acid, equilibrium is maintained. To push the reaction forward, a catalyst is used.
$$\ce{H2(g) + I2(g) <=> 2 HI(g)}$$
My question: Why is it so?