Discuss the consequence of high enthalpy of $H-H$ bond in terms of chemical reactivity of dihydrogen.

(N/A) The chemical behaviour of dihydrogen is determined,to a large extent,by its bond dissociation enthalpy.
The $H-H$ bond dissociation enthalpy is the highest for a single bond between two atoms of any element.
The dissociation of dihydrogen into its atoms is only $0.081 \%$ at $2000 \ K$,which increases to $95.5 \%$ at $5000 \ K$.
It is relatively inert at room temperature due to the high $H-H$ bond enthalpy. Thus,atomic hydrogen is typically produced at high temperatures in an electric arc or under ultraviolet radiations.
Since its $1s$ orbital is incomplete with $1s^{1}$ electronic configuration,it combines with almost all elements.
It accomplishes reactions by $(i)$ loss of the only electron to give $H^{+}$,$(ii)$ gain of an electron to form $H^{-}$,and $(iii)$ sharing electrons to form a single covalent bond.
Reaction with halogens: It reacts with halogens $(X_{2})$ to give hydrogen halides $(HX)$: $H_{2(g)} + X_{2(g)} \longrightarrow 2HX_{(g)}$ $(X = F, Cl, Br, I)$. While the reaction with fluorine occurs even in the dark,the reaction with iodine requires a catalyst.
Reaction with dioxygen: It reacts with dioxygen to form water. The reaction is highly exothermic: $2H_{2(g)} + O_{2(g)} \xrightarrow{\text{Catalyst or heat}} 2H_{2}O_{(l)}$,$\Delta H^{\ominus} = -285.9 \ kJ \ mol^{-1}$.
Reaction with dinitrogen: With dinitrogen,it forms ammonia: $3H_{2(g)} + N_{2(g)} \xrightarrow{673 \ K, 200 \ atm, Fe} 2NH_{3(g)}$,$\Delta H^{\ominus} = -92.6 \ kJ \ mol^{-1}$. This is the method for the manufacture of ammonia by the Haber process.